Numerical Study of Laser Interaction With Solid Materials

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Numerical Study of Laser Interaction with Solid Materials. University of Sulaimani, College of Science, Physics Department, 2011

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Kurdistan Iraqi Region Ministry of Higher Education University of Sulaimani College of Science Physics Department

Numerical Study of Laser Interaction with

Solid Materials

Prepared by

Chia HQadr Shiraz Q Ghafur Hewar AAbdul

Supervised by

Dr Omed Ghareb Abdullah

2010 - 2011

ii

Acknowledgments

First of all we would like to say Alhamdulillah for giving us the

strength and health to do this project work until it done and not forgotten

to our family for providing with everything that are related to this project

work and their advice They also supported us and encouraged us to

complete this task so that we will not procrastinate in doing it

Then we would like to thank our Supervisor Dr Omed Ghareb

Abdullah for guiding us throughout this project We had some difficulties

in doing this task but he taught us patiently until we knew what to do He

tried and tried to teach us until we understand what we supposed to do with

the project work

Last but not least those who were helping us in doing this project by

sharing ideas They were helpful that when we combined and discussed

together we had this task done

Chia Shiraz amp Hewar

iii

Contents

Chapter One Basic Concepts

11 Introduction

12 Definition of the Laser

13 Active laser medium or gain medium

14 A Survey of Laser Types

141 Gas Lasers

142 Solid Lasers

143 Molecular Lasers

144 Free-Electron Lasers

15 Pulsed operation

16 Heat and heat capacity

17 Thermal conductivity

18 Derivation in one dimension

19 Aim of present work

Chapter Two Theoretical Aspects

21 Introduction

22 One dimension laser heating equation

23 Numerical solution of Initial value problems

24 Finite Difference Method

241 First Order Forward Difference

iv

242 First Order Backward Difference

242 First Order Central Difference

25 Procedures

Chapter three Results and Discussion

31 Introduction

32 Numerical solution with constant laser power density and

constant thermal properties

33 Evaluation of function 119920119920(119957119957) of laser flux density

34 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and constant thermal properties

35 Evaluation the Thermal Conductivity as functions of

temperature

36 Evaluation the Specific heat as functions of temperature

37 Evaluation the Density as functions of temperature

38 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

119914119914(119931119931)120646120646 = 120646120646(119931119931)

39 Laser interaction with copper material

310 Conclusions

References

v

Abstract

In recent years much effort has gone into the understanding of the

interaction of short laser pulses with matter The present works have

typically involved studying the interaction of high intensity laser pulses with

high-density solid target In this study the NDYAG pulsed laser with

maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

laser energy with time as well as a function of laser intensity with time are

presented in this study

The finite difference method was used to calculate the temperature

distribution as a function of laser depth penetration in lead and copper

materials

The best polynomial fits for thermal conductivity specific heat capacity

and density of metals as a function of temperature was obtained using

Matlab software At the first all these properties were assumed to be

constants and then the influence of varying these properties with

temperature was tacked in to account The temperature gradient of lead

shows to be greater than that of copper this may be due to the high thermal

conductivity and high specific heat capacity of copper with that of lead

1

Chapter One

Basic concepts

11 Introduction

Laser is a mechanism for emitting light with in electromagnetic radiation

region of the spectrum with different output intensity Max Plank published

work in 1900 that provided the understanding that light is a form of

electromagnetic radiation without this understanding the laser would have

been invented The principle of the laser was first known in 1917 when Albert

Einstein describe the theory of stimulated emission and Theodor Maiman in

1960 invent the first laser using a lasing medium of ruby that was stimulated

by using high energy flash of intense light

We have four types of laser according to their gain medium which are

(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

So laser is provided a controlled source of atomic and electronic excitations

involving non equilibrium phenomena that lend themselves to processing of

novel material and structure because laser used in wide range application in

our life such as welding cutting drilling industrial and medical field Maiman

and other developer of laser weapons sighting system and powerful laser for

use in surgery and other areas where moderated powerful pinpoint source of

heat was needed And today laser are used in corrective eye surgery and

providing apprecise source of heat for cutting and cauterizing tissue

12 Definition of the Laser

The word laser is an acronym for Light Amplification by Stimulated Emission

of Radiation The laser makes use of processes that increase or amplify light

signals after those signals have been generated by other means These

processes include (1) stimulated emission a natural effect that was deduced

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser

    ii

    Acknowledgments

    First of all we would like to say Alhamdulillah for giving us the

    strength and health to do this project work until it done and not forgotten

    to our family for providing with everything that are related to this project

    work and their advice They also supported us and encouraged us to

    complete this task so that we will not procrastinate in doing it

    Then we would like to thank our Supervisor Dr Omed Ghareb

    Abdullah for guiding us throughout this project We had some difficulties

    in doing this task but he taught us patiently until we knew what to do He

    tried and tried to teach us until we understand what we supposed to do with

    the project work

    Last but not least those who were helping us in doing this project by

    sharing ideas They were helpful that when we combined and discussed

    together we had this task done

    Chia Shiraz amp Hewar

    iii

    Contents

    Chapter One Basic Concepts

    11 Introduction

    12 Definition of the Laser

    13 Active laser medium or gain medium

    14 A Survey of Laser Types

    141 Gas Lasers

    142 Solid Lasers

    143 Molecular Lasers

    144 Free-Electron Lasers

    15 Pulsed operation

    16 Heat and heat capacity

    17 Thermal conductivity

    18 Derivation in one dimension

    19 Aim of present work

    Chapter Two Theoretical Aspects

    21 Introduction

    22 One dimension laser heating equation

    23 Numerical solution of Initial value problems

    24 Finite Difference Method

    241 First Order Forward Difference

    iv

    242 First Order Backward Difference

    242 First Order Central Difference

    25 Procedures

    Chapter three Results and Discussion

    31 Introduction

    32 Numerical solution with constant laser power density and

    constant thermal properties

    33 Evaluation of function 119920119920(119957119957) of laser flux density

    34 Numerical solution with variable laser power density (

    119920119920 = 119920119920 (119957119957) ) and constant thermal properties

    35 Evaluation the Thermal Conductivity as functions of

    temperature

    36 Evaluation the Specific heat as functions of temperature

    37 Evaluation the Density as functions of temperature

    38 Numerical solution with variable laser power density (

    119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

    119914119914(119931119931)120646120646 = 120646120646(119931119931)

    39 Laser interaction with copper material

    310 Conclusions

    References

    v

    Abstract

    In recent years much effort has gone into the understanding of the

    interaction of short laser pulses with matter The present works have

    typically involved studying the interaction of high intensity laser pulses with

    high-density solid target In this study the NDYAG pulsed laser with

    maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

    laser energy with time as well as a function of laser intensity with time are

    presented in this study

    The finite difference method was used to calculate the temperature

    distribution as a function of laser depth penetration in lead and copper

    materials

    The best polynomial fits for thermal conductivity specific heat capacity

    and density of metals as a function of temperature was obtained using

    Matlab software At the first all these properties were assumed to be

    constants and then the influence of varying these properties with

    temperature was tacked in to account The temperature gradient of lead

    shows to be greater than that of copper this may be due to the high thermal

    conductivity and high specific heat capacity of copper with that of lead

    1

    Chapter One

    Basic concepts

    11 Introduction

    Laser is a mechanism for emitting light with in electromagnetic radiation

    region of the spectrum with different output intensity Max Plank published

    work in 1900 that provided the understanding that light is a form of

    electromagnetic radiation without this understanding the laser would have

    been invented The principle of the laser was first known in 1917 when Albert

    Einstein describe the theory of stimulated emission and Theodor Maiman in

    1960 invent the first laser using a lasing medium of ruby that was stimulated

    by using high energy flash of intense light

    We have four types of laser according to their gain medium which are

    (solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

    So laser is provided a controlled source of atomic and electronic excitations

    involving non equilibrium phenomena that lend themselves to processing of

    novel material and structure because laser used in wide range application in

    our life such as welding cutting drilling industrial and medical field Maiman

    and other developer of laser weapons sighting system and powerful laser for

    use in surgery and other areas where moderated powerful pinpoint source of

    heat was needed And today laser are used in corrective eye surgery and

    providing apprecise source of heat for cutting and cauterizing tissue

    12 Definition of the Laser

    The word laser is an acronym for Light Amplification by Stimulated Emission

    of Radiation The laser makes use of processes that increase or amplify light

    signals after those signals have been generated by other means These

    processes include (1) stimulated emission a natural effect that was deduced

    2

    by considerations relating to thermodynamic equilibrium and (2) optical

    feedback (present in most lasers) that is usually provided by mirrors

    Thus in its simplest form a laser consists of a gain or amplifying medium

    (where stimulated emission occurs) and a set of mirrors to feed the light back

    into the amplifier for continued growth of the developing beam as seen in

    Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

    more intense directional monochromatic and coherent Most lasers consist

    of a column of active material with a partly reflecting mirror at one end and a

    fully reflecting mirror at the other The active material can be solid (ruby

    crystal) liquid or gas (HeNe COR2R etc)

    Fig(11) Simplified schematic of typical laser

    13 Active laser medium or gain medium

    Laser medium is the heart of the laser system and is responsible for

    producing gain and subsequent generation of laser It can be a crystal solid

    liquid semiconductor or gas medium and can be pumped to a higher energy

    state The material should be of controlled purity size and shape and should

    have the suitable energy levels to support population inversion In other

    words it must have a metastable state to support stimulated emission Most

    lasers are based on 3 or 4 level energy level systems which depends on the

    lasing medium These systems are shown in Figs (12) and (13)

    3

    In case of a three-level laser the material is pumped from level 1 to level 3

    which decays rapidly to level 2 through spontaneous emission Level 2 is a

    metastable level and promotes stimulated emission from level 2 to level 1

    Fig(12) Energy states of Three-level active medium

    On the other hand in a four-level laser the material is pumped to level 4

    which is a fast decaying level and the atoms decay rapidly to level 3 which is

    a metastable level The stimulated emission takes place from level 3 to level 2

    from where the atoms decay back to level 1 Four level lasers is an

    improvement on a system based on three level systems In this case the laser

    transition takes place between the third and second excited states Since

    lower laser level 2 is a fast decaying level which ensures that it rapidly gets

    empty and as such always supports the population inversion condition

    Fig(13) Energy states of Four-level active medium

    4

    14 A Survey of Laser Types

    Laser technology is available to us since 1960rsquos and since then has been

    quite well developed Currently there is a great variety of lasers of different

    output power operating voltages sizes etc The major classes of lasers

    currently used are Gas Solid Molecular and Free Electron lasers Below we

    will cover some most popular representative types of lasers of each class and

    describe specific principles of operation construction and main highlights

    141 Gas Lasers

    1 Helium-Neon Laser

    The most common and inexpensive gas laser the helium-neon laser is

    usually constructed to operate in the red at 6328 nm It can also be

    constructed to produce laser action in the green at 5435 nm and in the

    infrared at 1523 nm

    One of the excited levels of helium at 2061 eV is very close to a level in

    neon at 2066 eV so close in fact that upon collision of a helium and a neon

    atom the energy can be transferred from the helium to the neon atom

    Fig (14) The components of a Hilium-Neon Laser

    5

    Fig(15) The lasing action of He-Ne laser

    Helium-Neon lasers are common in the introductory physics laboratories

    but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

    brightness equal to sunshine on a clear day (01 wattcmP

    2P) and is just as

    dangerous to stare at directly

    2- Carbon Dioxide Laser

    The carbon dioxide gas laser is capable of continuous output powers above

    10 kilowatts It is also capable of extremely high power pulse operation It

    exhibits laser action at several infrared frequencies but none in the visible

    spectrum Operating in a manner similar to the helium-neon laser it employs

    an electric discharge for pumping using a percentage of nitrogen gas as a

    pumping gas The COR2R laser is the most efficient laser capable of operating at

    more than 30 efficiency

    The carbon dioxide laser finds many applications in industry particularly for

    welding and Cutting

    6

    3- Argon Laser

    The argon ion laser can be operated as a continuous gas laser at about 25

    different wavelengths in the visible between (4089 - 6861) nm but is best

    known for its most efficient transitions in the green at 488 nm and 5145 nm

    Operating at much higher powers than the Helium-Neon gas laser it is not

    uncommon to achieve (30 ndash 100) watts of continuous power using several

    transitions This output is produced in hot plasma and takes extremely high

    power typically (9 ndash 12) kW so these are large and expensive devices

    142 Solid Lasers

    1 Ruby Laser

    The ruby laser is the first type of laser actually constructed first

    demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

    aluminum oxide with a small amount (about 005) of Chromium which gives

    it its characteristic pink or red color by absorbing green and blue light

    The ruby laser is used as a pulsed laser producing red light at 6943 nm

    After receiving a pumping flash from the flash tube the laser light emerges for

    as long as the excited atoms persist in the ruby rod which is typically about a

    millisecond

    A pulsed ruby laser was used for the famous laser ranging experiment which

    was conducted with a corner reflector placed on the Moon by the Apollo

    astronauts This determined the distance to the Moon with an accuracy of

    about 15 cm

    7

    Fig (16) Principle of operation of a Ruby laser

    2- Neodymium-YAG Laser

    An example of a solid-state laser the neodymium-YAG uses the NdP

    3+P ion to

    dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

    geometry which makes population inversion possible Neodymium-YAG lasers

    have become very important because they can be used to produce high

    powers Such lasers have been constructed to produce over a kilowatt of

    continuous laser power at 1065 nm and can achieve extremely high powers in

    a pulsed mode

    Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

    production of a series of very short pulses for research with femtosecond time

    resolution

    Fig(17) Construction of a Neodymium-YAG laser

    8

    3- Neodymium-Glass Lasers

    Neodymium glass lasers have emerged as the design choice for research in

    laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

    short as 10-12 seconds with peak powers of 109 kilowatts

    143 Molecular Lasers

    Eximer Lasers

    Eximer is a shortened form of excited dimer denoting the fact that the

    lasing medium in this type of laser is an excited diatomic molecule These

    lasers typically produce ultraviolet pulses They are under investigation for use

    in communicating with submarines by conversion to blue-green light and

    pulsing from overhead satellites through sea water to submarines below

    The eximers used are typically those formed by rare gases and halogens in

    electron excited Gas discharges Molecules like XeF are stable only in their

    excited states and quickly dissociate when they make the transition to their

    ground state This makes possible large population inversions because the

    ground state is depleted by this dissociation However the excited states are

    very short-lived compared to other laser metastable states and lasers like the

    XeF eximer laser require high pumping rates

    Eximer lasers typically produce high power pulse outputs in the blue or

    ultraviolet after excitation by fast electron-beam discharges

    The rare-gas xenon and the highly active fluorine seem unlikely to form a

    molecule but they do in the hot plasma environment of an electron-beam

    initiated gas discharge They are only stable in their excited states if stable

    can be used for molecules which undergo radioactive decay in 1 to 10

    nanoseconds This is long enough to achieve pulsed laser action in the blue-

    green over a band from 450 to 510 nm peaking at 486 nm Very high power

    9

    pulses can be achieved because the stimulated emission cross-sections of the

    laser transitions are relatively low allowing a large population inversion to

    build up The power is also enhanced by the fact that the ground state of XeF

    quickly dissociates so that there is little absorption to quench the laser pulse

    action

    144 Free-Electron Lasers

    The radiation from a free-electron laser is produced from free electrons

    which are forced to oscillate in a regular fashion by an applied field They are

    therefore more like synchrotron light sources or microwave tubes than like

    other lasers They are able to produce highly coherent collimated radiation

    over a wide range of frequencies The magnetic field arrangement which

    produces the alternating field is commonly called a wiggler magnet

    Fig(18) Principle of operation of Free-Electron laser

    The free-electron laser is a highly tunable device which has been used to

    generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

    this range they are the highest power source Applications of free-electron

    lasers are envisioned in isotope separation plasma heating for nuclear fusion

    long-range high resolution radar and particle acceleration in accelerators

    10

    15 Pulsed operation

    Pulsed operation of lasers refers to any laser not classified as continuous

    wave so that the optical power appears in pulses of some duration at some

    repetition rate This encompasses a wide range of technologies addressing a

    number of different motivations Some lasers are pulsed simply because they

    cannot be run in continuous mode

    In other cases the application requires the production of pulses having as

    large an energy as possible Since the pulse energy is equal to the average

    power divided by the repitition rate this goal can sometimes be satisfied by

    lowering the rate of pulses so that more energy can be built up in between

    pulses In laser ablation for example a small volume of material at the surface

    of a work piece can be evaporated if it is heated in a very short time whereas

    supplying the energy gradually would allow for the heat to be absorbed into

    the bulk of the piece never attaining a sufficiently high temperature at a

    particular point

    Other applications rely on the peak pulse power (rather than the energy in

    the pulse) especially in order to obtain nonlinear optical effects For a given

    pulse energy this requires creating pulses of the shortest possible duration

    utilizing techniques such as Q-switching

    16 Heat and heat capacity

    When a sample is heated meaning it receives thermal energy from an

    external source some of the introduced heat is converted into kinetic energy

    the rest to other forms of internal energy specific to the material The amount

    converted into kinetic energy causes the temperature of the material to rise

    The amount of the temperature increase depends on how much heat was

    added the size of the sample the original temperature of the sample and on

    how the heat was added The two obvious choices on how to add the heat are

    11

    to add it holding volume constant or to add it holding pressure constant

    (There may be other choices but they will not concern us)

    Lets assume for the moment that we are going to add heat to our sample

    holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

    subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

    be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

    amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

    (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

    lim119876119876119889119889rarr0

    119876119876119889119889∆120549120549119889119889

    = 120597120597119876119876120597120597120549120549119889119889

    = 119862119862119889119889 (11)

    We have given this derivative the symbol 119862119862119889119889 and we call it the heat

    capacity at constant volume Usually one quotes the molar heat capacity

    119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

    (12)

    We can rearrange Equation (11) as follows

    119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

    Then we can integrate this equation to find the heat involved in a finite

    change at constant volume

    119876119876119889119889 = 119862119862119889119889

    1205491205492

    1205491205491

    119889119889120549120549 (14)

    If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

    out of the integral and the heat at constant volume becomes

    119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

    Let us now go through the same sequence of steps except holding pressure

    constant instead of volume Our initial definition of the heat capacity at

    constant pressure 119862119862119875119875 R Rbecomes

    lim119876119876119875119875rarr0

    119876119876119875119875∆120549120549119875119875

    = 120597120597119876119876120597120597120549120549119875119875

    = 119862119862119875119875 (16)

    The analogous molar heat capacity is

    12

    119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

    (17)

    Equation (16) rearranges to

    119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

    which integrates to give

    119876119876119875119875 = 119862119862119875119875

    1205491205492

    1205491205491

    119889119889120549120549 (19)

    When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

    119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

    Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

    regarded as constant In these cases the heat capacity is fit to a polynomial (or

    similar function) in 120549120549 For example some tables give the heat capacity as

    119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

    where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

    dependent heat capacity the heat at constant pressure would integrate as

    follows

    119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

    1205491205492

    1205491205491

    119889119889120549120549 (112)

    119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

    2 minus 12054912054912 + 119899119899

    1205741205743

    12054912054923 minus 1205491205491

    3 (113)

    Occasionally one finds a different form for the temperature dependent heat

    capacity in the literature

    119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

    When you do calculations with temperature dependent heat capacities you

    must check to see which form is being used for 119862119862119875119875 We are using the

    convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

    be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

    13

    positive then heat was indeed absorbed by the system On the other hand if

    119876119876 is negative it means that the system gave up heat to the surroundings

    17 Thermal conductivity

    In physics thermal conductivity 119896119896 is the property of a material that indicates

    its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

    Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

    The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

    a piece of material The reciprocal of thermal conductivity is thermal

    resistivity

    18 Derivation in one dimension

    The heat equation is derived from Fouriers law and conservation of energy

    (Cannon 1984) By Fouriers law the flow rate of heat energy through a

    surface is proportional to the negative temperature gradient across the

    surface

    119902119902 = minus119896119896 120571120571120549120549 (115)

    where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

    dimension the gradient is an ordinary spatial derivative and so Fouriers law is

    119902119902 = minus119896119896 120549120549119909119909 (116)

    where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

    energy per unit volume in the material 120549120549119876119876 is proportional to the change in

    temperature 120549120549120549120549 That is

    ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

    where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

    material Choosing zero energy at absolute zero temperature this can be

    rewritten as

    ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

    14

    The increase in internal energy in a small spatial region of the material

    (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

    given by

    119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

    119909119909minus∆119909119909

    = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

    119889119889120577120577 119889119889120591120591119909119909+∆119909119909

    119909119909minus∆119909119909

    119905119905+∆119905119905

    119905119905minus∆119905119905

    (119)

    Where the fundamental theorem of calculus was used Additionally with no

    work done and absent any heat sources or sinks the change in internal energy

    in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

    across the boundaries By Fouriers law this is

    119896119896 120597120597120549120549120597120597119909119909

    (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

    (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

    119905119905minus∆119905119905

    = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

    119909119909+∆119909119909

    119909119909minus∆119909119909

    119905119905+∆119905119905

    119905119905minus∆119905119905

    (120)

    again by the fundamental theorem of calculus By conservation of energy

    119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

    119909119909minus∆119909119909

    119905119905+∆119905119905

    119905119905minus∆119905119905

    = 0 (121)

    This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

    Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

    Which can be rewritten as

    120549120549119905119905 =119896119896119888119888119875119875 120588120588

    120549120549119909119909119909119909 (122)

    or

    120597120597120549120549120597120597119905119905

    =119896119896119888119888119875119875 120588120588

    12059712059721205491205491205971205971199091199092 (123)

    15

    which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

    diffusivity and is often denoted 120572120572

    19 Aim of present work

    The goal of this study is to estimate the solution of partial differential

    equation that governs the laser-solid interaction using numerical methods

    The solution will been restricted into one dimensional situation in which we

    assume that both the laser power density and thermal properties are

    functions of time and temperature respectively In this project we attempt to

    investigate the laser interaction with both lead and copper materials by

    predicting the temperature gradient with the depth of the metals

    16

    Chapter Two

    Theoretical Aspects

    21 Introduction

    When a laser interacts with a solid surface a variety of processes can

    occur We are mainly interested in the interaction of pulsed lasers with a

    solid surface in first instance a metal When such a laser interacts with a

    copper surface the laser energy will be transformed into heat The

    temperature of the solid material will increase leading to melting and

    evaporation of the solid material

    The evaporated material (vapour atoms) will expand Depending on the

    applications this can happen in vacuum (or very low pressure) or in a

    background gas (helium argon air)

    22 One dimension laser heating equation

    In general the one dimension laser heating processes of opaque solid slab is

    represented as

    120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

    ( 119870119870 119879119879120597120597 ) (21)

    With boundary conditions and initial condition which represent the pre-

    vaporization stage

    minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

    minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

    17

    119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

    where

    119870119870 represents the thermal conductivity

    120588120588 represents the density

    119862119862 represents the specific heat

    119879119879 represents the temperature

    119879119879infin represents the ambient temperature

    119879119879119907119907 represents the front surface vaporization

    120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

    Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

    119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

    With the same boundary conditions as in equation (22)

    where 119889119889119889119889 = 119870119870120588120588119862119862

    which represents the thermal diffusion

    But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

    equation (21) with this assuming implies

    119879119879119905119905 = 1

    120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

    With the same boundary and initial conditions in equation (22) Where 119870119870

    represents the derivative of K with respect the temperature

    23 Numerical solution of Initial value problems

    An immense number of analytical solutions for conduction heat-transfer

    problems have been accumulated in literature over the past 100 years Even so

    in many practical situations the geometry or boundary conditions are such that an

    analytical solution has not been obtained at all or if the solution has been

    18

    developed it involves such a complex series solution that numerical evaluation

    becomes exceedingly difficult For such situation the most fruitful approach to

    the problem is numerical techniques the basic principles of which we shall

    outline in this section

    One way to guarantee accuracy in the solution of an initial values problems

    (IVP) is to solve the problem twice using step sizes h and h2 and compare

    answers at the mesh points corresponding to the larger step size But this requires

    a significant amount of computation for the smaller step size and must be

    repeated if it is determined that the agreement is not good enough

    24 Finite Difference Method

    The finite difference method is one of several techniques for obtaining

    numerical solutions to differential equations In all numerical solutions the

    continuous partial differential equation (PDE) is replaced with a discrete

    approximation In this context the word discrete means that the numerical

    solution is known only at a finite number of points in the physical domain The

    number of those points can be selected by the user of the numerical method In

    general increasing the number of points not only increases the resolution but

    also the accuracy of the numerical solution

    The discrete approximation results in a set of algebraic equations that are

    evaluated for the values of the discrete unknowns

    The mesh is the set of locations where the discrete solution is computed

    These points are called nodes and if one were to draw lines between adjacent

    nodes in the domain the resulting image would resemble a net or mesh Two key

    parameters of the mesh are ∆120597120597 the local distance between adjacent points in

    space and ∆119905119905 the local distance between adjacent time steps For the simple

    examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

    19

    The core idea of the finite-difference method is to replace continuous

    derivatives with so-called difference formulas that involve only the discrete

    values associated with positions on the mesh

    Applying the finite-difference method to a differential equation involves

    replacing all derivatives with difference formulas In the heat equation there are

    derivatives with respect to time and derivatives with respect to space Using

    different combinations of mesh points in the difference formulas results in

    different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

    numerical solution obtained with any useful scheme will approach the true

    solution to the original differential equation However the rate at which the

    numerical solution approaches the true solution varies with the scheme

    241 First Order Forward Difference

    Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

    empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

    +1205751205751205971205972

    2 1205971205972empty1205971205971205971205972

    1205971205971

    +1205751205751205971205973

    3 1205971205973empty1205971205971205971205973

    1205971205971

    + ⋯ (25)

    where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

    consider the value of empty at the location of the 120597120597119894119894+1 mesh line

    empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

    +∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120597120597119894119894

    +∆1205971205973

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    + ⋯ (26)

    Solve for (120597120597empty120597120597120597120597)120597120597119894119894

    120597120597empty120597120597120597120597120597120597119894119894

    =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

    ∆120597120597minus∆1205971205972

    1205971205972empty1205971205971205971205972

    120597120597119894119894

    minus∆1205971205972

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    minus ⋯ (27)

    Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

    hand side have been reduced by one

    20

    Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

    and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

    120597120597empty120597120597120597120597120597120597119894119894

    =empty119894119894+1 minus empty119894119894

    ∆120597120597minus∆1205971205972

    1205971205972empty1205971205971205971205972

    120597120597119894119894

    minus∆1205971205972

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    minus ⋯ (28)

    The mean value theorem can be used to replace the higher order derivatives

    ∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120597120597119894119894

    +∆1205971205973

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    + ⋯ =∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120585120585 (29)

    where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

    120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

    ∆120597120597+∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120585120585 (210)

    120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

    ∆120597120597asymp∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120585120585 (211)

    The term on the right hand side of previous equation is called the truncation

    error of the finite difference approximation

    In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

    unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

    truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

    analytical form) the big 119978119978 notation can be used to express the dependence of

    the truncation error on the mesh spacing Note that the right hand side of last

    equation contains the mesh parameter ∆120597120597 which is chosen by the person using

    the finite difference simulation Since this is the only parameter under the users

    control that determines the error the truncation error is simply written

    ∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120585120585= 119978119978(∆1205971205972) (212)

    The equals sign in this expression is true in the order of magnitude sense In

    other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

    21

    equality Rather the expression means that the left hand side is a product of an

    unknown constant and ∆1205971205972 Although the expression does not give us the exact

    magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

    approaches zero as ∆120597120597 is reduced

    Using big 119978119978 notation Equation (28) can be written

    120597120597empty120597120597120597120597120597120597119894119894

    =empty119894119894+1 minus empty119894119894

    ∆120597120597+ 119978119978(∆120597120597) (213)

    This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

    it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

    truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

    our control because we can choose the mesh size ∆120597120597 The part of the truncation

    error that is not under our control is |120597120597empty120597120597120597120597|120585120585

    242 First Order Backward Difference

    An alternative first order finite difference formula is obtained if the Taylor series

    like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

    variables in place of all the unknowns one obtains

    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

    +∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120597120597119894119894

    minus∆1205971205973

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    + ⋯ (214)

    Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

    to get

    120597120597empty120597120597120597120597120597120597119894119894

    =empty119894119894+1 minus empty119894119894

    ∆120597120597minus∆1205971205972

    1205971205972empty1205971205971205971205972

    120597120597119894119894

    minus∆1205971205972

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    minus ⋯ (215)

    Or using big 119978119978 notation

    120597120597empty120597120597120597120597120597120597119894119894

    =empty119894119894 minus empty119894119894minus1

    ∆120597120597+ 119978119978(∆120597120597) (216)

    22

    This is called the backward difference formula because it involves the values of

    empty at 120597120597119894119894 and 120597120597119894119894minus1

    The order of magnitude of the truncation error for the backward difference

    approximation is the same as that of the forward difference approximation Can

    we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

    truncation error The answer is yes

    242 First Order Central Difference

    Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

    empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

    +∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120597120597119894119894

    +∆1205971205973

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    + ⋯ (217)

    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

    +∆1205971205972

    2 1205971205972empty1205971205971205971205972

    120597120597119894119894

    minus∆1205971205973

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    + ⋯ (218)

    Subtracting Equation (10) from Equation (9) yields

    empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

    + 2∆1205971205973

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    + ⋯ (219)

    Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

    120597120597empty120597120597120597120597120597120597119894119894

    =empty119894119894+1 minus empty119894119894minus1

    2∆120597120597minus∆1205971205972

    3 1205971205973empty1205971205971205971205973

    120597120597119894119894

    minus ⋯ (220)

    or

    120597120597empty120597120597120597120597120597120597119894119894

    =empty119894119894+1 minus empty119894119894minus1

    2∆120597120597+ 119978119978(∆1205971205972) (221)

    This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

    approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

    the truncation error for the central difference approximation goes to zero much

    faster than the truncation error in forward and backward equations

    23

    25 Procedures

    The simple case in this investigation was assuming the constant thermal

    properties of the material First we assumed all the thermal properties of the

    materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

    point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

    assume the constant energy after that the pulse of special shapes was selected

    The numerical solution of equation (23) with boundary and initial conditions

    in equation (22) was investigated using Matlab program as shown in Appendix

    The equation of thermal conductivity and specific heat capacity of metal as a

    function of temperature was obtained by best fitting of polynomials using

    tabulated data in references

    24

    Chapter Three

    Results and Discursion

    31 Introduction

    The development of laser has been an exciting chapter in the history of

    science and engineering It has produced a new type of advice with potential for

    application in an extremely wide variety of fields Mach basic development in

    lasers were occurred during last 35 years The lasers interaction with metal and

    vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

    The status of laser development and application were still rather rudimentary

    The light emitted by laser is electro magnetic radiation this radiation has a wave

    nature the waves consists of vibrating electric and magnetic fields many studies

    have tried to find and solve models of laser interactions Some researchers

    proposed the mathematical model related to the laser - plasma interaction and

    the others have developed an analytical model to study the temperature

    distribution in Infrared optical materials heated by laser pulses Also an attempt

    have made to study the interaction of nanosecond pulsed lasers with material

    from point of view using experimental technique and theoretical approach of

    dimensional analysis

    In this study we have evaluate the solution of partial difference equation

    (PDE) that represent the laser interaction with solid situation in one dimension

    assuming that the power density of laser and thermal properties are functions

    with time and temperature respectively

    25

    32 Numerical solution with constant laser power density and constant

    thermal properties

    First we have taken the lead metal (Pb) with thermal properties

    119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

    119862119862 = 014016119869119869119892119892119870119870

    120588120588 = 10751 1198921198921198981198981198981198983

    119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

    119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

    and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

    represent the area under laser influence

    The numerical solution of equation (23) with boundary and initial conditions

    in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

    of lead metal by explicit method using Matlab program give us the results as

    shown in Fig (31)

    Fig(31) Depth dependence of the temperature with the laser power density

    1198681198680 = 76 times 106 119882119882119898119898119898119898 2

    26

    33 Evaluation of function 119920119920(119957119957) of laser flux density

    From following data that represent the energy (119869119869) with time (millie second)

    Time 0 001 01 02 03 04 05 06 07 08

    Energy 0 002 017 022 024 02 012 007 002 0

    By using Matlab program the best polynomial with deduced from above data

    was

    119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

    minus 10069 1198981198985 (31)

    As shown in Fig (32)

    Fig(32) Laser energy as a function of time

    Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

    normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

    maximum value (119864119864119898119898119907119907119898119898 )

    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

    (119864119864119898119898119907119907119898119898 ) (32)

    The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

    27

    Fig(33) Normalized laser energy as a function of time

    The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

    equal to 3 (total laser energy) ie

    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

    08

    00

    119899119899119898119898 = 3 (33)

    Therefore there exist a real number 119875119875 such that

    119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

    08

    00

    119899119899119898119898 = 3 (34)

    that implies 119875119875 = 68241 and

    119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

    08

    00

    119899119899119898119898 = 3 (35)

    The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

    08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

    119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

    00

    (36)

    28

    Where 119863119863119898119898 put to balance the units of equation (36)

    But integral

    119868119868 = 119864119864119860119860

    (37)

    and from equations (35) (36) and (37) we have

    119868119868 (119898119898)11989911989911989811989808

    00

    = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

    0800 119899119899119898119898

    119860119860 119863119863119898119898 (38)

    Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

    (38)

    There fore

    119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

    119860119860 119863119863119898119898 (39)

    As shown in Fig(34) Matlab program was used to obtain the best polynomial

    that agrees with result data

    119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

    + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

    Fig(34) Time dependence of laser intensity

    29

    34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

    constant thermal properties

    With all constant thermal properties of lead metal as in article (23) and

    119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

    equation (23) with boundary and initial condition as in equation (22) and the

    depth penetration is shown in Fig(35)

    Fig(35) Depth dependence of the temperature when laser intensity function

    of time and constant thermal properties of Lead

    35 Evaluation the Thermal Conductivity as functions of temperature

    The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

    temperature for Lead material was obtained by Matlab program using the

    experimental data tabulated in researches

    119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

    times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

    30

    119879119879 ( 119870119870) 119870119870 119869119869

    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

    300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

    The previous thermal conductivity data and the best fitting of the data are

    shown in Fig (36)

    Fig(36) The best fitting of thermal conductivity of Lead as a function of

    temperature

    31

    36 Evaluation the Specific heat as functions of temperature

    The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

    material was obtained from the following experimental data tacked from

    literatures

    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

    300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

    The best polynomial fitted for these data was

    119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

    + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

    times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

    The specific heat capacity data and the best polynomial fitting of the data are

    shown in Fig (37)

    32

    Fig(37) The best fitting of specific heat capacity of Lead as a function of

    temperature

    37 Evaluation the Density as functions of temperature

    The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

    was used to find the best polynomial fitting

    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

    300 11330 400 11230 500 11130 600 11010 800 10430

    1000 10190 1200 9940

    The best polynomial of this data was

    120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

    minus 45158 times 10minus12 1198791198794 (313)

    33

    The density of Lead as a function of temperature and the best polynomial fitting

    are shown in Fig (38)

    Fig(38) The best fitting of density of Lead as a function of temperature

    38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

    variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

    We have deduced the solution of equation (24) with initial and boundary

    condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

    and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

    respectively then by using Matlab program the depth penetration is shown in

    Fig (39)

    34

    Fig(39) Depth dependence of the temperature for pulse laser on Lead

    material

    39 Laser interaction with copper material

    The same time dependence of laser intensity as shown in Fig(34) with

    thermal properties of copper was used to calculate the temperature distribution as

    a function of depth penetration

    The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

    copper material was obtained from the experimental data tabulated in literary

    The Matlab program used to obtain the best polynomial equation that agrees

    with the above data

    119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

    minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

    times 10minus17 1198791198795 (314)

    35

    119879119879 ( 119870119870) 119870119870 119869119869

    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

    100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

    The previous thermal conductivity data and the best fitting of the data are

    shown in Fig (310)

    Fig(310) The best fitting of thermal conductivity of Copper as a function of

    temperature

    36

    The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

    material was obtained from the following experimental data tacked from

    literatures

    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

    100 0254

    200 0357

    273 0384

    298 0387

    400 0397

    600 0416

    800 0435

    1000 0454

    1100 0464

    1200 0474

    1300 0483

    The best polynomial fitted for these data was

    119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

    + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

    times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

    The specific heat capacity data and the best polynomial fitting of the data are

    shown in Fig (311)

    37

    Fig(311) The best fitting of specific heat capacity of Copper as a function of

    temperature

    The density of copper 120588120588(119879119879) as a function of temperature tacked from

    literature was used to find the best polynomial fitting

    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

    100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

    1000 8576 1100 8519 1200 8458 1300 8396

    38

    The best polynomial of this data was

    120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

    minus 76765 times 10minus14 1198791198794 (316)

    The density of copper as a function of temperature and the best polynomial

    fitting are shown in Fig (312)

    Fig(312) The best fitting of density of copper as a function of temperature

    The depth penetration of laser energy for copper metal was calculated using

    the polynomial equations of thermal conductivity specific heat capacity and

    density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

    (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

    function of time the result was shown in Fig (313)

    39

    The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

    for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

    copper metal so the depth penetration of laser energy of lead metal was smaller

    than that of copper metal this may be due to the high thermal conductivity and

    high specific heat capacity of copper with that of lead metal

    Fig(313) Depth dependence of the temperature for pulse laser on Copper

    material

    40

    310 Conclusions

    The Depth dependence of temperature for lead metal was investigated in two

    case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

    thermal properties (thermal conductivity specific heat) and density of metal are

    also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

    vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

    specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

    120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

    penetration depth in the first case is smaller than that of the second case about

    (190) times

    The temperature distribution as a function of depth dependence for copper

    metal was also investigated in the case when the laser intensity vary with time

    119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

    density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

    The depth penetration of laser energy of lead metal was found to be smaller

    than that of copper metal this may be due to the high thermal conductivity and

    high specific heat capacity of copper with that of lead metal

    41

    References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

    Sons Inc Hoboken New Jersey Canada (2003)

    [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

    [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

    [9]

    Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

    [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

    [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

    [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

    [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

    [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

    httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

    [10] httpwwwworldoflaserscomlaserprincipleshtm

    [11] httpenwikipediaorgwikiLaserPulsed_operation

    [12] httpenwikipediaorgwikiThermal_conductivity

    [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

    [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

    [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

    42

    Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

    This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

    This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

    43

    E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

    This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

    This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

    This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

    44

    title(Dencity as a function of temperature) hold off

    This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

    45

    elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

    46

    r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

    47

    for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

    48

    for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

    49

    6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

    This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

    50

    u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

    51

    alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

    52

    715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

    • 01 Title
      • 0TDr Omed Ghareb Abdullah
        • Ch1-Laser
          • 16 Heat and heat capacity
          • 18 Derivation in one dimension
            • Ch2 laser2
            • Ch3 laser2
            • Appendix-Laser

      iii

      Contents

      Chapter One Basic Concepts

      11 Introduction

      12 Definition of the Laser

      13 Active laser medium or gain medium

      14 A Survey of Laser Types

      141 Gas Lasers

      142 Solid Lasers

      143 Molecular Lasers

      144 Free-Electron Lasers

      15 Pulsed operation

      16 Heat and heat capacity

      17 Thermal conductivity

      18 Derivation in one dimension

      19 Aim of present work

      Chapter Two Theoretical Aspects

      21 Introduction

      22 One dimension laser heating equation

      23 Numerical solution of Initial value problems

      24 Finite Difference Method

      241 First Order Forward Difference

      iv

      242 First Order Backward Difference

      242 First Order Central Difference

      25 Procedures

      Chapter three Results and Discussion

      31 Introduction

      32 Numerical solution with constant laser power density and

      constant thermal properties

      33 Evaluation of function 119920119920(119957119957) of laser flux density

      34 Numerical solution with variable laser power density (

      119920119920 = 119920119920 (119957119957) ) and constant thermal properties

      35 Evaluation the Thermal Conductivity as functions of

      temperature

      36 Evaluation the Specific heat as functions of temperature

      37 Evaluation the Density as functions of temperature

      38 Numerical solution with variable laser power density (

      119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

      119914119914(119931119931)120646120646 = 120646120646(119931119931)

      39 Laser interaction with copper material

      310 Conclusions

      References

      v

      Abstract

      In recent years much effort has gone into the understanding of the

      interaction of short laser pulses with matter The present works have

      typically involved studying the interaction of high intensity laser pulses with

      high-density solid target In this study the NDYAG pulsed laser with

      maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

      laser energy with time as well as a function of laser intensity with time are

      presented in this study

      The finite difference method was used to calculate the temperature

      distribution as a function of laser depth penetration in lead and copper

      materials

      The best polynomial fits for thermal conductivity specific heat capacity

      and density of metals as a function of temperature was obtained using

      Matlab software At the first all these properties were assumed to be

      constants and then the influence of varying these properties with

      temperature was tacked in to account The temperature gradient of lead

      shows to be greater than that of copper this may be due to the high thermal

      conductivity and high specific heat capacity of copper with that of lead

      1

      Chapter One

      Basic concepts

      11 Introduction

      Laser is a mechanism for emitting light with in electromagnetic radiation

      region of the spectrum with different output intensity Max Plank published

      work in 1900 that provided the understanding that light is a form of

      electromagnetic radiation without this understanding the laser would have

      been invented The principle of the laser was first known in 1917 when Albert

      Einstein describe the theory of stimulated emission and Theodor Maiman in

      1960 invent the first laser using a lasing medium of ruby that was stimulated

      by using high energy flash of intense light

      We have four types of laser according to their gain medium which are

      (solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

      So laser is provided a controlled source of atomic and electronic excitations

      involving non equilibrium phenomena that lend themselves to processing of

      novel material and structure because laser used in wide range application in

      our life such as welding cutting drilling industrial and medical field Maiman

      and other developer of laser weapons sighting system and powerful laser for

      use in surgery and other areas where moderated powerful pinpoint source of

      heat was needed And today laser are used in corrective eye surgery and

      providing apprecise source of heat for cutting and cauterizing tissue

      12 Definition of the Laser

      The word laser is an acronym for Light Amplification by Stimulated Emission

      of Radiation The laser makes use of processes that increase or amplify light

      signals after those signals have been generated by other means These

      processes include (1) stimulated emission a natural effect that was deduced

      2

      by considerations relating to thermodynamic equilibrium and (2) optical

      feedback (present in most lasers) that is usually provided by mirrors

      Thus in its simplest form a laser consists of a gain or amplifying medium

      (where stimulated emission occurs) and a set of mirrors to feed the light back

      into the amplifier for continued growth of the developing beam as seen in

      Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

      more intense directional monochromatic and coherent Most lasers consist

      of a column of active material with a partly reflecting mirror at one end and a

      fully reflecting mirror at the other The active material can be solid (ruby

      crystal) liquid or gas (HeNe COR2R etc)

      Fig(11) Simplified schematic of typical laser

      13 Active laser medium or gain medium

      Laser medium is the heart of the laser system and is responsible for

      producing gain and subsequent generation of laser It can be a crystal solid

      liquid semiconductor or gas medium and can be pumped to a higher energy

      state The material should be of controlled purity size and shape and should

      have the suitable energy levels to support population inversion In other

      words it must have a metastable state to support stimulated emission Most

      lasers are based on 3 or 4 level energy level systems which depends on the

      lasing medium These systems are shown in Figs (12) and (13)

      3

      In case of a three-level laser the material is pumped from level 1 to level 3

      which decays rapidly to level 2 through spontaneous emission Level 2 is a

      metastable level and promotes stimulated emission from level 2 to level 1

      Fig(12) Energy states of Three-level active medium

      On the other hand in a four-level laser the material is pumped to level 4

      which is a fast decaying level and the atoms decay rapidly to level 3 which is

      a metastable level The stimulated emission takes place from level 3 to level 2

      from where the atoms decay back to level 1 Four level lasers is an

      improvement on a system based on three level systems In this case the laser

      transition takes place between the third and second excited states Since

      lower laser level 2 is a fast decaying level which ensures that it rapidly gets

      empty and as such always supports the population inversion condition

      Fig(13) Energy states of Four-level active medium

      4

      14 A Survey of Laser Types

      Laser technology is available to us since 1960rsquos and since then has been

      quite well developed Currently there is a great variety of lasers of different

      output power operating voltages sizes etc The major classes of lasers

      currently used are Gas Solid Molecular and Free Electron lasers Below we

      will cover some most popular representative types of lasers of each class and

      describe specific principles of operation construction and main highlights

      141 Gas Lasers

      1 Helium-Neon Laser

      The most common and inexpensive gas laser the helium-neon laser is

      usually constructed to operate in the red at 6328 nm It can also be

      constructed to produce laser action in the green at 5435 nm and in the

      infrared at 1523 nm

      One of the excited levels of helium at 2061 eV is very close to a level in

      neon at 2066 eV so close in fact that upon collision of a helium and a neon

      atom the energy can be transferred from the helium to the neon atom

      Fig (14) The components of a Hilium-Neon Laser

      5

      Fig(15) The lasing action of He-Ne laser

      Helium-Neon lasers are common in the introductory physics laboratories

      but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

      brightness equal to sunshine on a clear day (01 wattcmP

      2P) and is just as

      dangerous to stare at directly

      2- Carbon Dioxide Laser

      The carbon dioxide gas laser is capable of continuous output powers above

      10 kilowatts It is also capable of extremely high power pulse operation It

      exhibits laser action at several infrared frequencies but none in the visible

      spectrum Operating in a manner similar to the helium-neon laser it employs

      an electric discharge for pumping using a percentage of nitrogen gas as a

      pumping gas The COR2R laser is the most efficient laser capable of operating at

      more than 30 efficiency

      The carbon dioxide laser finds many applications in industry particularly for

      welding and Cutting

      6

      3- Argon Laser

      The argon ion laser can be operated as a continuous gas laser at about 25

      different wavelengths in the visible between (4089 - 6861) nm but is best

      known for its most efficient transitions in the green at 488 nm and 5145 nm

      Operating at much higher powers than the Helium-Neon gas laser it is not

      uncommon to achieve (30 ndash 100) watts of continuous power using several

      transitions This output is produced in hot plasma and takes extremely high

      power typically (9 ndash 12) kW so these are large and expensive devices

      142 Solid Lasers

      1 Ruby Laser

      The ruby laser is the first type of laser actually constructed first

      demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

      aluminum oxide with a small amount (about 005) of Chromium which gives

      it its characteristic pink or red color by absorbing green and blue light

      The ruby laser is used as a pulsed laser producing red light at 6943 nm

      After receiving a pumping flash from the flash tube the laser light emerges for

      as long as the excited atoms persist in the ruby rod which is typically about a

      millisecond

      A pulsed ruby laser was used for the famous laser ranging experiment which

      was conducted with a corner reflector placed on the Moon by the Apollo

      astronauts This determined the distance to the Moon with an accuracy of

      about 15 cm

      7

      Fig (16) Principle of operation of a Ruby laser

      2- Neodymium-YAG Laser

      An example of a solid-state laser the neodymium-YAG uses the NdP

      3+P ion to

      dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

      geometry which makes population inversion possible Neodymium-YAG lasers

      have become very important because they can be used to produce high

      powers Such lasers have been constructed to produce over a kilowatt of

      continuous laser power at 1065 nm and can achieve extremely high powers in

      a pulsed mode

      Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

      production of a series of very short pulses for research with femtosecond time

      resolution

      Fig(17) Construction of a Neodymium-YAG laser

      8

      3- Neodymium-Glass Lasers

      Neodymium glass lasers have emerged as the design choice for research in

      laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

      short as 10-12 seconds with peak powers of 109 kilowatts

      143 Molecular Lasers

      Eximer Lasers

      Eximer is a shortened form of excited dimer denoting the fact that the

      lasing medium in this type of laser is an excited diatomic molecule These

      lasers typically produce ultraviolet pulses They are under investigation for use

      in communicating with submarines by conversion to blue-green light and

      pulsing from overhead satellites through sea water to submarines below

      The eximers used are typically those formed by rare gases and halogens in

      electron excited Gas discharges Molecules like XeF are stable only in their

      excited states and quickly dissociate when they make the transition to their

      ground state This makes possible large population inversions because the

      ground state is depleted by this dissociation However the excited states are

      very short-lived compared to other laser metastable states and lasers like the

      XeF eximer laser require high pumping rates

      Eximer lasers typically produce high power pulse outputs in the blue or

      ultraviolet after excitation by fast electron-beam discharges

      The rare-gas xenon and the highly active fluorine seem unlikely to form a

      molecule but they do in the hot plasma environment of an electron-beam

      initiated gas discharge They are only stable in their excited states if stable

      can be used for molecules which undergo radioactive decay in 1 to 10

      nanoseconds This is long enough to achieve pulsed laser action in the blue-

      green over a band from 450 to 510 nm peaking at 486 nm Very high power

      9

      pulses can be achieved because the stimulated emission cross-sections of the

      laser transitions are relatively low allowing a large population inversion to

      build up The power is also enhanced by the fact that the ground state of XeF

      quickly dissociates so that there is little absorption to quench the laser pulse

      action

      144 Free-Electron Lasers

      The radiation from a free-electron laser is produced from free electrons

      which are forced to oscillate in a regular fashion by an applied field They are

      therefore more like synchrotron light sources or microwave tubes than like

      other lasers They are able to produce highly coherent collimated radiation

      over a wide range of frequencies The magnetic field arrangement which

      produces the alternating field is commonly called a wiggler magnet

      Fig(18) Principle of operation of Free-Electron laser

      The free-electron laser is a highly tunable device which has been used to

      generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

      this range they are the highest power source Applications of free-electron

      lasers are envisioned in isotope separation plasma heating for nuclear fusion

      long-range high resolution radar and particle acceleration in accelerators

      10

      15 Pulsed operation

      Pulsed operation of lasers refers to any laser not classified as continuous

      wave so that the optical power appears in pulses of some duration at some

      repetition rate This encompasses a wide range of technologies addressing a

      number of different motivations Some lasers are pulsed simply because they

      cannot be run in continuous mode

      In other cases the application requires the production of pulses having as

      large an energy as possible Since the pulse energy is equal to the average

      power divided by the repitition rate this goal can sometimes be satisfied by

      lowering the rate of pulses so that more energy can be built up in between

      pulses In laser ablation for example a small volume of material at the surface

      of a work piece can be evaporated if it is heated in a very short time whereas

      supplying the energy gradually would allow for the heat to be absorbed into

      the bulk of the piece never attaining a sufficiently high temperature at a

      particular point

      Other applications rely on the peak pulse power (rather than the energy in

      the pulse) especially in order to obtain nonlinear optical effects For a given

      pulse energy this requires creating pulses of the shortest possible duration

      utilizing techniques such as Q-switching

      16 Heat and heat capacity

      When a sample is heated meaning it receives thermal energy from an

      external source some of the introduced heat is converted into kinetic energy

      the rest to other forms of internal energy specific to the material The amount

      converted into kinetic energy causes the temperature of the material to rise

      The amount of the temperature increase depends on how much heat was

      added the size of the sample the original temperature of the sample and on

      how the heat was added The two obvious choices on how to add the heat are

      11

      to add it holding volume constant or to add it holding pressure constant

      (There may be other choices but they will not concern us)

      Lets assume for the moment that we are going to add heat to our sample

      holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

      subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

      be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

      amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

      (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

      lim119876119876119889119889rarr0

      119876119876119889119889∆120549120549119889119889

      = 120597120597119876119876120597120597120549120549119889119889

      = 119862119862119889119889 (11)

      We have given this derivative the symbol 119862119862119889119889 and we call it the heat

      capacity at constant volume Usually one quotes the molar heat capacity

      119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

      (12)

      We can rearrange Equation (11) as follows

      119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

      Then we can integrate this equation to find the heat involved in a finite

      change at constant volume

      119876119876119889119889 = 119862119862119889119889

      1205491205492

      1205491205491

      119889119889120549120549 (14)

      If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

      out of the integral and the heat at constant volume becomes

      119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

      Let us now go through the same sequence of steps except holding pressure

      constant instead of volume Our initial definition of the heat capacity at

      constant pressure 119862119862119875119875 R Rbecomes

      lim119876119876119875119875rarr0

      119876119876119875119875∆120549120549119875119875

      = 120597120597119876119876120597120597120549120549119875119875

      = 119862119862119875119875 (16)

      The analogous molar heat capacity is

      12

      119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

      (17)

      Equation (16) rearranges to

      119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

      which integrates to give

      119876119876119875119875 = 119862119862119875119875

      1205491205492

      1205491205491

      119889119889120549120549 (19)

      When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

      119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

      Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

      regarded as constant In these cases the heat capacity is fit to a polynomial (or

      similar function) in 120549120549 For example some tables give the heat capacity as

      119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

      where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

      dependent heat capacity the heat at constant pressure would integrate as

      follows

      119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

      1205491205492

      1205491205491

      119889119889120549120549 (112)

      119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

      2 minus 12054912054912 + 119899119899

      1205741205743

      12054912054923 minus 1205491205491

      3 (113)

      Occasionally one finds a different form for the temperature dependent heat

      capacity in the literature

      119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

      When you do calculations with temperature dependent heat capacities you

      must check to see which form is being used for 119862119862119875119875 We are using the

      convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

      be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

      13

      positive then heat was indeed absorbed by the system On the other hand if

      119876119876 is negative it means that the system gave up heat to the surroundings

      17 Thermal conductivity

      In physics thermal conductivity 119896119896 is the property of a material that indicates

      its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

      Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

      The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

      a piece of material The reciprocal of thermal conductivity is thermal

      resistivity

      18 Derivation in one dimension

      The heat equation is derived from Fouriers law and conservation of energy

      (Cannon 1984) By Fouriers law the flow rate of heat energy through a

      surface is proportional to the negative temperature gradient across the

      surface

      119902119902 = minus119896119896 120571120571120549120549 (115)

      where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

      dimension the gradient is an ordinary spatial derivative and so Fouriers law is

      119902119902 = minus119896119896 120549120549119909119909 (116)

      where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

      energy per unit volume in the material 120549120549119876119876 is proportional to the change in

      temperature 120549120549120549120549 That is

      ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

      where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

      material Choosing zero energy at absolute zero temperature this can be

      rewritten as

      ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

      14

      The increase in internal energy in a small spatial region of the material

      (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

      given by

      119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

      119909119909minus∆119909119909

      = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

      119889119889120577120577 119889119889120591120591119909119909+∆119909119909

      119909119909minus∆119909119909

      119905119905+∆119905119905

      119905119905minus∆119905119905

      (119)

      Where the fundamental theorem of calculus was used Additionally with no

      work done and absent any heat sources or sinks the change in internal energy

      in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

      across the boundaries By Fouriers law this is

      119896119896 120597120597120549120549120597120597119909119909

      (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

      (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

      119905119905minus∆119905119905

      = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

      119909119909+∆119909119909

      119909119909minus∆119909119909

      119905119905+∆119905119905

      119905119905minus∆119905119905

      (120)

      again by the fundamental theorem of calculus By conservation of energy

      119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

      119909119909minus∆119909119909

      119905119905+∆119905119905

      119905119905minus∆119905119905

      = 0 (121)

      This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

      Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

      Which can be rewritten as

      120549120549119905119905 =119896119896119888119888119875119875 120588120588

      120549120549119909119909119909119909 (122)

      or

      120597120597120549120549120597120597119905119905

      =119896119896119888119888119875119875 120588120588

      12059712059721205491205491205971205971199091199092 (123)

      15

      which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

      diffusivity and is often denoted 120572120572

      19 Aim of present work

      The goal of this study is to estimate the solution of partial differential

      equation that governs the laser-solid interaction using numerical methods

      The solution will been restricted into one dimensional situation in which we

      assume that both the laser power density and thermal properties are

      functions of time and temperature respectively In this project we attempt to

      investigate the laser interaction with both lead and copper materials by

      predicting the temperature gradient with the depth of the metals

      16

      Chapter Two

      Theoretical Aspects

      21 Introduction

      When a laser interacts with a solid surface a variety of processes can

      occur We are mainly interested in the interaction of pulsed lasers with a

      solid surface in first instance a metal When such a laser interacts with a

      copper surface the laser energy will be transformed into heat The

      temperature of the solid material will increase leading to melting and

      evaporation of the solid material

      The evaporated material (vapour atoms) will expand Depending on the

      applications this can happen in vacuum (or very low pressure) or in a

      background gas (helium argon air)

      22 One dimension laser heating equation

      In general the one dimension laser heating processes of opaque solid slab is

      represented as

      120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

      ( 119870119870 119879119879120597120597 ) (21)

      With boundary conditions and initial condition which represent the pre-

      vaporization stage

      minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

      minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

      17

      119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

      where

      119870119870 represents the thermal conductivity

      120588120588 represents the density

      119862119862 represents the specific heat

      119879119879 represents the temperature

      119879119879infin represents the ambient temperature

      119879119879119907119907 represents the front surface vaporization

      120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

      Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

      119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

      With the same boundary conditions as in equation (22)

      where 119889119889119889119889 = 119870119870120588120588119862119862

      which represents the thermal diffusion

      But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

      equation (21) with this assuming implies

      119879119879119905119905 = 1

      120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

      With the same boundary and initial conditions in equation (22) Where 119870119870

      represents the derivative of K with respect the temperature

      23 Numerical solution of Initial value problems

      An immense number of analytical solutions for conduction heat-transfer

      problems have been accumulated in literature over the past 100 years Even so

      in many practical situations the geometry or boundary conditions are such that an

      analytical solution has not been obtained at all or if the solution has been

      18

      developed it involves such a complex series solution that numerical evaluation

      becomes exceedingly difficult For such situation the most fruitful approach to

      the problem is numerical techniques the basic principles of which we shall

      outline in this section

      One way to guarantee accuracy in the solution of an initial values problems

      (IVP) is to solve the problem twice using step sizes h and h2 and compare

      answers at the mesh points corresponding to the larger step size But this requires

      a significant amount of computation for the smaller step size and must be

      repeated if it is determined that the agreement is not good enough

      24 Finite Difference Method

      The finite difference method is one of several techniques for obtaining

      numerical solutions to differential equations In all numerical solutions the

      continuous partial differential equation (PDE) is replaced with a discrete

      approximation In this context the word discrete means that the numerical

      solution is known only at a finite number of points in the physical domain The

      number of those points can be selected by the user of the numerical method In

      general increasing the number of points not only increases the resolution but

      also the accuracy of the numerical solution

      The discrete approximation results in a set of algebraic equations that are

      evaluated for the values of the discrete unknowns

      The mesh is the set of locations where the discrete solution is computed

      These points are called nodes and if one were to draw lines between adjacent

      nodes in the domain the resulting image would resemble a net or mesh Two key

      parameters of the mesh are ∆120597120597 the local distance between adjacent points in

      space and ∆119905119905 the local distance between adjacent time steps For the simple

      examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

      19

      The core idea of the finite-difference method is to replace continuous

      derivatives with so-called difference formulas that involve only the discrete

      values associated with positions on the mesh

      Applying the finite-difference method to a differential equation involves

      replacing all derivatives with difference formulas In the heat equation there are

      derivatives with respect to time and derivatives with respect to space Using

      different combinations of mesh points in the difference formulas results in

      different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

      numerical solution obtained with any useful scheme will approach the true

      solution to the original differential equation However the rate at which the

      numerical solution approaches the true solution varies with the scheme

      241 First Order Forward Difference

      Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

      empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

      +1205751205751205971205972

      2 1205971205972empty1205971205971205971205972

      1205971205971

      +1205751205751205971205973

      3 1205971205973empty1205971205971205971205973

      1205971205971

      + ⋯ (25)

      where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

      consider the value of empty at the location of the 120597120597119894119894+1 mesh line

      empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

      +∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120597120597119894119894

      +∆1205971205973

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      + ⋯ (26)

      Solve for (120597120597empty120597120597120597120597)120597120597119894119894

      120597120597empty120597120597120597120597120597120597119894119894

      =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

      ∆120597120597minus∆1205971205972

      1205971205972empty1205971205971205971205972

      120597120597119894119894

      minus∆1205971205972

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      minus ⋯ (27)

      Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

      hand side have been reduced by one

      20

      Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

      and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

      120597120597empty120597120597120597120597120597120597119894119894

      =empty119894119894+1 minus empty119894119894

      ∆120597120597minus∆1205971205972

      1205971205972empty1205971205971205971205972

      120597120597119894119894

      minus∆1205971205972

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      minus ⋯ (28)

      The mean value theorem can be used to replace the higher order derivatives

      ∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120597120597119894119894

      +∆1205971205973

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      + ⋯ =∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120585120585 (29)

      where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

      120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

      ∆120597120597+∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120585120585 (210)

      120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

      ∆120597120597asymp∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120585120585 (211)

      The term on the right hand side of previous equation is called the truncation

      error of the finite difference approximation

      In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

      unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

      truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

      analytical form) the big 119978119978 notation can be used to express the dependence of

      the truncation error on the mesh spacing Note that the right hand side of last

      equation contains the mesh parameter ∆120597120597 which is chosen by the person using

      the finite difference simulation Since this is the only parameter under the users

      control that determines the error the truncation error is simply written

      ∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120585120585= 119978119978(∆1205971205972) (212)

      The equals sign in this expression is true in the order of magnitude sense In

      other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

      21

      equality Rather the expression means that the left hand side is a product of an

      unknown constant and ∆1205971205972 Although the expression does not give us the exact

      magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

      approaches zero as ∆120597120597 is reduced

      Using big 119978119978 notation Equation (28) can be written

      120597120597empty120597120597120597120597120597120597119894119894

      =empty119894119894+1 minus empty119894119894

      ∆120597120597+ 119978119978(∆120597120597) (213)

      This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

      it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

      truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

      our control because we can choose the mesh size ∆120597120597 The part of the truncation

      error that is not under our control is |120597120597empty120597120597120597120597|120585120585

      242 First Order Backward Difference

      An alternative first order finite difference formula is obtained if the Taylor series

      like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

      variables in place of all the unknowns one obtains

      empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

      +∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120597120597119894119894

      minus∆1205971205973

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      + ⋯ (214)

      Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

      to get

      120597120597empty120597120597120597120597120597120597119894119894

      =empty119894119894+1 minus empty119894119894

      ∆120597120597minus∆1205971205972

      1205971205972empty1205971205971205971205972

      120597120597119894119894

      minus∆1205971205972

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      minus ⋯ (215)

      Or using big 119978119978 notation

      120597120597empty120597120597120597120597120597120597119894119894

      =empty119894119894 minus empty119894119894minus1

      ∆120597120597+ 119978119978(∆120597120597) (216)

      22

      This is called the backward difference formula because it involves the values of

      empty at 120597120597119894119894 and 120597120597119894119894minus1

      The order of magnitude of the truncation error for the backward difference

      approximation is the same as that of the forward difference approximation Can

      we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

      truncation error The answer is yes

      242 First Order Central Difference

      Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

      empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

      +∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120597120597119894119894

      +∆1205971205973

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      + ⋯ (217)

      empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

      +∆1205971205972

      2 1205971205972empty1205971205971205971205972

      120597120597119894119894

      minus∆1205971205973

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      + ⋯ (218)

      Subtracting Equation (10) from Equation (9) yields

      empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

      + 2∆1205971205973

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      + ⋯ (219)

      Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

      120597120597empty120597120597120597120597120597120597119894119894

      =empty119894119894+1 minus empty119894119894minus1

      2∆120597120597minus∆1205971205972

      3 1205971205973empty1205971205971205971205973

      120597120597119894119894

      minus ⋯ (220)

      or

      120597120597empty120597120597120597120597120597120597119894119894

      =empty119894119894+1 minus empty119894119894minus1

      2∆120597120597+ 119978119978(∆1205971205972) (221)

      This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

      approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

      the truncation error for the central difference approximation goes to zero much

      faster than the truncation error in forward and backward equations

      23

      25 Procedures

      The simple case in this investigation was assuming the constant thermal

      properties of the material First we assumed all the thermal properties of the

      materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

      point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

      assume the constant energy after that the pulse of special shapes was selected

      The numerical solution of equation (23) with boundary and initial conditions

      in equation (22) was investigated using Matlab program as shown in Appendix

      The equation of thermal conductivity and specific heat capacity of metal as a

      function of temperature was obtained by best fitting of polynomials using

      tabulated data in references

      24

      Chapter Three

      Results and Discursion

      31 Introduction

      The development of laser has been an exciting chapter in the history of

      science and engineering It has produced a new type of advice with potential for

      application in an extremely wide variety of fields Mach basic development in

      lasers were occurred during last 35 years The lasers interaction with metal and

      vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

      The status of laser development and application were still rather rudimentary

      The light emitted by laser is electro magnetic radiation this radiation has a wave

      nature the waves consists of vibrating electric and magnetic fields many studies

      have tried to find and solve models of laser interactions Some researchers

      proposed the mathematical model related to the laser - plasma interaction and

      the others have developed an analytical model to study the temperature

      distribution in Infrared optical materials heated by laser pulses Also an attempt

      have made to study the interaction of nanosecond pulsed lasers with material

      from point of view using experimental technique and theoretical approach of

      dimensional analysis

      In this study we have evaluate the solution of partial difference equation

      (PDE) that represent the laser interaction with solid situation in one dimension

      assuming that the power density of laser and thermal properties are functions

      with time and temperature respectively

      25

      32 Numerical solution with constant laser power density and constant

      thermal properties

      First we have taken the lead metal (Pb) with thermal properties

      119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

      119862119862 = 014016119869119869119892119892119870119870

      120588120588 = 10751 1198921198921198981198981198981198983

      119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

      119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

      and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

      represent the area under laser influence

      The numerical solution of equation (23) with boundary and initial conditions

      in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

      of lead metal by explicit method using Matlab program give us the results as

      shown in Fig (31)

      Fig(31) Depth dependence of the temperature with the laser power density

      1198681198680 = 76 times 106 119882119882119898119898119898119898 2

      26

      33 Evaluation of function 119920119920(119957119957) of laser flux density

      From following data that represent the energy (119869119869) with time (millie second)

      Time 0 001 01 02 03 04 05 06 07 08

      Energy 0 002 017 022 024 02 012 007 002 0

      By using Matlab program the best polynomial with deduced from above data

      was

      119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

      minus 10069 1198981198985 (31)

      As shown in Fig (32)

      Fig(32) Laser energy as a function of time

      Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

      normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

      maximum value (119864119864119898119898119907119907119898119898 )

      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

      (119864119864119898119898119907119907119898119898 ) (32)

      The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

      27

      Fig(33) Normalized laser energy as a function of time

      The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

      equal to 3 (total laser energy) ie

      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

      08

      00

      119899119899119898119898 = 3 (33)

      Therefore there exist a real number 119875119875 such that

      119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

      08

      00

      119899119899119898119898 = 3 (34)

      that implies 119875119875 = 68241 and

      119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

      08

      00

      119899119899119898119898 = 3 (35)

      The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

      08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

      119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

      00

      (36)

      28

      Where 119863119863119898119898 put to balance the units of equation (36)

      But integral

      119868119868 = 119864119864119860119860

      (37)

      and from equations (35) (36) and (37) we have

      119868119868 (119898119898)11989911989911989811989808

      00

      = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

      0800 119899119899119898119898

      119860119860 119863119863119898119898 (38)

      Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

      (38)

      There fore

      119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

      119860119860 119863119863119898119898 (39)

      As shown in Fig(34) Matlab program was used to obtain the best polynomial

      that agrees with result data

      119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

      + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

      Fig(34) Time dependence of laser intensity

      29

      34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

      constant thermal properties

      With all constant thermal properties of lead metal as in article (23) and

      119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

      equation (23) with boundary and initial condition as in equation (22) and the

      depth penetration is shown in Fig(35)

      Fig(35) Depth dependence of the temperature when laser intensity function

      of time and constant thermal properties of Lead

      35 Evaluation the Thermal Conductivity as functions of temperature

      The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

      temperature for Lead material was obtained by Matlab program using the

      experimental data tabulated in researches

      119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

      times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

      30

      119879119879 ( 119870119870) 119870119870 119869119869

      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

      300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

      The previous thermal conductivity data and the best fitting of the data are

      shown in Fig (36)

      Fig(36) The best fitting of thermal conductivity of Lead as a function of

      temperature

      31

      36 Evaluation the Specific heat as functions of temperature

      The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

      material was obtained from the following experimental data tacked from

      literatures

      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

      300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

      The best polynomial fitted for these data was

      119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

      + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

      times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

      The specific heat capacity data and the best polynomial fitting of the data are

      shown in Fig (37)

      32

      Fig(37) The best fitting of specific heat capacity of Lead as a function of

      temperature

      37 Evaluation the Density as functions of temperature

      The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

      was used to find the best polynomial fitting

      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

      300 11330 400 11230 500 11130 600 11010 800 10430

      1000 10190 1200 9940

      The best polynomial of this data was

      120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

      minus 45158 times 10minus12 1198791198794 (313)

      33

      The density of Lead as a function of temperature and the best polynomial fitting

      are shown in Fig (38)

      Fig(38) The best fitting of density of Lead as a function of temperature

      38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

      variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

      We have deduced the solution of equation (24) with initial and boundary

      condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

      and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

      respectively then by using Matlab program the depth penetration is shown in

      Fig (39)

      34

      Fig(39) Depth dependence of the temperature for pulse laser on Lead

      material

      39 Laser interaction with copper material

      The same time dependence of laser intensity as shown in Fig(34) with

      thermal properties of copper was used to calculate the temperature distribution as

      a function of depth penetration

      The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

      copper material was obtained from the experimental data tabulated in literary

      The Matlab program used to obtain the best polynomial equation that agrees

      with the above data

      119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

      minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

      times 10minus17 1198791198795 (314)

      35

      119879119879 ( 119870119870) 119870119870 119869119869

      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

      100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

      The previous thermal conductivity data and the best fitting of the data are

      shown in Fig (310)

      Fig(310) The best fitting of thermal conductivity of Copper as a function of

      temperature

      36

      The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

      material was obtained from the following experimental data tacked from

      literatures

      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

      100 0254

      200 0357

      273 0384

      298 0387

      400 0397

      600 0416

      800 0435

      1000 0454

      1100 0464

      1200 0474

      1300 0483

      The best polynomial fitted for these data was

      119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

      + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

      times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

      The specific heat capacity data and the best polynomial fitting of the data are

      shown in Fig (311)

      37

      Fig(311) The best fitting of specific heat capacity of Copper as a function of

      temperature

      The density of copper 120588120588(119879119879) as a function of temperature tacked from

      literature was used to find the best polynomial fitting

      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

      100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

      1000 8576 1100 8519 1200 8458 1300 8396

      38

      The best polynomial of this data was

      120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

      minus 76765 times 10minus14 1198791198794 (316)

      The density of copper as a function of temperature and the best polynomial

      fitting are shown in Fig (312)

      Fig(312) The best fitting of density of copper as a function of temperature

      The depth penetration of laser energy for copper metal was calculated using

      the polynomial equations of thermal conductivity specific heat capacity and

      density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

      (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

      function of time the result was shown in Fig (313)

      39

      The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

      for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

      copper metal so the depth penetration of laser energy of lead metal was smaller

      than that of copper metal this may be due to the high thermal conductivity and

      high specific heat capacity of copper with that of lead metal

      Fig(313) Depth dependence of the temperature for pulse laser on Copper

      material

      40

      310 Conclusions

      The Depth dependence of temperature for lead metal was investigated in two

      case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

      thermal properties (thermal conductivity specific heat) and density of metal are

      also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

      vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

      specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

      120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

      penetration depth in the first case is smaller than that of the second case about

      (190) times

      The temperature distribution as a function of depth dependence for copper

      metal was also investigated in the case when the laser intensity vary with time

      119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

      density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

      The depth penetration of laser energy of lead metal was found to be smaller

      than that of copper metal this may be due to the high thermal conductivity and

      high specific heat capacity of copper with that of lead metal

      41

      References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

      Sons Inc Hoboken New Jersey Canada (2003)

      [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

      [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

      [9]

      Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

      [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

      [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

      [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

      [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

      [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

      httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

      [10] httpwwwworldoflaserscomlaserprincipleshtm

      [11] httpenwikipediaorgwikiLaserPulsed_operation

      [12] httpenwikipediaorgwikiThermal_conductivity

      [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

      [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

      [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

      42

      Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

      This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

      This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

      43

      E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

      This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

      This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

      This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

      44

      title(Dencity as a function of temperature) hold off

      This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

      45

      elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

      46

      r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

      47

      for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

      48

      for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

      49

      6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

      This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

      50

      u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

      51

      alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

      52

      715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

      • 01 Title
        • 0TDr Omed Ghareb Abdullah
          • Ch1-Laser
            • 16 Heat and heat capacity
            • 18 Derivation in one dimension
              • Ch2 laser2
              • Ch3 laser2
              • Appendix-Laser

        iv

        242 First Order Backward Difference

        242 First Order Central Difference

        25 Procedures

        Chapter three Results and Discussion

        31 Introduction

        32 Numerical solution with constant laser power density and

        constant thermal properties

        33 Evaluation of function 119920119920(119957119957) of laser flux density

        34 Numerical solution with variable laser power density (

        119920119920 = 119920119920 (119957119957) ) and constant thermal properties

        35 Evaluation the Thermal Conductivity as functions of

        temperature

        36 Evaluation the Specific heat as functions of temperature

        37 Evaluation the Density as functions of temperature

        38 Numerical solution with variable laser power density (

        119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

        119914119914(119931119931)120646120646 = 120646120646(119931119931)

        39 Laser interaction with copper material

        310 Conclusions

        References

        v

        Abstract

        In recent years much effort has gone into the understanding of the

        interaction of short laser pulses with matter The present works have

        typically involved studying the interaction of high intensity laser pulses with

        high-density solid target In this study the NDYAG pulsed laser with

        maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

        laser energy with time as well as a function of laser intensity with time are

        presented in this study

        The finite difference method was used to calculate the temperature

        distribution as a function of laser depth penetration in lead and copper

        materials

        The best polynomial fits for thermal conductivity specific heat capacity

        and density of metals as a function of temperature was obtained using

        Matlab software At the first all these properties were assumed to be

        constants and then the influence of varying these properties with

        temperature was tacked in to account The temperature gradient of lead

        shows to be greater than that of copper this may be due to the high thermal

        conductivity and high specific heat capacity of copper with that of lead

        1

        Chapter One

        Basic concepts

        11 Introduction

        Laser is a mechanism for emitting light with in electromagnetic radiation

        region of the spectrum with different output intensity Max Plank published

        work in 1900 that provided the understanding that light is a form of

        electromagnetic radiation without this understanding the laser would have

        been invented The principle of the laser was first known in 1917 when Albert

        Einstein describe the theory of stimulated emission and Theodor Maiman in

        1960 invent the first laser using a lasing medium of ruby that was stimulated

        by using high energy flash of intense light

        We have four types of laser according to their gain medium which are

        (solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

        So laser is provided a controlled source of atomic and electronic excitations

        involving non equilibrium phenomena that lend themselves to processing of

        novel material and structure because laser used in wide range application in

        our life such as welding cutting drilling industrial and medical field Maiman

        and other developer of laser weapons sighting system and powerful laser for

        use in surgery and other areas where moderated powerful pinpoint source of

        heat was needed And today laser are used in corrective eye surgery and

        providing apprecise source of heat for cutting and cauterizing tissue

        12 Definition of the Laser

        The word laser is an acronym for Light Amplification by Stimulated Emission

        of Radiation The laser makes use of processes that increase or amplify light

        signals after those signals have been generated by other means These

        processes include (1) stimulated emission a natural effect that was deduced

        2

        by considerations relating to thermodynamic equilibrium and (2) optical

        feedback (present in most lasers) that is usually provided by mirrors

        Thus in its simplest form a laser consists of a gain or amplifying medium

        (where stimulated emission occurs) and a set of mirrors to feed the light back

        into the amplifier for continued growth of the developing beam as seen in

        Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

        more intense directional monochromatic and coherent Most lasers consist

        of a column of active material with a partly reflecting mirror at one end and a

        fully reflecting mirror at the other The active material can be solid (ruby

        crystal) liquid or gas (HeNe COR2R etc)

        Fig(11) Simplified schematic of typical laser

        13 Active laser medium or gain medium

        Laser medium is the heart of the laser system and is responsible for

        producing gain and subsequent generation of laser It can be a crystal solid

        liquid semiconductor or gas medium and can be pumped to a higher energy

        state The material should be of controlled purity size and shape and should

        have the suitable energy levels to support population inversion In other

        words it must have a metastable state to support stimulated emission Most

        lasers are based on 3 or 4 level energy level systems which depends on the

        lasing medium These systems are shown in Figs (12) and (13)

        3

        In case of a three-level laser the material is pumped from level 1 to level 3

        which decays rapidly to level 2 through spontaneous emission Level 2 is a

        metastable level and promotes stimulated emission from level 2 to level 1

        Fig(12) Energy states of Three-level active medium

        On the other hand in a four-level laser the material is pumped to level 4

        which is a fast decaying level and the atoms decay rapidly to level 3 which is

        a metastable level The stimulated emission takes place from level 3 to level 2

        from where the atoms decay back to level 1 Four level lasers is an

        improvement on a system based on three level systems In this case the laser

        transition takes place between the third and second excited states Since

        lower laser level 2 is a fast decaying level which ensures that it rapidly gets

        empty and as such always supports the population inversion condition

        Fig(13) Energy states of Four-level active medium

        4

        14 A Survey of Laser Types

        Laser technology is available to us since 1960rsquos and since then has been

        quite well developed Currently there is a great variety of lasers of different

        output power operating voltages sizes etc The major classes of lasers

        currently used are Gas Solid Molecular and Free Electron lasers Below we

        will cover some most popular representative types of lasers of each class and

        describe specific principles of operation construction and main highlights

        141 Gas Lasers

        1 Helium-Neon Laser

        The most common and inexpensive gas laser the helium-neon laser is

        usually constructed to operate in the red at 6328 nm It can also be

        constructed to produce laser action in the green at 5435 nm and in the

        infrared at 1523 nm

        One of the excited levels of helium at 2061 eV is very close to a level in

        neon at 2066 eV so close in fact that upon collision of a helium and a neon

        atom the energy can be transferred from the helium to the neon atom

        Fig (14) The components of a Hilium-Neon Laser

        5

        Fig(15) The lasing action of He-Ne laser

        Helium-Neon lasers are common in the introductory physics laboratories

        but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

        brightness equal to sunshine on a clear day (01 wattcmP

        2P) and is just as

        dangerous to stare at directly

        2- Carbon Dioxide Laser

        The carbon dioxide gas laser is capable of continuous output powers above

        10 kilowatts It is also capable of extremely high power pulse operation It

        exhibits laser action at several infrared frequencies but none in the visible

        spectrum Operating in a manner similar to the helium-neon laser it employs

        an electric discharge for pumping using a percentage of nitrogen gas as a

        pumping gas The COR2R laser is the most efficient laser capable of operating at

        more than 30 efficiency

        The carbon dioxide laser finds many applications in industry particularly for

        welding and Cutting

        6

        3- Argon Laser

        The argon ion laser can be operated as a continuous gas laser at about 25

        different wavelengths in the visible between (4089 - 6861) nm but is best

        known for its most efficient transitions in the green at 488 nm and 5145 nm

        Operating at much higher powers than the Helium-Neon gas laser it is not

        uncommon to achieve (30 ndash 100) watts of continuous power using several

        transitions This output is produced in hot plasma and takes extremely high

        power typically (9 ndash 12) kW so these are large and expensive devices

        142 Solid Lasers

        1 Ruby Laser

        The ruby laser is the first type of laser actually constructed first

        demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

        aluminum oxide with a small amount (about 005) of Chromium which gives

        it its characteristic pink or red color by absorbing green and blue light

        The ruby laser is used as a pulsed laser producing red light at 6943 nm

        After receiving a pumping flash from the flash tube the laser light emerges for

        as long as the excited atoms persist in the ruby rod which is typically about a

        millisecond

        A pulsed ruby laser was used for the famous laser ranging experiment which

        was conducted with a corner reflector placed on the Moon by the Apollo

        astronauts This determined the distance to the Moon with an accuracy of

        about 15 cm

        7

        Fig (16) Principle of operation of a Ruby laser

        2- Neodymium-YAG Laser

        An example of a solid-state laser the neodymium-YAG uses the NdP

        3+P ion to

        dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

        geometry which makes population inversion possible Neodymium-YAG lasers

        have become very important because they can be used to produce high

        powers Such lasers have been constructed to produce over a kilowatt of

        continuous laser power at 1065 nm and can achieve extremely high powers in

        a pulsed mode

        Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

        production of a series of very short pulses for research with femtosecond time

        resolution

        Fig(17) Construction of a Neodymium-YAG laser

        8

        3- Neodymium-Glass Lasers

        Neodymium glass lasers have emerged as the design choice for research in

        laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

        short as 10-12 seconds with peak powers of 109 kilowatts

        143 Molecular Lasers

        Eximer Lasers

        Eximer is a shortened form of excited dimer denoting the fact that the

        lasing medium in this type of laser is an excited diatomic molecule These

        lasers typically produce ultraviolet pulses They are under investigation for use

        in communicating with submarines by conversion to blue-green light and

        pulsing from overhead satellites through sea water to submarines below

        The eximers used are typically those formed by rare gases and halogens in

        electron excited Gas discharges Molecules like XeF are stable only in their

        excited states and quickly dissociate when they make the transition to their

        ground state This makes possible large population inversions because the

        ground state is depleted by this dissociation However the excited states are

        very short-lived compared to other laser metastable states and lasers like the

        XeF eximer laser require high pumping rates

        Eximer lasers typically produce high power pulse outputs in the blue or

        ultraviolet after excitation by fast electron-beam discharges

        The rare-gas xenon and the highly active fluorine seem unlikely to form a

        molecule but they do in the hot plasma environment of an electron-beam

        initiated gas discharge They are only stable in their excited states if stable

        can be used for molecules which undergo radioactive decay in 1 to 10

        nanoseconds This is long enough to achieve pulsed laser action in the blue-

        green over a band from 450 to 510 nm peaking at 486 nm Very high power

        9

        pulses can be achieved because the stimulated emission cross-sections of the

        laser transitions are relatively low allowing a large population inversion to

        build up The power is also enhanced by the fact that the ground state of XeF

        quickly dissociates so that there is little absorption to quench the laser pulse

        action

        144 Free-Electron Lasers

        The radiation from a free-electron laser is produced from free electrons

        which are forced to oscillate in a regular fashion by an applied field They are

        therefore more like synchrotron light sources or microwave tubes than like

        other lasers They are able to produce highly coherent collimated radiation

        over a wide range of frequencies The magnetic field arrangement which

        produces the alternating field is commonly called a wiggler magnet

        Fig(18) Principle of operation of Free-Electron laser

        The free-electron laser is a highly tunable device which has been used to

        generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

        this range they are the highest power source Applications of free-electron

        lasers are envisioned in isotope separation plasma heating for nuclear fusion

        long-range high resolution radar and particle acceleration in accelerators

        10

        15 Pulsed operation

        Pulsed operation of lasers refers to any laser not classified as continuous

        wave so that the optical power appears in pulses of some duration at some

        repetition rate This encompasses a wide range of technologies addressing a

        number of different motivations Some lasers are pulsed simply because they

        cannot be run in continuous mode

        In other cases the application requires the production of pulses having as

        large an energy as possible Since the pulse energy is equal to the average

        power divided by the repitition rate this goal can sometimes be satisfied by

        lowering the rate of pulses so that more energy can be built up in between

        pulses In laser ablation for example a small volume of material at the surface

        of a work piece can be evaporated if it is heated in a very short time whereas

        supplying the energy gradually would allow for the heat to be absorbed into

        the bulk of the piece never attaining a sufficiently high temperature at a

        particular point

        Other applications rely on the peak pulse power (rather than the energy in

        the pulse) especially in order to obtain nonlinear optical effects For a given

        pulse energy this requires creating pulses of the shortest possible duration

        utilizing techniques such as Q-switching

        16 Heat and heat capacity

        When a sample is heated meaning it receives thermal energy from an

        external source some of the introduced heat is converted into kinetic energy

        the rest to other forms of internal energy specific to the material The amount

        converted into kinetic energy causes the temperature of the material to rise

        The amount of the temperature increase depends on how much heat was

        added the size of the sample the original temperature of the sample and on

        how the heat was added The two obvious choices on how to add the heat are

        11

        to add it holding volume constant or to add it holding pressure constant

        (There may be other choices but they will not concern us)

        Lets assume for the moment that we are going to add heat to our sample

        holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

        subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

        be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

        amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

        (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

        lim119876119876119889119889rarr0

        119876119876119889119889∆120549120549119889119889

        = 120597120597119876119876120597120597120549120549119889119889

        = 119862119862119889119889 (11)

        We have given this derivative the symbol 119862119862119889119889 and we call it the heat

        capacity at constant volume Usually one quotes the molar heat capacity

        119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

        (12)

        We can rearrange Equation (11) as follows

        119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

        Then we can integrate this equation to find the heat involved in a finite

        change at constant volume

        119876119876119889119889 = 119862119862119889119889

        1205491205492

        1205491205491

        119889119889120549120549 (14)

        If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

        out of the integral and the heat at constant volume becomes

        119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

        Let us now go through the same sequence of steps except holding pressure

        constant instead of volume Our initial definition of the heat capacity at

        constant pressure 119862119862119875119875 R Rbecomes

        lim119876119876119875119875rarr0

        119876119876119875119875∆120549120549119875119875

        = 120597120597119876119876120597120597120549120549119875119875

        = 119862119862119875119875 (16)

        The analogous molar heat capacity is

        12

        119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

        (17)

        Equation (16) rearranges to

        119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

        which integrates to give

        119876119876119875119875 = 119862119862119875119875

        1205491205492

        1205491205491

        119889119889120549120549 (19)

        When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

        119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

        Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

        regarded as constant In these cases the heat capacity is fit to a polynomial (or

        similar function) in 120549120549 For example some tables give the heat capacity as

        119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

        where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

        dependent heat capacity the heat at constant pressure would integrate as

        follows

        119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

        1205491205492

        1205491205491

        119889119889120549120549 (112)

        119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

        2 minus 12054912054912 + 119899119899

        1205741205743

        12054912054923 minus 1205491205491

        3 (113)

        Occasionally one finds a different form for the temperature dependent heat

        capacity in the literature

        119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

        When you do calculations with temperature dependent heat capacities you

        must check to see which form is being used for 119862119862119875119875 We are using the

        convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

        be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

        13

        positive then heat was indeed absorbed by the system On the other hand if

        119876119876 is negative it means that the system gave up heat to the surroundings

        17 Thermal conductivity

        In physics thermal conductivity 119896119896 is the property of a material that indicates

        its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

        Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

        The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

        a piece of material The reciprocal of thermal conductivity is thermal

        resistivity

        18 Derivation in one dimension

        The heat equation is derived from Fouriers law and conservation of energy

        (Cannon 1984) By Fouriers law the flow rate of heat energy through a

        surface is proportional to the negative temperature gradient across the

        surface

        119902119902 = minus119896119896 120571120571120549120549 (115)

        where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

        dimension the gradient is an ordinary spatial derivative and so Fouriers law is

        119902119902 = minus119896119896 120549120549119909119909 (116)

        where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

        energy per unit volume in the material 120549120549119876119876 is proportional to the change in

        temperature 120549120549120549120549 That is

        ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

        where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

        material Choosing zero energy at absolute zero temperature this can be

        rewritten as

        ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

        14

        The increase in internal energy in a small spatial region of the material

        (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

        given by

        119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

        119909119909minus∆119909119909

        = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

        119889119889120577120577 119889119889120591120591119909119909+∆119909119909

        119909119909minus∆119909119909

        119905119905+∆119905119905

        119905119905minus∆119905119905

        (119)

        Where the fundamental theorem of calculus was used Additionally with no

        work done and absent any heat sources or sinks the change in internal energy

        in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

        across the boundaries By Fouriers law this is

        119896119896 120597120597120549120549120597120597119909119909

        (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

        (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

        119905119905minus∆119905119905

        = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

        119909119909+∆119909119909

        119909119909minus∆119909119909

        119905119905+∆119905119905

        119905119905minus∆119905119905

        (120)

        again by the fundamental theorem of calculus By conservation of energy

        119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

        119909119909minus∆119909119909

        119905119905+∆119905119905

        119905119905minus∆119905119905

        = 0 (121)

        This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

        Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

        Which can be rewritten as

        120549120549119905119905 =119896119896119888119888119875119875 120588120588

        120549120549119909119909119909119909 (122)

        or

        120597120597120549120549120597120597119905119905

        =119896119896119888119888119875119875 120588120588

        12059712059721205491205491205971205971199091199092 (123)

        15

        which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

        diffusivity and is often denoted 120572120572

        19 Aim of present work

        The goal of this study is to estimate the solution of partial differential

        equation that governs the laser-solid interaction using numerical methods

        The solution will been restricted into one dimensional situation in which we

        assume that both the laser power density and thermal properties are

        functions of time and temperature respectively In this project we attempt to

        investigate the laser interaction with both lead and copper materials by

        predicting the temperature gradient with the depth of the metals

        16

        Chapter Two

        Theoretical Aspects

        21 Introduction

        When a laser interacts with a solid surface a variety of processes can

        occur We are mainly interested in the interaction of pulsed lasers with a

        solid surface in first instance a metal When such a laser interacts with a

        copper surface the laser energy will be transformed into heat The

        temperature of the solid material will increase leading to melting and

        evaporation of the solid material

        The evaporated material (vapour atoms) will expand Depending on the

        applications this can happen in vacuum (or very low pressure) or in a

        background gas (helium argon air)

        22 One dimension laser heating equation

        In general the one dimension laser heating processes of opaque solid slab is

        represented as

        120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

        ( 119870119870 119879119879120597120597 ) (21)

        With boundary conditions and initial condition which represent the pre-

        vaporization stage

        minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

        minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

        17

        119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

        where

        119870119870 represents the thermal conductivity

        120588120588 represents the density

        119862119862 represents the specific heat

        119879119879 represents the temperature

        119879119879infin represents the ambient temperature

        119879119879119907119907 represents the front surface vaporization

        120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

        Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

        119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

        With the same boundary conditions as in equation (22)

        where 119889119889119889119889 = 119870119870120588120588119862119862

        which represents the thermal diffusion

        But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

        equation (21) with this assuming implies

        119879119879119905119905 = 1

        120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

        With the same boundary and initial conditions in equation (22) Where 119870119870

        represents the derivative of K with respect the temperature

        23 Numerical solution of Initial value problems

        An immense number of analytical solutions for conduction heat-transfer

        problems have been accumulated in literature over the past 100 years Even so

        in many practical situations the geometry or boundary conditions are such that an

        analytical solution has not been obtained at all or if the solution has been

        18

        developed it involves such a complex series solution that numerical evaluation

        becomes exceedingly difficult For such situation the most fruitful approach to

        the problem is numerical techniques the basic principles of which we shall

        outline in this section

        One way to guarantee accuracy in the solution of an initial values problems

        (IVP) is to solve the problem twice using step sizes h and h2 and compare

        answers at the mesh points corresponding to the larger step size But this requires

        a significant amount of computation for the smaller step size and must be

        repeated if it is determined that the agreement is not good enough

        24 Finite Difference Method

        The finite difference method is one of several techniques for obtaining

        numerical solutions to differential equations In all numerical solutions the

        continuous partial differential equation (PDE) is replaced with a discrete

        approximation In this context the word discrete means that the numerical

        solution is known only at a finite number of points in the physical domain The

        number of those points can be selected by the user of the numerical method In

        general increasing the number of points not only increases the resolution but

        also the accuracy of the numerical solution

        The discrete approximation results in a set of algebraic equations that are

        evaluated for the values of the discrete unknowns

        The mesh is the set of locations where the discrete solution is computed

        These points are called nodes and if one were to draw lines between adjacent

        nodes in the domain the resulting image would resemble a net or mesh Two key

        parameters of the mesh are ∆120597120597 the local distance between adjacent points in

        space and ∆119905119905 the local distance between adjacent time steps For the simple

        examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

        19

        The core idea of the finite-difference method is to replace continuous

        derivatives with so-called difference formulas that involve only the discrete

        values associated with positions on the mesh

        Applying the finite-difference method to a differential equation involves

        replacing all derivatives with difference formulas In the heat equation there are

        derivatives with respect to time and derivatives with respect to space Using

        different combinations of mesh points in the difference formulas results in

        different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

        numerical solution obtained with any useful scheme will approach the true

        solution to the original differential equation However the rate at which the

        numerical solution approaches the true solution varies with the scheme

        241 First Order Forward Difference

        Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

        empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

        +1205751205751205971205972

        2 1205971205972empty1205971205971205971205972

        1205971205971

        +1205751205751205971205973

        3 1205971205973empty1205971205971205971205973

        1205971205971

        + ⋯ (25)

        where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

        consider the value of empty at the location of the 120597120597119894119894+1 mesh line

        empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

        +∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120597120597119894119894

        +∆1205971205973

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        + ⋯ (26)

        Solve for (120597120597empty120597120597120597120597)120597120597119894119894

        120597120597empty120597120597120597120597120597120597119894119894

        =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

        ∆120597120597minus∆1205971205972

        1205971205972empty1205971205971205971205972

        120597120597119894119894

        minus∆1205971205972

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        minus ⋯ (27)

        Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

        hand side have been reduced by one

        20

        Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

        and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

        120597120597empty120597120597120597120597120597120597119894119894

        =empty119894119894+1 minus empty119894119894

        ∆120597120597minus∆1205971205972

        1205971205972empty1205971205971205971205972

        120597120597119894119894

        minus∆1205971205972

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        minus ⋯ (28)

        The mean value theorem can be used to replace the higher order derivatives

        ∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120597120597119894119894

        +∆1205971205973

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        + ⋯ =∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120585120585 (29)

        where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

        120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

        ∆120597120597+∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120585120585 (210)

        120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

        ∆120597120597asymp∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120585120585 (211)

        The term on the right hand side of previous equation is called the truncation

        error of the finite difference approximation

        In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

        unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

        truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

        analytical form) the big 119978119978 notation can be used to express the dependence of

        the truncation error on the mesh spacing Note that the right hand side of last

        equation contains the mesh parameter ∆120597120597 which is chosen by the person using

        the finite difference simulation Since this is the only parameter under the users

        control that determines the error the truncation error is simply written

        ∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120585120585= 119978119978(∆1205971205972) (212)

        The equals sign in this expression is true in the order of magnitude sense In

        other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

        21

        equality Rather the expression means that the left hand side is a product of an

        unknown constant and ∆1205971205972 Although the expression does not give us the exact

        magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

        approaches zero as ∆120597120597 is reduced

        Using big 119978119978 notation Equation (28) can be written

        120597120597empty120597120597120597120597120597120597119894119894

        =empty119894119894+1 minus empty119894119894

        ∆120597120597+ 119978119978(∆120597120597) (213)

        This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

        it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

        truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

        our control because we can choose the mesh size ∆120597120597 The part of the truncation

        error that is not under our control is |120597120597empty120597120597120597120597|120585120585

        242 First Order Backward Difference

        An alternative first order finite difference formula is obtained if the Taylor series

        like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

        variables in place of all the unknowns one obtains

        empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

        +∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120597120597119894119894

        minus∆1205971205973

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        + ⋯ (214)

        Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

        to get

        120597120597empty120597120597120597120597120597120597119894119894

        =empty119894119894+1 minus empty119894119894

        ∆120597120597minus∆1205971205972

        1205971205972empty1205971205971205971205972

        120597120597119894119894

        minus∆1205971205972

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        minus ⋯ (215)

        Or using big 119978119978 notation

        120597120597empty120597120597120597120597120597120597119894119894

        =empty119894119894 minus empty119894119894minus1

        ∆120597120597+ 119978119978(∆120597120597) (216)

        22

        This is called the backward difference formula because it involves the values of

        empty at 120597120597119894119894 and 120597120597119894119894minus1

        The order of magnitude of the truncation error for the backward difference

        approximation is the same as that of the forward difference approximation Can

        we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

        truncation error The answer is yes

        242 First Order Central Difference

        Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

        empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

        +∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120597120597119894119894

        +∆1205971205973

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        + ⋯ (217)

        empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

        +∆1205971205972

        2 1205971205972empty1205971205971205971205972

        120597120597119894119894

        minus∆1205971205973

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        + ⋯ (218)

        Subtracting Equation (10) from Equation (9) yields

        empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

        + 2∆1205971205973

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        + ⋯ (219)

        Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

        120597120597empty120597120597120597120597120597120597119894119894

        =empty119894119894+1 minus empty119894119894minus1

        2∆120597120597minus∆1205971205972

        3 1205971205973empty1205971205971205971205973

        120597120597119894119894

        minus ⋯ (220)

        or

        120597120597empty120597120597120597120597120597120597119894119894

        =empty119894119894+1 minus empty119894119894minus1

        2∆120597120597+ 119978119978(∆1205971205972) (221)

        This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

        approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

        the truncation error for the central difference approximation goes to zero much

        faster than the truncation error in forward and backward equations

        23

        25 Procedures

        The simple case in this investigation was assuming the constant thermal

        properties of the material First we assumed all the thermal properties of the

        materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

        point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

        assume the constant energy after that the pulse of special shapes was selected

        The numerical solution of equation (23) with boundary and initial conditions

        in equation (22) was investigated using Matlab program as shown in Appendix

        The equation of thermal conductivity and specific heat capacity of metal as a

        function of temperature was obtained by best fitting of polynomials using

        tabulated data in references

        24

        Chapter Three

        Results and Discursion

        31 Introduction

        The development of laser has been an exciting chapter in the history of

        science and engineering It has produced a new type of advice with potential for

        application in an extremely wide variety of fields Mach basic development in

        lasers were occurred during last 35 years The lasers interaction with metal and

        vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

        The status of laser development and application were still rather rudimentary

        The light emitted by laser is electro magnetic radiation this radiation has a wave

        nature the waves consists of vibrating electric and magnetic fields many studies

        have tried to find and solve models of laser interactions Some researchers

        proposed the mathematical model related to the laser - plasma interaction and

        the others have developed an analytical model to study the temperature

        distribution in Infrared optical materials heated by laser pulses Also an attempt

        have made to study the interaction of nanosecond pulsed lasers with material

        from point of view using experimental technique and theoretical approach of

        dimensional analysis

        In this study we have evaluate the solution of partial difference equation

        (PDE) that represent the laser interaction with solid situation in one dimension

        assuming that the power density of laser and thermal properties are functions

        with time and temperature respectively

        25

        32 Numerical solution with constant laser power density and constant

        thermal properties

        First we have taken the lead metal (Pb) with thermal properties

        119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

        119862119862 = 014016119869119869119892119892119870119870

        120588120588 = 10751 1198921198921198981198981198981198983

        119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

        119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

        and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

        represent the area under laser influence

        The numerical solution of equation (23) with boundary and initial conditions

        in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

        of lead metal by explicit method using Matlab program give us the results as

        shown in Fig (31)

        Fig(31) Depth dependence of the temperature with the laser power density

        1198681198680 = 76 times 106 119882119882119898119898119898119898 2

        26

        33 Evaluation of function 119920119920(119957119957) of laser flux density

        From following data that represent the energy (119869119869) with time (millie second)

        Time 0 001 01 02 03 04 05 06 07 08

        Energy 0 002 017 022 024 02 012 007 002 0

        By using Matlab program the best polynomial with deduced from above data

        was

        119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

        minus 10069 1198981198985 (31)

        As shown in Fig (32)

        Fig(32) Laser energy as a function of time

        Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

        normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

        maximum value (119864119864119898119898119907119907119898119898 )

        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

        (119864119864119898119898119907119907119898119898 ) (32)

        The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

        27

        Fig(33) Normalized laser energy as a function of time

        The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

        equal to 3 (total laser energy) ie

        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

        08

        00

        119899119899119898119898 = 3 (33)

        Therefore there exist a real number 119875119875 such that

        119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

        08

        00

        119899119899119898119898 = 3 (34)

        that implies 119875119875 = 68241 and

        119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

        08

        00

        119899119899119898119898 = 3 (35)

        The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

        08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

        119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

        00

        (36)

        28

        Where 119863119863119898119898 put to balance the units of equation (36)

        But integral

        119868119868 = 119864119864119860119860

        (37)

        and from equations (35) (36) and (37) we have

        119868119868 (119898119898)11989911989911989811989808

        00

        = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

        0800 119899119899119898119898

        119860119860 119863119863119898119898 (38)

        Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

        (38)

        There fore

        119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

        119860119860 119863119863119898119898 (39)

        As shown in Fig(34) Matlab program was used to obtain the best polynomial

        that agrees with result data

        119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

        + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

        Fig(34) Time dependence of laser intensity

        29

        34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

        constant thermal properties

        With all constant thermal properties of lead metal as in article (23) and

        119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

        equation (23) with boundary and initial condition as in equation (22) and the

        depth penetration is shown in Fig(35)

        Fig(35) Depth dependence of the temperature when laser intensity function

        of time and constant thermal properties of Lead

        35 Evaluation the Thermal Conductivity as functions of temperature

        The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

        temperature for Lead material was obtained by Matlab program using the

        experimental data tabulated in researches

        119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

        times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

        30

        119879119879 ( 119870119870) 119870119870 119869119869

        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

        300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

        The previous thermal conductivity data and the best fitting of the data are

        shown in Fig (36)

        Fig(36) The best fitting of thermal conductivity of Lead as a function of

        temperature

        31

        36 Evaluation the Specific heat as functions of temperature

        The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

        material was obtained from the following experimental data tacked from

        literatures

        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

        300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

        The best polynomial fitted for these data was

        119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

        + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

        times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

        The specific heat capacity data and the best polynomial fitting of the data are

        shown in Fig (37)

        32

        Fig(37) The best fitting of specific heat capacity of Lead as a function of

        temperature

        37 Evaluation the Density as functions of temperature

        The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

        was used to find the best polynomial fitting

        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

        300 11330 400 11230 500 11130 600 11010 800 10430

        1000 10190 1200 9940

        The best polynomial of this data was

        120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

        minus 45158 times 10minus12 1198791198794 (313)

        33

        The density of Lead as a function of temperature and the best polynomial fitting

        are shown in Fig (38)

        Fig(38) The best fitting of density of Lead as a function of temperature

        38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

        variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

        We have deduced the solution of equation (24) with initial and boundary

        condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

        and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

        respectively then by using Matlab program the depth penetration is shown in

        Fig (39)

        34

        Fig(39) Depth dependence of the temperature for pulse laser on Lead

        material

        39 Laser interaction with copper material

        The same time dependence of laser intensity as shown in Fig(34) with

        thermal properties of copper was used to calculate the temperature distribution as

        a function of depth penetration

        The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

        copper material was obtained from the experimental data tabulated in literary

        The Matlab program used to obtain the best polynomial equation that agrees

        with the above data

        119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

        minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

        times 10minus17 1198791198795 (314)

        35

        119879119879 ( 119870119870) 119870119870 119869119869

        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

        100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

        The previous thermal conductivity data and the best fitting of the data are

        shown in Fig (310)

        Fig(310) The best fitting of thermal conductivity of Copper as a function of

        temperature

        36

        The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

        material was obtained from the following experimental data tacked from

        literatures

        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

        100 0254

        200 0357

        273 0384

        298 0387

        400 0397

        600 0416

        800 0435

        1000 0454

        1100 0464

        1200 0474

        1300 0483

        The best polynomial fitted for these data was

        119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

        + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

        times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

        The specific heat capacity data and the best polynomial fitting of the data are

        shown in Fig (311)

        37

        Fig(311) The best fitting of specific heat capacity of Copper as a function of

        temperature

        The density of copper 120588120588(119879119879) as a function of temperature tacked from

        literature was used to find the best polynomial fitting

        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

        100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

        1000 8576 1100 8519 1200 8458 1300 8396

        38

        The best polynomial of this data was

        120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

        minus 76765 times 10minus14 1198791198794 (316)

        The density of copper as a function of temperature and the best polynomial

        fitting are shown in Fig (312)

        Fig(312) The best fitting of density of copper as a function of temperature

        The depth penetration of laser energy for copper metal was calculated using

        the polynomial equations of thermal conductivity specific heat capacity and

        density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

        (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

        function of time the result was shown in Fig (313)

        39

        The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

        for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

        copper metal so the depth penetration of laser energy of lead metal was smaller

        than that of copper metal this may be due to the high thermal conductivity and

        high specific heat capacity of copper with that of lead metal

        Fig(313) Depth dependence of the temperature for pulse laser on Copper

        material

        40

        310 Conclusions

        The Depth dependence of temperature for lead metal was investigated in two

        case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

        thermal properties (thermal conductivity specific heat) and density of metal are

        also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

        vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

        specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

        120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

        penetration depth in the first case is smaller than that of the second case about

        (190) times

        The temperature distribution as a function of depth dependence for copper

        metal was also investigated in the case when the laser intensity vary with time

        119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

        density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

        The depth penetration of laser energy of lead metal was found to be smaller

        than that of copper metal this may be due to the high thermal conductivity and

        high specific heat capacity of copper with that of lead metal

        41

        References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

        Sons Inc Hoboken New Jersey Canada (2003)

        [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

        [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

        [9]

        Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

        [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

        [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

        [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

        [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

        [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

        httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

        [10] httpwwwworldoflaserscomlaserprincipleshtm

        [11] httpenwikipediaorgwikiLaserPulsed_operation

        [12] httpenwikipediaorgwikiThermal_conductivity

        [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

        [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

        [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

        42

        Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

        This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

        This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

        43

        E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

        This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

        This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

        This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

        44

        title(Dencity as a function of temperature) hold off

        This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

        45

        elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

        46

        r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

        47

        for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

        48

        for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

        49

        6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

        This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

        50

        u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

        51

        alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

        52

        715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

        • 01 Title
          • 0TDr Omed Ghareb Abdullah
            • Ch1-Laser
              • 16 Heat and heat capacity
              • 18 Derivation in one dimension
                • Ch2 laser2
                • Ch3 laser2
                • Appendix-Laser

          v

          Abstract

          In recent years much effort has gone into the understanding of the

          interaction of short laser pulses with matter The present works have

          typically involved studying the interaction of high intensity laser pulses with

          high-density solid target In this study the NDYAG pulsed laser with

          maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

          laser energy with time as well as a function of laser intensity with time are

          presented in this study

          The finite difference method was used to calculate the temperature

          distribution as a function of laser depth penetration in lead and copper

          materials

          The best polynomial fits for thermal conductivity specific heat capacity

          and density of metals as a function of temperature was obtained using

          Matlab software At the first all these properties were assumed to be

          constants and then the influence of varying these properties with

          temperature was tacked in to account The temperature gradient of lead

          shows to be greater than that of copper this may be due to the high thermal

          conductivity and high specific heat capacity of copper with that of lead

          1

          Chapter One

          Basic concepts

          11 Introduction

          Laser is a mechanism for emitting light with in electromagnetic radiation

          region of the spectrum with different output intensity Max Plank published

          work in 1900 that provided the understanding that light is a form of

          electromagnetic radiation without this understanding the laser would have

          been invented The principle of the laser was first known in 1917 when Albert

          Einstein describe the theory of stimulated emission and Theodor Maiman in

          1960 invent the first laser using a lasing medium of ruby that was stimulated

          by using high energy flash of intense light

          We have four types of laser according to their gain medium which are

          (solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

          So laser is provided a controlled source of atomic and electronic excitations

          involving non equilibrium phenomena that lend themselves to processing of

          novel material and structure because laser used in wide range application in

          our life such as welding cutting drilling industrial and medical field Maiman

          and other developer of laser weapons sighting system and powerful laser for

          use in surgery and other areas where moderated powerful pinpoint source of

          heat was needed And today laser are used in corrective eye surgery and

          providing apprecise source of heat for cutting and cauterizing tissue

          12 Definition of the Laser

          The word laser is an acronym for Light Amplification by Stimulated Emission

          of Radiation The laser makes use of processes that increase or amplify light

          signals after those signals have been generated by other means These

          processes include (1) stimulated emission a natural effect that was deduced

          2

          by considerations relating to thermodynamic equilibrium and (2) optical

          feedback (present in most lasers) that is usually provided by mirrors

          Thus in its simplest form a laser consists of a gain or amplifying medium

          (where stimulated emission occurs) and a set of mirrors to feed the light back

          into the amplifier for continued growth of the developing beam as seen in

          Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

          more intense directional monochromatic and coherent Most lasers consist

          of a column of active material with a partly reflecting mirror at one end and a

          fully reflecting mirror at the other The active material can be solid (ruby

          crystal) liquid or gas (HeNe COR2R etc)

          Fig(11) Simplified schematic of typical laser

          13 Active laser medium or gain medium

          Laser medium is the heart of the laser system and is responsible for

          producing gain and subsequent generation of laser It can be a crystal solid

          liquid semiconductor or gas medium and can be pumped to a higher energy

          state The material should be of controlled purity size and shape and should

          have the suitable energy levels to support population inversion In other

          words it must have a metastable state to support stimulated emission Most

          lasers are based on 3 or 4 level energy level systems which depends on the

          lasing medium These systems are shown in Figs (12) and (13)

          3

          In case of a three-level laser the material is pumped from level 1 to level 3

          which decays rapidly to level 2 through spontaneous emission Level 2 is a

          metastable level and promotes stimulated emission from level 2 to level 1

          Fig(12) Energy states of Three-level active medium

          On the other hand in a four-level laser the material is pumped to level 4

          which is a fast decaying level and the atoms decay rapidly to level 3 which is

          a metastable level The stimulated emission takes place from level 3 to level 2

          from where the atoms decay back to level 1 Four level lasers is an

          improvement on a system based on three level systems In this case the laser

          transition takes place between the third and second excited states Since

          lower laser level 2 is a fast decaying level which ensures that it rapidly gets

          empty and as such always supports the population inversion condition

          Fig(13) Energy states of Four-level active medium

          4

          14 A Survey of Laser Types

          Laser technology is available to us since 1960rsquos and since then has been

          quite well developed Currently there is a great variety of lasers of different

          output power operating voltages sizes etc The major classes of lasers

          currently used are Gas Solid Molecular and Free Electron lasers Below we

          will cover some most popular representative types of lasers of each class and

          describe specific principles of operation construction and main highlights

          141 Gas Lasers

          1 Helium-Neon Laser

          The most common and inexpensive gas laser the helium-neon laser is

          usually constructed to operate in the red at 6328 nm It can also be

          constructed to produce laser action in the green at 5435 nm and in the

          infrared at 1523 nm

          One of the excited levels of helium at 2061 eV is very close to a level in

          neon at 2066 eV so close in fact that upon collision of a helium and a neon

          atom the energy can be transferred from the helium to the neon atom

          Fig (14) The components of a Hilium-Neon Laser

          5

          Fig(15) The lasing action of He-Ne laser

          Helium-Neon lasers are common in the introductory physics laboratories

          but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

          brightness equal to sunshine on a clear day (01 wattcmP

          2P) and is just as

          dangerous to stare at directly

          2- Carbon Dioxide Laser

          The carbon dioxide gas laser is capable of continuous output powers above

          10 kilowatts It is also capable of extremely high power pulse operation It

          exhibits laser action at several infrared frequencies but none in the visible

          spectrum Operating in a manner similar to the helium-neon laser it employs

          an electric discharge for pumping using a percentage of nitrogen gas as a

          pumping gas The COR2R laser is the most efficient laser capable of operating at

          more than 30 efficiency

          The carbon dioxide laser finds many applications in industry particularly for

          welding and Cutting

          6

          3- Argon Laser

          The argon ion laser can be operated as a continuous gas laser at about 25

          different wavelengths in the visible between (4089 - 6861) nm but is best

          known for its most efficient transitions in the green at 488 nm and 5145 nm

          Operating at much higher powers than the Helium-Neon gas laser it is not

          uncommon to achieve (30 ndash 100) watts of continuous power using several

          transitions This output is produced in hot plasma and takes extremely high

          power typically (9 ndash 12) kW so these are large and expensive devices

          142 Solid Lasers

          1 Ruby Laser

          The ruby laser is the first type of laser actually constructed first

          demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

          aluminum oxide with a small amount (about 005) of Chromium which gives

          it its characteristic pink or red color by absorbing green and blue light

          The ruby laser is used as a pulsed laser producing red light at 6943 nm

          After receiving a pumping flash from the flash tube the laser light emerges for

          as long as the excited atoms persist in the ruby rod which is typically about a

          millisecond

          A pulsed ruby laser was used for the famous laser ranging experiment which

          was conducted with a corner reflector placed on the Moon by the Apollo

          astronauts This determined the distance to the Moon with an accuracy of

          about 15 cm

          7

          Fig (16) Principle of operation of a Ruby laser

          2- Neodymium-YAG Laser

          An example of a solid-state laser the neodymium-YAG uses the NdP

          3+P ion to

          dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

          geometry which makes population inversion possible Neodymium-YAG lasers

          have become very important because they can be used to produce high

          powers Such lasers have been constructed to produce over a kilowatt of

          continuous laser power at 1065 nm and can achieve extremely high powers in

          a pulsed mode

          Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

          production of a series of very short pulses for research with femtosecond time

          resolution

          Fig(17) Construction of a Neodymium-YAG laser

          8

          3- Neodymium-Glass Lasers

          Neodymium glass lasers have emerged as the design choice for research in

          laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

          short as 10-12 seconds with peak powers of 109 kilowatts

          143 Molecular Lasers

          Eximer Lasers

          Eximer is a shortened form of excited dimer denoting the fact that the

          lasing medium in this type of laser is an excited diatomic molecule These

          lasers typically produce ultraviolet pulses They are under investigation for use

          in communicating with submarines by conversion to blue-green light and

          pulsing from overhead satellites through sea water to submarines below

          The eximers used are typically those formed by rare gases and halogens in

          electron excited Gas discharges Molecules like XeF are stable only in their

          excited states and quickly dissociate when they make the transition to their

          ground state This makes possible large population inversions because the

          ground state is depleted by this dissociation However the excited states are

          very short-lived compared to other laser metastable states and lasers like the

          XeF eximer laser require high pumping rates

          Eximer lasers typically produce high power pulse outputs in the blue or

          ultraviolet after excitation by fast electron-beam discharges

          The rare-gas xenon and the highly active fluorine seem unlikely to form a

          molecule but they do in the hot plasma environment of an electron-beam

          initiated gas discharge They are only stable in their excited states if stable

          can be used for molecules which undergo radioactive decay in 1 to 10

          nanoseconds This is long enough to achieve pulsed laser action in the blue-

          green over a band from 450 to 510 nm peaking at 486 nm Very high power

          9

          pulses can be achieved because the stimulated emission cross-sections of the

          laser transitions are relatively low allowing a large population inversion to

          build up The power is also enhanced by the fact that the ground state of XeF

          quickly dissociates so that there is little absorption to quench the laser pulse

          action

          144 Free-Electron Lasers

          The radiation from a free-electron laser is produced from free electrons

          which are forced to oscillate in a regular fashion by an applied field They are

          therefore more like synchrotron light sources or microwave tubes than like

          other lasers They are able to produce highly coherent collimated radiation

          over a wide range of frequencies The magnetic field arrangement which

          produces the alternating field is commonly called a wiggler magnet

          Fig(18) Principle of operation of Free-Electron laser

          The free-electron laser is a highly tunable device which has been used to

          generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

          this range they are the highest power source Applications of free-electron

          lasers are envisioned in isotope separation plasma heating for nuclear fusion

          long-range high resolution radar and particle acceleration in accelerators

          10

          15 Pulsed operation

          Pulsed operation of lasers refers to any laser not classified as continuous

          wave so that the optical power appears in pulses of some duration at some

          repetition rate This encompasses a wide range of technologies addressing a

          number of different motivations Some lasers are pulsed simply because they

          cannot be run in continuous mode

          In other cases the application requires the production of pulses having as

          large an energy as possible Since the pulse energy is equal to the average

          power divided by the repitition rate this goal can sometimes be satisfied by

          lowering the rate of pulses so that more energy can be built up in between

          pulses In laser ablation for example a small volume of material at the surface

          of a work piece can be evaporated if it is heated in a very short time whereas

          supplying the energy gradually would allow for the heat to be absorbed into

          the bulk of the piece never attaining a sufficiently high temperature at a

          particular point

          Other applications rely on the peak pulse power (rather than the energy in

          the pulse) especially in order to obtain nonlinear optical effects For a given

          pulse energy this requires creating pulses of the shortest possible duration

          utilizing techniques such as Q-switching

          16 Heat and heat capacity

          When a sample is heated meaning it receives thermal energy from an

          external source some of the introduced heat is converted into kinetic energy

          the rest to other forms of internal energy specific to the material The amount

          converted into kinetic energy causes the temperature of the material to rise

          The amount of the temperature increase depends on how much heat was

          added the size of the sample the original temperature of the sample and on

          how the heat was added The two obvious choices on how to add the heat are

          11

          to add it holding volume constant or to add it holding pressure constant

          (There may be other choices but they will not concern us)

          Lets assume for the moment that we are going to add heat to our sample

          holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

          subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

          be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

          amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

          (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

          lim119876119876119889119889rarr0

          119876119876119889119889∆120549120549119889119889

          = 120597120597119876119876120597120597120549120549119889119889

          = 119862119862119889119889 (11)

          We have given this derivative the symbol 119862119862119889119889 and we call it the heat

          capacity at constant volume Usually one quotes the molar heat capacity

          119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

          (12)

          We can rearrange Equation (11) as follows

          119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

          Then we can integrate this equation to find the heat involved in a finite

          change at constant volume

          119876119876119889119889 = 119862119862119889119889

          1205491205492

          1205491205491

          119889119889120549120549 (14)

          If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

          out of the integral and the heat at constant volume becomes

          119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

          Let us now go through the same sequence of steps except holding pressure

          constant instead of volume Our initial definition of the heat capacity at

          constant pressure 119862119862119875119875 R Rbecomes

          lim119876119876119875119875rarr0

          119876119876119875119875∆120549120549119875119875

          = 120597120597119876119876120597120597120549120549119875119875

          = 119862119862119875119875 (16)

          The analogous molar heat capacity is

          12

          119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

          (17)

          Equation (16) rearranges to

          119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

          which integrates to give

          119876119876119875119875 = 119862119862119875119875

          1205491205492

          1205491205491

          119889119889120549120549 (19)

          When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

          119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

          Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

          regarded as constant In these cases the heat capacity is fit to a polynomial (or

          similar function) in 120549120549 For example some tables give the heat capacity as

          119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

          where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

          dependent heat capacity the heat at constant pressure would integrate as

          follows

          119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

          1205491205492

          1205491205491

          119889119889120549120549 (112)

          119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

          2 minus 12054912054912 + 119899119899

          1205741205743

          12054912054923 minus 1205491205491

          3 (113)

          Occasionally one finds a different form for the temperature dependent heat

          capacity in the literature

          119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

          When you do calculations with temperature dependent heat capacities you

          must check to see which form is being used for 119862119862119875119875 We are using the

          convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

          be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

          13

          positive then heat was indeed absorbed by the system On the other hand if

          119876119876 is negative it means that the system gave up heat to the surroundings

          17 Thermal conductivity

          In physics thermal conductivity 119896119896 is the property of a material that indicates

          its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

          Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

          The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

          a piece of material The reciprocal of thermal conductivity is thermal

          resistivity

          18 Derivation in one dimension

          The heat equation is derived from Fouriers law and conservation of energy

          (Cannon 1984) By Fouriers law the flow rate of heat energy through a

          surface is proportional to the negative temperature gradient across the

          surface

          119902119902 = minus119896119896 120571120571120549120549 (115)

          where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

          dimension the gradient is an ordinary spatial derivative and so Fouriers law is

          119902119902 = minus119896119896 120549120549119909119909 (116)

          where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

          energy per unit volume in the material 120549120549119876119876 is proportional to the change in

          temperature 120549120549120549120549 That is

          ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

          where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

          material Choosing zero energy at absolute zero temperature this can be

          rewritten as

          ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

          14

          The increase in internal energy in a small spatial region of the material

          (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

          given by

          119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

          119909119909minus∆119909119909

          = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

          119889119889120577120577 119889119889120591120591119909119909+∆119909119909

          119909119909minus∆119909119909

          119905119905+∆119905119905

          119905119905minus∆119905119905

          (119)

          Where the fundamental theorem of calculus was used Additionally with no

          work done and absent any heat sources or sinks the change in internal energy

          in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

          across the boundaries By Fouriers law this is

          119896119896 120597120597120549120549120597120597119909119909

          (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

          (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

          119905119905minus∆119905119905

          = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

          119909119909+∆119909119909

          119909119909minus∆119909119909

          119905119905+∆119905119905

          119905119905minus∆119905119905

          (120)

          again by the fundamental theorem of calculus By conservation of energy

          119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

          119909119909minus∆119909119909

          119905119905+∆119905119905

          119905119905minus∆119905119905

          = 0 (121)

          This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

          Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

          Which can be rewritten as

          120549120549119905119905 =119896119896119888119888119875119875 120588120588

          120549120549119909119909119909119909 (122)

          or

          120597120597120549120549120597120597119905119905

          =119896119896119888119888119875119875 120588120588

          12059712059721205491205491205971205971199091199092 (123)

          15

          which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

          diffusivity and is often denoted 120572120572

          19 Aim of present work

          The goal of this study is to estimate the solution of partial differential

          equation that governs the laser-solid interaction using numerical methods

          The solution will been restricted into one dimensional situation in which we

          assume that both the laser power density and thermal properties are

          functions of time and temperature respectively In this project we attempt to

          investigate the laser interaction with both lead and copper materials by

          predicting the temperature gradient with the depth of the metals

          16

          Chapter Two

          Theoretical Aspects

          21 Introduction

          When a laser interacts with a solid surface a variety of processes can

          occur We are mainly interested in the interaction of pulsed lasers with a

          solid surface in first instance a metal When such a laser interacts with a

          copper surface the laser energy will be transformed into heat The

          temperature of the solid material will increase leading to melting and

          evaporation of the solid material

          The evaporated material (vapour atoms) will expand Depending on the

          applications this can happen in vacuum (or very low pressure) or in a

          background gas (helium argon air)

          22 One dimension laser heating equation

          In general the one dimension laser heating processes of opaque solid slab is

          represented as

          120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

          ( 119870119870 119879119879120597120597 ) (21)

          With boundary conditions and initial condition which represent the pre-

          vaporization stage

          minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

          minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

          17

          119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

          where

          119870119870 represents the thermal conductivity

          120588120588 represents the density

          119862119862 represents the specific heat

          119879119879 represents the temperature

          119879119879infin represents the ambient temperature

          119879119879119907119907 represents the front surface vaporization

          120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

          Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

          119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

          With the same boundary conditions as in equation (22)

          where 119889119889119889119889 = 119870119870120588120588119862119862

          which represents the thermal diffusion

          But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

          equation (21) with this assuming implies

          119879119879119905119905 = 1

          120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

          With the same boundary and initial conditions in equation (22) Where 119870119870

          represents the derivative of K with respect the temperature

          23 Numerical solution of Initial value problems

          An immense number of analytical solutions for conduction heat-transfer

          problems have been accumulated in literature over the past 100 years Even so

          in many practical situations the geometry or boundary conditions are such that an

          analytical solution has not been obtained at all or if the solution has been

          18

          developed it involves such a complex series solution that numerical evaluation

          becomes exceedingly difficult For such situation the most fruitful approach to

          the problem is numerical techniques the basic principles of which we shall

          outline in this section

          One way to guarantee accuracy in the solution of an initial values problems

          (IVP) is to solve the problem twice using step sizes h and h2 and compare

          answers at the mesh points corresponding to the larger step size But this requires

          a significant amount of computation for the smaller step size and must be

          repeated if it is determined that the agreement is not good enough

          24 Finite Difference Method

          The finite difference method is one of several techniques for obtaining

          numerical solutions to differential equations In all numerical solutions the

          continuous partial differential equation (PDE) is replaced with a discrete

          approximation In this context the word discrete means that the numerical

          solution is known only at a finite number of points in the physical domain The

          number of those points can be selected by the user of the numerical method In

          general increasing the number of points not only increases the resolution but

          also the accuracy of the numerical solution

          The discrete approximation results in a set of algebraic equations that are

          evaluated for the values of the discrete unknowns

          The mesh is the set of locations where the discrete solution is computed

          These points are called nodes and if one were to draw lines between adjacent

          nodes in the domain the resulting image would resemble a net or mesh Two key

          parameters of the mesh are ∆120597120597 the local distance between adjacent points in

          space and ∆119905119905 the local distance between adjacent time steps For the simple

          examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

          19

          The core idea of the finite-difference method is to replace continuous

          derivatives with so-called difference formulas that involve only the discrete

          values associated with positions on the mesh

          Applying the finite-difference method to a differential equation involves

          replacing all derivatives with difference formulas In the heat equation there are

          derivatives with respect to time and derivatives with respect to space Using

          different combinations of mesh points in the difference formulas results in

          different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

          numerical solution obtained with any useful scheme will approach the true

          solution to the original differential equation However the rate at which the

          numerical solution approaches the true solution varies with the scheme

          241 First Order Forward Difference

          Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

          empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

          +1205751205751205971205972

          2 1205971205972empty1205971205971205971205972

          1205971205971

          +1205751205751205971205973

          3 1205971205973empty1205971205971205971205973

          1205971205971

          + ⋯ (25)

          where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

          consider the value of empty at the location of the 120597120597119894119894+1 mesh line

          empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

          +∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120597120597119894119894

          +∆1205971205973

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          + ⋯ (26)

          Solve for (120597120597empty120597120597120597120597)120597120597119894119894

          120597120597empty120597120597120597120597120597120597119894119894

          =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

          ∆120597120597minus∆1205971205972

          1205971205972empty1205971205971205971205972

          120597120597119894119894

          minus∆1205971205972

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          minus ⋯ (27)

          Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

          hand side have been reduced by one

          20

          Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

          and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

          120597120597empty120597120597120597120597120597120597119894119894

          =empty119894119894+1 minus empty119894119894

          ∆120597120597minus∆1205971205972

          1205971205972empty1205971205971205971205972

          120597120597119894119894

          minus∆1205971205972

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          minus ⋯ (28)

          The mean value theorem can be used to replace the higher order derivatives

          ∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120597120597119894119894

          +∆1205971205973

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          + ⋯ =∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120585120585 (29)

          where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

          120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

          ∆120597120597+∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120585120585 (210)

          120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

          ∆120597120597asymp∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120585120585 (211)

          The term on the right hand side of previous equation is called the truncation

          error of the finite difference approximation

          In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

          unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

          truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

          analytical form) the big 119978119978 notation can be used to express the dependence of

          the truncation error on the mesh spacing Note that the right hand side of last

          equation contains the mesh parameter ∆120597120597 which is chosen by the person using

          the finite difference simulation Since this is the only parameter under the users

          control that determines the error the truncation error is simply written

          ∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120585120585= 119978119978(∆1205971205972) (212)

          The equals sign in this expression is true in the order of magnitude sense In

          other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

          21

          equality Rather the expression means that the left hand side is a product of an

          unknown constant and ∆1205971205972 Although the expression does not give us the exact

          magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

          approaches zero as ∆120597120597 is reduced

          Using big 119978119978 notation Equation (28) can be written

          120597120597empty120597120597120597120597120597120597119894119894

          =empty119894119894+1 minus empty119894119894

          ∆120597120597+ 119978119978(∆120597120597) (213)

          This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

          it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

          truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

          our control because we can choose the mesh size ∆120597120597 The part of the truncation

          error that is not under our control is |120597120597empty120597120597120597120597|120585120585

          242 First Order Backward Difference

          An alternative first order finite difference formula is obtained if the Taylor series

          like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

          variables in place of all the unknowns one obtains

          empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

          +∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120597120597119894119894

          minus∆1205971205973

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          + ⋯ (214)

          Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

          to get

          120597120597empty120597120597120597120597120597120597119894119894

          =empty119894119894+1 minus empty119894119894

          ∆120597120597minus∆1205971205972

          1205971205972empty1205971205971205971205972

          120597120597119894119894

          minus∆1205971205972

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          minus ⋯ (215)

          Or using big 119978119978 notation

          120597120597empty120597120597120597120597120597120597119894119894

          =empty119894119894 minus empty119894119894minus1

          ∆120597120597+ 119978119978(∆120597120597) (216)

          22

          This is called the backward difference formula because it involves the values of

          empty at 120597120597119894119894 and 120597120597119894119894minus1

          The order of magnitude of the truncation error for the backward difference

          approximation is the same as that of the forward difference approximation Can

          we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

          truncation error The answer is yes

          242 First Order Central Difference

          Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

          empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

          +∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120597120597119894119894

          +∆1205971205973

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          + ⋯ (217)

          empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

          +∆1205971205972

          2 1205971205972empty1205971205971205971205972

          120597120597119894119894

          minus∆1205971205973

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          + ⋯ (218)

          Subtracting Equation (10) from Equation (9) yields

          empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

          + 2∆1205971205973

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          + ⋯ (219)

          Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

          120597120597empty120597120597120597120597120597120597119894119894

          =empty119894119894+1 minus empty119894119894minus1

          2∆120597120597minus∆1205971205972

          3 1205971205973empty1205971205971205971205973

          120597120597119894119894

          minus ⋯ (220)

          or

          120597120597empty120597120597120597120597120597120597119894119894

          =empty119894119894+1 minus empty119894119894minus1

          2∆120597120597+ 119978119978(∆1205971205972) (221)

          This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

          approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

          the truncation error for the central difference approximation goes to zero much

          faster than the truncation error in forward and backward equations

          23

          25 Procedures

          The simple case in this investigation was assuming the constant thermal

          properties of the material First we assumed all the thermal properties of the

          materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

          point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

          assume the constant energy after that the pulse of special shapes was selected

          The numerical solution of equation (23) with boundary and initial conditions

          in equation (22) was investigated using Matlab program as shown in Appendix

          The equation of thermal conductivity and specific heat capacity of metal as a

          function of temperature was obtained by best fitting of polynomials using

          tabulated data in references

          24

          Chapter Three

          Results and Discursion

          31 Introduction

          The development of laser has been an exciting chapter in the history of

          science and engineering It has produced a new type of advice with potential for

          application in an extremely wide variety of fields Mach basic development in

          lasers were occurred during last 35 years The lasers interaction with metal and

          vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

          The status of laser development and application were still rather rudimentary

          The light emitted by laser is electro magnetic radiation this radiation has a wave

          nature the waves consists of vibrating electric and magnetic fields many studies

          have tried to find and solve models of laser interactions Some researchers

          proposed the mathematical model related to the laser - plasma interaction and

          the others have developed an analytical model to study the temperature

          distribution in Infrared optical materials heated by laser pulses Also an attempt

          have made to study the interaction of nanosecond pulsed lasers with material

          from point of view using experimental technique and theoretical approach of

          dimensional analysis

          In this study we have evaluate the solution of partial difference equation

          (PDE) that represent the laser interaction with solid situation in one dimension

          assuming that the power density of laser and thermal properties are functions

          with time and temperature respectively

          25

          32 Numerical solution with constant laser power density and constant

          thermal properties

          First we have taken the lead metal (Pb) with thermal properties

          119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

          119862119862 = 014016119869119869119892119892119870119870

          120588120588 = 10751 1198921198921198981198981198981198983

          119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

          119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

          and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

          represent the area under laser influence

          The numerical solution of equation (23) with boundary and initial conditions

          in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

          of lead metal by explicit method using Matlab program give us the results as

          shown in Fig (31)

          Fig(31) Depth dependence of the temperature with the laser power density

          1198681198680 = 76 times 106 119882119882119898119898119898119898 2

          26

          33 Evaluation of function 119920119920(119957119957) of laser flux density

          From following data that represent the energy (119869119869) with time (millie second)

          Time 0 001 01 02 03 04 05 06 07 08

          Energy 0 002 017 022 024 02 012 007 002 0

          By using Matlab program the best polynomial with deduced from above data

          was

          119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

          minus 10069 1198981198985 (31)

          As shown in Fig (32)

          Fig(32) Laser energy as a function of time

          Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

          normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

          maximum value (119864119864119898119898119907119907119898119898 )

          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

          (119864119864119898119898119907119907119898119898 ) (32)

          The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

          27

          Fig(33) Normalized laser energy as a function of time

          The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

          equal to 3 (total laser energy) ie

          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

          08

          00

          119899119899119898119898 = 3 (33)

          Therefore there exist a real number 119875119875 such that

          119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

          08

          00

          119899119899119898119898 = 3 (34)

          that implies 119875119875 = 68241 and

          119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

          08

          00

          119899119899119898119898 = 3 (35)

          The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

          08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

          119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

          00

          (36)

          28

          Where 119863119863119898119898 put to balance the units of equation (36)

          But integral

          119868119868 = 119864119864119860119860

          (37)

          and from equations (35) (36) and (37) we have

          119868119868 (119898119898)11989911989911989811989808

          00

          = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

          0800 119899119899119898119898

          119860119860 119863119863119898119898 (38)

          Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

          (38)

          There fore

          119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

          119860119860 119863119863119898119898 (39)

          As shown in Fig(34) Matlab program was used to obtain the best polynomial

          that agrees with result data

          119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

          + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

          Fig(34) Time dependence of laser intensity

          29

          34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

          constant thermal properties

          With all constant thermal properties of lead metal as in article (23) and

          119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

          equation (23) with boundary and initial condition as in equation (22) and the

          depth penetration is shown in Fig(35)

          Fig(35) Depth dependence of the temperature when laser intensity function

          of time and constant thermal properties of Lead

          35 Evaluation the Thermal Conductivity as functions of temperature

          The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

          temperature for Lead material was obtained by Matlab program using the

          experimental data tabulated in researches

          119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

          times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

          30

          119879119879 ( 119870119870) 119870119870 119869119869

          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

          300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

          The previous thermal conductivity data and the best fitting of the data are

          shown in Fig (36)

          Fig(36) The best fitting of thermal conductivity of Lead as a function of

          temperature

          31

          36 Evaluation the Specific heat as functions of temperature

          The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

          material was obtained from the following experimental data tacked from

          literatures

          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

          300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

          The best polynomial fitted for these data was

          119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

          + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

          times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

          The specific heat capacity data and the best polynomial fitting of the data are

          shown in Fig (37)

          32

          Fig(37) The best fitting of specific heat capacity of Lead as a function of

          temperature

          37 Evaluation the Density as functions of temperature

          The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

          was used to find the best polynomial fitting

          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

          300 11330 400 11230 500 11130 600 11010 800 10430

          1000 10190 1200 9940

          The best polynomial of this data was

          120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

          minus 45158 times 10minus12 1198791198794 (313)

          33

          The density of Lead as a function of temperature and the best polynomial fitting

          are shown in Fig (38)

          Fig(38) The best fitting of density of Lead as a function of temperature

          38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

          variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

          We have deduced the solution of equation (24) with initial and boundary

          condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

          and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

          respectively then by using Matlab program the depth penetration is shown in

          Fig (39)

          34

          Fig(39) Depth dependence of the temperature for pulse laser on Lead

          material

          39 Laser interaction with copper material

          The same time dependence of laser intensity as shown in Fig(34) with

          thermal properties of copper was used to calculate the temperature distribution as

          a function of depth penetration

          The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

          copper material was obtained from the experimental data tabulated in literary

          The Matlab program used to obtain the best polynomial equation that agrees

          with the above data

          119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

          minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

          times 10minus17 1198791198795 (314)

          35

          119879119879 ( 119870119870) 119870119870 119869119869

          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

          100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

          The previous thermal conductivity data and the best fitting of the data are

          shown in Fig (310)

          Fig(310) The best fitting of thermal conductivity of Copper as a function of

          temperature

          36

          The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

          material was obtained from the following experimental data tacked from

          literatures

          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

          100 0254

          200 0357

          273 0384

          298 0387

          400 0397

          600 0416

          800 0435

          1000 0454

          1100 0464

          1200 0474

          1300 0483

          The best polynomial fitted for these data was

          119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

          + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

          times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

          The specific heat capacity data and the best polynomial fitting of the data are

          shown in Fig (311)

          37

          Fig(311) The best fitting of specific heat capacity of Copper as a function of

          temperature

          The density of copper 120588120588(119879119879) as a function of temperature tacked from

          literature was used to find the best polynomial fitting

          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

          100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

          1000 8576 1100 8519 1200 8458 1300 8396

          38

          The best polynomial of this data was

          120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

          minus 76765 times 10minus14 1198791198794 (316)

          The density of copper as a function of temperature and the best polynomial

          fitting are shown in Fig (312)

          Fig(312) The best fitting of density of copper as a function of temperature

          The depth penetration of laser energy for copper metal was calculated using

          the polynomial equations of thermal conductivity specific heat capacity and

          density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

          (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

          function of time the result was shown in Fig (313)

          39

          The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

          for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

          copper metal so the depth penetration of laser energy of lead metal was smaller

          than that of copper metal this may be due to the high thermal conductivity and

          high specific heat capacity of copper with that of lead metal

          Fig(313) Depth dependence of the temperature for pulse laser on Copper

          material

          40

          310 Conclusions

          The Depth dependence of temperature for lead metal was investigated in two

          case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

          thermal properties (thermal conductivity specific heat) and density of metal are

          also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

          vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

          specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

          120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

          penetration depth in the first case is smaller than that of the second case about

          (190) times

          The temperature distribution as a function of depth dependence for copper

          metal was also investigated in the case when the laser intensity vary with time

          119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

          density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

          The depth penetration of laser energy of lead metal was found to be smaller

          than that of copper metal this may be due to the high thermal conductivity and

          high specific heat capacity of copper with that of lead metal

          41

          References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

          Sons Inc Hoboken New Jersey Canada (2003)

          [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

          [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

          [9]

          Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

          [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

          [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

          [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

          [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

          [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

          httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

          [10] httpwwwworldoflaserscomlaserprincipleshtm

          [11] httpenwikipediaorgwikiLaserPulsed_operation

          [12] httpenwikipediaorgwikiThermal_conductivity

          [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

          [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

          [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

          42

          Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

          This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

          This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

          43

          E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

          This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

          This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

          This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

          44

          title(Dencity as a function of temperature) hold off

          This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

          45

          elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

          46

          r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

          47

          for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

          48

          for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

          49

          6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

          This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

          50

          u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

          51

          alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

          52

          715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

          • 01 Title
            • 0TDr Omed Ghareb Abdullah
              • Ch1-Laser
                • 16 Heat and heat capacity
                • 18 Derivation in one dimension
                  • Ch2 laser2
                  • Ch3 laser2
                  • Appendix-Laser

            1

            Chapter One

            Basic concepts

            11 Introduction

            Laser is a mechanism for emitting light with in electromagnetic radiation

            region of the spectrum with different output intensity Max Plank published

            work in 1900 that provided the understanding that light is a form of

            electromagnetic radiation without this understanding the laser would have

            been invented The principle of the laser was first known in 1917 when Albert

            Einstein describe the theory of stimulated emission and Theodor Maiman in

            1960 invent the first laser using a lasing medium of ruby that was stimulated

            by using high energy flash of intense light

            We have four types of laser according to their gain medium which are

            (solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

            So laser is provided a controlled source of atomic and electronic excitations

            involving non equilibrium phenomena that lend themselves to processing of

            novel material and structure because laser used in wide range application in

            our life such as welding cutting drilling industrial and medical field Maiman

            and other developer of laser weapons sighting system and powerful laser for

            use in surgery and other areas where moderated powerful pinpoint source of

            heat was needed And today laser are used in corrective eye surgery and

            providing apprecise source of heat for cutting and cauterizing tissue

            12 Definition of the Laser

            The word laser is an acronym for Light Amplification by Stimulated Emission

            of Radiation The laser makes use of processes that increase or amplify light

            signals after those signals have been generated by other means These

            processes include (1) stimulated emission a natural effect that was deduced

            2

            by considerations relating to thermodynamic equilibrium and (2) optical

            feedback (present in most lasers) that is usually provided by mirrors

            Thus in its simplest form a laser consists of a gain or amplifying medium

            (where stimulated emission occurs) and a set of mirrors to feed the light back

            into the amplifier for continued growth of the developing beam as seen in

            Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

            more intense directional monochromatic and coherent Most lasers consist

            of a column of active material with a partly reflecting mirror at one end and a

            fully reflecting mirror at the other The active material can be solid (ruby

            crystal) liquid or gas (HeNe COR2R etc)

            Fig(11) Simplified schematic of typical laser

            13 Active laser medium or gain medium

            Laser medium is the heart of the laser system and is responsible for

            producing gain and subsequent generation of laser It can be a crystal solid

            liquid semiconductor or gas medium and can be pumped to a higher energy

            state The material should be of controlled purity size and shape and should

            have the suitable energy levels to support population inversion In other

            words it must have a metastable state to support stimulated emission Most

            lasers are based on 3 or 4 level energy level systems which depends on the

            lasing medium These systems are shown in Figs (12) and (13)

            3

            In case of a three-level laser the material is pumped from level 1 to level 3

            which decays rapidly to level 2 through spontaneous emission Level 2 is a

            metastable level and promotes stimulated emission from level 2 to level 1

            Fig(12) Energy states of Three-level active medium

            On the other hand in a four-level laser the material is pumped to level 4

            which is a fast decaying level and the atoms decay rapidly to level 3 which is

            a metastable level The stimulated emission takes place from level 3 to level 2

            from where the atoms decay back to level 1 Four level lasers is an

            improvement on a system based on three level systems In this case the laser

            transition takes place between the third and second excited states Since

            lower laser level 2 is a fast decaying level which ensures that it rapidly gets

            empty and as such always supports the population inversion condition

            Fig(13) Energy states of Four-level active medium

            4

            14 A Survey of Laser Types

            Laser technology is available to us since 1960rsquos and since then has been

            quite well developed Currently there is a great variety of lasers of different

            output power operating voltages sizes etc The major classes of lasers

            currently used are Gas Solid Molecular and Free Electron lasers Below we

            will cover some most popular representative types of lasers of each class and

            describe specific principles of operation construction and main highlights

            141 Gas Lasers

            1 Helium-Neon Laser

            The most common and inexpensive gas laser the helium-neon laser is

            usually constructed to operate in the red at 6328 nm It can also be

            constructed to produce laser action in the green at 5435 nm and in the

            infrared at 1523 nm

            One of the excited levels of helium at 2061 eV is very close to a level in

            neon at 2066 eV so close in fact that upon collision of a helium and a neon

            atom the energy can be transferred from the helium to the neon atom

            Fig (14) The components of a Hilium-Neon Laser

            5

            Fig(15) The lasing action of He-Ne laser

            Helium-Neon lasers are common in the introductory physics laboratories

            but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

            brightness equal to sunshine on a clear day (01 wattcmP

            2P) and is just as

            dangerous to stare at directly

            2- Carbon Dioxide Laser

            The carbon dioxide gas laser is capable of continuous output powers above

            10 kilowatts It is also capable of extremely high power pulse operation It

            exhibits laser action at several infrared frequencies but none in the visible

            spectrum Operating in a manner similar to the helium-neon laser it employs

            an electric discharge for pumping using a percentage of nitrogen gas as a

            pumping gas The COR2R laser is the most efficient laser capable of operating at

            more than 30 efficiency

            The carbon dioxide laser finds many applications in industry particularly for

            welding and Cutting

            6

            3- Argon Laser

            The argon ion laser can be operated as a continuous gas laser at about 25

            different wavelengths in the visible between (4089 - 6861) nm but is best

            known for its most efficient transitions in the green at 488 nm and 5145 nm

            Operating at much higher powers than the Helium-Neon gas laser it is not

            uncommon to achieve (30 ndash 100) watts of continuous power using several

            transitions This output is produced in hot plasma and takes extremely high

            power typically (9 ndash 12) kW so these are large and expensive devices

            142 Solid Lasers

            1 Ruby Laser

            The ruby laser is the first type of laser actually constructed first

            demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

            aluminum oxide with a small amount (about 005) of Chromium which gives

            it its characteristic pink or red color by absorbing green and blue light

            The ruby laser is used as a pulsed laser producing red light at 6943 nm

            After receiving a pumping flash from the flash tube the laser light emerges for

            as long as the excited atoms persist in the ruby rod which is typically about a

            millisecond

            A pulsed ruby laser was used for the famous laser ranging experiment which

            was conducted with a corner reflector placed on the Moon by the Apollo

            astronauts This determined the distance to the Moon with an accuracy of

            about 15 cm

            7

            Fig (16) Principle of operation of a Ruby laser

            2- Neodymium-YAG Laser

            An example of a solid-state laser the neodymium-YAG uses the NdP

            3+P ion to

            dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

            geometry which makes population inversion possible Neodymium-YAG lasers

            have become very important because they can be used to produce high

            powers Such lasers have been constructed to produce over a kilowatt of

            continuous laser power at 1065 nm and can achieve extremely high powers in

            a pulsed mode

            Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

            production of a series of very short pulses for research with femtosecond time

            resolution

            Fig(17) Construction of a Neodymium-YAG laser

            8

            3- Neodymium-Glass Lasers

            Neodymium glass lasers have emerged as the design choice for research in

            laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

            short as 10-12 seconds with peak powers of 109 kilowatts

            143 Molecular Lasers

            Eximer Lasers

            Eximer is a shortened form of excited dimer denoting the fact that the

            lasing medium in this type of laser is an excited diatomic molecule These

            lasers typically produce ultraviolet pulses They are under investigation for use

            in communicating with submarines by conversion to blue-green light and

            pulsing from overhead satellites through sea water to submarines below

            The eximers used are typically those formed by rare gases and halogens in

            electron excited Gas discharges Molecules like XeF are stable only in their

            excited states and quickly dissociate when they make the transition to their

            ground state This makes possible large population inversions because the

            ground state is depleted by this dissociation However the excited states are

            very short-lived compared to other laser metastable states and lasers like the

            XeF eximer laser require high pumping rates

            Eximer lasers typically produce high power pulse outputs in the blue or

            ultraviolet after excitation by fast electron-beam discharges

            The rare-gas xenon and the highly active fluorine seem unlikely to form a

            molecule but they do in the hot plasma environment of an electron-beam

            initiated gas discharge They are only stable in their excited states if stable

            can be used for molecules which undergo radioactive decay in 1 to 10

            nanoseconds This is long enough to achieve pulsed laser action in the blue-

            green over a band from 450 to 510 nm peaking at 486 nm Very high power

            9

            pulses can be achieved because the stimulated emission cross-sections of the

            laser transitions are relatively low allowing a large population inversion to

            build up The power is also enhanced by the fact that the ground state of XeF

            quickly dissociates so that there is little absorption to quench the laser pulse

            action

            144 Free-Electron Lasers

            The radiation from a free-electron laser is produced from free electrons

            which are forced to oscillate in a regular fashion by an applied field They are

            therefore more like synchrotron light sources or microwave tubes than like

            other lasers They are able to produce highly coherent collimated radiation

            over a wide range of frequencies The magnetic field arrangement which

            produces the alternating field is commonly called a wiggler magnet

            Fig(18) Principle of operation of Free-Electron laser

            The free-electron laser is a highly tunable device which has been used to

            generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

            this range they are the highest power source Applications of free-electron

            lasers are envisioned in isotope separation plasma heating for nuclear fusion

            long-range high resolution radar and particle acceleration in accelerators

            10

            15 Pulsed operation

            Pulsed operation of lasers refers to any laser not classified as continuous

            wave so that the optical power appears in pulses of some duration at some

            repetition rate This encompasses a wide range of technologies addressing a

            number of different motivations Some lasers are pulsed simply because they

            cannot be run in continuous mode

            In other cases the application requires the production of pulses having as

            large an energy as possible Since the pulse energy is equal to the average

            power divided by the repitition rate this goal can sometimes be satisfied by

            lowering the rate of pulses so that more energy can be built up in between

            pulses In laser ablation for example a small volume of material at the surface

            of a work piece can be evaporated if it is heated in a very short time whereas

            supplying the energy gradually would allow for the heat to be absorbed into

            the bulk of the piece never attaining a sufficiently high temperature at a

            particular point

            Other applications rely on the peak pulse power (rather than the energy in

            the pulse) especially in order to obtain nonlinear optical effects For a given

            pulse energy this requires creating pulses of the shortest possible duration

            utilizing techniques such as Q-switching

            16 Heat and heat capacity

            When a sample is heated meaning it receives thermal energy from an

            external source some of the introduced heat is converted into kinetic energy

            the rest to other forms of internal energy specific to the material The amount

            converted into kinetic energy causes the temperature of the material to rise

            The amount of the temperature increase depends on how much heat was

            added the size of the sample the original temperature of the sample and on

            how the heat was added The two obvious choices on how to add the heat are

            11

            to add it holding volume constant or to add it holding pressure constant

            (There may be other choices but they will not concern us)

            Lets assume for the moment that we are going to add heat to our sample

            holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

            subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

            be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

            amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

            (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

            lim119876119876119889119889rarr0

            119876119876119889119889∆120549120549119889119889

            = 120597120597119876119876120597120597120549120549119889119889

            = 119862119862119889119889 (11)

            We have given this derivative the symbol 119862119862119889119889 and we call it the heat

            capacity at constant volume Usually one quotes the molar heat capacity

            119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

            (12)

            We can rearrange Equation (11) as follows

            119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

            Then we can integrate this equation to find the heat involved in a finite

            change at constant volume

            119876119876119889119889 = 119862119862119889119889

            1205491205492

            1205491205491

            119889119889120549120549 (14)

            If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

            out of the integral and the heat at constant volume becomes

            119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

            Let us now go through the same sequence of steps except holding pressure

            constant instead of volume Our initial definition of the heat capacity at

            constant pressure 119862119862119875119875 R Rbecomes

            lim119876119876119875119875rarr0

            119876119876119875119875∆120549120549119875119875

            = 120597120597119876119876120597120597120549120549119875119875

            = 119862119862119875119875 (16)

            The analogous molar heat capacity is

            12

            119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

            (17)

            Equation (16) rearranges to

            119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

            which integrates to give

            119876119876119875119875 = 119862119862119875119875

            1205491205492

            1205491205491

            119889119889120549120549 (19)

            When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

            119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

            Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

            regarded as constant In these cases the heat capacity is fit to a polynomial (or

            similar function) in 120549120549 For example some tables give the heat capacity as

            119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

            where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

            dependent heat capacity the heat at constant pressure would integrate as

            follows

            119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

            1205491205492

            1205491205491

            119889119889120549120549 (112)

            119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

            2 minus 12054912054912 + 119899119899

            1205741205743

            12054912054923 minus 1205491205491

            3 (113)

            Occasionally one finds a different form for the temperature dependent heat

            capacity in the literature

            119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

            When you do calculations with temperature dependent heat capacities you

            must check to see which form is being used for 119862119862119875119875 We are using the

            convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

            be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

            13

            positive then heat was indeed absorbed by the system On the other hand if

            119876119876 is negative it means that the system gave up heat to the surroundings

            17 Thermal conductivity

            In physics thermal conductivity 119896119896 is the property of a material that indicates

            its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

            Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

            The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

            a piece of material The reciprocal of thermal conductivity is thermal

            resistivity

            18 Derivation in one dimension

            The heat equation is derived from Fouriers law and conservation of energy

            (Cannon 1984) By Fouriers law the flow rate of heat energy through a

            surface is proportional to the negative temperature gradient across the

            surface

            119902119902 = minus119896119896 120571120571120549120549 (115)

            where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

            dimension the gradient is an ordinary spatial derivative and so Fouriers law is

            119902119902 = minus119896119896 120549120549119909119909 (116)

            where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

            energy per unit volume in the material 120549120549119876119876 is proportional to the change in

            temperature 120549120549120549120549 That is

            ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

            where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

            material Choosing zero energy at absolute zero temperature this can be

            rewritten as

            ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

            14

            The increase in internal energy in a small spatial region of the material

            (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

            given by

            119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

            119909119909minus∆119909119909

            = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

            119889119889120577120577 119889119889120591120591119909119909+∆119909119909

            119909119909minus∆119909119909

            119905119905+∆119905119905

            119905119905minus∆119905119905

            (119)

            Where the fundamental theorem of calculus was used Additionally with no

            work done and absent any heat sources or sinks the change in internal energy

            in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

            across the boundaries By Fouriers law this is

            119896119896 120597120597120549120549120597120597119909119909

            (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

            (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

            119905119905minus∆119905119905

            = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

            119909119909+∆119909119909

            119909119909minus∆119909119909

            119905119905+∆119905119905

            119905119905minus∆119905119905

            (120)

            again by the fundamental theorem of calculus By conservation of energy

            119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

            119909119909minus∆119909119909

            119905119905+∆119905119905

            119905119905minus∆119905119905

            = 0 (121)

            This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

            Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

            Which can be rewritten as

            120549120549119905119905 =119896119896119888119888119875119875 120588120588

            120549120549119909119909119909119909 (122)

            or

            120597120597120549120549120597120597119905119905

            =119896119896119888119888119875119875 120588120588

            12059712059721205491205491205971205971199091199092 (123)

            15

            which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

            diffusivity and is often denoted 120572120572

            19 Aim of present work

            The goal of this study is to estimate the solution of partial differential

            equation that governs the laser-solid interaction using numerical methods

            The solution will been restricted into one dimensional situation in which we

            assume that both the laser power density and thermal properties are

            functions of time and temperature respectively In this project we attempt to

            investigate the laser interaction with both lead and copper materials by

            predicting the temperature gradient with the depth of the metals

            16

            Chapter Two

            Theoretical Aspects

            21 Introduction

            When a laser interacts with a solid surface a variety of processes can

            occur We are mainly interested in the interaction of pulsed lasers with a

            solid surface in first instance a metal When such a laser interacts with a

            copper surface the laser energy will be transformed into heat The

            temperature of the solid material will increase leading to melting and

            evaporation of the solid material

            The evaporated material (vapour atoms) will expand Depending on the

            applications this can happen in vacuum (or very low pressure) or in a

            background gas (helium argon air)

            22 One dimension laser heating equation

            In general the one dimension laser heating processes of opaque solid slab is

            represented as

            120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

            ( 119870119870 119879119879120597120597 ) (21)

            With boundary conditions and initial condition which represent the pre-

            vaporization stage

            minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

            minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

            17

            119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

            where

            119870119870 represents the thermal conductivity

            120588120588 represents the density

            119862119862 represents the specific heat

            119879119879 represents the temperature

            119879119879infin represents the ambient temperature

            119879119879119907119907 represents the front surface vaporization

            120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

            Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

            119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

            With the same boundary conditions as in equation (22)

            where 119889119889119889119889 = 119870119870120588120588119862119862

            which represents the thermal diffusion

            But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

            equation (21) with this assuming implies

            119879119879119905119905 = 1

            120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

            With the same boundary and initial conditions in equation (22) Where 119870119870

            represents the derivative of K with respect the temperature

            23 Numerical solution of Initial value problems

            An immense number of analytical solutions for conduction heat-transfer

            problems have been accumulated in literature over the past 100 years Even so

            in many practical situations the geometry or boundary conditions are such that an

            analytical solution has not been obtained at all or if the solution has been

            18

            developed it involves such a complex series solution that numerical evaluation

            becomes exceedingly difficult For such situation the most fruitful approach to

            the problem is numerical techniques the basic principles of which we shall

            outline in this section

            One way to guarantee accuracy in the solution of an initial values problems

            (IVP) is to solve the problem twice using step sizes h and h2 and compare

            answers at the mesh points corresponding to the larger step size But this requires

            a significant amount of computation for the smaller step size and must be

            repeated if it is determined that the agreement is not good enough

            24 Finite Difference Method

            The finite difference method is one of several techniques for obtaining

            numerical solutions to differential equations In all numerical solutions the

            continuous partial differential equation (PDE) is replaced with a discrete

            approximation In this context the word discrete means that the numerical

            solution is known only at a finite number of points in the physical domain The

            number of those points can be selected by the user of the numerical method In

            general increasing the number of points not only increases the resolution but

            also the accuracy of the numerical solution

            The discrete approximation results in a set of algebraic equations that are

            evaluated for the values of the discrete unknowns

            The mesh is the set of locations where the discrete solution is computed

            These points are called nodes and if one were to draw lines between adjacent

            nodes in the domain the resulting image would resemble a net or mesh Two key

            parameters of the mesh are ∆120597120597 the local distance between adjacent points in

            space and ∆119905119905 the local distance between adjacent time steps For the simple

            examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

            19

            The core idea of the finite-difference method is to replace continuous

            derivatives with so-called difference formulas that involve only the discrete

            values associated with positions on the mesh

            Applying the finite-difference method to a differential equation involves

            replacing all derivatives with difference formulas In the heat equation there are

            derivatives with respect to time and derivatives with respect to space Using

            different combinations of mesh points in the difference formulas results in

            different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

            numerical solution obtained with any useful scheme will approach the true

            solution to the original differential equation However the rate at which the

            numerical solution approaches the true solution varies with the scheme

            241 First Order Forward Difference

            Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

            empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

            +1205751205751205971205972

            2 1205971205972empty1205971205971205971205972

            1205971205971

            +1205751205751205971205973

            3 1205971205973empty1205971205971205971205973

            1205971205971

            + ⋯ (25)

            where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

            consider the value of empty at the location of the 120597120597119894119894+1 mesh line

            empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

            +∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120597120597119894119894

            +∆1205971205973

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            + ⋯ (26)

            Solve for (120597120597empty120597120597120597120597)120597120597119894119894

            120597120597empty120597120597120597120597120597120597119894119894

            =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

            ∆120597120597minus∆1205971205972

            1205971205972empty1205971205971205971205972

            120597120597119894119894

            minus∆1205971205972

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            minus ⋯ (27)

            Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

            hand side have been reduced by one

            20

            Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

            and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

            120597120597empty120597120597120597120597120597120597119894119894

            =empty119894119894+1 minus empty119894119894

            ∆120597120597minus∆1205971205972

            1205971205972empty1205971205971205971205972

            120597120597119894119894

            minus∆1205971205972

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            minus ⋯ (28)

            The mean value theorem can be used to replace the higher order derivatives

            ∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120597120597119894119894

            +∆1205971205973

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            + ⋯ =∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120585120585 (29)

            where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

            120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

            ∆120597120597+∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120585120585 (210)

            120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

            ∆120597120597asymp∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120585120585 (211)

            The term on the right hand side of previous equation is called the truncation

            error of the finite difference approximation

            In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

            unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

            truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

            analytical form) the big 119978119978 notation can be used to express the dependence of

            the truncation error on the mesh spacing Note that the right hand side of last

            equation contains the mesh parameter ∆120597120597 which is chosen by the person using

            the finite difference simulation Since this is the only parameter under the users

            control that determines the error the truncation error is simply written

            ∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120585120585= 119978119978(∆1205971205972) (212)

            The equals sign in this expression is true in the order of magnitude sense In

            other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

            21

            equality Rather the expression means that the left hand side is a product of an

            unknown constant and ∆1205971205972 Although the expression does not give us the exact

            magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

            approaches zero as ∆120597120597 is reduced

            Using big 119978119978 notation Equation (28) can be written

            120597120597empty120597120597120597120597120597120597119894119894

            =empty119894119894+1 minus empty119894119894

            ∆120597120597+ 119978119978(∆120597120597) (213)

            This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

            it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

            truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

            our control because we can choose the mesh size ∆120597120597 The part of the truncation

            error that is not under our control is |120597120597empty120597120597120597120597|120585120585

            242 First Order Backward Difference

            An alternative first order finite difference formula is obtained if the Taylor series

            like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

            variables in place of all the unknowns one obtains

            empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

            +∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120597120597119894119894

            minus∆1205971205973

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            + ⋯ (214)

            Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

            to get

            120597120597empty120597120597120597120597120597120597119894119894

            =empty119894119894+1 minus empty119894119894

            ∆120597120597minus∆1205971205972

            1205971205972empty1205971205971205971205972

            120597120597119894119894

            minus∆1205971205972

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            minus ⋯ (215)

            Or using big 119978119978 notation

            120597120597empty120597120597120597120597120597120597119894119894

            =empty119894119894 minus empty119894119894minus1

            ∆120597120597+ 119978119978(∆120597120597) (216)

            22

            This is called the backward difference formula because it involves the values of

            empty at 120597120597119894119894 and 120597120597119894119894minus1

            The order of magnitude of the truncation error for the backward difference

            approximation is the same as that of the forward difference approximation Can

            we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

            truncation error The answer is yes

            242 First Order Central Difference

            Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

            empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

            +∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120597120597119894119894

            +∆1205971205973

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            + ⋯ (217)

            empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

            +∆1205971205972

            2 1205971205972empty1205971205971205971205972

            120597120597119894119894

            minus∆1205971205973

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            + ⋯ (218)

            Subtracting Equation (10) from Equation (9) yields

            empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

            + 2∆1205971205973

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            + ⋯ (219)

            Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

            120597120597empty120597120597120597120597120597120597119894119894

            =empty119894119894+1 minus empty119894119894minus1

            2∆120597120597minus∆1205971205972

            3 1205971205973empty1205971205971205971205973

            120597120597119894119894

            minus ⋯ (220)

            or

            120597120597empty120597120597120597120597120597120597119894119894

            =empty119894119894+1 minus empty119894119894minus1

            2∆120597120597+ 119978119978(∆1205971205972) (221)

            This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

            approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

            the truncation error for the central difference approximation goes to zero much

            faster than the truncation error in forward and backward equations

            23

            25 Procedures

            The simple case in this investigation was assuming the constant thermal

            properties of the material First we assumed all the thermal properties of the

            materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

            point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

            assume the constant energy after that the pulse of special shapes was selected

            The numerical solution of equation (23) with boundary and initial conditions

            in equation (22) was investigated using Matlab program as shown in Appendix

            The equation of thermal conductivity and specific heat capacity of metal as a

            function of temperature was obtained by best fitting of polynomials using

            tabulated data in references

            24

            Chapter Three

            Results and Discursion

            31 Introduction

            The development of laser has been an exciting chapter in the history of

            science and engineering It has produced a new type of advice with potential for

            application in an extremely wide variety of fields Mach basic development in

            lasers were occurred during last 35 years The lasers interaction with metal and

            vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

            The status of laser development and application were still rather rudimentary

            The light emitted by laser is electro magnetic radiation this radiation has a wave

            nature the waves consists of vibrating electric and magnetic fields many studies

            have tried to find and solve models of laser interactions Some researchers

            proposed the mathematical model related to the laser - plasma interaction and

            the others have developed an analytical model to study the temperature

            distribution in Infrared optical materials heated by laser pulses Also an attempt

            have made to study the interaction of nanosecond pulsed lasers with material

            from point of view using experimental technique and theoretical approach of

            dimensional analysis

            In this study we have evaluate the solution of partial difference equation

            (PDE) that represent the laser interaction with solid situation in one dimension

            assuming that the power density of laser and thermal properties are functions

            with time and temperature respectively

            25

            32 Numerical solution with constant laser power density and constant

            thermal properties

            First we have taken the lead metal (Pb) with thermal properties

            119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

            119862119862 = 014016119869119869119892119892119870119870

            120588120588 = 10751 1198921198921198981198981198981198983

            119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

            119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

            and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

            represent the area under laser influence

            The numerical solution of equation (23) with boundary and initial conditions

            in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

            of lead metal by explicit method using Matlab program give us the results as

            shown in Fig (31)

            Fig(31) Depth dependence of the temperature with the laser power density

            1198681198680 = 76 times 106 119882119882119898119898119898119898 2

            26

            33 Evaluation of function 119920119920(119957119957) of laser flux density

            From following data that represent the energy (119869119869) with time (millie second)

            Time 0 001 01 02 03 04 05 06 07 08

            Energy 0 002 017 022 024 02 012 007 002 0

            By using Matlab program the best polynomial with deduced from above data

            was

            119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

            minus 10069 1198981198985 (31)

            As shown in Fig (32)

            Fig(32) Laser energy as a function of time

            Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

            normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

            maximum value (119864119864119898119898119907119907119898119898 )

            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

            (119864119864119898119898119907119907119898119898 ) (32)

            The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

            27

            Fig(33) Normalized laser energy as a function of time

            The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

            equal to 3 (total laser energy) ie

            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

            08

            00

            119899119899119898119898 = 3 (33)

            Therefore there exist a real number 119875119875 such that

            119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

            08

            00

            119899119899119898119898 = 3 (34)

            that implies 119875119875 = 68241 and

            119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

            08

            00

            119899119899119898119898 = 3 (35)

            The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

            08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

            119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

            00

            (36)

            28

            Where 119863119863119898119898 put to balance the units of equation (36)

            But integral

            119868119868 = 119864119864119860119860

            (37)

            and from equations (35) (36) and (37) we have

            119868119868 (119898119898)11989911989911989811989808

            00

            = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

            0800 119899119899119898119898

            119860119860 119863119863119898119898 (38)

            Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

            (38)

            There fore

            119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

            119860119860 119863119863119898119898 (39)

            As shown in Fig(34) Matlab program was used to obtain the best polynomial

            that agrees with result data

            119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

            + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

            Fig(34) Time dependence of laser intensity

            29

            34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

            constant thermal properties

            With all constant thermal properties of lead metal as in article (23) and

            119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

            equation (23) with boundary and initial condition as in equation (22) and the

            depth penetration is shown in Fig(35)

            Fig(35) Depth dependence of the temperature when laser intensity function

            of time and constant thermal properties of Lead

            35 Evaluation the Thermal Conductivity as functions of temperature

            The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

            temperature for Lead material was obtained by Matlab program using the

            experimental data tabulated in researches

            119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

            times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

            30

            119879119879 ( 119870119870) 119870119870 119869119869

            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

            300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

            The previous thermal conductivity data and the best fitting of the data are

            shown in Fig (36)

            Fig(36) The best fitting of thermal conductivity of Lead as a function of

            temperature

            31

            36 Evaluation the Specific heat as functions of temperature

            The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

            material was obtained from the following experimental data tacked from

            literatures

            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

            300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

            The best polynomial fitted for these data was

            119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

            + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

            times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

            The specific heat capacity data and the best polynomial fitting of the data are

            shown in Fig (37)

            32

            Fig(37) The best fitting of specific heat capacity of Lead as a function of

            temperature

            37 Evaluation the Density as functions of temperature

            The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

            was used to find the best polynomial fitting

            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

            300 11330 400 11230 500 11130 600 11010 800 10430

            1000 10190 1200 9940

            The best polynomial of this data was

            120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

            minus 45158 times 10minus12 1198791198794 (313)

            33

            The density of Lead as a function of temperature and the best polynomial fitting

            are shown in Fig (38)

            Fig(38) The best fitting of density of Lead as a function of temperature

            38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

            variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

            We have deduced the solution of equation (24) with initial and boundary

            condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

            and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

            respectively then by using Matlab program the depth penetration is shown in

            Fig (39)

            34

            Fig(39) Depth dependence of the temperature for pulse laser on Lead

            material

            39 Laser interaction with copper material

            The same time dependence of laser intensity as shown in Fig(34) with

            thermal properties of copper was used to calculate the temperature distribution as

            a function of depth penetration

            The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

            copper material was obtained from the experimental data tabulated in literary

            The Matlab program used to obtain the best polynomial equation that agrees

            with the above data

            119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

            minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

            times 10minus17 1198791198795 (314)

            35

            119879119879 ( 119870119870) 119870119870 119869119869

            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

            100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

            The previous thermal conductivity data and the best fitting of the data are

            shown in Fig (310)

            Fig(310) The best fitting of thermal conductivity of Copper as a function of

            temperature

            36

            The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

            material was obtained from the following experimental data tacked from

            literatures

            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

            100 0254

            200 0357

            273 0384

            298 0387

            400 0397

            600 0416

            800 0435

            1000 0454

            1100 0464

            1200 0474

            1300 0483

            The best polynomial fitted for these data was

            119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

            + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

            times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

            The specific heat capacity data and the best polynomial fitting of the data are

            shown in Fig (311)

            37

            Fig(311) The best fitting of specific heat capacity of Copper as a function of

            temperature

            The density of copper 120588120588(119879119879) as a function of temperature tacked from

            literature was used to find the best polynomial fitting

            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

            100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

            1000 8576 1100 8519 1200 8458 1300 8396

            38

            The best polynomial of this data was

            120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

            minus 76765 times 10minus14 1198791198794 (316)

            The density of copper as a function of temperature and the best polynomial

            fitting are shown in Fig (312)

            Fig(312) The best fitting of density of copper as a function of temperature

            The depth penetration of laser energy for copper metal was calculated using

            the polynomial equations of thermal conductivity specific heat capacity and

            density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

            (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

            function of time the result was shown in Fig (313)

            39

            The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

            for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

            copper metal so the depth penetration of laser energy of lead metal was smaller

            than that of copper metal this may be due to the high thermal conductivity and

            high specific heat capacity of copper with that of lead metal

            Fig(313) Depth dependence of the temperature for pulse laser on Copper

            material

            40

            310 Conclusions

            The Depth dependence of temperature for lead metal was investigated in two

            case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

            thermal properties (thermal conductivity specific heat) and density of metal are

            also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

            vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

            specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

            120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

            penetration depth in the first case is smaller than that of the second case about

            (190) times

            The temperature distribution as a function of depth dependence for copper

            metal was also investigated in the case when the laser intensity vary with time

            119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

            density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

            The depth penetration of laser energy of lead metal was found to be smaller

            than that of copper metal this may be due to the high thermal conductivity and

            high specific heat capacity of copper with that of lead metal

            41

            References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

            Sons Inc Hoboken New Jersey Canada (2003)

            [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

            [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

            [9]

            Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

            [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

            [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

            [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

            [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

            [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

            httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

            [10] httpwwwworldoflaserscomlaserprincipleshtm

            [11] httpenwikipediaorgwikiLaserPulsed_operation

            [12] httpenwikipediaorgwikiThermal_conductivity

            [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

            [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

            [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

            42

            Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

            This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

            This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

            43

            E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

            This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

            This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

            This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

            44

            title(Dencity as a function of temperature) hold off

            This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

            45

            elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

            46

            r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

            47

            for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

            48

            for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

            49

            6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

            This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

            50

            u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

            51

            alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

            52

            715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

            • 01 Title
              • 0TDr Omed Ghareb Abdullah
                • Ch1-Laser
                  • 16 Heat and heat capacity
                  • 18 Derivation in one dimension
                    • Ch2 laser2
                    • Ch3 laser2
                    • Appendix-Laser

              2

              by considerations relating to thermodynamic equilibrium and (2) optical

              feedback (present in most lasers) that is usually provided by mirrors

              Thus in its simplest form a laser consists of a gain or amplifying medium

              (where stimulated emission occurs) and a set of mirrors to feed the light back

              into the amplifier for continued growth of the developing beam as seen in

              Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

              more intense directional monochromatic and coherent Most lasers consist

              of a column of active material with a partly reflecting mirror at one end and a

              fully reflecting mirror at the other The active material can be solid (ruby

              crystal) liquid or gas (HeNe COR2R etc)

              Fig(11) Simplified schematic of typical laser

              13 Active laser medium or gain medium

              Laser medium is the heart of the laser system and is responsible for

              producing gain and subsequent generation of laser It can be a crystal solid

              liquid semiconductor or gas medium and can be pumped to a higher energy

              state The material should be of controlled purity size and shape and should

              have the suitable energy levels to support population inversion In other

              words it must have a metastable state to support stimulated emission Most

              lasers are based on 3 or 4 level energy level systems which depends on the

              lasing medium These systems are shown in Figs (12) and (13)

              3

              In case of a three-level laser the material is pumped from level 1 to level 3

              which decays rapidly to level 2 through spontaneous emission Level 2 is a

              metastable level and promotes stimulated emission from level 2 to level 1

              Fig(12) Energy states of Three-level active medium

              On the other hand in a four-level laser the material is pumped to level 4

              which is a fast decaying level and the atoms decay rapidly to level 3 which is

              a metastable level The stimulated emission takes place from level 3 to level 2

              from where the atoms decay back to level 1 Four level lasers is an

              improvement on a system based on three level systems In this case the laser

              transition takes place between the third and second excited states Since

              lower laser level 2 is a fast decaying level which ensures that it rapidly gets

              empty and as such always supports the population inversion condition

              Fig(13) Energy states of Four-level active medium

              4

              14 A Survey of Laser Types

              Laser technology is available to us since 1960rsquos and since then has been

              quite well developed Currently there is a great variety of lasers of different

              output power operating voltages sizes etc The major classes of lasers

              currently used are Gas Solid Molecular and Free Electron lasers Below we

              will cover some most popular representative types of lasers of each class and

              describe specific principles of operation construction and main highlights

              141 Gas Lasers

              1 Helium-Neon Laser

              The most common and inexpensive gas laser the helium-neon laser is

              usually constructed to operate in the red at 6328 nm It can also be

              constructed to produce laser action in the green at 5435 nm and in the

              infrared at 1523 nm

              One of the excited levels of helium at 2061 eV is very close to a level in

              neon at 2066 eV so close in fact that upon collision of a helium and a neon

              atom the energy can be transferred from the helium to the neon atom

              Fig (14) The components of a Hilium-Neon Laser

              5

              Fig(15) The lasing action of He-Ne laser

              Helium-Neon lasers are common in the introductory physics laboratories

              but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

              brightness equal to sunshine on a clear day (01 wattcmP

              2P) and is just as

              dangerous to stare at directly

              2- Carbon Dioxide Laser

              The carbon dioxide gas laser is capable of continuous output powers above

              10 kilowatts It is also capable of extremely high power pulse operation It

              exhibits laser action at several infrared frequencies but none in the visible

              spectrum Operating in a manner similar to the helium-neon laser it employs

              an electric discharge for pumping using a percentage of nitrogen gas as a

              pumping gas The COR2R laser is the most efficient laser capable of operating at

              more than 30 efficiency

              The carbon dioxide laser finds many applications in industry particularly for

              welding and Cutting

              6

              3- Argon Laser

              The argon ion laser can be operated as a continuous gas laser at about 25

              different wavelengths in the visible between (4089 - 6861) nm but is best

              known for its most efficient transitions in the green at 488 nm and 5145 nm

              Operating at much higher powers than the Helium-Neon gas laser it is not

              uncommon to achieve (30 ndash 100) watts of continuous power using several

              transitions This output is produced in hot plasma and takes extremely high

              power typically (9 ndash 12) kW so these are large and expensive devices

              142 Solid Lasers

              1 Ruby Laser

              The ruby laser is the first type of laser actually constructed first

              demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

              aluminum oxide with a small amount (about 005) of Chromium which gives

              it its characteristic pink or red color by absorbing green and blue light

              The ruby laser is used as a pulsed laser producing red light at 6943 nm

              After receiving a pumping flash from the flash tube the laser light emerges for

              as long as the excited atoms persist in the ruby rod which is typically about a

              millisecond

              A pulsed ruby laser was used for the famous laser ranging experiment which

              was conducted with a corner reflector placed on the Moon by the Apollo

              astronauts This determined the distance to the Moon with an accuracy of

              about 15 cm

              7

              Fig (16) Principle of operation of a Ruby laser

              2- Neodymium-YAG Laser

              An example of a solid-state laser the neodymium-YAG uses the NdP

              3+P ion to

              dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

              geometry which makes population inversion possible Neodymium-YAG lasers

              have become very important because they can be used to produce high

              powers Such lasers have been constructed to produce over a kilowatt of

              continuous laser power at 1065 nm and can achieve extremely high powers in

              a pulsed mode

              Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

              production of a series of very short pulses for research with femtosecond time

              resolution

              Fig(17) Construction of a Neodymium-YAG laser

              8

              3- Neodymium-Glass Lasers

              Neodymium glass lasers have emerged as the design choice for research in

              laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

              short as 10-12 seconds with peak powers of 109 kilowatts

              143 Molecular Lasers

              Eximer Lasers

              Eximer is a shortened form of excited dimer denoting the fact that the

              lasing medium in this type of laser is an excited diatomic molecule These

              lasers typically produce ultraviolet pulses They are under investigation for use

              in communicating with submarines by conversion to blue-green light and

              pulsing from overhead satellites through sea water to submarines below

              The eximers used are typically those formed by rare gases and halogens in

              electron excited Gas discharges Molecules like XeF are stable only in their

              excited states and quickly dissociate when they make the transition to their

              ground state This makes possible large population inversions because the

              ground state is depleted by this dissociation However the excited states are

              very short-lived compared to other laser metastable states and lasers like the

              XeF eximer laser require high pumping rates

              Eximer lasers typically produce high power pulse outputs in the blue or

              ultraviolet after excitation by fast electron-beam discharges

              The rare-gas xenon and the highly active fluorine seem unlikely to form a

              molecule but they do in the hot plasma environment of an electron-beam

              initiated gas discharge They are only stable in their excited states if stable

              can be used for molecules which undergo radioactive decay in 1 to 10

              nanoseconds This is long enough to achieve pulsed laser action in the blue-

              green over a band from 450 to 510 nm peaking at 486 nm Very high power

              9

              pulses can be achieved because the stimulated emission cross-sections of the

              laser transitions are relatively low allowing a large population inversion to

              build up The power is also enhanced by the fact that the ground state of XeF

              quickly dissociates so that there is little absorption to quench the laser pulse

              action

              144 Free-Electron Lasers

              The radiation from a free-electron laser is produced from free electrons

              which are forced to oscillate in a regular fashion by an applied field They are

              therefore more like synchrotron light sources or microwave tubes than like

              other lasers They are able to produce highly coherent collimated radiation

              over a wide range of frequencies The magnetic field arrangement which

              produces the alternating field is commonly called a wiggler magnet

              Fig(18) Principle of operation of Free-Electron laser

              The free-electron laser is a highly tunable device which has been used to

              generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

              this range they are the highest power source Applications of free-electron

              lasers are envisioned in isotope separation plasma heating for nuclear fusion

              long-range high resolution radar and particle acceleration in accelerators

              10

              15 Pulsed operation

              Pulsed operation of lasers refers to any laser not classified as continuous

              wave so that the optical power appears in pulses of some duration at some

              repetition rate This encompasses a wide range of technologies addressing a

              number of different motivations Some lasers are pulsed simply because they

              cannot be run in continuous mode

              In other cases the application requires the production of pulses having as

              large an energy as possible Since the pulse energy is equal to the average

              power divided by the repitition rate this goal can sometimes be satisfied by

              lowering the rate of pulses so that more energy can be built up in between

              pulses In laser ablation for example a small volume of material at the surface

              of a work piece can be evaporated if it is heated in a very short time whereas

              supplying the energy gradually would allow for the heat to be absorbed into

              the bulk of the piece never attaining a sufficiently high temperature at a

              particular point

              Other applications rely on the peak pulse power (rather than the energy in

              the pulse) especially in order to obtain nonlinear optical effects For a given

              pulse energy this requires creating pulses of the shortest possible duration

              utilizing techniques such as Q-switching

              16 Heat and heat capacity

              When a sample is heated meaning it receives thermal energy from an

              external source some of the introduced heat is converted into kinetic energy

              the rest to other forms of internal energy specific to the material The amount

              converted into kinetic energy causes the temperature of the material to rise

              The amount of the temperature increase depends on how much heat was

              added the size of the sample the original temperature of the sample and on

              how the heat was added The two obvious choices on how to add the heat are

              11

              to add it holding volume constant or to add it holding pressure constant

              (There may be other choices but they will not concern us)

              Lets assume for the moment that we are going to add heat to our sample

              holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

              subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

              be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

              amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

              (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

              lim119876119876119889119889rarr0

              119876119876119889119889∆120549120549119889119889

              = 120597120597119876119876120597120597120549120549119889119889

              = 119862119862119889119889 (11)

              We have given this derivative the symbol 119862119862119889119889 and we call it the heat

              capacity at constant volume Usually one quotes the molar heat capacity

              119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

              (12)

              We can rearrange Equation (11) as follows

              119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

              Then we can integrate this equation to find the heat involved in a finite

              change at constant volume

              119876119876119889119889 = 119862119862119889119889

              1205491205492

              1205491205491

              119889119889120549120549 (14)

              If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

              out of the integral and the heat at constant volume becomes

              119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

              Let us now go through the same sequence of steps except holding pressure

              constant instead of volume Our initial definition of the heat capacity at

              constant pressure 119862119862119875119875 R Rbecomes

              lim119876119876119875119875rarr0

              119876119876119875119875∆120549120549119875119875

              = 120597120597119876119876120597120597120549120549119875119875

              = 119862119862119875119875 (16)

              The analogous molar heat capacity is

              12

              119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

              (17)

              Equation (16) rearranges to

              119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

              which integrates to give

              119876119876119875119875 = 119862119862119875119875

              1205491205492

              1205491205491

              119889119889120549120549 (19)

              When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

              119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

              Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

              regarded as constant In these cases the heat capacity is fit to a polynomial (or

              similar function) in 120549120549 For example some tables give the heat capacity as

              119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

              where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

              dependent heat capacity the heat at constant pressure would integrate as

              follows

              119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

              1205491205492

              1205491205491

              119889119889120549120549 (112)

              119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

              2 minus 12054912054912 + 119899119899

              1205741205743

              12054912054923 minus 1205491205491

              3 (113)

              Occasionally one finds a different form for the temperature dependent heat

              capacity in the literature

              119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

              When you do calculations with temperature dependent heat capacities you

              must check to see which form is being used for 119862119862119875119875 We are using the

              convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

              be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

              13

              positive then heat was indeed absorbed by the system On the other hand if

              119876119876 is negative it means that the system gave up heat to the surroundings

              17 Thermal conductivity

              In physics thermal conductivity 119896119896 is the property of a material that indicates

              its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

              Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

              The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

              a piece of material The reciprocal of thermal conductivity is thermal

              resistivity

              18 Derivation in one dimension

              The heat equation is derived from Fouriers law and conservation of energy

              (Cannon 1984) By Fouriers law the flow rate of heat energy through a

              surface is proportional to the negative temperature gradient across the

              surface

              119902119902 = minus119896119896 120571120571120549120549 (115)

              where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

              dimension the gradient is an ordinary spatial derivative and so Fouriers law is

              119902119902 = minus119896119896 120549120549119909119909 (116)

              where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

              energy per unit volume in the material 120549120549119876119876 is proportional to the change in

              temperature 120549120549120549120549 That is

              ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

              where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

              material Choosing zero energy at absolute zero temperature this can be

              rewritten as

              ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

              14

              The increase in internal energy in a small spatial region of the material

              (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

              given by

              119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

              119909119909minus∆119909119909

              = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

              119889119889120577120577 119889119889120591120591119909119909+∆119909119909

              119909119909minus∆119909119909

              119905119905+∆119905119905

              119905119905minus∆119905119905

              (119)

              Where the fundamental theorem of calculus was used Additionally with no

              work done and absent any heat sources or sinks the change in internal energy

              in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

              across the boundaries By Fouriers law this is

              119896119896 120597120597120549120549120597120597119909119909

              (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

              (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

              119905119905minus∆119905119905

              = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

              119909119909+∆119909119909

              119909119909minus∆119909119909

              119905119905+∆119905119905

              119905119905minus∆119905119905

              (120)

              again by the fundamental theorem of calculus By conservation of energy

              119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

              119909119909minus∆119909119909

              119905119905+∆119905119905

              119905119905minus∆119905119905

              = 0 (121)

              This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

              Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

              Which can be rewritten as

              120549120549119905119905 =119896119896119888119888119875119875 120588120588

              120549120549119909119909119909119909 (122)

              or

              120597120597120549120549120597120597119905119905

              =119896119896119888119888119875119875 120588120588

              12059712059721205491205491205971205971199091199092 (123)

              15

              which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

              diffusivity and is often denoted 120572120572

              19 Aim of present work

              The goal of this study is to estimate the solution of partial differential

              equation that governs the laser-solid interaction using numerical methods

              The solution will been restricted into one dimensional situation in which we

              assume that both the laser power density and thermal properties are

              functions of time and temperature respectively In this project we attempt to

              investigate the laser interaction with both lead and copper materials by

              predicting the temperature gradient with the depth of the metals

              16

              Chapter Two

              Theoretical Aspects

              21 Introduction

              When a laser interacts with a solid surface a variety of processes can

              occur We are mainly interested in the interaction of pulsed lasers with a

              solid surface in first instance a metal When such a laser interacts with a

              copper surface the laser energy will be transformed into heat The

              temperature of the solid material will increase leading to melting and

              evaporation of the solid material

              The evaporated material (vapour atoms) will expand Depending on the

              applications this can happen in vacuum (or very low pressure) or in a

              background gas (helium argon air)

              22 One dimension laser heating equation

              In general the one dimension laser heating processes of opaque solid slab is

              represented as

              120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

              ( 119870119870 119879119879120597120597 ) (21)

              With boundary conditions and initial condition which represent the pre-

              vaporization stage

              minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

              minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

              17

              119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

              where

              119870119870 represents the thermal conductivity

              120588120588 represents the density

              119862119862 represents the specific heat

              119879119879 represents the temperature

              119879119879infin represents the ambient temperature

              119879119879119907119907 represents the front surface vaporization

              120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

              Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

              119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

              With the same boundary conditions as in equation (22)

              where 119889119889119889119889 = 119870119870120588120588119862119862

              which represents the thermal diffusion

              But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

              equation (21) with this assuming implies

              119879119879119905119905 = 1

              120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

              With the same boundary and initial conditions in equation (22) Where 119870119870

              represents the derivative of K with respect the temperature

              23 Numerical solution of Initial value problems

              An immense number of analytical solutions for conduction heat-transfer

              problems have been accumulated in literature over the past 100 years Even so

              in many practical situations the geometry or boundary conditions are such that an

              analytical solution has not been obtained at all or if the solution has been

              18

              developed it involves such a complex series solution that numerical evaluation

              becomes exceedingly difficult For such situation the most fruitful approach to

              the problem is numerical techniques the basic principles of which we shall

              outline in this section

              One way to guarantee accuracy in the solution of an initial values problems

              (IVP) is to solve the problem twice using step sizes h and h2 and compare

              answers at the mesh points corresponding to the larger step size But this requires

              a significant amount of computation for the smaller step size and must be

              repeated if it is determined that the agreement is not good enough

              24 Finite Difference Method

              The finite difference method is one of several techniques for obtaining

              numerical solutions to differential equations In all numerical solutions the

              continuous partial differential equation (PDE) is replaced with a discrete

              approximation In this context the word discrete means that the numerical

              solution is known only at a finite number of points in the physical domain The

              number of those points can be selected by the user of the numerical method In

              general increasing the number of points not only increases the resolution but

              also the accuracy of the numerical solution

              The discrete approximation results in a set of algebraic equations that are

              evaluated for the values of the discrete unknowns

              The mesh is the set of locations where the discrete solution is computed

              These points are called nodes and if one were to draw lines between adjacent

              nodes in the domain the resulting image would resemble a net or mesh Two key

              parameters of the mesh are ∆120597120597 the local distance between adjacent points in

              space and ∆119905119905 the local distance between adjacent time steps For the simple

              examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

              19

              The core idea of the finite-difference method is to replace continuous

              derivatives with so-called difference formulas that involve only the discrete

              values associated with positions on the mesh

              Applying the finite-difference method to a differential equation involves

              replacing all derivatives with difference formulas In the heat equation there are

              derivatives with respect to time and derivatives with respect to space Using

              different combinations of mesh points in the difference formulas results in

              different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

              numerical solution obtained with any useful scheme will approach the true

              solution to the original differential equation However the rate at which the

              numerical solution approaches the true solution varies with the scheme

              241 First Order Forward Difference

              Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

              empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

              +1205751205751205971205972

              2 1205971205972empty1205971205971205971205972

              1205971205971

              +1205751205751205971205973

              3 1205971205973empty1205971205971205971205973

              1205971205971

              + ⋯ (25)

              where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

              consider the value of empty at the location of the 120597120597119894119894+1 mesh line

              empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

              +∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120597120597119894119894

              +∆1205971205973

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              + ⋯ (26)

              Solve for (120597120597empty120597120597120597120597)120597120597119894119894

              120597120597empty120597120597120597120597120597120597119894119894

              =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

              ∆120597120597minus∆1205971205972

              1205971205972empty1205971205971205971205972

              120597120597119894119894

              minus∆1205971205972

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              minus ⋯ (27)

              Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

              hand side have been reduced by one

              20

              Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

              and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

              120597120597empty120597120597120597120597120597120597119894119894

              =empty119894119894+1 minus empty119894119894

              ∆120597120597minus∆1205971205972

              1205971205972empty1205971205971205971205972

              120597120597119894119894

              minus∆1205971205972

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              minus ⋯ (28)

              The mean value theorem can be used to replace the higher order derivatives

              ∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120597120597119894119894

              +∆1205971205973

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              + ⋯ =∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120585120585 (29)

              where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

              120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

              ∆120597120597+∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120585120585 (210)

              120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

              ∆120597120597asymp∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120585120585 (211)

              The term on the right hand side of previous equation is called the truncation

              error of the finite difference approximation

              In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

              unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

              truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

              analytical form) the big 119978119978 notation can be used to express the dependence of

              the truncation error on the mesh spacing Note that the right hand side of last

              equation contains the mesh parameter ∆120597120597 which is chosen by the person using

              the finite difference simulation Since this is the only parameter under the users

              control that determines the error the truncation error is simply written

              ∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120585120585= 119978119978(∆1205971205972) (212)

              The equals sign in this expression is true in the order of magnitude sense In

              other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

              21

              equality Rather the expression means that the left hand side is a product of an

              unknown constant and ∆1205971205972 Although the expression does not give us the exact

              magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

              approaches zero as ∆120597120597 is reduced

              Using big 119978119978 notation Equation (28) can be written

              120597120597empty120597120597120597120597120597120597119894119894

              =empty119894119894+1 minus empty119894119894

              ∆120597120597+ 119978119978(∆120597120597) (213)

              This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

              it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

              truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

              our control because we can choose the mesh size ∆120597120597 The part of the truncation

              error that is not under our control is |120597120597empty120597120597120597120597|120585120585

              242 First Order Backward Difference

              An alternative first order finite difference formula is obtained if the Taylor series

              like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

              variables in place of all the unknowns one obtains

              empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

              +∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120597120597119894119894

              minus∆1205971205973

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              + ⋯ (214)

              Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

              to get

              120597120597empty120597120597120597120597120597120597119894119894

              =empty119894119894+1 minus empty119894119894

              ∆120597120597minus∆1205971205972

              1205971205972empty1205971205971205971205972

              120597120597119894119894

              minus∆1205971205972

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              minus ⋯ (215)

              Or using big 119978119978 notation

              120597120597empty120597120597120597120597120597120597119894119894

              =empty119894119894 minus empty119894119894minus1

              ∆120597120597+ 119978119978(∆120597120597) (216)

              22

              This is called the backward difference formula because it involves the values of

              empty at 120597120597119894119894 and 120597120597119894119894minus1

              The order of magnitude of the truncation error for the backward difference

              approximation is the same as that of the forward difference approximation Can

              we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

              truncation error The answer is yes

              242 First Order Central Difference

              Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

              empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

              +∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120597120597119894119894

              +∆1205971205973

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              + ⋯ (217)

              empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

              +∆1205971205972

              2 1205971205972empty1205971205971205971205972

              120597120597119894119894

              minus∆1205971205973

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              + ⋯ (218)

              Subtracting Equation (10) from Equation (9) yields

              empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

              + 2∆1205971205973

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              + ⋯ (219)

              Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

              120597120597empty120597120597120597120597120597120597119894119894

              =empty119894119894+1 minus empty119894119894minus1

              2∆120597120597minus∆1205971205972

              3 1205971205973empty1205971205971205971205973

              120597120597119894119894

              minus ⋯ (220)

              or

              120597120597empty120597120597120597120597120597120597119894119894

              =empty119894119894+1 minus empty119894119894minus1

              2∆120597120597+ 119978119978(∆1205971205972) (221)

              This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

              approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

              the truncation error for the central difference approximation goes to zero much

              faster than the truncation error in forward and backward equations

              23

              25 Procedures

              The simple case in this investigation was assuming the constant thermal

              properties of the material First we assumed all the thermal properties of the

              materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

              point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

              assume the constant energy after that the pulse of special shapes was selected

              The numerical solution of equation (23) with boundary and initial conditions

              in equation (22) was investigated using Matlab program as shown in Appendix

              The equation of thermal conductivity and specific heat capacity of metal as a

              function of temperature was obtained by best fitting of polynomials using

              tabulated data in references

              24

              Chapter Three

              Results and Discursion

              31 Introduction

              The development of laser has been an exciting chapter in the history of

              science and engineering It has produced a new type of advice with potential for

              application in an extremely wide variety of fields Mach basic development in

              lasers were occurred during last 35 years The lasers interaction with metal and

              vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

              The status of laser development and application were still rather rudimentary

              The light emitted by laser is electro magnetic radiation this radiation has a wave

              nature the waves consists of vibrating electric and magnetic fields many studies

              have tried to find and solve models of laser interactions Some researchers

              proposed the mathematical model related to the laser - plasma interaction and

              the others have developed an analytical model to study the temperature

              distribution in Infrared optical materials heated by laser pulses Also an attempt

              have made to study the interaction of nanosecond pulsed lasers with material

              from point of view using experimental technique and theoretical approach of

              dimensional analysis

              In this study we have evaluate the solution of partial difference equation

              (PDE) that represent the laser interaction with solid situation in one dimension

              assuming that the power density of laser and thermal properties are functions

              with time and temperature respectively

              25

              32 Numerical solution with constant laser power density and constant

              thermal properties

              First we have taken the lead metal (Pb) with thermal properties

              119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

              119862119862 = 014016119869119869119892119892119870119870

              120588120588 = 10751 1198921198921198981198981198981198983

              119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

              119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

              and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

              represent the area under laser influence

              The numerical solution of equation (23) with boundary and initial conditions

              in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

              of lead metal by explicit method using Matlab program give us the results as

              shown in Fig (31)

              Fig(31) Depth dependence of the temperature with the laser power density

              1198681198680 = 76 times 106 119882119882119898119898119898119898 2

              26

              33 Evaluation of function 119920119920(119957119957) of laser flux density

              From following data that represent the energy (119869119869) with time (millie second)

              Time 0 001 01 02 03 04 05 06 07 08

              Energy 0 002 017 022 024 02 012 007 002 0

              By using Matlab program the best polynomial with deduced from above data

              was

              119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

              minus 10069 1198981198985 (31)

              As shown in Fig (32)

              Fig(32) Laser energy as a function of time

              Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

              normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

              maximum value (119864119864119898119898119907119907119898119898 )

              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

              (119864119864119898119898119907119907119898119898 ) (32)

              The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

              27

              Fig(33) Normalized laser energy as a function of time

              The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

              equal to 3 (total laser energy) ie

              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

              08

              00

              119899119899119898119898 = 3 (33)

              Therefore there exist a real number 119875119875 such that

              119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

              08

              00

              119899119899119898119898 = 3 (34)

              that implies 119875119875 = 68241 and

              119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

              08

              00

              119899119899119898119898 = 3 (35)

              The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

              08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

              119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

              00

              (36)

              28

              Where 119863119863119898119898 put to balance the units of equation (36)

              But integral

              119868119868 = 119864119864119860119860

              (37)

              and from equations (35) (36) and (37) we have

              119868119868 (119898119898)11989911989911989811989808

              00

              = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

              0800 119899119899119898119898

              119860119860 119863119863119898119898 (38)

              Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

              (38)

              There fore

              119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

              119860119860 119863119863119898119898 (39)

              As shown in Fig(34) Matlab program was used to obtain the best polynomial

              that agrees with result data

              119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

              + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

              Fig(34) Time dependence of laser intensity

              29

              34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

              constant thermal properties

              With all constant thermal properties of lead metal as in article (23) and

              119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

              equation (23) with boundary and initial condition as in equation (22) and the

              depth penetration is shown in Fig(35)

              Fig(35) Depth dependence of the temperature when laser intensity function

              of time and constant thermal properties of Lead

              35 Evaluation the Thermal Conductivity as functions of temperature

              The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

              temperature for Lead material was obtained by Matlab program using the

              experimental data tabulated in researches

              119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

              times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

              30

              119879119879 ( 119870119870) 119870119870 119869119869

              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

              300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

              The previous thermal conductivity data and the best fitting of the data are

              shown in Fig (36)

              Fig(36) The best fitting of thermal conductivity of Lead as a function of

              temperature

              31

              36 Evaluation the Specific heat as functions of temperature

              The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

              material was obtained from the following experimental data tacked from

              literatures

              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

              300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

              The best polynomial fitted for these data was

              119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

              + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

              times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

              The specific heat capacity data and the best polynomial fitting of the data are

              shown in Fig (37)

              32

              Fig(37) The best fitting of specific heat capacity of Lead as a function of

              temperature

              37 Evaluation the Density as functions of temperature

              The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

              was used to find the best polynomial fitting

              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

              300 11330 400 11230 500 11130 600 11010 800 10430

              1000 10190 1200 9940

              The best polynomial of this data was

              120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

              minus 45158 times 10minus12 1198791198794 (313)

              33

              The density of Lead as a function of temperature and the best polynomial fitting

              are shown in Fig (38)

              Fig(38) The best fitting of density of Lead as a function of temperature

              38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

              variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

              We have deduced the solution of equation (24) with initial and boundary

              condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

              and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

              respectively then by using Matlab program the depth penetration is shown in

              Fig (39)

              34

              Fig(39) Depth dependence of the temperature for pulse laser on Lead

              material

              39 Laser interaction with copper material

              The same time dependence of laser intensity as shown in Fig(34) with

              thermal properties of copper was used to calculate the temperature distribution as

              a function of depth penetration

              The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

              copper material was obtained from the experimental data tabulated in literary

              The Matlab program used to obtain the best polynomial equation that agrees

              with the above data

              119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

              minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

              times 10minus17 1198791198795 (314)

              35

              119879119879 ( 119870119870) 119870119870 119869119869

              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

              100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

              The previous thermal conductivity data and the best fitting of the data are

              shown in Fig (310)

              Fig(310) The best fitting of thermal conductivity of Copper as a function of

              temperature

              36

              The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

              material was obtained from the following experimental data tacked from

              literatures

              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

              100 0254

              200 0357

              273 0384

              298 0387

              400 0397

              600 0416

              800 0435

              1000 0454

              1100 0464

              1200 0474

              1300 0483

              The best polynomial fitted for these data was

              119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

              + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

              times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

              The specific heat capacity data and the best polynomial fitting of the data are

              shown in Fig (311)

              37

              Fig(311) The best fitting of specific heat capacity of Copper as a function of

              temperature

              The density of copper 120588120588(119879119879) as a function of temperature tacked from

              literature was used to find the best polynomial fitting

              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

              100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

              1000 8576 1100 8519 1200 8458 1300 8396

              38

              The best polynomial of this data was

              120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

              minus 76765 times 10minus14 1198791198794 (316)

              The density of copper as a function of temperature and the best polynomial

              fitting are shown in Fig (312)

              Fig(312) The best fitting of density of copper as a function of temperature

              The depth penetration of laser energy for copper metal was calculated using

              the polynomial equations of thermal conductivity specific heat capacity and

              density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

              (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

              function of time the result was shown in Fig (313)

              39

              The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

              for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

              copper metal so the depth penetration of laser energy of lead metal was smaller

              than that of copper metal this may be due to the high thermal conductivity and

              high specific heat capacity of copper with that of lead metal

              Fig(313) Depth dependence of the temperature for pulse laser on Copper

              material

              40

              310 Conclusions

              The Depth dependence of temperature for lead metal was investigated in two

              case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

              thermal properties (thermal conductivity specific heat) and density of metal are

              also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

              vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

              specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

              120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

              penetration depth in the first case is smaller than that of the second case about

              (190) times

              The temperature distribution as a function of depth dependence for copper

              metal was also investigated in the case when the laser intensity vary with time

              119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

              density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

              The depth penetration of laser energy of lead metal was found to be smaller

              than that of copper metal this may be due to the high thermal conductivity and

              high specific heat capacity of copper with that of lead metal

              41

              References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

              Sons Inc Hoboken New Jersey Canada (2003)

              [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

              [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

              [9]

              Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

              [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

              [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

              [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

              [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

              [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

              httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

              [10] httpwwwworldoflaserscomlaserprincipleshtm

              [11] httpenwikipediaorgwikiLaserPulsed_operation

              [12] httpenwikipediaorgwikiThermal_conductivity

              [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

              [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

              [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

              42

              Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

              This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

              This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

              43

              E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

              This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

              This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

              This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

              44

              title(Dencity as a function of temperature) hold off

              This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

              45

              elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

              46

              r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

              47

              for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

              48

              for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

              49

              6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

              This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

              50

              u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

              51

              alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

              52

              715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

              • 01 Title
                • 0TDr Omed Ghareb Abdullah
                  • Ch1-Laser
                    • 16 Heat and heat capacity
                    • 18 Derivation in one dimension
                      • Ch2 laser2
                      • Ch3 laser2
                      • Appendix-Laser

                3

                In case of a three-level laser the material is pumped from level 1 to level 3

                which decays rapidly to level 2 through spontaneous emission Level 2 is a

                metastable level and promotes stimulated emission from level 2 to level 1

                Fig(12) Energy states of Three-level active medium

                On the other hand in a four-level laser the material is pumped to level 4

                which is a fast decaying level and the atoms decay rapidly to level 3 which is

                a metastable level The stimulated emission takes place from level 3 to level 2

                from where the atoms decay back to level 1 Four level lasers is an

                improvement on a system based on three level systems In this case the laser

                transition takes place between the third and second excited states Since

                lower laser level 2 is a fast decaying level which ensures that it rapidly gets

                empty and as such always supports the population inversion condition

                Fig(13) Energy states of Four-level active medium

                4

                14 A Survey of Laser Types

                Laser technology is available to us since 1960rsquos and since then has been

                quite well developed Currently there is a great variety of lasers of different

                output power operating voltages sizes etc The major classes of lasers

                currently used are Gas Solid Molecular and Free Electron lasers Below we

                will cover some most popular representative types of lasers of each class and

                describe specific principles of operation construction and main highlights

                141 Gas Lasers

                1 Helium-Neon Laser

                The most common and inexpensive gas laser the helium-neon laser is

                usually constructed to operate in the red at 6328 nm It can also be

                constructed to produce laser action in the green at 5435 nm and in the

                infrared at 1523 nm

                One of the excited levels of helium at 2061 eV is very close to a level in

                neon at 2066 eV so close in fact that upon collision of a helium and a neon

                atom the energy can be transferred from the helium to the neon atom

                Fig (14) The components of a Hilium-Neon Laser

                5

                Fig(15) The lasing action of He-Ne laser

                Helium-Neon lasers are common in the introductory physics laboratories

                but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

                brightness equal to sunshine on a clear day (01 wattcmP

                2P) and is just as

                dangerous to stare at directly

                2- Carbon Dioxide Laser

                The carbon dioxide gas laser is capable of continuous output powers above

                10 kilowatts It is also capable of extremely high power pulse operation It

                exhibits laser action at several infrared frequencies but none in the visible

                spectrum Operating in a manner similar to the helium-neon laser it employs

                an electric discharge for pumping using a percentage of nitrogen gas as a

                pumping gas The COR2R laser is the most efficient laser capable of operating at

                more than 30 efficiency

                The carbon dioxide laser finds many applications in industry particularly for

                welding and Cutting

                6

                3- Argon Laser

                The argon ion laser can be operated as a continuous gas laser at about 25

                different wavelengths in the visible between (4089 - 6861) nm but is best

                known for its most efficient transitions in the green at 488 nm and 5145 nm

                Operating at much higher powers than the Helium-Neon gas laser it is not

                uncommon to achieve (30 ndash 100) watts of continuous power using several

                transitions This output is produced in hot plasma and takes extremely high

                power typically (9 ndash 12) kW so these are large and expensive devices

                142 Solid Lasers

                1 Ruby Laser

                The ruby laser is the first type of laser actually constructed first

                demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

                aluminum oxide with a small amount (about 005) of Chromium which gives

                it its characteristic pink or red color by absorbing green and blue light

                The ruby laser is used as a pulsed laser producing red light at 6943 nm

                After receiving a pumping flash from the flash tube the laser light emerges for

                as long as the excited atoms persist in the ruby rod which is typically about a

                millisecond

                A pulsed ruby laser was used for the famous laser ranging experiment which

                was conducted with a corner reflector placed on the Moon by the Apollo

                astronauts This determined the distance to the Moon with an accuracy of

                about 15 cm

                7

                Fig (16) Principle of operation of a Ruby laser

                2- Neodymium-YAG Laser

                An example of a solid-state laser the neodymium-YAG uses the NdP

                3+P ion to

                dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

                geometry which makes population inversion possible Neodymium-YAG lasers

                have become very important because they can be used to produce high

                powers Such lasers have been constructed to produce over a kilowatt of

                continuous laser power at 1065 nm and can achieve extremely high powers in

                a pulsed mode

                Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

                production of a series of very short pulses for research with femtosecond time

                resolution

                Fig(17) Construction of a Neodymium-YAG laser

                8

                3- Neodymium-Glass Lasers

                Neodymium glass lasers have emerged as the design choice for research in

                laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

                short as 10-12 seconds with peak powers of 109 kilowatts

                143 Molecular Lasers

                Eximer Lasers

                Eximer is a shortened form of excited dimer denoting the fact that the

                lasing medium in this type of laser is an excited diatomic molecule These

                lasers typically produce ultraviolet pulses They are under investigation for use

                in communicating with submarines by conversion to blue-green light and

                pulsing from overhead satellites through sea water to submarines below

                The eximers used are typically those formed by rare gases and halogens in

                electron excited Gas discharges Molecules like XeF are stable only in their

                excited states and quickly dissociate when they make the transition to their

                ground state This makes possible large population inversions because the

                ground state is depleted by this dissociation However the excited states are

                very short-lived compared to other laser metastable states and lasers like the

                XeF eximer laser require high pumping rates

                Eximer lasers typically produce high power pulse outputs in the blue or

                ultraviolet after excitation by fast electron-beam discharges

                The rare-gas xenon and the highly active fluorine seem unlikely to form a

                molecule but they do in the hot plasma environment of an electron-beam

                initiated gas discharge They are only stable in their excited states if stable

                can be used for molecules which undergo radioactive decay in 1 to 10

                nanoseconds This is long enough to achieve pulsed laser action in the blue-

                green over a band from 450 to 510 nm peaking at 486 nm Very high power

                9

                pulses can be achieved because the stimulated emission cross-sections of the

                laser transitions are relatively low allowing a large population inversion to

                build up The power is also enhanced by the fact that the ground state of XeF

                quickly dissociates so that there is little absorption to quench the laser pulse

                action

                144 Free-Electron Lasers

                The radiation from a free-electron laser is produced from free electrons

                which are forced to oscillate in a regular fashion by an applied field They are

                therefore more like synchrotron light sources or microwave tubes than like

                other lasers They are able to produce highly coherent collimated radiation

                over a wide range of frequencies The magnetic field arrangement which

                produces the alternating field is commonly called a wiggler magnet

                Fig(18) Principle of operation of Free-Electron laser

                The free-electron laser is a highly tunable device which has been used to

                generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

                this range they are the highest power source Applications of free-electron

                lasers are envisioned in isotope separation plasma heating for nuclear fusion

                long-range high resolution radar and particle acceleration in accelerators

                10

                15 Pulsed operation

                Pulsed operation of lasers refers to any laser not classified as continuous

                wave so that the optical power appears in pulses of some duration at some

                repetition rate This encompasses a wide range of technologies addressing a

                number of different motivations Some lasers are pulsed simply because they

                cannot be run in continuous mode

                In other cases the application requires the production of pulses having as

                large an energy as possible Since the pulse energy is equal to the average

                power divided by the repitition rate this goal can sometimes be satisfied by

                lowering the rate of pulses so that more energy can be built up in between

                pulses In laser ablation for example a small volume of material at the surface

                of a work piece can be evaporated if it is heated in a very short time whereas

                supplying the energy gradually would allow for the heat to be absorbed into

                the bulk of the piece never attaining a sufficiently high temperature at a

                particular point

                Other applications rely on the peak pulse power (rather than the energy in

                the pulse) especially in order to obtain nonlinear optical effects For a given

                pulse energy this requires creating pulses of the shortest possible duration

                utilizing techniques such as Q-switching

                16 Heat and heat capacity

                When a sample is heated meaning it receives thermal energy from an

                external source some of the introduced heat is converted into kinetic energy

                the rest to other forms of internal energy specific to the material The amount

                converted into kinetic energy causes the temperature of the material to rise

                The amount of the temperature increase depends on how much heat was

                added the size of the sample the original temperature of the sample and on

                how the heat was added The two obvious choices on how to add the heat are

                11

                to add it holding volume constant or to add it holding pressure constant

                (There may be other choices but they will not concern us)

                Lets assume for the moment that we are going to add heat to our sample

                holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                lim119876119876119889119889rarr0

                119876119876119889119889∆120549120549119889119889

                = 120597120597119876119876120597120597120549120549119889119889

                = 119862119862119889119889 (11)

                We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                capacity at constant volume Usually one quotes the molar heat capacity

                119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                (12)

                We can rearrange Equation (11) as follows

                119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                Then we can integrate this equation to find the heat involved in a finite

                change at constant volume

                119876119876119889119889 = 119862119862119889119889

                1205491205492

                1205491205491

                119889119889120549120549 (14)

                If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                out of the integral and the heat at constant volume becomes

                119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                Let us now go through the same sequence of steps except holding pressure

                constant instead of volume Our initial definition of the heat capacity at

                constant pressure 119862119862119875119875 R Rbecomes

                lim119876119876119875119875rarr0

                119876119876119875119875∆120549120549119875119875

                = 120597120597119876119876120597120597120549120549119875119875

                = 119862119862119875119875 (16)

                The analogous molar heat capacity is

                12

                119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                (17)

                Equation (16) rearranges to

                119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                which integrates to give

                119876119876119875119875 = 119862119862119875119875

                1205491205492

                1205491205491

                119889119889120549120549 (19)

                When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                regarded as constant In these cases the heat capacity is fit to a polynomial (or

                similar function) in 120549120549 For example some tables give the heat capacity as

                119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                dependent heat capacity the heat at constant pressure would integrate as

                follows

                119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                1205491205492

                1205491205491

                119889119889120549120549 (112)

                119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                2 minus 12054912054912 + 119899119899

                1205741205743

                12054912054923 minus 1205491205491

                3 (113)

                Occasionally one finds a different form for the temperature dependent heat

                capacity in the literature

                119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                When you do calculations with temperature dependent heat capacities you

                must check to see which form is being used for 119862119862119875119875 We are using the

                convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                13

                positive then heat was indeed absorbed by the system On the other hand if

                119876119876 is negative it means that the system gave up heat to the surroundings

                17 Thermal conductivity

                In physics thermal conductivity 119896119896 is the property of a material that indicates

                its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                a piece of material The reciprocal of thermal conductivity is thermal

                resistivity

                18 Derivation in one dimension

                The heat equation is derived from Fouriers law and conservation of energy

                (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                surface is proportional to the negative temperature gradient across the

                surface

                119902119902 = minus119896119896 120571120571120549120549 (115)

                where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                119902119902 = minus119896119896 120549120549119909119909 (116)

                where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                temperature 120549120549120549120549 That is

                ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                material Choosing zero energy at absolute zero temperature this can be

                rewritten as

                ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                14

                The increase in internal energy in a small spatial region of the material

                (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                given by

                119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                119909119909minus∆119909119909

                = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                119909119909minus∆119909119909

                119905119905+∆119905119905

                119905119905minus∆119905119905

                (119)

                Where the fundamental theorem of calculus was used Additionally with no

                work done and absent any heat sources or sinks the change in internal energy

                in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                across the boundaries By Fouriers law this is

                119896119896 120597120597120549120549120597120597119909119909

                (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                119905119905minus∆119905119905

                = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                119909119909+∆119909119909

                119909119909minus∆119909119909

                119905119905+∆119905119905

                119905119905minus∆119905119905

                (120)

                again by the fundamental theorem of calculus By conservation of energy

                119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                119909119909minus∆119909119909

                119905119905+∆119905119905

                119905119905minus∆119905119905

                = 0 (121)

                This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                Which can be rewritten as

                120549120549119905119905 =119896119896119888119888119875119875 120588120588

                120549120549119909119909119909119909 (122)

                or

                120597120597120549120549120597120597119905119905

                =119896119896119888119888119875119875 120588120588

                12059712059721205491205491205971205971199091199092 (123)

                15

                which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                diffusivity and is often denoted 120572120572

                19 Aim of present work

                The goal of this study is to estimate the solution of partial differential

                equation that governs the laser-solid interaction using numerical methods

                The solution will been restricted into one dimensional situation in which we

                assume that both the laser power density and thermal properties are

                functions of time and temperature respectively In this project we attempt to

                investigate the laser interaction with both lead and copper materials by

                predicting the temperature gradient with the depth of the metals

                16

                Chapter Two

                Theoretical Aspects

                21 Introduction

                When a laser interacts with a solid surface a variety of processes can

                occur We are mainly interested in the interaction of pulsed lasers with a

                solid surface in first instance a metal When such a laser interacts with a

                copper surface the laser energy will be transformed into heat The

                temperature of the solid material will increase leading to melting and

                evaporation of the solid material

                The evaporated material (vapour atoms) will expand Depending on the

                applications this can happen in vacuum (or very low pressure) or in a

                background gas (helium argon air)

                22 One dimension laser heating equation

                In general the one dimension laser heating processes of opaque solid slab is

                represented as

                120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                ( 119870119870 119879119879120597120597 ) (21)

                With boundary conditions and initial condition which represent the pre-

                vaporization stage

                minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                17

                119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                where

                119870119870 represents the thermal conductivity

                120588120588 represents the density

                119862119862 represents the specific heat

                119879119879 represents the temperature

                119879119879infin represents the ambient temperature

                119879119879119907119907 represents the front surface vaporization

                120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                With the same boundary conditions as in equation (22)

                where 119889119889119889119889 = 119870119870120588120588119862119862

                which represents the thermal diffusion

                But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                equation (21) with this assuming implies

                119879119879119905119905 = 1

                120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                With the same boundary and initial conditions in equation (22) Where 119870119870

                represents the derivative of K with respect the temperature

                23 Numerical solution of Initial value problems

                An immense number of analytical solutions for conduction heat-transfer

                problems have been accumulated in literature over the past 100 years Even so

                in many practical situations the geometry or boundary conditions are such that an

                analytical solution has not been obtained at all or if the solution has been

                18

                developed it involves such a complex series solution that numerical evaluation

                becomes exceedingly difficult For such situation the most fruitful approach to

                the problem is numerical techniques the basic principles of which we shall

                outline in this section

                One way to guarantee accuracy in the solution of an initial values problems

                (IVP) is to solve the problem twice using step sizes h and h2 and compare

                answers at the mesh points corresponding to the larger step size But this requires

                a significant amount of computation for the smaller step size and must be

                repeated if it is determined that the agreement is not good enough

                24 Finite Difference Method

                The finite difference method is one of several techniques for obtaining

                numerical solutions to differential equations In all numerical solutions the

                continuous partial differential equation (PDE) is replaced with a discrete

                approximation In this context the word discrete means that the numerical

                solution is known only at a finite number of points in the physical domain The

                number of those points can be selected by the user of the numerical method In

                general increasing the number of points not only increases the resolution but

                also the accuracy of the numerical solution

                The discrete approximation results in a set of algebraic equations that are

                evaluated for the values of the discrete unknowns

                The mesh is the set of locations where the discrete solution is computed

                These points are called nodes and if one were to draw lines between adjacent

                nodes in the domain the resulting image would resemble a net or mesh Two key

                parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                space and ∆119905119905 the local distance between adjacent time steps For the simple

                examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                19

                The core idea of the finite-difference method is to replace continuous

                derivatives with so-called difference formulas that involve only the discrete

                values associated with positions on the mesh

                Applying the finite-difference method to a differential equation involves

                replacing all derivatives with difference formulas In the heat equation there are

                derivatives with respect to time and derivatives with respect to space Using

                different combinations of mesh points in the difference formulas results in

                different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                numerical solution obtained with any useful scheme will approach the true

                solution to the original differential equation However the rate at which the

                numerical solution approaches the true solution varies with the scheme

                241 First Order Forward Difference

                Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                +1205751205751205971205972

                2 1205971205972empty1205971205971205971205972

                1205971205971

                +1205751205751205971205973

                3 1205971205973empty1205971205971205971205973

                1205971205971

                + ⋯ (25)

                where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                +∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120597120597119894119894

                +∆1205971205973

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                + ⋯ (26)

                Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                120597120597empty120597120597120597120597120597120597119894119894

                =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                ∆120597120597minus∆1205971205972

                1205971205972empty1205971205971205971205972

                120597120597119894119894

                minus∆1205971205972

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                minus ⋯ (27)

                Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                hand side have been reduced by one

                20

                Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                120597120597empty120597120597120597120597120597120597119894119894

                =empty119894119894+1 minus empty119894119894

                ∆120597120597minus∆1205971205972

                1205971205972empty1205971205971205971205972

                120597120597119894119894

                minus∆1205971205972

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                minus ⋯ (28)

                The mean value theorem can be used to replace the higher order derivatives

                ∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120597120597119894119894

                +∆1205971205973

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                + ⋯ =∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120585120585 (29)

                where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                ∆120597120597+∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120585120585 (210)

                120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                ∆120597120597asymp∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120585120585 (211)

                The term on the right hand side of previous equation is called the truncation

                error of the finite difference approximation

                In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                analytical form) the big 119978119978 notation can be used to express the dependence of

                the truncation error on the mesh spacing Note that the right hand side of last

                equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                the finite difference simulation Since this is the only parameter under the users

                control that determines the error the truncation error is simply written

                ∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120585120585= 119978119978(∆1205971205972) (212)

                The equals sign in this expression is true in the order of magnitude sense In

                other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                21

                equality Rather the expression means that the left hand side is a product of an

                unknown constant and ∆1205971205972 Although the expression does not give us the exact

                magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                approaches zero as ∆120597120597 is reduced

                Using big 119978119978 notation Equation (28) can be written

                120597120597empty120597120597120597120597120597120597119894119894

                =empty119894119894+1 minus empty119894119894

                ∆120597120597+ 119978119978(∆120597120597) (213)

                This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                our control because we can choose the mesh size ∆120597120597 The part of the truncation

                error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                242 First Order Backward Difference

                An alternative first order finite difference formula is obtained if the Taylor series

                like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                variables in place of all the unknowns one obtains

                empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                +∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120597120597119894119894

                minus∆1205971205973

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                + ⋯ (214)

                Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                to get

                120597120597empty120597120597120597120597120597120597119894119894

                =empty119894119894+1 minus empty119894119894

                ∆120597120597minus∆1205971205972

                1205971205972empty1205971205971205971205972

                120597120597119894119894

                minus∆1205971205972

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                minus ⋯ (215)

                Or using big 119978119978 notation

                120597120597empty120597120597120597120597120597120597119894119894

                =empty119894119894 minus empty119894119894minus1

                ∆120597120597+ 119978119978(∆120597120597) (216)

                22

                This is called the backward difference formula because it involves the values of

                empty at 120597120597119894119894 and 120597120597119894119894minus1

                The order of magnitude of the truncation error for the backward difference

                approximation is the same as that of the forward difference approximation Can

                we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                truncation error The answer is yes

                242 First Order Central Difference

                Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                +∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120597120597119894119894

                +∆1205971205973

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                + ⋯ (217)

                empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                +∆1205971205972

                2 1205971205972empty1205971205971205971205972

                120597120597119894119894

                minus∆1205971205973

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                + ⋯ (218)

                Subtracting Equation (10) from Equation (9) yields

                empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                + 2∆1205971205973

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                + ⋯ (219)

                Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                120597120597empty120597120597120597120597120597120597119894119894

                =empty119894119894+1 minus empty119894119894minus1

                2∆120597120597minus∆1205971205972

                3 1205971205973empty1205971205971205971205973

                120597120597119894119894

                minus ⋯ (220)

                or

                120597120597empty120597120597120597120597120597120597119894119894

                =empty119894119894+1 minus empty119894119894minus1

                2∆120597120597+ 119978119978(∆1205971205972) (221)

                This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                the truncation error for the central difference approximation goes to zero much

                faster than the truncation error in forward and backward equations

                23

                25 Procedures

                The simple case in this investigation was assuming the constant thermal

                properties of the material First we assumed all the thermal properties of the

                materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                assume the constant energy after that the pulse of special shapes was selected

                The numerical solution of equation (23) with boundary and initial conditions

                in equation (22) was investigated using Matlab program as shown in Appendix

                The equation of thermal conductivity and specific heat capacity of metal as a

                function of temperature was obtained by best fitting of polynomials using

                tabulated data in references

                24

                Chapter Three

                Results and Discursion

                31 Introduction

                The development of laser has been an exciting chapter in the history of

                science and engineering It has produced a new type of advice with potential for

                application in an extremely wide variety of fields Mach basic development in

                lasers were occurred during last 35 years The lasers interaction with metal and

                vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                The status of laser development and application were still rather rudimentary

                The light emitted by laser is electro magnetic radiation this radiation has a wave

                nature the waves consists of vibrating electric and magnetic fields many studies

                have tried to find and solve models of laser interactions Some researchers

                proposed the mathematical model related to the laser - plasma interaction and

                the others have developed an analytical model to study the temperature

                distribution in Infrared optical materials heated by laser pulses Also an attempt

                have made to study the interaction of nanosecond pulsed lasers with material

                from point of view using experimental technique and theoretical approach of

                dimensional analysis

                In this study we have evaluate the solution of partial difference equation

                (PDE) that represent the laser interaction with solid situation in one dimension

                assuming that the power density of laser and thermal properties are functions

                with time and temperature respectively

                25

                32 Numerical solution with constant laser power density and constant

                thermal properties

                First we have taken the lead metal (Pb) with thermal properties

                119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                119862119862 = 014016119869119869119892119892119870119870

                120588120588 = 10751 1198921198921198981198981198981198983

                119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                represent the area under laser influence

                The numerical solution of equation (23) with boundary and initial conditions

                in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                of lead metal by explicit method using Matlab program give us the results as

                shown in Fig (31)

                Fig(31) Depth dependence of the temperature with the laser power density

                1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                26

                33 Evaluation of function 119920119920(119957119957) of laser flux density

                From following data that represent the energy (119869119869) with time (millie second)

                Time 0 001 01 02 03 04 05 06 07 08

                Energy 0 002 017 022 024 02 012 007 002 0

                By using Matlab program the best polynomial with deduced from above data

                was

                119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                minus 10069 1198981198985 (31)

                As shown in Fig (32)

                Fig(32) Laser energy as a function of time

                Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                maximum value (119864119864119898119898119907119907119898119898 )

                119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                (119864119864119898119898119907119907119898119898 ) (32)

                The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                27

                Fig(33) Normalized laser energy as a function of time

                The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                equal to 3 (total laser energy) ie

                119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                08

                00

                119899119899119898119898 = 3 (33)

                Therefore there exist a real number 119875119875 such that

                119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                08

                00

                119899119899119898119898 = 3 (34)

                that implies 119875119875 = 68241 and

                119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                08

                00

                119899119899119898119898 = 3 (35)

                The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                00

                (36)

                28

                Where 119863119863119898119898 put to balance the units of equation (36)

                But integral

                119868119868 = 119864119864119860119860

                (37)

                and from equations (35) (36) and (37) we have

                119868119868 (119898119898)11989911989911989811989808

                00

                = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                0800 119899119899119898119898

                119860119860 119863119863119898119898 (38)

                Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                (38)

                There fore

                119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                119860119860 119863119863119898119898 (39)

                As shown in Fig(34) Matlab program was used to obtain the best polynomial

                that agrees with result data

                119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                Fig(34) Time dependence of laser intensity

                29

                34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                constant thermal properties

                With all constant thermal properties of lead metal as in article (23) and

                119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                equation (23) with boundary and initial condition as in equation (22) and the

                depth penetration is shown in Fig(35)

                Fig(35) Depth dependence of the temperature when laser intensity function

                of time and constant thermal properties of Lead

                35 Evaluation the Thermal Conductivity as functions of temperature

                The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                temperature for Lead material was obtained by Matlab program using the

                experimental data tabulated in researches

                119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                30

                119879119879 ( 119870119870) 119870119870 119869119869

                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                The previous thermal conductivity data and the best fitting of the data are

                shown in Fig (36)

                Fig(36) The best fitting of thermal conductivity of Lead as a function of

                temperature

                31

                36 Evaluation the Specific heat as functions of temperature

                The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                material was obtained from the following experimental data tacked from

                literatures

                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                The best polynomial fitted for these data was

                119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                The specific heat capacity data and the best polynomial fitting of the data are

                shown in Fig (37)

                32

                Fig(37) The best fitting of specific heat capacity of Lead as a function of

                temperature

                37 Evaluation the Density as functions of temperature

                The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                was used to find the best polynomial fitting

                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                300 11330 400 11230 500 11130 600 11010 800 10430

                1000 10190 1200 9940

                The best polynomial of this data was

                120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                minus 45158 times 10minus12 1198791198794 (313)

                33

                The density of Lead as a function of temperature and the best polynomial fitting

                are shown in Fig (38)

                Fig(38) The best fitting of density of Lead as a function of temperature

                38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                We have deduced the solution of equation (24) with initial and boundary

                condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                respectively then by using Matlab program the depth penetration is shown in

                Fig (39)

                34

                Fig(39) Depth dependence of the temperature for pulse laser on Lead

                material

                39 Laser interaction with copper material

                The same time dependence of laser intensity as shown in Fig(34) with

                thermal properties of copper was used to calculate the temperature distribution as

                a function of depth penetration

                The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                copper material was obtained from the experimental data tabulated in literary

                The Matlab program used to obtain the best polynomial equation that agrees

                with the above data

                119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                times 10minus17 1198791198795 (314)

                35

                119879119879 ( 119870119870) 119870119870 119869119869

                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                The previous thermal conductivity data and the best fitting of the data are

                shown in Fig (310)

                Fig(310) The best fitting of thermal conductivity of Copper as a function of

                temperature

                36

                The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                material was obtained from the following experimental data tacked from

                literatures

                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                100 0254

                200 0357

                273 0384

                298 0387

                400 0397

                600 0416

                800 0435

                1000 0454

                1100 0464

                1200 0474

                1300 0483

                The best polynomial fitted for these data was

                119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                The specific heat capacity data and the best polynomial fitting of the data are

                shown in Fig (311)

                37

                Fig(311) The best fitting of specific heat capacity of Copper as a function of

                temperature

                The density of copper 120588120588(119879119879) as a function of temperature tacked from

                literature was used to find the best polynomial fitting

                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                1000 8576 1100 8519 1200 8458 1300 8396

                38

                The best polynomial of this data was

                120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                minus 76765 times 10minus14 1198791198794 (316)

                The density of copper as a function of temperature and the best polynomial

                fitting are shown in Fig (312)

                Fig(312) The best fitting of density of copper as a function of temperature

                The depth penetration of laser energy for copper metal was calculated using

                the polynomial equations of thermal conductivity specific heat capacity and

                density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                function of time the result was shown in Fig (313)

                39

                The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                copper metal so the depth penetration of laser energy of lead metal was smaller

                than that of copper metal this may be due to the high thermal conductivity and

                high specific heat capacity of copper with that of lead metal

                Fig(313) Depth dependence of the temperature for pulse laser on Copper

                material

                40

                310 Conclusions

                The Depth dependence of temperature for lead metal was investigated in two

                case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                thermal properties (thermal conductivity specific heat) and density of metal are

                also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                penetration depth in the first case is smaller than that of the second case about

                (190) times

                The temperature distribution as a function of depth dependence for copper

                metal was also investigated in the case when the laser intensity vary with time

                119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                The depth penetration of laser energy of lead metal was found to be smaller

                than that of copper metal this may be due to the high thermal conductivity and

                high specific heat capacity of copper with that of lead metal

                41

                References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                Sons Inc Hoboken New Jersey Canada (2003)

                [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                [9]

                Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                [10] httpwwwworldoflaserscomlaserprincipleshtm

                [11] httpenwikipediaorgwikiLaserPulsed_operation

                [12] httpenwikipediaorgwikiThermal_conductivity

                [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                42

                Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                43

                E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                44

                title(Dencity as a function of temperature) hold off

                This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                45

                elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                46

                r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                47

                for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                48

                for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                49

                6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                50

                u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                51

                alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                52

                715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                • 01 Title
                  • 0TDr Omed Ghareb Abdullah
                    • Ch1-Laser
                      • 16 Heat and heat capacity
                      • 18 Derivation in one dimension
                        • Ch2 laser2
                        • Ch3 laser2
                        • Appendix-Laser

                  4

                  14 A Survey of Laser Types

                  Laser technology is available to us since 1960rsquos and since then has been

                  quite well developed Currently there is a great variety of lasers of different

                  output power operating voltages sizes etc The major classes of lasers

                  currently used are Gas Solid Molecular and Free Electron lasers Below we

                  will cover some most popular representative types of lasers of each class and

                  describe specific principles of operation construction and main highlights

                  141 Gas Lasers

                  1 Helium-Neon Laser

                  The most common and inexpensive gas laser the helium-neon laser is

                  usually constructed to operate in the red at 6328 nm It can also be

                  constructed to produce laser action in the green at 5435 nm and in the

                  infrared at 1523 nm

                  One of the excited levels of helium at 2061 eV is very close to a level in

                  neon at 2066 eV so close in fact that upon collision of a helium and a neon

                  atom the energy can be transferred from the helium to the neon atom

                  Fig (14) The components of a Hilium-Neon Laser

                  5

                  Fig(15) The lasing action of He-Ne laser

                  Helium-Neon lasers are common in the introductory physics laboratories

                  but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

                  brightness equal to sunshine on a clear day (01 wattcmP

                  2P) and is just as

                  dangerous to stare at directly

                  2- Carbon Dioxide Laser

                  The carbon dioxide gas laser is capable of continuous output powers above

                  10 kilowatts It is also capable of extremely high power pulse operation It

                  exhibits laser action at several infrared frequencies but none in the visible

                  spectrum Operating in a manner similar to the helium-neon laser it employs

                  an electric discharge for pumping using a percentage of nitrogen gas as a

                  pumping gas The COR2R laser is the most efficient laser capable of operating at

                  more than 30 efficiency

                  The carbon dioxide laser finds many applications in industry particularly for

                  welding and Cutting

                  6

                  3- Argon Laser

                  The argon ion laser can be operated as a continuous gas laser at about 25

                  different wavelengths in the visible between (4089 - 6861) nm but is best

                  known for its most efficient transitions in the green at 488 nm and 5145 nm

                  Operating at much higher powers than the Helium-Neon gas laser it is not

                  uncommon to achieve (30 ndash 100) watts of continuous power using several

                  transitions This output is produced in hot plasma and takes extremely high

                  power typically (9 ndash 12) kW so these are large and expensive devices

                  142 Solid Lasers

                  1 Ruby Laser

                  The ruby laser is the first type of laser actually constructed first

                  demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

                  aluminum oxide with a small amount (about 005) of Chromium which gives

                  it its characteristic pink or red color by absorbing green and blue light

                  The ruby laser is used as a pulsed laser producing red light at 6943 nm

                  After receiving a pumping flash from the flash tube the laser light emerges for

                  as long as the excited atoms persist in the ruby rod which is typically about a

                  millisecond

                  A pulsed ruby laser was used for the famous laser ranging experiment which

                  was conducted with a corner reflector placed on the Moon by the Apollo

                  astronauts This determined the distance to the Moon with an accuracy of

                  about 15 cm

                  7

                  Fig (16) Principle of operation of a Ruby laser

                  2- Neodymium-YAG Laser

                  An example of a solid-state laser the neodymium-YAG uses the NdP

                  3+P ion to

                  dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

                  geometry which makes population inversion possible Neodymium-YAG lasers

                  have become very important because they can be used to produce high

                  powers Such lasers have been constructed to produce over a kilowatt of

                  continuous laser power at 1065 nm and can achieve extremely high powers in

                  a pulsed mode

                  Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

                  production of a series of very short pulses for research with femtosecond time

                  resolution

                  Fig(17) Construction of a Neodymium-YAG laser

                  8

                  3- Neodymium-Glass Lasers

                  Neodymium glass lasers have emerged as the design choice for research in

                  laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

                  short as 10-12 seconds with peak powers of 109 kilowatts

                  143 Molecular Lasers

                  Eximer Lasers

                  Eximer is a shortened form of excited dimer denoting the fact that the

                  lasing medium in this type of laser is an excited diatomic molecule These

                  lasers typically produce ultraviolet pulses They are under investigation for use

                  in communicating with submarines by conversion to blue-green light and

                  pulsing from overhead satellites through sea water to submarines below

                  The eximers used are typically those formed by rare gases and halogens in

                  electron excited Gas discharges Molecules like XeF are stable only in their

                  excited states and quickly dissociate when they make the transition to their

                  ground state This makes possible large population inversions because the

                  ground state is depleted by this dissociation However the excited states are

                  very short-lived compared to other laser metastable states and lasers like the

                  XeF eximer laser require high pumping rates

                  Eximer lasers typically produce high power pulse outputs in the blue or

                  ultraviolet after excitation by fast electron-beam discharges

                  The rare-gas xenon and the highly active fluorine seem unlikely to form a

                  molecule but they do in the hot plasma environment of an electron-beam

                  initiated gas discharge They are only stable in their excited states if stable

                  can be used for molecules which undergo radioactive decay in 1 to 10

                  nanoseconds This is long enough to achieve pulsed laser action in the blue-

                  green over a band from 450 to 510 nm peaking at 486 nm Very high power

                  9

                  pulses can be achieved because the stimulated emission cross-sections of the

                  laser transitions are relatively low allowing a large population inversion to

                  build up The power is also enhanced by the fact that the ground state of XeF

                  quickly dissociates so that there is little absorption to quench the laser pulse

                  action

                  144 Free-Electron Lasers

                  The radiation from a free-electron laser is produced from free electrons

                  which are forced to oscillate in a regular fashion by an applied field They are

                  therefore more like synchrotron light sources or microwave tubes than like

                  other lasers They are able to produce highly coherent collimated radiation

                  over a wide range of frequencies The magnetic field arrangement which

                  produces the alternating field is commonly called a wiggler magnet

                  Fig(18) Principle of operation of Free-Electron laser

                  The free-electron laser is a highly tunable device which has been used to

                  generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

                  this range they are the highest power source Applications of free-electron

                  lasers are envisioned in isotope separation plasma heating for nuclear fusion

                  long-range high resolution radar and particle acceleration in accelerators

                  10

                  15 Pulsed operation

                  Pulsed operation of lasers refers to any laser not classified as continuous

                  wave so that the optical power appears in pulses of some duration at some

                  repetition rate This encompasses a wide range of technologies addressing a

                  number of different motivations Some lasers are pulsed simply because they

                  cannot be run in continuous mode

                  In other cases the application requires the production of pulses having as

                  large an energy as possible Since the pulse energy is equal to the average

                  power divided by the repitition rate this goal can sometimes be satisfied by

                  lowering the rate of pulses so that more energy can be built up in between

                  pulses In laser ablation for example a small volume of material at the surface

                  of a work piece can be evaporated if it is heated in a very short time whereas

                  supplying the energy gradually would allow for the heat to be absorbed into

                  the bulk of the piece never attaining a sufficiently high temperature at a

                  particular point

                  Other applications rely on the peak pulse power (rather than the energy in

                  the pulse) especially in order to obtain nonlinear optical effects For a given

                  pulse energy this requires creating pulses of the shortest possible duration

                  utilizing techniques such as Q-switching

                  16 Heat and heat capacity

                  When a sample is heated meaning it receives thermal energy from an

                  external source some of the introduced heat is converted into kinetic energy

                  the rest to other forms of internal energy specific to the material The amount

                  converted into kinetic energy causes the temperature of the material to rise

                  The amount of the temperature increase depends on how much heat was

                  added the size of the sample the original temperature of the sample and on

                  how the heat was added The two obvious choices on how to add the heat are

                  11

                  to add it holding volume constant or to add it holding pressure constant

                  (There may be other choices but they will not concern us)

                  Lets assume for the moment that we are going to add heat to our sample

                  holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                  subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                  be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                  amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                  (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                  lim119876119876119889119889rarr0

                  119876119876119889119889∆120549120549119889119889

                  = 120597120597119876119876120597120597120549120549119889119889

                  = 119862119862119889119889 (11)

                  We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                  capacity at constant volume Usually one quotes the molar heat capacity

                  119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                  (12)

                  We can rearrange Equation (11) as follows

                  119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                  Then we can integrate this equation to find the heat involved in a finite

                  change at constant volume

                  119876119876119889119889 = 119862119862119889119889

                  1205491205492

                  1205491205491

                  119889119889120549120549 (14)

                  If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                  out of the integral and the heat at constant volume becomes

                  119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                  Let us now go through the same sequence of steps except holding pressure

                  constant instead of volume Our initial definition of the heat capacity at

                  constant pressure 119862119862119875119875 R Rbecomes

                  lim119876119876119875119875rarr0

                  119876119876119875119875∆120549120549119875119875

                  = 120597120597119876119876120597120597120549120549119875119875

                  = 119862119862119875119875 (16)

                  The analogous molar heat capacity is

                  12

                  119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                  (17)

                  Equation (16) rearranges to

                  119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                  which integrates to give

                  119876119876119875119875 = 119862119862119875119875

                  1205491205492

                  1205491205491

                  119889119889120549120549 (19)

                  When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                  119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                  Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                  regarded as constant In these cases the heat capacity is fit to a polynomial (or

                  similar function) in 120549120549 For example some tables give the heat capacity as

                  119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                  where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                  dependent heat capacity the heat at constant pressure would integrate as

                  follows

                  119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                  1205491205492

                  1205491205491

                  119889119889120549120549 (112)

                  119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                  2 minus 12054912054912 + 119899119899

                  1205741205743

                  12054912054923 minus 1205491205491

                  3 (113)

                  Occasionally one finds a different form for the temperature dependent heat

                  capacity in the literature

                  119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                  When you do calculations with temperature dependent heat capacities you

                  must check to see which form is being used for 119862119862119875119875 We are using the

                  convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                  be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                  13

                  positive then heat was indeed absorbed by the system On the other hand if

                  119876119876 is negative it means that the system gave up heat to the surroundings

                  17 Thermal conductivity

                  In physics thermal conductivity 119896119896 is the property of a material that indicates

                  its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                  Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                  The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                  a piece of material The reciprocal of thermal conductivity is thermal

                  resistivity

                  18 Derivation in one dimension

                  The heat equation is derived from Fouriers law and conservation of energy

                  (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                  surface is proportional to the negative temperature gradient across the

                  surface

                  119902119902 = minus119896119896 120571120571120549120549 (115)

                  where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                  dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                  119902119902 = minus119896119896 120549120549119909119909 (116)

                  where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                  energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                  temperature 120549120549120549120549 That is

                  ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                  where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                  material Choosing zero energy at absolute zero temperature this can be

                  rewritten as

                  ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                  14

                  The increase in internal energy in a small spatial region of the material

                  (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                  given by

                  119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                  119909119909minus∆119909119909

                  = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                  119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                  119909119909minus∆119909119909

                  119905119905+∆119905119905

                  119905119905minus∆119905119905

                  (119)

                  Where the fundamental theorem of calculus was used Additionally with no

                  work done and absent any heat sources or sinks the change in internal energy

                  in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                  across the boundaries By Fouriers law this is

                  119896119896 120597120597120549120549120597120597119909119909

                  (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                  (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                  119905119905minus∆119905119905

                  = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                  119909119909+∆119909119909

                  119909119909minus∆119909119909

                  119905119905+∆119905119905

                  119905119905minus∆119905119905

                  (120)

                  again by the fundamental theorem of calculus By conservation of energy

                  119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                  119909119909minus∆119909119909

                  119905119905+∆119905119905

                  119905119905minus∆119905119905

                  = 0 (121)

                  This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                  Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                  Which can be rewritten as

                  120549120549119905119905 =119896119896119888119888119875119875 120588120588

                  120549120549119909119909119909119909 (122)

                  or

                  120597120597120549120549120597120597119905119905

                  =119896119896119888119888119875119875 120588120588

                  12059712059721205491205491205971205971199091199092 (123)

                  15

                  which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                  diffusivity and is often denoted 120572120572

                  19 Aim of present work

                  The goal of this study is to estimate the solution of partial differential

                  equation that governs the laser-solid interaction using numerical methods

                  The solution will been restricted into one dimensional situation in which we

                  assume that both the laser power density and thermal properties are

                  functions of time and temperature respectively In this project we attempt to

                  investigate the laser interaction with both lead and copper materials by

                  predicting the temperature gradient with the depth of the metals

                  16

                  Chapter Two

                  Theoretical Aspects

                  21 Introduction

                  When a laser interacts with a solid surface a variety of processes can

                  occur We are mainly interested in the interaction of pulsed lasers with a

                  solid surface in first instance a metal When such a laser interacts with a

                  copper surface the laser energy will be transformed into heat The

                  temperature of the solid material will increase leading to melting and

                  evaporation of the solid material

                  The evaporated material (vapour atoms) will expand Depending on the

                  applications this can happen in vacuum (or very low pressure) or in a

                  background gas (helium argon air)

                  22 One dimension laser heating equation

                  In general the one dimension laser heating processes of opaque solid slab is

                  represented as

                  120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                  ( 119870119870 119879119879120597120597 ) (21)

                  With boundary conditions and initial condition which represent the pre-

                  vaporization stage

                  minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                  minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                  17

                  119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                  where

                  119870119870 represents the thermal conductivity

                  120588120588 represents the density

                  119862119862 represents the specific heat

                  119879119879 represents the temperature

                  119879119879infin represents the ambient temperature

                  119879119879119907119907 represents the front surface vaporization

                  120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                  Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                  119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                  With the same boundary conditions as in equation (22)

                  where 119889119889119889119889 = 119870119870120588120588119862119862

                  which represents the thermal diffusion

                  But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                  equation (21) with this assuming implies

                  119879119879119905119905 = 1

                  120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                  With the same boundary and initial conditions in equation (22) Where 119870119870

                  represents the derivative of K with respect the temperature

                  23 Numerical solution of Initial value problems

                  An immense number of analytical solutions for conduction heat-transfer

                  problems have been accumulated in literature over the past 100 years Even so

                  in many practical situations the geometry or boundary conditions are such that an

                  analytical solution has not been obtained at all or if the solution has been

                  18

                  developed it involves such a complex series solution that numerical evaluation

                  becomes exceedingly difficult For such situation the most fruitful approach to

                  the problem is numerical techniques the basic principles of which we shall

                  outline in this section

                  One way to guarantee accuracy in the solution of an initial values problems

                  (IVP) is to solve the problem twice using step sizes h and h2 and compare

                  answers at the mesh points corresponding to the larger step size But this requires

                  a significant amount of computation for the smaller step size and must be

                  repeated if it is determined that the agreement is not good enough

                  24 Finite Difference Method

                  The finite difference method is one of several techniques for obtaining

                  numerical solutions to differential equations In all numerical solutions the

                  continuous partial differential equation (PDE) is replaced with a discrete

                  approximation In this context the word discrete means that the numerical

                  solution is known only at a finite number of points in the physical domain The

                  number of those points can be selected by the user of the numerical method In

                  general increasing the number of points not only increases the resolution but

                  also the accuracy of the numerical solution

                  The discrete approximation results in a set of algebraic equations that are

                  evaluated for the values of the discrete unknowns

                  The mesh is the set of locations where the discrete solution is computed

                  These points are called nodes and if one were to draw lines between adjacent

                  nodes in the domain the resulting image would resemble a net or mesh Two key

                  parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                  space and ∆119905119905 the local distance between adjacent time steps For the simple

                  examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                  19

                  The core idea of the finite-difference method is to replace continuous

                  derivatives with so-called difference formulas that involve only the discrete

                  values associated with positions on the mesh

                  Applying the finite-difference method to a differential equation involves

                  replacing all derivatives with difference formulas In the heat equation there are

                  derivatives with respect to time and derivatives with respect to space Using

                  different combinations of mesh points in the difference formulas results in

                  different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                  numerical solution obtained with any useful scheme will approach the true

                  solution to the original differential equation However the rate at which the

                  numerical solution approaches the true solution varies with the scheme

                  241 First Order Forward Difference

                  Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                  empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                  +1205751205751205971205972

                  2 1205971205972empty1205971205971205971205972

                  1205971205971

                  +1205751205751205971205973

                  3 1205971205973empty1205971205971205971205973

                  1205971205971

                  + ⋯ (25)

                  where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                  consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                  empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                  +∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  +∆1205971205973

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  + ⋯ (26)

                  Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                  120597120597empty120597120597120597120597120597120597119894119894

                  =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                  ∆120597120597minus∆1205971205972

                  1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  minus∆1205971205972

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  minus ⋯ (27)

                  Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                  hand side have been reduced by one

                  20

                  Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                  and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                  120597120597empty120597120597120597120597120597120597119894119894

                  =empty119894119894+1 minus empty119894119894

                  ∆120597120597minus∆1205971205972

                  1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  minus∆1205971205972

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  minus ⋯ (28)

                  The mean value theorem can be used to replace the higher order derivatives

                  ∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  +∆1205971205973

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  + ⋯ =∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120585120585 (29)

                  where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                  120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                  ∆120597120597+∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120585120585 (210)

                  120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                  ∆120597120597asymp∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120585120585 (211)

                  The term on the right hand side of previous equation is called the truncation

                  error of the finite difference approximation

                  In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                  unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                  truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                  analytical form) the big 119978119978 notation can be used to express the dependence of

                  the truncation error on the mesh spacing Note that the right hand side of last

                  equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                  the finite difference simulation Since this is the only parameter under the users

                  control that determines the error the truncation error is simply written

                  ∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120585120585= 119978119978(∆1205971205972) (212)

                  The equals sign in this expression is true in the order of magnitude sense In

                  other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                  21

                  equality Rather the expression means that the left hand side is a product of an

                  unknown constant and ∆1205971205972 Although the expression does not give us the exact

                  magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                  approaches zero as ∆120597120597 is reduced

                  Using big 119978119978 notation Equation (28) can be written

                  120597120597empty120597120597120597120597120597120597119894119894

                  =empty119894119894+1 minus empty119894119894

                  ∆120597120597+ 119978119978(∆120597120597) (213)

                  This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                  it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                  truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                  our control because we can choose the mesh size ∆120597120597 The part of the truncation

                  error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                  242 First Order Backward Difference

                  An alternative first order finite difference formula is obtained if the Taylor series

                  like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                  variables in place of all the unknowns one obtains

                  empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                  +∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  minus∆1205971205973

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  + ⋯ (214)

                  Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                  to get

                  120597120597empty120597120597120597120597120597120597119894119894

                  =empty119894119894+1 minus empty119894119894

                  ∆120597120597minus∆1205971205972

                  1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  minus∆1205971205972

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  minus ⋯ (215)

                  Or using big 119978119978 notation

                  120597120597empty120597120597120597120597120597120597119894119894

                  =empty119894119894 minus empty119894119894minus1

                  ∆120597120597+ 119978119978(∆120597120597) (216)

                  22

                  This is called the backward difference formula because it involves the values of

                  empty at 120597120597119894119894 and 120597120597119894119894minus1

                  The order of magnitude of the truncation error for the backward difference

                  approximation is the same as that of the forward difference approximation Can

                  we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                  truncation error The answer is yes

                  242 First Order Central Difference

                  Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                  empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                  +∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  +∆1205971205973

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  + ⋯ (217)

                  empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                  +∆1205971205972

                  2 1205971205972empty1205971205971205971205972

                  120597120597119894119894

                  minus∆1205971205973

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  + ⋯ (218)

                  Subtracting Equation (10) from Equation (9) yields

                  empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                  + 2∆1205971205973

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  + ⋯ (219)

                  Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                  120597120597empty120597120597120597120597120597120597119894119894

                  =empty119894119894+1 minus empty119894119894minus1

                  2∆120597120597minus∆1205971205972

                  3 1205971205973empty1205971205971205971205973

                  120597120597119894119894

                  minus ⋯ (220)

                  or

                  120597120597empty120597120597120597120597120597120597119894119894

                  =empty119894119894+1 minus empty119894119894minus1

                  2∆120597120597+ 119978119978(∆1205971205972) (221)

                  This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                  approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                  the truncation error for the central difference approximation goes to zero much

                  faster than the truncation error in forward and backward equations

                  23

                  25 Procedures

                  The simple case in this investigation was assuming the constant thermal

                  properties of the material First we assumed all the thermal properties of the

                  materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                  point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                  assume the constant energy after that the pulse of special shapes was selected

                  The numerical solution of equation (23) with boundary and initial conditions

                  in equation (22) was investigated using Matlab program as shown in Appendix

                  The equation of thermal conductivity and specific heat capacity of metal as a

                  function of temperature was obtained by best fitting of polynomials using

                  tabulated data in references

                  24

                  Chapter Three

                  Results and Discursion

                  31 Introduction

                  The development of laser has been an exciting chapter in the history of

                  science and engineering It has produced a new type of advice with potential for

                  application in an extremely wide variety of fields Mach basic development in

                  lasers were occurred during last 35 years The lasers interaction with metal and

                  vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                  The status of laser development and application were still rather rudimentary

                  The light emitted by laser is electro magnetic radiation this radiation has a wave

                  nature the waves consists of vibrating electric and magnetic fields many studies

                  have tried to find and solve models of laser interactions Some researchers

                  proposed the mathematical model related to the laser - plasma interaction and

                  the others have developed an analytical model to study the temperature

                  distribution in Infrared optical materials heated by laser pulses Also an attempt

                  have made to study the interaction of nanosecond pulsed lasers with material

                  from point of view using experimental technique and theoretical approach of

                  dimensional analysis

                  In this study we have evaluate the solution of partial difference equation

                  (PDE) that represent the laser interaction with solid situation in one dimension

                  assuming that the power density of laser and thermal properties are functions

                  with time and temperature respectively

                  25

                  32 Numerical solution with constant laser power density and constant

                  thermal properties

                  First we have taken the lead metal (Pb) with thermal properties

                  119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                  119862119862 = 014016119869119869119892119892119870119870

                  120588120588 = 10751 1198921198921198981198981198981198983

                  119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                  119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                  and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                  represent the area under laser influence

                  The numerical solution of equation (23) with boundary and initial conditions

                  in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                  of lead metal by explicit method using Matlab program give us the results as

                  shown in Fig (31)

                  Fig(31) Depth dependence of the temperature with the laser power density

                  1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                  26

                  33 Evaluation of function 119920119920(119957119957) of laser flux density

                  From following data that represent the energy (119869119869) with time (millie second)

                  Time 0 001 01 02 03 04 05 06 07 08

                  Energy 0 002 017 022 024 02 012 007 002 0

                  By using Matlab program the best polynomial with deduced from above data

                  was

                  119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                  minus 10069 1198981198985 (31)

                  As shown in Fig (32)

                  Fig(32) Laser energy as a function of time

                  Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                  normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                  maximum value (119864119864119898119898119907119907119898119898 )

                  119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                  (119864119864119898119898119907119907119898119898 ) (32)

                  The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                  27

                  Fig(33) Normalized laser energy as a function of time

                  The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                  equal to 3 (total laser energy) ie

                  119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                  08

                  00

                  119899119899119898119898 = 3 (33)

                  Therefore there exist a real number 119875119875 such that

                  119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                  08

                  00

                  119899119899119898119898 = 3 (34)

                  that implies 119875119875 = 68241 and

                  119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                  08

                  00

                  119899119899119898119898 = 3 (35)

                  The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                  08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                  119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                  00

                  (36)

                  28

                  Where 119863119863119898119898 put to balance the units of equation (36)

                  But integral

                  119868119868 = 119864119864119860119860

                  (37)

                  and from equations (35) (36) and (37) we have

                  119868119868 (119898119898)11989911989911989811989808

                  00

                  = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                  0800 119899119899119898119898

                  119860119860 119863119863119898119898 (38)

                  Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                  (38)

                  There fore

                  119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                  119860119860 119863119863119898119898 (39)

                  As shown in Fig(34) Matlab program was used to obtain the best polynomial

                  that agrees with result data

                  119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                  + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                  Fig(34) Time dependence of laser intensity

                  29

                  34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                  constant thermal properties

                  With all constant thermal properties of lead metal as in article (23) and

                  119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                  equation (23) with boundary and initial condition as in equation (22) and the

                  depth penetration is shown in Fig(35)

                  Fig(35) Depth dependence of the temperature when laser intensity function

                  of time and constant thermal properties of Lead

                  35 Evaluation the Thermal Conductivity as functions of temperature

                  The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                  temperature for Lead material was obtained by Matlab program using the

                  experimental data tabulated in researches

                  119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                  times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                  30

                  119879119879 ( 119870119870) 119870119870 119869119869

                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                  300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                  The previous thermal conductivity data and the best fitting of the data are

                  shown in Fig (36)

                  Fig(36) The best fitting of thermal conductivity of Lead as a function of

                  temperature

                  31

                  36 Evaluation the Specific heat as functions of temperature

                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                  material was obtained from the following experimental data tacked from

                  literatures

                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                  300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                  The best polynomial fitted for these data was

                  119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                  + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                  times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                  The specific heat capacity data and the best polynomial fitting of the data are

                  shown in Fig (37)

                  32

                  Fig(37) The best fitting of specific heat capacity of Lead as a function of

                  temperature

                  37 Evaluation the Density as functions of temperature

                  The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                  was used to find the best polynomial fitting

                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                  300 11330 400 11230 500 11130 600 11010 800 10430

                  1000 10190 1200 9940

                  The best polynomial of this data was

                  120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                  minus 45158 times 10minus12 1198791198794 (313)

                  33

                  The density of Lead as a function of temperature and the best polynomial fitting

                  are shown in Fig (38)

                  Fig(38) The best fitting of density of Lead as a function of temperature

                  38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                  variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                  We have deduced the solution of equation (24) with initial and boundary

                  condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                  and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                  respectively then by using Matlab program the depth penetration is shown in

                  Fig (39)

                  34

                  Fig(39) Depth dependence of the temperature for pulse laser on Lead

                  material

                  39 Laser interaction with copper material

                  The same time dependence of laser intensity as shown in Fig(34) with

                  thermal properties of copper was used to calculate the temperature distribution as

                  a function of depth penetration

                  The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                  copper material was obtained from the experimental data tabulated in literary

                  The Matlab program used to obtain the best polynomial equation that agrees

                  with the above data

                  119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                  minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                  times 10minus17 1198791198795 (314)

                  35

                  119879119879 ( 119870119870) 119870119870 119869119869

                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                  100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                  The previous thermal conductivity data and the best fitting of the data are

                  shown in Fig (310)

                  Fig(310) The best fitting of thermal conductivity of Copper as a function of

                  temperature

                  36

                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                  material was obtained from the following experimental data tacked from

                  literatures

                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                  100 0254

                  200 0357

                  273 0384

                  298 0387

                  400 0397

                  600 0416

                  800 0435

                  1000 0454

                  1100 0464

                  1200 0474

                  1300 0483

                  The best polynomial fitted for these data was

                  119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                  + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                  times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                  The specific heat capacity data and the best polynomial fitting of the data are

                  shown in Fig (311)

                  37

                  Fig(311) The best fitting of specific heat capacity of Copper as a function of

                  temperature

                  The density of copper 120588120588(119879119879) as a function of temperature tacked from

                  literature was used to find the best polynomial fitting

                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                  100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                  1000 8576 1100 8519 1200 8458 1300 8396

                  38

                  The best polynomial of this data was

                  120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                  minus 76765 times 10minus14 1198791198794 (316)

                  The density of copper as a function of temperature and the best polynomial

                  fitting are shown in Fig (312)

                  Fig(312) The best fitting of density of copper as a function of temperature

                  The depth penetration of laser energy for copper metal was calculated using

                  the polynomial equations of thermal conductivity specific heat capacity and

                  density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                  (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                  function of time the result was shown in Fig (313)

                  39

                  The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                  for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                  copper metal so the depth penetration of laser energy of lead metal was smaller

                  than that of copper metal this may be due to the high thermal conductivity and

                  high specific heat capacity of copper with that of lead metal

                  Fig(313) Depth dependence of the temperature for pulse laser on Copper

                  material

                  40

                  310 Conclusions

                  The Depth dependence of temperature for lead metal was investigated in two

                  case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                  thermal properties (thermal conductivity specific heat) and density of metal are

                  also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                  vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                  specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                  120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                  penetration depth in the first case is smaller than that of the second case about

                  (190) times

                  The temperature distribution as a function of depth dependence for copper

                  metal was also investigated in the case when the laser intensity vary with time

                  119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                  density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                  The depth penetration of laser energy of lead metal was found to be smaller

                  than that of copper metal this may be due to the high thermal conductivity and

                  high specific heat capacity of copper with that of lead metal

                  41

                  References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                  Sons Inc Hoboken New Jersey Canada (2003)

                  [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                  [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                  [9]

                  Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                  [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                  [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                  [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                  [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                  [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                  httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                  [10] httpwwwworldoflaserscomlaserprincipleshtm

                  [11] httpenwikipediaorgwikiLaserPulsed_operation

                  [12] httpenwikipediaorgwikiThermal_conductivity

                  [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                  [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                  [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                  42

                  Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                  This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                  This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                  43

                  E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                  This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                  This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                  This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                  44

                  title(Dencity as a function of temperature) hold off

                  This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                  45

                  elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                  46

                  r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                  47

                  for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                  48

                  for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                  49

                  6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                  This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                  50

                  u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                  51

                  alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                  52

                  715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                  • 01 Title
                    • 0TDr Omed Ghareb Abdullah
                      • Ch1-Laser
                        • 16 Heat and heat capacity
                        • 18 Derivation in one dimension
                          • Ch2 laser2
                          • Ch3 laser2
                          • Appendix-Laser

                    5

                    Fig(15) The lasing action of He-Ne laser

                    Helium-Neon lasers are common in the introductory physics laboratories

                    but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

                    brightness equal to sunshine on a clear day (01 wattcmP

                    2P) and is just as

                    dangerous to stare at directly

                    2- Carbon Dioxide Laser

                    The carbon dioxide gas laser is capable of continuous output powers above

                    10 kilowatts It is also capable of extremely high power pulse operation It

                    exhibits laser action at several infrared frequencies but none in the visible

                    spectrum Operating in a manner similar to the helium-neon laser it employs

                    an electric discharge for pumping using a percentage of nitrogen gas as a

                    pumping gas The COR2R laser is the most efficient laser capable of operating at

                    more than 30 efficiency

                    The carbon dioxide laser finds many applications in industry particularly for

                    welding and Cutting

                    6

                    3- Argon Laser

                    The argon ion laser can be operated as a continuous gas laser at about 25

                    different wavelengths in the visible between (4089 - 6861) nm but is best

                    known for its most efficient transitions in the green at 488 nm and 5145 nm

                    Operating at much higher powers than the Helium-Neon gas laser it is not

                    uncommon to achieve (30 ndash 100) watts of continuous power using several

                    transitions This output is produced in hot plasma and takes extremely high

                    power typically (9 ndash 12) kW so these are large and expensive devices

                    142 Solid Lasers

                    1 Ruby Laser

                    The ruby laser is the first type of laser actually constructed first

                    demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

                    aluminum oxide with a small amount (about 005) of Chromium which gives

                    it its characteristic pink or red color by absorbing green and blue light

                    The ruby laser is used as a pulsed laser producing red light at 6943 nm

                    After receiving a pumping flash from the flash tube the laser light emerges for

                    as long as the excited atoms persist in the ruby rod which is typically about a

                    millisecond

                    A pulsed ruby laser was used for the famous laser ranging experiment which

                    was conducted with a corner reflector placed on the Moon by the Apollo

                    astronauts This determined the distance to the Moon with an accuracy of

                    about 15 cm

                    7

                    Fig (16) Principle of operation of a Ruby laser

                    2- Neodymium-YAG Laser

                    An example of a solid-state laser the neodymium-YAG uses the NdP

                    3+P ion to

                    dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

                    geometry which makes population inversion possible Neodymium-YAG lasers

                    have become very important because they can be used to produce high

                    powers Such lasers have been constructed to produce over a kilowatt of

                    continuous laser power at 1065 nm and can achieve extremely high powers in

                    a pulsed mode

                    Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

                    production of a series of very short pulses for research with femtosecond time

                    resolution

                    Fig(17) Construction of a Neodymium-YAG laser

                    8

                    3- Neodymium-Glass Lasers

                    Neodymium glass lasers have emerged as the design choice for research in

                    laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

                    short as 10-12 seconds with peak powers of 109 kilowatts

                    143 Molecular Lasers

                    Eximer Lasers

                    Eximer is a shortened form of excited dimer denoting the fact that the

                    lasing medium in this type of laser is an excited diatomic molecule These

                    lasers typically produce ultraviolet pulses They are under investigation for use

                    in communicating with submarines by conversion to blue-green light and

                    pulsing from overhead satellites through sea water to submarines below

                    The eximers used are typically those formed by rare gases and halogens in

                    electron excited Gas discharges Molecules like XeF are stable only in their

                    excited states and quickly dissociate when they make the transition to their

                    ground state This makes possible large population inversions because the

                    ground state is depleted by this dissociation However the excited states are

                    very short-lived compared to other laser metastable states and lasers like the

                    XeF eximer laser require high pumping rates

                    Eximer lasers typically produce high power pulse outputs in the blue or

                    ultraviolet after excitation by fast electron-beam discharges

                    The rare-gas xenon and the highly active fluorine seem unlikely to form a

                    molecule but they do in the hot plasma environment of an electron-beam

                    initiated gas discharge They are only stable in their excited states if stable

                    can be used for molecules which undergo radioactive decay in 1 to 10

                    nanoseconds This is long enough to achieve pulsed laser action in the blue-

                    green over a band from 450 to 510 nm peaking at 486 nm Very high power

                    9

                    pulses can be achieved because the stimulated emission cross-sections of the

                    laser transitions are relatively low allowing a large population inversion to

                    build up The power is also enhanced by the fact that the ground state of XeF

                    quickly dissociates so that there is little absorption to quench the laser pulse

                    action

                    144 Free-Electron Lasers

                    The radiation from a free-electron laser is produced from free electrons

                    which are forced to oscillate in a regular fashion by an applied field They are

                    therefore more like synchrotron light sources or microwave tubes than like

                    other lasers They are able to produce highly coherent collimated radiation

                    over a wide range of frequencies The magnetic field arrangement which

                    produces the alternating field is commonly called a wiggler magnet

                    Fig(18) Principle of operation of Free-Electron laser

                    The free-electron laser is a highly tunable device which has been used to

                    generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

                    this range they are the highest power source Applications of free-electron

                    lasers are envisioned in isotope separation plasma heating for nuclear fusion

                    long-range high resolution radar and particle acceleration in accelerators

                    10

                    15 Pulsed operation

                    Pulsed operation of lasers refers to any laser not classified as continuous

                    wave so that the optical power appears in pulses of some duration at some

                    repetition rate This encompasses a wide range of technologies addressing a

                    number of different motivations Some lasers are pulsed simply because they

                    cannot be run in continuous mode

                    In other cases the application requires the production of pulses having as

                    large an energy as possible Since the pulse energy is equal to the average

                    power divided by the repitition rate this goal can sometimes be satisfied by

                    lowering the rate of pulses so that more energy can be built up in between

                    pulses In laser ablation for example a small volume of material at the surface

                    of a work piece can be evaporated if it is heated in a very short time whereas

                    supplying the energy gradually would allow for the heat to be absorbed into

                    the bulk of the piece never attaining a sufficiently high temperature at a

                    particular point

                    Other applications rely on the peak pulse power (rather than the energy in

                    the pulse) especially in order to obtain nonlinear optical effects For a given

                    pulse energy this requires creating pulses of the shortest possible duration

                    utilizing techniques such as Q-switching

                    16 Heat and heat capacity

                    When a sample is heated meaning it receives thermal energy from an

                    external source some of the introduced heat is converted into kinetic energy

                    the rest to other forms of internal energy specific to the material The amount

                    converted into kinetic energy causes the temperature of the material to rise

                    The amount of the temperature increase depends on how much heat was

                    added the size of the sample the original temperature of the sample and on

                    how the heat was added The two obvious choices on how to add the heat are

                    11

                    to add it holding volume constant or to add it holding pressure constant

                    (There may be other choices but they will not concern us)

                    Lets assume for the moment that we are going to add heat to our sample

                    holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                    subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                    be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                    amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                    (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                    lim119876119876119889119889rarr0

                    119876119876119889119889∆120549120549119889119889

                    = 120597120597119876119876120597120597120549120549119889119889

                    = 119862119862119889119889 (11)

                    We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                    capacity at constant volume Usually one quotes the molar heat capacity

                    119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                    (12)

                    We can rearrange Equation (11) as follows

                    119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                    Then we can integrate this equation to find the heat involved in a finite

                    change at constant volume

                    119876119876119889119889 = 119862119862119889119889

                    1205491205492

                    1205491205491

                    119889119889120549120549 (14)

                    If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                    out of the integral and the heat at constant volume becomes

                    119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                    Let us now go through the same sequence of steps except holding pressure

                    constant instead of volume Our initial definition of the heat capacity at

                    constant pressure 119862119862119875119875 R Rbecomes

                    lim119876119876119875119875rarr0

                    119876119876119875119875∆120549120549119875119875

                    = 120597120597119876119876120597120597120549120549119875119875

                    = 119862119862119875119875 (16)

                    The analogous molar heat capacity is

                    12

                    119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                    (17)

                    Equation (16) rearranges to

                    119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                    which integrates to give

                    119876119876119875119875 = 119862119862119875119875

                    1205491205492

                    1205491205491

                    119889119889120549120549 (19)

                    When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                    119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                    Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                    regarded as constant In these cases the heat capacity is fit to a polynomial (or

                    similar function) in 120549120549 For example some tables give the heat capacity as

                    119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                    where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                    dependent heat capacity the heat at constant pressure would integrate as

                    follows

                    119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                    1205491205492

                    1205491205491

                    119889119889120549120549 (112)

                    119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                    2 minus 12054912054912 + 119899119899

                    1205741205743

                    12054912054923 minus 1205491205491

                    3 (113)

                    Occasionally one finds a different form for the temperature dependent heat

                    capacity in the literature

                    119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                    When you do calculations with temperature dependent heat capacities you

                    must check to see which form is being used for 119862119862119875119875 We are using the

                    convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                    be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                    13

                    positive then heat was indeed absorbed by the system On the other hand if

                    119876119876 is negative it means that the system gave up heat to the surroundings

                    17 Thermal conductivity

                    In physics thermal conductivity 119896119896 is the property of a material that indicates

                    its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                    Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                    The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                    a piece of material The reciprocal of thermal conductivity is thermal

                    resistivity

                    18 Derivation in one dimension

                    The heat equation is derived from Fouriers law and conservation of energy

                    (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                    surface is proportional to the negative temperature gradient across the

                    surface

                    119902119902 = minus119896119896 120571120571120549120549 (115)

                    where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                    dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                    119902119902 = minus119896119896 120549120549119909119909 (116)

                    where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                    energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                    temperature 120549120549120549120549 That is

                    ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                    where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                    material Choosing zero energy at absolute zero temperature this can be

                    rewritten as

                    ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                    14

                    The increase in internal energy in a small spatial region of the material

                    (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                    given by

                    119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                    119909119909minus∆119909119909

                    = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                    119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                    119909119909minus∆119909119909

                    119905119905+∆119905119905

                    119905119905minus∆119905119905

                    (119)

                    Where the fundamental theorem of calculus was used Additionally with no

                    work done and absent any heat sources or sinks the change in internal energy

                    in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                    across the boundaries By Fouriers law this is

                    119896119896 120597120597120549120549120597120597119909119909

                    (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                    (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                    119905119905minus∆119905119905

                    = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                    119909119909+∆119909119909

                    119909119909minus∆119909119909

                    119905119905+∆119905119905

                    119905119905minus∆119905119905

                    (120)

                    again by the fundamental theorem of calculus By conservation of energy

                    119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                    119909119909minus∆119909119909

                    119905119905+∆119905119905

                    119905119905minus∆119905119905

                    = 0 (121)

                    This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                    Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                    Which can be rewritten as

                    120549120549119905119905 =119896119896119888119888119875119875 120588120588

                    120549120549119909119909119909119909 (122)

                    or

                    120597120597120549120549120597120597119905119905

                    =119896119896119888119888119875119875 120588120588

                    12059712059721205491205491205971205971199091199092 (123)

                    15

                    which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                    diffusivity and is often denoted 120572120572

                    19 Aim of present work

                    The goal of this study is to estimate the solution of partial differential

                    equation that governs the laser-solid interaction using numerical methods

                    The solution will been restricted into one dimensional situation in which we

                    assume that both the laser power density and thermal properties are

                    functions of time and temperature respectively In this project we attempt to

                    investigate the laser interaction with both lead and copper materials by

                    predicting the temperature gradient with the depth of the metals

                    16

                    Chapter Two

                    Theoretical Aspects

                    21 Introduction

                    When a laser interacts with a solid surface a variety of processes can

                    occur We are mainly interested in the interaction of pulsed lasers with a

                    solid surface in first instance a metal When such a laser interacts with a

                    copper surface the laser energy will be transformed into heat The

                    temperature of the solid material will increase leading to melting and

                    evaporation of the solid material

                    The evaporated material (vapour atoms) will expand Depending on the

                    applications this can happen in vacuum (or very low pressure) or in a

                    background gas (helium argon air)

                    22 One dimension laser heating equation

                    In general the one dimension laser heating processes of opaque solid slab is

                    represented as

                    120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                    ( 119870119870 119879119879120597120597 ) (21)

                    With boundary conditions and initial condition which represent the pre-

                    vaporization stage

                    minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                    minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                    17

                    119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                    where

                    119870119870 represents the thermal conductivity

                    120588120588 represents the density

                    119862119862 represents the specific heat

                    119879119879 represents the temperature

                    119879119879infin represents the ambient temperature

                    119879119879119907119907 represents the front surface vaporization

                    120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                    Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                    119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                    With the same boundary conditions as in equation (22)

                    where 119889119889119889119889 = 119870119870120588120588119862119862

                    which represents the thermal diffusion

                    But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                    equation (21) with this assuming implies

                    119879119879119905119905 = 1

                    120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                    With the same boundary and initial conditions in equation (22) Where 119870119870

                    represents the derivative of K with respect the temperature

                    23 Numerical solution of Initial value problems

                    An immense number of analytical solutions for conduction heat-transfer

                    problems have been accumulated in literature over the past 100 years Even so

                    in many practical situations the geometry or boundary conditions are such that an

                    analytical solution has not been obtained at all or if the solution has been

                    18

                    developed it involves such a complex series solution that numerical evaluation

                    becomes exceedingly difficult For such situation the most fruitful approach to

                    the problem is numerical techniques the basic principles of which we shall

                    outline in this section

                    One way to guarantee accuracy in the solution of an initial values problems

                    (IVP) is to solve the problem twice using step sizes h and h2 and compare

                    answers at the mesh points corresponding to the larger step size But this requires

                    a significant amount of computation for the smaller step size and must be

                    repeated if it is determined that the agreement is not good enough

                    24 Finite Difference Method

                    The finite difference method is one of several techniques for obtaining

                    numerical solutions to differential equations In all numerical solutions the

                    continuous partial differential equation (PDE) is replaced with a discrete

                    approximation In this context the word discrete means that the numerical

                    solution is known only at a finite number of points in the physical domain The

                    number of those points can be selected by the user of the numerical method In

                    general increasing the number of points not only increases the resolution but

                    also the accuracy of the numerical solution

                    The discrete approximation results in a set of algebraic equations that are

                    evaluated for the values of the discrete unknowns

                    The mesh is the set of locations where the discrete solution is computed

                    These points are called nodes and if one were to draw lines between adjacent

                    nodes in the domain the resulting image would resemble a net or mesh Two key

                    parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                    space and ∆119905119905 the local distance between adjacent time steps For the simple

                    examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                    19

                    The core idea of the finite-difference method is to replace continuous

                    derivatives with so-called difference formulas that involve only the discrete

                    values associated with positions on the mesh

                    Applying the finite-difference method to a differential equation involves

                    replacing all derivatives with difference formulas In the heat equation there are

                    derivatives with respect to time and derivatives with respect to space Using

                    different combinations of mesh points in the difference formulas results in

                    different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                    numerical solution obtained with any useful scheme will approach the true

                    solution to the original differential equation However the rate at which the

                    numerical solution approaches the true solution varies with the scheme

                    241 First Order Forward Difference

                    Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                    empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                    +1205751205751205971205972

                    2 1205971205972empty1205971205971205971205972

                    1205971205971

                    +1205751205751205971205973

                    3 1205971205973empty1205971205971205971205973

                    1205971205971

                    + ⋯ (25)

                    where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                    consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                    empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                    +∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    +∆1205971205973

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    + ⋯ (26)

                    Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                    120597120597empty120597120597120597120597120597120597119894119894

                    =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                    ∆120597120597minus∆1205971205972

                    1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    minus∆1205971205972

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    minus ⋯ (27)

                    Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                    hand side have been reduced by one

                    20

                    Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                    and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                    120597120597empty120597120597120597120597120597120597119894119894

                    =empty119894119894+1 minus empty119894119894

                    ∆120597120597minus∆1205971205972

                    1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    minus∆1205971205972

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    minus ⋯ (28)

                    The mean value theorem can be used to replace the higher order derivatives

                    ∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    +∆1205971205973

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    + ⋯ =∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120585120585 (29)

                    where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                    120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                    ∆120597120597+∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120585120585 (210)

                    120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                    ∆120597120597asymp∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120585120585 (211)

                    The term on the right hand side of previous equation is called the truncation

                    error of the finite difference approximation

                    In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                    unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                    truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                    analytical form) the big 119978119978 notation can be used to express the dependence of

                    the truncation error on the mesh spacing Note that the right hand side of last

                    equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                    the finite difference simulation Since this is the only parameter under the users

                    control that determines the error the truncation error is simply written

                    ∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120585120585= 119978119978(∆1205971205972) (212)

                    The equals sign in this expression is true in the order of magnitude sense In

                    other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                    21

                    equality Rather the expression means that the left hand side is a product of an

                    unknown constant and ∆1205971205972 Although the expression does not give us the exact

                    magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                    approaches zero as ∆120597120597 is reduced

                    Using big 119978119978 notation Equation (28) can be written

                    120597120597empty120597120597120597120597120597120597119894119894

                    =empty119894119894+1 minus empty119894119894

                    ∆120597120597+ 119978119978(∆120597120597) (213)

                    This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                    it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                    truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                    our control because we can choose the mesh size ∆120597120597 The part of the truncation

                    error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                    242 First Order Backward Difference

                    An alternative first order finite difference formula is obtained if the Taylor series

                    like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                    variables in place of all the unknowns one obtains

                    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                    +∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    minus∆1205971205973

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    + ⋯ (214)

                    Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                    to get

                    120597120597empty120597120597120597120597120597120597119894119894

                    =empty119894119894+1 minus empty119894119894

                    ∆120597120597minus∆1205971205972

                    1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    minus∆1205971205972

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    minus ⋯ (215)

                    Or using big 119978119978 notation

                    120597120597empty120597120597120597120597120597120597119894119894

                    =empty119894119894 minus empty119894119894minus1

                    ∆120597120597+ 119978119978(∆120597120597) (216)

                    22

                    This is called the backward difference formula because it involves the values of

                    empty at 120597120597119894119894 and 120597120597119894119894minus1

                    The order of magnitude of the truncation error for the backward difference

                    approximation is the same as that of the forward difference approximation Can

                    we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                    truncation error The answer is yes

                    242 First Order Central Difference

                    Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                    empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                    +∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    +∆1205971205973

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    + ⋯ (217)

                    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                    +∆1205971205972

                    2 1205971205972empty1205971205971205971205972

                    120597120597119894119894

                    minus∆1205971205973

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    + ⋯ (218)

                    Subtracting Equation (10) from Equation (9) yields

                    empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                    + 2∆1205971205973

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    + ⋯ (219)

                    Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                    120597120597empty120597120597120597120597120597120597119894119894

                    =empty119894119894+1 minus empty119894119894minus1

                    2∆120597120597minus∆1205971205972

                    3 1205971205973empty1205971205971205971205973

                    120597120597119894119894

                    minus ⋯ (220)

                    or

                    120597120597empty120597120597120597120597120597120597119894119894

                    =empty119894119894+1 minus empty119894119894minus1

                    2∆120597120597+ 119978119978(∆1205971205972) (221)

                    This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                    approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                    the truncation error for the central difference approximation goes to zero much

                    faster than the truncation error in forward and backward equations

                    23

                    25 Procedures

                    The simple case in this investigation was assuming the constant thermal

                    properties of the material First we assumed all the thermal properties of the

                    materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                    point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                    assume the constant energy after that the pulse of special shapes was selected

                    The numerical solution of equation (23) with boundary and initial conditions

                    in equation (22) was investigated using Matlab program as shown in Appendix

                    The equation of thermal conductivity and specific heat capacity of metal as a

                    function of temperature was obtained by best fitting of polynomials using

                    tabulated data in references

                    24

                    Chapter Three

                    Results and Discursion

                    31 Introduction

                    The development of laser has been an exciting chapter in the history of

                    science and engineering It has produced a new type of advice with potential for

                    application in an extremely wide variety of fields Mach basic development in

                    lasers were occurred during last 35 years The lasers interaction with metal and

                    vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                    The status of laser development and application were still rather rudimentary

                    The light emitted by laser is electro magnetic radiation this radiation has a wave

                    nature the waves consists of vibrating electric and magnetic fields many studies

                    have tried to find and solve models of laser interactions Some researchers

                    proposed the mathematical model related to the laser - plasma interaction and

                    the others have developed an analytical model to study the temperature

                    distribution in Infrared optical materials heated by laser pulses Also an attempt

                    have made to study the interaction of nanosecond pulsed lasers with material

                    from point of view using experimental technique and theoretical approach of

                    dimensional analysis

                    In this study we have evaluate the solution of partial difference equation

                    (PDE) that represent the laser interaction with solid situation in one dimension

                    assuming that the power density of laser and thermal properties are functions

                    with time and temperature respectively

                    25

                    32 Numerical solution with constant laser power density and constant

                    thermal properties

                    First we have taken the lead metal (Pb) with thermal properties

                    119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                    119862119862 = 014016119869119869119892119892119870119870

                    120588120588 = 10751 1198921198921198981198981198981198983

                    119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                    119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                    and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                    represent the area under laser influence

                    The numerical solution of equation (23) with boundary and initial conditions

                    in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                    of lead metal by explicit method using Matlab program give us the results as

                    shown in Fig (31)

                    Fig(31) Depth dependence of the temperature with the laser power density

                    1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                    26

                    33 Evaluation of function 119920119920(119957119957) of laser flux density

                    From following data that represent the energy (119869119869) with time (millie second)

                    Time 0 001 01 02 03 04 05 06 07 08

                    Energy 0 002 017 022 024 02 012 007 002 0

                    By using Matlab program the best polynomial with deduced from above data

                    was

                    119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                    minus 10069 1198981198985 (31)

                    As shown in Fig (32)

                    Fig(32) Laser energy as a function of time

                    Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                    normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                    maximum value (119864119864119898119898119907119907119898119898 )

                    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                    (119864119864119898119898119907119907119898119898 ) (32)

                    The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                    27

                    Fig(33) Normalized laser energy as a function of time

                    The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                    equal to 3 (total laser energy) ie

                    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                    08

                    00

                    119899119899119898119898 = 3 (33)

                    Therefore there exist a real number 119875119875 such that

                    119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                    08

                    00

                    119899119899119898119898 = 3 (34)

                    that implies 119875119875 = 68241 and

                    119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                    08

                    00

                    119899119899119898119898 = 3 (35)

                    The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                    08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                    119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                    00

                    (36)

                    28

                    Where 119863119863119898119898 put to balance the units of equation (36)

                    But integral

                    119868119868 = 119864119864119860119860

                    (37)

                    and from equations (35) (36) and (37) we have

                    119868119868 (119898119898)11989911989911989811989808

                    00

                    = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                    0800 119899119899119898119898

                    119860119860 119863119863119898119898 (38)

                    Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                    (38)

                    There fore

                    119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                    119860119860 119863119863119898119898 (39)

                    As shown in Fig(34) Matlab program was used to obtain the best polynomial

                    that agrees with result data

                    119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                    + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                    Fig(34) Time dependence of laser intensity

                    29

                    34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                    constant thermal properties

                    With all constant thermal properties of lead metal as in article (23) and

                    119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                    equation (23) with boundary and initial condition as in equation (22) and the

                    depth penetration is shown in Fig(35)

                    Fig(35) Depth dependence of the temperature when laser intensity function

                    of time and constant thermal properties of Lead

                    35 Evaluation the Thermal Conductivity as functions of temperature

                    The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                    temperature for Lead material was obtained by Matlab program using the

                    experimental data tabulated in researches

                    119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                    times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                    30

                    119879119879 ( 119870119870) 119870119870 119869119869

                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                    300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                    The previous thermal conductivity data and the best fitting of the data are

                    shown in Fig (36)

                    Fig(36) The best fitting of thermal conductivity of Lead as a function of

                    temperature

                    31

                    36 Evaluation the Specific heat as functions of temperature

                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                    material was obtained from the following experimental data tacked from

                    literatures

                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                    300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                    The best polynomial fitted for these data was

                    119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                    + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                    times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                    The specific heat capacity data and the best polynomial fitting of the data are

                    shown in Fig (37)

                    32

                    Fig(37) The best fitting of specific heat capacity of Lead as a function of

                    temperature

                    37 Evaluation the Density as functions of temperature

                    The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                    was used to find the best polynomial fitting

                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                    300 11330 400 11230 500 11130 600 11010 800 10430

                    1000 10190 1200 9940

                    The best polynomial of this data was

                    120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                    minus 45158 times 10minus12 1198791198794 (313)

                    33

                    The density of Lead as a function of temperature and the best polynomial fitting

                    are shown in Fig (38)

                    Fig(38) The best fitting of density of Lead as a function of temperature

                    38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                    variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                    We have deduced the solution of equation (24) with initial and boundary

                    condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                    and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                    respectively then by using Matlab program the depth penetration is shown in

                    Fig (39)

                    34

                    Fig(39) Depth dependence of the temperature for pulse laser on Lead

                    material

                    39 Laser interaction with copper material

                    The same time dependence of laser intensity as shown in Fig(34) with

                    thermal properties of copper was used to calculate the temperature distribution as

                    a function of depth penetration

                    The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                    copper material was obtained from the experimental data tabulated in literary

                    The Matlab program used to obtain the best polynomial equation that agrees

                    with the above data

                    119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                    minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                    times 10minus17 1198791198795 (314)

                    35

                    119879119879 ( 119870119870) 119870119870 119869119869

                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                    100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                    The previous thermal conductivity data and the best fitting of the data are

                    shown in Fig (310)

                    Fig(310) The best fitting of thermal conductivity of Copper as a function of

                    temperature

                    36

                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                    material was obtained from the following experimental data tacked from

                    literatures

                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                    100 0254

                    200 0357

                    273 0384

                    298 0387

                    400 0397

                    600 0416

                    800 0435

                    1000 0454

                    1100 0464

                    1200 0474

                    1300 0483

                    The best polynomial fitted for these data was

                    119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                    + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                    times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                    The specific heat capacity data and the best polynomial fitting of the data are

                    shown in Fig (311)

                    37

                    Fig(311) The best fitting of specific heat capacity of Copper as a function of

                    temperature

                    The density of copper 120588120588(119879119879) as a function of temperature tacked from

                    literature was used to find the best polynomial fitting

                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                    100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                    1000 8576 1100 8519 1200 8458 1300 8396

                    38

                    The best polynomial of this data was

                    120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                    minus 76765 times 10minus14 1198791198794 (316)

                    The density of copper as a function of temperature and the best polynomial

                    fitting are shown in Fig (312)

                    Fig(312) The best fitting of density of copper as a function of temperature

                    The depth penetration of laser energy for copper metal was calculated using

                    the polynomial equations of thermal conductivity specific heat capacity and

                    density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                    (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                    function of time the result was shown in Fig (313)

                    39

                    The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                    for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                    copper metal so the depth penetration of laser energy of lead metal was smaller

                    than that of copper metal this may be due to the high thermal conductivity and

                    high specific heat capacity of copper with that of lead metal

                    Fig(313) Depth dependence of the temperature for pulse laser on Copper

                    material

                    40

                    310 Conclusions

                    The Depth dependence of temperature for lead metal was investigated in two

                    case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                    thermal properties (thermal conductivity specific heat) and density of metal are

                    also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                    vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                    specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                    120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                    penetration depth in the first case is smaller than that of the second case about

                    (190) times

                    The temperature distribution as a function of depth dependence for copper

                    metal was also investigated in the case when the laser intensity vary with time

                    119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                    density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                    The depth penetration of laser energy of lead metal was found to be smaller

                    than that of copper metal this may be due to the high thermal conductivity and

                    high specific heat capacity of copper with that of lead metal

                    41

                    References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                    Sons Inc Hoboken New Jersey Canada (2003)

                    [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                    [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                    [9]

                    Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                    [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                    [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                    [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                    [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                    [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                    httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                    [10] httpwwwworldoflaserscomlaserprincipleshtm

                    [11] httpenwikipediaorgwikiLaserPulsed_operation

                    [12] httpenwikipediaorgwikiThermal_conductivity

                    [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                    [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                    [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                    42

                    Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                    This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                    This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                    43

                    E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                    This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                    This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                    This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                    44

                    title(Dencity as a function of temperature) hold off

                    This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                    45

                    elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                    46

                    r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                    47

                    for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                    48

                    for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                    49

                    6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                    This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                    50

                    u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                    51

                    alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                    52

                    715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                    • 01 Title
                      • 0TDr Omed Ghareb Abdullah
                        • Ch1-Laser
                          • 16 Heat and heat capacity
                          • 18 Derivation in one dimension
                            • Ch2 laser2
                            • Ch3 laser2
                            • Appendix-Laser

                      6

                      3- Argon Laser

                      The argon ion laser can be operated as a continuous gas laser at about 25

                      different wavelengths in the visible between (4089 - 6861) nm but is best

                      known for its most efficient transitions in the green at 488 nm and 5145 nm

                      Operating at much higher powers than the Helium-Neon gas laser it is not

                      uncommon to achieve (30 ndash 100) watts of continuous power using several

                      transitions This output is produced in hot plasma and takes extremely high

                      power typically (9 ndash 12) kW so these are large and expensive devices

                      142 Solid Lasers

                      1 Ruby Laser

                      The ruby laser is the first type of laser actually constructed first

                      demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

                      aluminum oxide with a small amount (about 005) of Chromium which gives

                      it its characteristic pink or red color by absorbing green and blue light

                      The ruby laser is used as a pulsed laser producing red light at 6943 nm

                      After receiving a pumping flash from the flash tube the laser light emerges for

                      as long as the excited atoms persist in the ruby rod which is typically about a

                      millisecond

                      A pulsed ruby laser was used for the famous laser ranging experiment which

                      was conducted with a corner reflector placed on the Moon by the Apollo

                      astronauts This determined the distance to the Moon with an accuracy of

                      about 15 cm

                      7

                      Fig (16) Principle of operation of a Ruby laser

                      2- Neodymium-YAG Laser

                      An example of a solid-state laser the neodymium-YAG uses the NdP

                      3+P ion to

                      dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

                      geometry which makes population inversion possible Neodymium-YAG lasers

                      have become very important because they can be used to produce high

                      powers Such lasers have been constructed to produce over a kilowatt of

                      continuous laser power at 1065 nm and can achieve extremely high powers in

                      a pulsed mode

                      Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

                      production of a series of very short pulses for research with femtosecond time

                      resolution

                      Fig(17) Construction of a Neodymium-YAG laser

                      8

                      3- Neodymium-Glass Lasers

                      Neodymium glass lasers have emerged as the design choice for research in

                      laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

                      short as 10-12 seconds with peak powers of 109 kilowatts

                      143 Molecular Lasers

                      Eximer Lasers

                      Eximer is a shortened form of excited dimer denoting the fact that the

                      lasing medium in this type of laser is an excited diatomic molecule These

                      lasers typically produce ultraviolet pulses They are under investigation for use

                      in communicating with submarines by conversion to blue-green light and

                      pulsing from overhead satellites through sea water to submarines below

                      The eximers used are typically those formed by rare gases and halogens in

                      electron excited Gas discharges Molecules like XeF are stable only in their

                      excited states and quickly dissociate when they make the transition to their

                      ground state This makes possible large population inversions because the

                      ground state is depleted by this dissociation However the excited states are

                      very short-lived compared to other laser metastable states and lasers like the

                      XeF eximer laser require high pumping rates

                      Eximer lasers typically produce high power pulse outputs in the blue or

                      ultraviolet after excitation by fast electron-beam discharges

                      The rare-gas xenon and the highly active fluorine seem unlikely to form a

                      molecule but they do in the hot plasma environment of an electron-beam

                      initiated gas discharge They are only stable in their excited states if stable

                      can be used for molecules which undergo radioactive decay in 1 to 10

                      nanoseconds This is long enough to achieve pulsed laser action in the blue-

                      green over a band from 450 to 510 nm peaking at 486 nm Very high power

                      9

                      pulses can be achieved because the stimulated emission cross-sections of the

                      laser transitions are relatively low allowing a large population inversion to

                      build up The power is also enhanced by the fact that the ground state of XeF

                      quickly dissociates so that there is little absorption to quench the laser pulse

                      action

                      144 Free-Electron Lasers

                      The radiation from a free-electron laser is produced from free electrons

                      which are forced to oscillate in a regular fashion by an applied field They are

                      therefore more like synchrotron light sources or microwave tubes than like

                      other lasers They are able to produce highly coherent collimated radiation

                      over a wide range of frequencies The magnetic field arrangement which

                      produces the alternating field is commonly called a wiggler magnet

                      Fig(18) Principle of operation of Free-Electron laser

                      The free-electron laser is a highly tunable device which has been used to

                      generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

                      this range they are the highest power source Applications of free-electron

                      lasers are envisioned in isotope separation plasma heating for nuclear fusion

                      long-range high resolution radar and particle acceleration in accelerators

                      10

                      15 Pulsed operation

                      Pulsed operation of lasers refers to any laser not classified as continuous

                      wave so that the optical power appears in pulses of some duration at some

                      repetition rate This encompasses a wide range of technologies addressing a

                      number of different motivations Some lasers are pulsed simply because they

                      cannot be run in continuous mode

                      In other cases the application requires the production of pulses having as

                      large an energy as possible Since the pulse energy is equal to the average

                      power divided by the repitition rate this goal can sometimes be satisfied by

                      lowering the rate of pulses so that more energy can be built up in between

                      pulses In laser ablation for example a small volume of material at the surface

                      of a work piece can be evaporated if it is heated in a very short time whereas

                      supplying the energy gradually would allow for the heat to be absorbed into

                      the bulk of the piece never attaining a sufficiently high temperature at a

                      particular point

                      Other applications rely on the peak pulse power (rather than the energy in

                      the pulse) especially in order to obtain nonlinear optical effects For a given

                      pulse energy this requires creating pulses of the shortest possible duration

                      utilizing techniques such as Q-switching

                      16 Heat and heat capacity

                      When a sample is heated meaning it receives thermal energy from an

                      external source some of the introduced heat is converted into kinetic energy

                      the rest to other forms of internal energy specific to the material The amount

                      converted into kinetic energy causes the temperature of the material to rise

                      The amount of the temperature increase depends on how much heat was

                      added the size of the sample the original temperature of the sample and on

                      how the heat was added The two obvious choices on how to add the heat are

                      11

                      to add it holding volume constant or to add it holding pressure constant

                      (There may be other choices but they will not concern us)

                      Lets assume for the moment that we are going to add heat to our sample

                      holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                      subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                      be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                      amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                      (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                      lim119876119876119889119889rarr0

                      119876119876119889119889∆120549120549119889119889

                      = 120597120597119876119876120597120597120549120549119889119889

                      = 119862119862119889119889 (11)

                      We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                      capacity at constant volume Usually one quotes the molar heat capacity

                      119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                      (12)

                      We can rearrange Equation (11) as follows

                      119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                      Then we can integrate this equation to find the heat involved in a finite

                      change at constant volume

                      119876119876119889119889 = 119862119862119889119889

                      1205491205492

                      1205491205491

                      119889119889120549120549 (14)

                      If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                      out of the integral and the heat at constant volume becomes

                      119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                      Let us now go through the same sequence of steps except holding pressure

                      constant instead of volume Our initial definition of the heat capacity at

                      constant pressure 119862119862119875119875 R Rbecomes

                      lim119876119876119875119875rarr0

                      119876119876119875119875∆120549120549119875119875

                      = 120597120597119876119876120597120597120549120549119875119875

                      = 119862119862119875119875 (16)

                      The analogous molar heat capacity is

                      12

                      119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                      (17)

                      Equation (16) rearranges to

                      119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                      which integrates to give

                      119876119876119875119875 = 119862119862119875119875

                      1205491205492

                      1205491205491

                      119889119889120549120549 (19)

                      When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                      119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                      Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                      regarded as constant In these cases the heat capacity is fit to a polynomial (or

                      similar function) in 120549120549 For example some tables give the heat capacity as

                      119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                      where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                      dependent heat capacity the heat at constant pressure would integrate as

                      follows

                      119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                      1205491205492

                      1205491205491

                      119889119889120549120549 (112)

                      119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                      2 minus 12054912054912 + 119899119899

                      1205741205743

                      12054912054923 minus 1205491205491

                      3 (113)

                      Occasionally one finds a different form for the temperature dependent heat

                      capacity in the literature

                      119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                      When you do calculations with temperature dependent heat capacities you

                      must check to see which form is being used for 119862119862119875119875 We are using the

                      convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                      be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                      13

                      positive then heat was indeed absorbed by the system On the other hand if

                      119876119876 is negative it means that the system gave up heat to the surroundings

                      17 Thermal conductivity

                      In physics thermal conductivity 119896119896 is the property of a material that indicates

                      its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                      Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                      The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                      a piece of material The reciprocal of thermal conductivity is thermal

                      resistivity

                      18 Derivation in one dimension

                      The heat equation is derived from Fouriers law and conservation of energy

                      (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                      surface is proportional to the negative temperature gradient across the

                      surface

                      119902119902 = minus119896119896 120571120571120549120549 (115)

                      where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                      dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                      119902119902 = minus119896119896 120549120549119909119909 (116)

                      where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                      energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                      temperature 120549120549120549120549 That is

                      ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                      where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                      material Choosing zero energy at absolute zero temperature this can be

                      rewritten as

                      ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                      14

                      The increase in internal energy in a small spatial region of the material

                      (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                      given by

                      119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                      119909119909minus∆119909119909

                      = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                      119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                      119909119909minus∆119909119909

                      119905119905+∆119905119905

                      119905119905minus∆119905119905

                      (119)

                      Where the fundamental theorem of calculus was used Additionally with no

                      work done and absent any heat sources or sinks the change in internal energy

                      in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                      across the boundaries By Fouriers law this is

                      119896119896 120597120597120549120549120597120597119909119909

                      (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                      (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                      119905119905minus∆119905119905

                      = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                      119909119909+∆119909119909

                      119909119909minus∆119909119909

                      119905119905+∆119905119905

                      119905119905minus∆119905119905

                      (120)

                      again by the fundamental theorem of calculus By conservation of energy

                      119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                      119909119909minus∆119909119909

                      119905119905+∆119905119905

                      119905119905minus∆119905119905

                      = 0 (121)

                      This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                      Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                      Which can be rewritten as

                      120549120549119905119905 =119896119896119888119888119875119875 120588120588

                      120549120549119909119909119909119909 (122)

                      or

                      120597120597120549120549120597120597119905119905

                      =119896119896119888119888119875119875 120588120588

                      12059712059721205491205491205971205971199091199092 (123)

                      15

                      which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                      diffusivity and is often denoted 120572120572

                      19 Aim of present work

                      The goal of this study is to estimate the solution of partial differential

                      equation that governs the laser-solid interaction using numerical methods

                      The solution will been restricted into one dimensional situation in which we

                      assume that both the laser power density and thermal properties are

                      functions of time and temperature respectively In this project we attempt to

                      investigate the laser interaction with both lead and copper materials by

                      predicting the temperature gradient with the depth of the metals

                      16

                      Chapter Two

                      Theoretical Aspects

                      21 Introduction

                      When a laser interacts with a solid surface a variety of processes can

                      occur We are mainly interested in the interaction of pulsed lasers with a

                      solid surface in first instance a metal When such a laser interacts with a

                      copper surface the laser energy will be transformed into heat The

                      temperature of the solid material will increase leading to melting and

                      evaporation of the solid material

                      The evaporated material (vapour atoms) will expand Depending on the

                      applications this can happen in vacuum (or very low pressure) or in a

                      background gas (helium argon air)

                      22 One dimension laser heating equation

                      In general the one dimension laser heating processes of opaque solid slab is

                      represented as

                      120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                      ( 119870119870 119879119879120597120597 ) (21)

                      With boundary conditions and initial condition which represent the pre-

                      vaporization stage

                      minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                      minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                      17

                      119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                      where

                      119870119870 represents the thermal conductivity

                      120588120588 represents the density

                      119862119862 represents the specific heat

                      119879119879 represents the temperature

                      119879119879infin represents the ambient temperature

                      119879119879119907119907 represents the front surface vaporization

                      120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                      Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                      119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                      With the same boundary conditions as in equation (22)

                      where 119889119889119889119889 = 119870119870120588120588119862119862

                      which represents the thermal diffusion

                      But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                      equation (21) with this assuming implies

                      119879119879119905119905 = 1

                      120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                      With the same boundary and initial conditions in equation (22) Where 119870119870

                      represents the derivative of K with respect the temperature

                      23 Numerical solution of Initial value problems

                      An immense number of analytical solutions for conduction heat-transfer

                      problems have been accumulated in literature over the past 100 years Even so

                      in many practical situations the geometry or boundary conditions are such that an

                      analytical solution has not been obtained at all or if the solution has been

                      18

                      developed it involves such a complex series solution that numerical evaluation

                      becomes exceedingly difficult For such situation the most fruitful approach to

                      the problem is numerical techniques the basic principles of which we shall

                      outline in this section

                      One way to guarantee accuracy in the solution of an initial values problems

                      (IVP) is to solve the problem twice using step sizes h and h2 and compare

                      answers at the mesh points corresponding to the larger step size But this requires

                      a significant amount of computation for the smaller step size and must be

                      repeated if it is determined that the agreement is not good enough

                      24 Finite Difference Method

                      The finite difference method is one of several techniques for obtaining

                      numerical solutions to differential equations In all numerical solutions the

                      continuous partial differential equation (PDE) is replaced with a discrete

                      approximation In this context the word discrete means that the numerical

                      solution is known only at a finite number of points in the physical domain The

                      number of those points can be selected by the user of the numerical method In

                      general increasing the number of points not only increases the resolution but

                      also the accuracy of the numerical solution

                      The discrete approximation results in a set of algebraic equations that are

                      evaluated for the values of the discrete unknowns

                      The mesh is the set of locations where the discrete solution is computed

                      These points are called nodes and if one were to draw lines between adjacent

                      nodes in the domain the resulting image would resemble a net or mesh Two key

                      parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                      space and ∆119905119905 the local distance between adjacent time steps For the simple

                      examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                      19

                      The core idea of the finite-difference method is to replace continuous

                      derivatives with so-called difference formulas that involve only the discrete

                      values associated with positions on the mesh

                      Applying the finite-difference method to a differential equation involves

                      replacing all derivatives with difference formulas In the heat equation there are

                      derivatives with respect to time and derivatives with respect to space Using

                      different combinations of mesh points in the difference formulas results in

                      different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                      numerical solution obtained with any useful scheme will approach the true

                      solution to the original differential equation However the rate at which the

                      numerical solution approaches the true solution varies with the scheme

                      241 First Order Forward Difference

                      Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                      empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                      +1205751205751205971205972

                      2 1205971205972empty1205971205971205971205972

                      1205971205971

                      +1205751205751205971205973

                      3 1205971205973empty1205971205971205971205973

                      1205971205971

                      + ⋯ (25)

                      where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                      consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                      empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                      +∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      +∆1205971205973

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      + ⋯ (26)

                      Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                      120597120597empty120597120597120597120597120597120597119894119894

                      =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                      ∆120597120597minus∆1205971205972

                      1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      minus∆1205971205972

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      minus ⋯ (27)

                      Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                      hand side have been reduced by one

                      20

                      Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                      and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                      120597120597empty120597120597120597120597120597120597119894119894

                      =empty119894119894+1 minus empty119894119894

                      ∆120597120597minus∆1205971205972

                      1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      minus∆1205971205972

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      minus ⋯ (28)

                      The mean value theorem can be used to replace the higher order derivatives

                      ∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      +∆1205971205973

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      + ⋯ =∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120585120585 (29)

                      where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                      120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                      ∆120597120597+∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120585120585 (210)

                      120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                      ∆120597120597asymp∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120585120585 (211)

                      The term on the right hand side of previous equation is called the truncation

                      error of the finite difference approximation

                      In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                      unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                      truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                      analytical form) the big 119978119978 notation can be used to express the dependence of

                      the truncation error on the mesh spacing Note that the right hand side of last

                      equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                      the finite difference simulation Since this is the only parameter under the users

                      control that determines the error the truncation error is simply written

                      ∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120585120585= 119978119978(∆1205971205972) (212)

                      The equals sign in this expression is true in the order of magnitude sense In

                      other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                      21

                      equality Rather the expression means that the left hand side is a product of an

                      unknown constant and ∆1205971205972 Although the expression does not give us the exact

                      magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                      approaches zero as ∆120597120597 is reduced

                      Using big 119978119978 notation Equation (28) can be written

                      120597120597empty120597120597120597120597120597120597119894119894

                      =empty119894119894+1 minus empty119894119894

                      ∆120597120597+ 119978119978(∆120597120597) (213)

                      This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                      it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                      truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                      our control because we can choose the mesh size ∆120597120597 The part of the truncation

                      error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                      242 First Order Backward Difference

                      An alternative first order finite difference formula is obtained if the Taylor series

                      like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                      variables in place of all the unknowns one obtains

                      empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                      +∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      minus∆1205971205973

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      + ⋯ (214)

                      Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                      to get

                      120597120597empty120597120597120597120597120597120597119894119894

                      =empty119894119894+1 minus empty119894119894

                      ∆120597120597minus∆1205971205972

                      1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      minus∆1205971205972

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      minus ⋯ (215)

                      Or using big 119978119978 notation

                      120597120597empty120597120597120597120597120597120597119894119894

                      =empty119894119894 minus empty119894119894minus1

                      ∆120597120597+ 119978119978(∆120597120597) (216)

                      22

                      This is called the backward difference formula because it involves the values of

                      empty at 120597120597119894119894 and 120597120597119894119894minus1

                      The order of magnitude of the truncation error for the backward difference

                      approximation is the same as that of the forward difference approximation Can

                      we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                      truncation error The answer is yes

                      242 First Order Central Difference

                      Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                      empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                      +∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      +∆1205971205973

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      + ⋯ (217)

                      empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                      +∆1205971205972

                      2 1205971205972empty1205971205971205971205972

                      120597120597119894119894

                      minus∆1205971205973

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      + ⋯ (218)

                      Subtracting Equation (10) from Equation (9) yields

                      empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                      + 2∆1205971205973

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      + ⋯ (219)

                      Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                      120597120597empty120597120597120597120597120597120597119894119894

                      =empty119894119894+1 minus empty119894119894minus1

                      2∆120597120597minus∆1205971205972

                      3 1205971205973empty1205971205971205971205973

                      120597120597119894119894

                      minus ⋯ (220)

                      or

                      120597120597empty120597120597120597120597120597120597119894119894

                      =empty119894119894+1 minus empty119894119894minus1

                      2∆120597120597+ 119978119978(∆1205971205972) (221)

                      This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                      approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                      the truncation error for the central difference approximation goes to zero much

                      faster than the truncation error in forward and backward equations

                      23

                      25 Procedures

                      The simple case in this investigation was assuming the constant thermal

                      properties of the material First we assumed all the thermal properties of the

                      materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                      point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                      assume the constant energy after that the pulse of special shapes was selected

                      The numerical solution of equation (23) with boundary and initial conditions

                      in equation (22) was investigated using Matlab program as shown in Appendix

                      The equation of thermal conductivity and specific heat capacity of metal as a

                      function of temperature was obtained by best fitting of polynomials using

                      tabulated data in references

                      24

                      Chapter Three

                      Results and Discursion

                      31 Introduction

                      The development of laser has been an exciting chapter in the history of

                      science and engineering It has produced a new type of advice with potential for

                      application in an extremely wide variety of fields Mach basic development in

                      lasers were occurred during last 35 years The lasers interaction with metal and

                      vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                      The status of laser development and application were still rather rudimentary

                      The light emitted by laser is electro magnetic radiation this radiation has a wave

                      nature the waves consists of vibrating electric and magnetic fields many studies

                      have tried to find and solve models of laser interactions Some researchers

                      proposed the mathematical model related to the laser - plasma interaction and

                      the others have developed an analytical model to study the temperature

                      distribution in Infrared optical materials heated by laser pulses Also an attempt

                      have made to study the interaction of nanosecond pulsed lasers with material

                      from point of view using experimental technique and theoretical approach of

                      dimensional analysis

                      In this study we have evaluate the solution of partial difference equation

                      (PDE) that represent the laser interaction with solid situation in one dimension

                      assuming that the power density of laser and thermal properties are functions

                      with time and temperature respectively

                      25

                      32 Numerical solution with constant laser power density and constant

                      thermal properties

                      First we have taken the lead metal (Pb) with thermal properties

                      119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                      119862119862 = 014016119869119869119892119892119870119870

                      120588120588 = 10751 1198921198921198981198981198981198983

                      119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                      119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                      and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                      represent the area under laser influence

                      The numerical solution of equation (23) with boundary and initial conditions

                      in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                      of lead metal by explicit method using Matlab program give us the results as

                      shown in Fig (31)

                      Fig(31) Depth dependence of the temperature with the laser power density

                      1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                      26

                      33 Evaluation of function 119920119920(119957119957) of laser flux density

                      From following data that represent the energy (119869119869) with time (millie second)

                      Time 0 001 01 02 03 04 05 06 07 08

                      Energy 0 002 017 022 024 02 012 007 002 0

                      By using Matlab program the best polynomial with deduced from above data

                      was

                      119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                      minus 10069 1198981198985 (31)

                      As shown in Fig (32)

                      Fig(32) Laser energy as a function of time

                      Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                      normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                      maximum value (119864119864119898119898119907119907119898119898 )

                      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                      (119864119864119898119898119907119907119898119898 ) (32)

                      The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                      27

                      Fig(33) Normalized laser energy as a function of time

                      The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                      equal to 3 (total laser energy) ie

                      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                      08

                      00

                      119899119899119898119898 = 3 (33)

                      Therefore there exist a real number 119875119875 such that

                      119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                      08

                      00

                      119899119899119898119898 = 3 (34)

                      that implies 119875119875 = 68241 and

                      119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                      08

                      00

                      119899119899119898119898 = 3 (35)

                      The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                      08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                      119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                      00

                      (36)

                      28

                      Where 119863119863119898119898 put to balance the units of equation (36)

                      But integral

                      119868119868 = 119864119864119860119860

                      (37)

                      and from equations (35) (36) and (37) we have

                      119868119868 (119898119898)11989911989911989811989808

                      00

                      = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                      0800 119899119899119898119898

                      119860119860 119863119863119898119898 (38)

                      Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                      (38)

                      There fore

                      119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                      119860119860 119863119863119898119898 (39)

                      As shown in Fig(34) Matlab program was used to obtain the best polynomial

                      that agrees with result data

                      119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                      + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                      Fig(34) Time dependence of laser intensity

                      29

                      34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                      constant thermal properties

                      With all constant thermal properties of lead metal as in article (23) and

                      119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                      equation (23) with boundary and initial condition as in equation (22) and the

                      depth penetration is shown in Fig(35)

                      Fig(35) Depth dependence of the temperature when laser intensity function

                      of time and constant thermal properties of Lead

                      35 Evaluation the Thermal Conductivity as functions of temperature

                      The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                      temperature for Lead material was obtained by Matlab program using the

                      experimental data tabulated in researches

                      119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                      times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                      30

                      119879119879 ( 119870119870) 119870119870 119869119869

                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                      300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                      The previous thermal conductivity data and the best fitting of the data are

                      shown in Fig (36)

                      Fig(36) The best fitting of thermal conductivity of Lead as a function of

                      temperature

                      31

                      36 Evaluation the Specific heat as functions of temperature

                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                      material was obtained from the following experimental data tacked from

                      literatures

                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                      300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                      The best polynomial fitted for these data was

                      119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                      + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                      times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                      The specific heat capacity data and the best polynomial fitting of the data are

                      shown in Fig (37)

                      32

                      Fig(37) The best fitting of specific heat capacity of Lead as a function of

                      temperature

                      37 Evaluation the Density as functions of temperature

                      The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                      was used to find the best polynomial fitting

                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                      300 11330 400 11230 500 11130 600 11010 800 10430

                      1000 10190 1200 9940

                      The best polynomial of this data was

                      120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                      minus 45158 times 10minus12 1198791198794 (313)

                      33

                      The density of Lead as a function of temperature and the best polynomial fitting

                      are shown in Fig (38)

                      Fig(38) The best fitting of density of Lead as a function of temperature

                      38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                      variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                      We have deduced the solution of equation (24) with initial and boundary

                      condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                      and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                      respectively then by using Matlab program the depth penetration is shown in

                      Fig (39)

                      34

                      Fig(39) Depth dependence of the temperature for pulse laser on Lead

                      material

                      39 Laser interaction with copper material

                      The same time dependence of laser intensity as shown in Fig(34) with

                      thermal properties of copper was used to calculate the temperature distribution as

                      a function of depth penetration

                      The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                      copper material was obtained from the experimental data tabulated in literary

                      The Matlab program used to obtain the best polynomial equation that agrees

                      with the above data

                      119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                      minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                      times 10minus17 1198791198795 (314)

                      35

                      119879119879 ( 119870119870) 119870119870 119869119869

                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                      100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                      The previous thermal conductivity data and the best fitting of the data are

                      shown in Fig (310)

                      Fig(310) The best fitting of thermal conductivity of Copper as a function of

                      temperature

                      36

                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                      material was obtained from the following experimental data tacked from

                      literatures

                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                      100 0254

                      200 0357

                      273 0384

                      298 0387

                      400 0397

                      600 0416

                      800 0435

                      1000 0454

                      1100 0464

                      1200 0474

                      1300 0483

                      The best polynomial fitted for these data was

                      119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                      + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                      times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                      The specific heat capacity data and the best polynomial fitting of the data are

                      shown in Fig (311)

                      37

                      Fig(311) The best fitting of specific heat capacity of Copper as a function of

                      temperature

                      The density of copper 120588120588(119879119879) as a function of temperature tacked from

                      literature was used to find the best polynomial fitting

                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                      100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                      1000 8576 1100 8519 1200 8458 1300 8396

                      38

                      The best polynomial of this data was

                      120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                      minus 76765 times 10minus14 1198791198794 (316)

                      The density of copper as a function of temperature and the best polynomial

                      fitting are shown in Fig (312)

                      Fig(312) The best fitting of density of copper as a function of temperature

                      The depth penetration of laser energy for copper metal was calculated using

                      the polynomial equations of thermal conductivity specific heat capacity and

                      density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                      (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                      function of time the result was shown in Fig (313)

                      39

                      The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                      for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                      copper metal so the depth penetration of laser energy of lead metal was smaller

                      than that of copper metal this may be due to the high thermal conductivity and

                      high specific heat capacity of copper with that of lead metal

                      Fig(313) Depth dependence of the temperature for pulse laser on Copper

                      material

                      40

                      310 Conclusions

                      The Depth dependence of temperature for lead metal was investigated in two

                      case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                      thermal properties (thermal conductivity specific heat) and density of metal are

                      also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                      vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                      specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                      120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                      penetration depth in the first case is smaller than that of the second case about

                      (190) times

                      The temperature distribution as a function of depth dependence for copper

                      metal was also investigated in the case when the laser intensity vary with time

                      119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                      density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                      The depth penetration of laser energy of lead metal was found to be smaller

                      than that of copper metal this may be due to the high thermal conductivity and

                      high specific heat capacity of copper with that of lead metal

                      41

                      References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                      Sons Inc Hoboken New Jersey Canada (2003)

                      [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                      [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                      [9]

                      Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                      [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                      [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                      [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                      [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                      [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                      httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                      [10] httpwwwworldoflaserscomlaserprincipleshtm

                      [11] httpenwikipediaorgwikiLaserPulsed_operation

                      [12] httpenwikipediaorgwikiThermal_conductivity

                      [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                      [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                      [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                      42

                      Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                      This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                      This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                      43

                      E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                      This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                      This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                      This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                      44

                      title(Dencity as a function of temperature) hold off

                      This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                      45

                      elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                      46

                      r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                      47

                      for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                      48

                      for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                      49

                      6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                      This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                      50

                      u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                      51

                      alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                      52

                      715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                      • 01 Title
                        • 0TDr Omed Ghareb Abdullah
                          • Ch1-Laser
                            • 16 Heat and heat capacity
                            • 18 Derivation in one dimension
                              • Ch2 laser2
                              • Ch3 laser2
                              • Appendix-Laser

                        7

                        Fig (16) Principle of operation of a Ruby laser

                        2- Neodymium-YAG Laser

                        An example of a solid-state laser the neodymium-YAG uses the NdP

                        3+P ion to

                        dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

                        geometry which makes population inversion possible Neodymium-YAG lasers

                        have become very important because they can be used to produce high

                        powers Such lasers have been constructed to produce over a kilowatt of

                        continuous laser power at 1065 nm and can achieve extremely high powers in

                        a pulsed mode

                        Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

                        production of a series of very short pulses for research with femtosecond time

                        resolution

                        Fig(17) Construction of a Neodymium-YAG laser

                        8

                        3- Neodymium-Glass Lasers

                        Neodymium glass lasers have emerged as the design choice for research in

                        laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

                        short as 10-12 seconds with peak powers of 109 kilowatts

                        143 Molecular Lasers

                        Eximer Lasers

                        Eximer is a shortened form of excited dimer denoting the fact that the

                        lasing medium in this type of laser is an excited diatomic molecule These

                        lasers typically produce ultraviolet pulses They are under investigation for use

                        in communicating with submarines by conversion to blue-green light and

                        pulsing from overhead satellites through sea water to submarines below

                        The eximers used are typically those formed by rare gases and halogens in

                        electron excited Gas discharges Molecules like XeF are stable only in their

                        excited states and quickly dissociate when they make the transition to their

                        ground state This makes possible large population inversions because the

                        ground state is depleted by this dissociation However the excited states are

                        very short-lived compared to other laser metastable states and lasers like the

                        XeF eximer laser require high pumping rates

                        Eximer lasers typically produce high power pulse outputs in the blue or

                        ultraviolet after excitation by fast electron-beam discharges

                        The rare-gas xenon and the highly active fluorine seem unlikely to form a

                        molecule but they do in the hot plasma environment of an electron-beam

                        initiated gas discharge They are only stable in their excited states if stable

                        can be used for molecules which undergo radioactive decay in 1 to 10

                        nanoseconds This is long enough to achieve pulsed laser action in the blue-

                        green over a band from 450 to 510 nm peaking at 486 nm Very high power

                        9

                        pulses can be achieved because the stimulated emission cross-sections of the

                        laser transitions are relatively low allowing a large population inversion to

                        build up The power is also enhanced by the fact that the ground state of XeF

                        quickly dissociates so that there is little absorption to quench the laser pulse

                        action

                        144 Free-Electron Lasers

                        The radiation from a free-electron laser is produced from free electrons

                        which are forced to oscillate in a regular fashion by an applied field They are

                        therefore more like synchrotron light sources or microwave tubes than like

                        other lasers They are able to produce highly coherent collimated radiation

                        over a wide range of frequencies The magnetic field arrangement which

                        produces the alternating field is commonly called a wiggler magnet

                        Fig(18) Principle of operation of Free-Electron laser

                        The free-electron laser is a highly tunable device which has been used to

                        generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

                        this range they are the highest power source Applications of free-electron

                        lasers are envisioned in isotope separation plasma heating for nuclear fusion

                        long-range high resolution radar and particle acceleration in accelerators

                        10

                        15 Pulsed operation

                        Pulsed operation of lasers refers to any laser not classified as continuous

                        wave so that the optical power appears in pulses of some duration at some

                        repetition rate This encompasses a wide range of technologies addressing a

                        number of different motivations Some lasers are pulsed simply because they

                        cannot be run in continuous mode

                        In other cases the application requires the production of pulses having as

                        large an energy as possible Since the pulse energy is equal to the average

                        power divided by the repitition rate this goal can sometimes be satisfied by

                        lowering the rate of pulses so that more energy can be built up in between

                        pulses In laser ablation for example a small volume of material at the surface

                        of a work piece can be evaporated if it is heated in a very short time whereas

                        supplying the energy gradually would allow for the heat to be absorbed into

                        the bulk of the piece never attaining a sufficiently high temperature at a

                        particular point

                        Other applications rely on the peak pulse power (rather than the energy in

                        the pulse) especially in order to obtain nonlinear optical effects For a given

                        pulse energy this requires creating pulses of the shortest possible duration

                        utilizing techniques such as Q-switching

                        16 Heat and heat capacity

                        When a sample is heated meaning it receives thermal energy from an

                        external source some of the introduced heat is converted into kinetic energy

                        the rest to other forms of internal energy specific to the material The amount

                        converted into kinetic energy causes the temperature of the material to rise

                        The amount of the temperature increase depends on how much heat was

                        added the size of the sample the original temperature of the sample and on

                        how the heat was added The two obvious choices on how to add the heat are

                        11

                        to add it holding volume constant or to add it holding pressure constant

                        (There may be other choices but they will not concern us)

                        Lets assume for the moment that we are going to add heat to our sample

                        holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                        subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                        be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                        amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                        (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                        lim119876119876119889119889rarr0

                        119876119876119889119889∆120549120549119889119889

                        = 120597120597119876119876120597120597120549120549119889119889

                        = 119862119862119889119889 (11)

                        We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                        capacity at constant volume Usually one quotes the molar heat capacity

                        119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                        (12)

                        We can rearrange Equation (11) as follows

                        119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                        Then we can integrate this equation to find the heat involved in a finite

                        change at constant volume

                        119876119876119889119889 = 119862119862119889119889

                        1205491205492

                        1205491205491

                        119889119889120549120549 (14)

                        If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                        out of the integral and the heat at constant volume becomes

                        119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                        Let us now go through the same sequence of steps except holding pressure

                        constant instead of volume Our initial definition of the heat capacity at

                        constant pressure 119862119862119875119875 R Rbecomes

                        lim119876119876119875119875rarr0

                        119876119876119875119875∆120549120549119875119875

                        = 120597120597119876119876120597120597120549120549119875119875

                        = 119862119862119875119875 (16)

                        The analogous molar heat capacity is

                        12

                        119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                        (17)

                        Equation (16) rearranges to

                        119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                        which integrates to give

                        119876119876119875119875 = 119862119862119875119875

                        1205491205492

                        1205491205491

                        119889119889120549120549 (19)

                        When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                        119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                        Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                        regarded as constant In these cases the heat capacity is fit to a polynomial (or

                        similar function) in 120549120549 For example some tables give the heat capacity as

                        119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                        where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                        dependent heat capacity the heat at constant pressure would integrate as

                        follows

                        119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                        1205491205492

                        1205491205491

                        119889119889120549120549 (112)

                        119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                        2 minus 12054912054912 + 119899119899

                        1205741205743

                        12054912054923 minus 1205491205491

                        3 (113)

                        Occasionally one finds a different form for the temperature dependent heat

                        capacity in the literature

                        119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                        When you do calculations with temperature dependent heat capacities you

                        must check to see which form is being used for 119862119862119875119875 We are using the

                        convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                        be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                        13

                        positive then heat was indeed absorbed by the system On the other hand if

                        119876119876 is negative it means that the system gave up heat to the surroundings

                        17 Thermal conductivity

                        In physics thermal conductivity 119896119896 is the property of a material that indicates

                        its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                        Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                        The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                        a piece of material The reciprocal of thermal conductivity is thermal

                        resistivity

                        18 Derivation in one dimension

                        The heat equation is derived from Fouriers law and conservation of energy

                        (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                        surface is proportional to the negative temperature gradient across the

                        surface

                        119902119902 = minus119896119896 120571120571120549120549 (115)

                        where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                        dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                        119902119902 = minus119896119896 120549120549119909119909 (116)

                        where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                        energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                        temperature 120549120549120549120549 That is

                        ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                        where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                        material Choosing zero energy at absolute zero temperature this can be

                        rewritten as

                        ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                        14

                        The increase in internal energy in a small spatial region of the material

                        (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                        given by

                        119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                        119909119909minus∆119909119909

                        = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                        119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                        119909119909minus∆119909119909

                        119905119905+∆119905119905

                        119905119905minus∆119905119905

                        (119)

                        Where the fundamental theorem of calculus was used Additionally with no

                        work done and absent any heat sources or sinks the change in internal energy

                        in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                        across the boundaries By Fouriers law this is

                        119896119896 120597120597120549120549120597120597119909119909

                        (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                        (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                        119905119905minus∆119905119905

                        = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                        119909119909+∆119909119909

                        119909119909minus∆119909119909

                        119905119905+∆119905119905

                        119905119905minus∆119905119905

                        (120)

                        again by the fundamental theorem of calculus By conservation of energy

                        119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                        119909119909minus∆119909119909

                        119905119905+∆119905119905

                        119905119905minus∆119905119905

                        = 0 (121)

                        This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                        Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                        Which can be rewritten as

                        120549120549119905119905 =119896119896119888119888119875119875 120588120588

                        120549120549119909119909119909119909 (122)

                        or

                        120597120597120549120549120597120597119905119905

                        =119896119896119888119888119875119875 120588120588

                        12059712059721205491205491205971205971199091199092 (123)

                        15

                        which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                        diffusivity and is often denoted 120572120572

                        19 Aim of present work

                        The goal of this study is to estimate the solution of partial differential

                        equation that governs the laser-solid interaction using numerical methods

                        The solution will been restricted into one dimensional situation in which we

                        assume that both the laser power density and thermal properties are

                        functions of time and temperature respectively In this project we attempt to

                        investigate the laser interaction with both lead and copper materials by

                        predicting the temperature gradient with the depth of the metals

                        16

                        Chapter Two

                        Theoretical Aspects

                        21 Introduction

                        When a laser interacts with a solid surface a variety of processes can

                        occur We are mainly interested in the interaction of pulsed lasers with a

                        solid surface in first instance a metal When such a laser interacts with a

                        copper surface the laser energy will be transformed into heat The

                        temperature of the solid material will increase leading to melting and

                        evaporation of the solid material

                        The evaporated material (vapour atoms) will expand Depending on the

                        applications this can happen in vacuum (or very low pressure) or in a

                        background gas (helium argon air)

                        22 One dimension laser heating equation

                        In general the one dimension laser heating processes of opaque solid slab is

                        represented as

                        120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                        ( 119870119870 119879119879120597120597 ) (21)

                        With boundary conditions and initial condition which represent the pre-

                        vaporization stage

                        minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                        minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                        17

                        119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                        where

                        119870119870 represents the thermal conductivity

                        120588120588 represents the density

                        119862119862 represents the specific heat

                        119879119879 represents the temperature

                        119879119879infin represents the ambient temperature

                        119879119879119907119907 represents the front surface vaporization

                        120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                        Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                        119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                        With the same boundary conditions as in equation (22)

                        where 119889119889119889119889 = 119870119870120588120588119862119862

                        which represents the thermal diffusion

                        But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                        equation (21) with this assuming implies

                        119879119879119905119905 = 1

                        120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                        With the same boundary and initial conditions in equation (22) Where 119870119870

                        represents the derivative of K with respect the temperature

                        23 Numerical solution of Initial value problems

                        An immense number of analytical solutions for conduction heat-transfer

                        problems have been accumulated in literature over the past 100 years Even so

                        in many practical situations the geometry or boundary conditions are such that an

                        analytical solution has not been obtained at all or if the solution has been

                        18

                        developed it involves such a complex series solution that numerical evaluation

                        becomes exceedingly difficult For such situation the most fruitful approach to

                        the problem is numerical techniques the basic principles of which we shall

                        outline in this section

                        One way to guarantee accuracy in the solution of an initial values problems

                        (IVP) is to solve the problem twice using step sizes h and h2 and compare

                        answers at the mesh points corresponding to the larger step size But this requires

                        a significant amount of computation for the smaller step size and must be

                        repeated if it is determined that the agreement is not good enough

                        24 Finite Difference Method

                        The finite difference method is one of several techniques for obtaining

                        numerical solutions to differential equations In all numerical solutions the

                        continuous partial differential equation (PDE) is replaced with a discrete

                        approximation In this context the word discrete means that the numerical

                        solution is known only at a finite number of points in the physical domain The

                        number of those points can be selected by the user of the numerical method In

                        general increasing the number of points not only increases the resolution but

                        also the accuracy of the numerical solution

                        The discrete approximation results in a set of algebraic equations that are

                        evaluated for the values of the discrete unknowns

                        The mesh is the set of locations where the discrete solution is computed

                        These points are called nodes and if one were to draw lines between adjacent

                        nodes in the domain the resulting image would resemble a net or mesh Two key

                        parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                        space and ∆119905119905 the local distance between adjacent time steps For the simple

                        examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                        19

                        The core idea of the finite-difference method is to replace continuous

                        derivatives with so-called difference formulas that involve only the discrete

                        values associated with positions on the mesh

                        Applying the finite-difference method to a differential equation involves

                        replacing all derivatives with difference formulas In the heat equation there are

                        derivatives with respect to time and derivatives with respect to space Using

                        different combinations of mesh points in the difference formulas results in

                        different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                        numerical solution obtained with any useful scheme will approach the true

                        solution to the original differential equation However the rate at which the

                        numerical solution approaches the true solution varies with the scheme

                        241 First Order Forward Difference

                        Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                        empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                        +1205751205751205971205972

                        2 1205971205972empty1205971205971205971205972

                        1205971205971

                        +1205751205751205971205973

                        3 1205971205973empty1205971205971205971205973

                        1205971205971

                        + ⋯ (25)

                        where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                        consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                        empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                        +∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        +∆1205971205973

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        + ⋯ (26)

                        Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                        120597120597empty120597120597120597120597120597120597119894119894

                        =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                        ∆120597120597minus∆1205971205972

                        1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        minus∆1205971205972

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        minus ⋯ (27)

                        Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                        hand side have been reduced by one

                        20

                        Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                        and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                        120597120597empty120597120597120597120597120597120597119894119894

                        =empty119894119894+1 minus empty119894119894

                        ∆120597120597minus∆1205971205972

                        1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        minus∆1205971205972

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        minus ⋯ (28)

                        The mean value theorem can be used to replace the higher order derivatives

                        ∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        +∆1205971205973

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        + ⋯ =∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120585120585 (29)

                        where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                        120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                        ∆120597120597+∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120585120585 (210)

                        120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                        ∆120597120597asymp∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120585120585 (211)

                        The term on the right hand side of previous equation is called the truncation

                        error of the finite difference approximation

                        In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                        unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                        truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                        analytical form) the big 119978119978 notation can be used to express the dependence of

                        the truncation error on the mesh spacing Note that the right hand side of last

                        equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                        the finite difference simulation Since this is the only parameter under the users

                        control that determines the error the truncation error is simply written

                        ∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120585120585= 119978119978(∆1205971205972) (212)

                        The equals sign in this expression is true in the order of magnitude sense In

                        other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                        21

                        equality Rather the expression means that the left hand side is a product of an

                        unknown constant and ∆1205971205972 Although the expression does not give us the exact

                        magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                        approaches zero as ∆120597120597 is reduced

                        Using big 119978119978 notation Equation (28) can be written

                        120597120597empty120597120597120597120597120597120597119894119894

                        =empty119894119894+1 minus empty119894119894

                        ∆120597120597+ 119978119978(∆120597120597) (213)

                        This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                        it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                        truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                        our control because we can choose the mesh size ∆120597120597 The part of the truncation

                        error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                        242 First Order Backward Difference

                        An alternative first order finite difference formula is obtained if the Taylor series

                        like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                        variables in place of all the unknowns one obtains

                        empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                        +∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        minus∆1205971205973

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        + ⋯ (214)

                        Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                        to get

                        120597120597empty120597120597120597120597120597120597119894119894

                        =empty119894119894+1 minus empty119894119894

                        ∆120597120597minus∆1205971205972

                        1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        minus∆1205971205972

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        minus ⋯ (215)

                        Or using big 119978119978 notation

                        120597120597empty120597120597120597120597120597120597119894119894

                        =empty119894119894 minus empty119894119894minus1

                        ∆120597120597+ 119978119978(∆120597120597) (216)

                        22

                        This is called the backward difference formula because it involves the values of

                        empty at 120597120597119894119894 and 120597120597119894119894minus1

                        The order of magnitude of the truncation error for the backward difference

                        approximation is the same as that of the forward difference approximation Can

                        we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                        truncation error The answer is yes

                        242 First Order Central Difference

                        Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                        empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                        +∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        +∆1205971205973

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        + ⋯ (217)

                        empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                        +∆1205971205972

                        2 1205971205972empty1205971205971205971205972

                        120597120597119894119894

                        minus∆1205971205973

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        + ⋯ (218)

                        Subtracting Equation (10) from Equation (9) yields

                        empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                        + 2∆1205971205973

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        + ⋯ (219)

                        Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                        120597120597empty120597120597120597120597120597120597119894119894

                        =empty119894119894+1 minus empty119894119894minus1

                        2∆120597120597minus∆1205971205972

                        3 1205971205973empty1205971205971205971205973

                        120597120597119894119894

                        minus ⋯ (220)

                        or

                        120597120597empty120597120597120597120597120597120597119894119894

                        =empty119894119894+1 minus empty119894119894minus1

                        2∆120597120597+ 119978119978(∆1205971205972) (221)

                        This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                        approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                        the truncation error for the central difference approximation goes to zero much

                        faster than the truncation error in forward and backward equations

                        23

                        25 Procedures

                        The simple case in this investigation was assuming the constant thermal

                        properties of the material First we assumed all the thermal properties of the

                        materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                        point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                        assume the constant energy after that the pulse of special shapes was selected

                        The numerical solution of equation (23) with boundary and initial conditions

                        in equation (22) was investigated using Matlab program as shown in Appendix

                        The equation of thermal conductivity and specific heat capacity of metal as a

                        function of temperature was obtained by best fitting of polynomials using

                        tabulated data in references

                        24

                        Chapter Three

                        Results and Discursion

                        31 Introduction

                        The development of laser has been an exciting chapter in the history of

                        science and engineering It has produced a new type of advice with potential for

                        application in an extremely wide variety of fields Mach basic development in

                        lasers were occurred during last 35 years The lasers interaction with metal and

                        vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                        The status of laser development and application were still rather rudimentary

                        The light emitted by laser is electro magnetic radiation this radiation has a wave

                        nature the waves consists of vibrating electric and magnetic fields many studies

                        have tried to find and solve models of laser interactions Some researchers

                        proposed the mathematical model related to the laser - plasma interaction and

                        the others have developed an analytical model to study the temperature

                        distribution in Infrared optical materials heated by laser pulses Also an attempt

                        have made to study the interaction of nanosecond pulsed lasers with material

                        from point of view using experimental technique and theoretical approach of

                        dimensional analysis

                        In this study we have evaluate the solution of partial difference equation

                        (PDE) that represent the laser interaction with solid situation in one dimension

                        assuming that the power density of laser and thermal properties are functions

                        with time and temperature respectively

                        25

                        32 Numerical solution with constant laser power density and constant

                        thermal properties

                        First we have taken the lead metal (Pb) with thermal properties

                        119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                        119862119862 = 014016119869119869119892119892119870119870

                        120588120588 = 10751 1198921198921198981198981198981198983

                        119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                        119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                        and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                        represent the area under laser influence

                        The numerical solution of equation (23) with boundary and initial conditions

                        in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                        of lead metal by explicit method using Matlab program give us the results as

                        shown in Fig (31)

                        Fig(31) Depth dependence of the temperature with the laser power density

                        1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                        26

                        33 Evaluation of function 119920119920(119957119957) of laser flux density

                        From following data that represent the energy (119869119869) with time (millie second)

                        Time 0 001 01 02 03 04 05 06 07 08

                        Energy 0 002 017 022 024 02 012 007 002 0

                        By using Matlab program the best polynomial with deduced from above data

                        was

                        119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                        minus 10069 1198981198985 (31)

                        As shown in Fig (32)

                        Fig(32) Laser energy as a function of time

                        Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                        normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                        maximum value (119864119864119898119898119907119907119898119898 )

                        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                        (119864119864119898119898119907119907119898119898 ) (32)

                        The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                        27

                        Fig(33) Normalized laser energy as a function of time

                        The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                        equal to 3 (total laser energy) ie

                        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                        08

                        00

                        119899119899119898119898 = 3 (33)

                        Therefore there exist a real number 119875119875 such that

                        119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                        08

                        00

                        119899119899119898119898 = 3 (34)

                        that implies 119875119875 = 68241 and

                        119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                        08

                        00

                        119899119899119898119898 = 3 (35)

                        The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                        08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                        119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                        00

                        (36)

                        28

                        Where 119863119863119898119898 put to balance the units of equation (36)

                        But integral

                        119868119868 = 119864119864119860119860

                        (37)

                        and from equations (35) (36) and (37) we have

                        119868119868 (119898119898)11989911989911989811989808

                        00

                        = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                        0800 119899119899119898119898

                        119860119860 119863119863119898119898 (38)

                        Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                        (38)

                        There fore

                        119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                        119860119860 119863119863119898119898 (39)

                        As shown in Fig(34) Matlab program was used to obtain the best polynomial

                        that agrees with result data

                        119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                        + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                        Fig(34) Time dependence of laser intensity

                        29

                        34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                        constant thermal properties

                        With all constant thermal properties of lead metal as in article (23) and

                        119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                        equation (23) with boundary and initial condition as in equation (22) and the

                        depth penetration is shown in Fig(35)

                        Fig(35) Depth dependence of the temperature when laser intensity function

                        of time and constant thermal properties of Lead

                        35 Evaluation the Thermal Conductivity as functions of temperature

                        The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                        temperature for Lead material was obtained by Matlab program using the

                        experimental data tabulated in researches

                        119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                        times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                        30

                        119879119879 ( 119870119870) 119870119870 119869119869

                        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                        300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                        The previous thermal conductivity data and the best fitting of the data are

                        shown in Fig (36)

                        Fig(36) The best fitting of thermal conductivity of Lead as a function of

                        temperature

                        31

                        36 Evaluation the Specific heat as functions of temperature

                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                        material was obtained from the following experimental data tacked from

                        literatures

                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                        300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                        The best polynomial fitted for these data was

                        119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                        + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                        times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                        The specific heat capacity data and the best polynomial fitting of the data are

                        shown in Fig (37)

                        32

                        Fig(37) The best fitting of specific heat capacity of Lead as a function of

                        temperature

                        37 Evaluation the Density as functions of temperature

                        The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                        was used to find the best polynomial fitting

                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                        300 11330 400 11230 500 11130 600 11010 800 10430

                        1000 10190 1200 9940

                        The best polynomial of this data was

                        120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                        minus 45158 times 10minus12 1198791198794 (313)

                        33

                        The density of Lead as a function of temperature and the best polynomial fitting

                        are shown in Fig (38)

                        Fig(38) The best fitting of density of Lead as a function of temperature

                        38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                        variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                        We have deduced the solution of equation (24) with initial and boundary

                        condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                        and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                        respectively then by using Matlab program the depth penetration is shown in

                        Fig (39)

                        34

                        Fig(39) Depth dependence of the temperature for pulse laser on Lead

                        material

                        39 Laser interaction with copper material

                        The same time dependence of laser intensity as shown in Fig(34) with

                        thermal properties of copper was used to calculate the temperature distribution as

                        a function of depth penetration

                        The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                        copper material was obtained from the experimental data tabulated in literary

                        The Matlab program used to obtain the best polynomial equation that agrees

                        with the above data

                        119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                        minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                        times 10minus17 1198791198795 (314)

                        35

                        119879119879 ( 119870119870) 119870119870 119869119869

                        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                        100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                        The previous thermal conductivity data and the best fitting of the data are

                        shown in Fig (310)

                        Fig(310) The best fitting of thermal conductivity of Copper as a function of

                        temperature

                        36

                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                        material was obtained from the following experimental data tacked from

                        literatures

                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                        100 0254

                        200 0357

                        273 0384

                        298 0387

                        400 0397

                        600 0416

                        800 0435

                        1000 0454

                        1100 0464

                        1200 0474

                        1300 0483

                        The best polynomial fitted for these data was

                        119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                        + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                        times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                        The specific heat capacity data and the best polynomial fitting of the data are

                        shown in Fig (311)

                        37

                        Fig(311) The best fitting of specific heat capacity of Copper as a function of

                        temperature

                        The density of copper 120588120588(119879119879) as a function of temperature tacked from

                        literature was used to find the best polynomial fitting

                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                        100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                        1000 8576 1100 8519 1200 8458 1300 8396

                        38

                        The best polynomial of this data was

                        120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                        minus 76765 times 10minus14 1198791198794 (316)

                        The density of copper as a function of temperature and the best polynomial

                        fitting are shown in Fig (312)

                        Fig(312) The best fitting of density of copper as a function of temperature

                        The depth penetration of laser energy for copper metal was calculated using

                        the polynomial equations of thermal conductivity specific heat capacity and

                        density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                        (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                        function of time the result was shown in Fig (313)

                        39

                        The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                        for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                        copper metal so the depth penetration of laser energy of lead metal was smaller

                        than that of copper metal this may be due to the high thermal conductivity and

                        high specific heat capacity of copper with that of lead metal

                        Fig(313) Depth dependence of the temperature for pulse laser on Copper

                        material

                        40

                        310 Conclusions

                        The Depth dependence of temperature for lead metal was investigated in two

                        case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                        thermal properties (thermal conductivity specific heat) and density of metal are

                        also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                        vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                        specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                        120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                        penetration depth in the first case is smaller than that of the second case about

                        (190) times

                        The temperature distribution as a function of depth dependence for copper

                        metal was also investigated in the case when the laser intensity vary with time

                        119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                        density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                        The depth penetration of laser energy of lead metal was found to be smaller

                        than that of copper metal this may be due to the high thermal conductivity and

                        high specific heat capacity of copper with that of lead metal

                        41

                        References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                        Sons Inc Hoboken New Jersey Canada (2003)

                        [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                        [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                        [9]

                        Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                        [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                        [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                        [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                        [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                        [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                        httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                        [10] httpwwwworldoflaserscomlaserprincipleshtm

                        [11] httpenwikipediaorgwikiLaserPulsed_operation

                        [12] httpenwikipediaorgwikiThermal_conductivity

                        [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                        [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                        [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                        42

                        Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                        This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                        This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                        43

                        E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                        This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                        This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                        This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                        44

                        title(Dencity as a function of temperature) hold off

                        This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                        45

                        elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                        46

                        r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                        47

                        for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                        48

                        for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                        49

                        6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                        This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                        50

                        u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                        51

                        alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                        52

                        715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                        • 01 Title
                          • 0TDr Omed Ghareb Abdullah
                            • Ch1-Laser
                              • 16 Heat and heat capacity
                              • 18 Derivation in one dimension
                                • Ch2 laser2
                                • Ch3 laser2
                                • Appendix-Laser

                          8

                          3- Neodymium-Glass Lasers

                          Neodymium glass lasers have emerged as the design choice for research in

                          laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

                          short as 10-12 seconds with peak powers of 109 kilowatts

                          143 Molecular Lasers

                          Eximer Lasers

                          Eximer is a shortened form of excited dimer denoting the fact that the

                          lasing medium in this type of laser is an excited diatomic molecule These

                          lasers typically produce ultraviolet pulses They are under investigation for use

                          in communicating with submarines by conversion to blue-green light and

                          pulsing from overhead satellites through sea water to submarines below

                          The eximers used are typically those formed by rare gases and halogens in

                          electron excited Gas discharges Molecules like XeF are stable only in their

                          excited states and quickly dissociate when they make the transition to their

                          ground state This makes possible large population inversions because the

                          ground state is depleted by this dissociation However the excited states are

                          very short-lived compared to other laser metastable states and lasers like the

                          XeF eximer laser require high pumping rates

                          Eximer lasers typically produce high power pulse outputs in the blue or

                          ultraviolet after excitation by fast electron-beam discharges

                          The rare-gas xenon and the highly active fluorine seem unlikely to form a

                          molecule but they do in the hot plasma environment of an electron-beam

                          initiated gas discharge They are only stable in their excited states if stable

                          can be used for molecules which undergo radioactive decay in 1 to 10

                          nanoseconds This is long enough to achieve pulsed laser action in the blue-

                          green over a band from 450 to 510 nm peaking at 486 nm Very high power

                          9

                          pulses can be achieved because the stimulated emission cross-sections of the

                          laser transitions are relatively low allowing a large population inversion to

                          build up The power is also enhanced by the fact that the ground state of XeF

                          quickly dissociates so that there is little absorption to quench the laser pulse

                          action

                          144 Free-Electron Lasers

                          The radiation from a free-electron laser is produced from free electrons

                          which are forced to oscillate in a regular fashion by an applied field They are

                          therefore more like synchrotron light sources or microwave tubes than like

                          other lasers They are able to produce highly coherent collimated radiation

                          over a wide range of frequencies The magnetic field arrangement which

                          produces the alternating field is commonly called a wiggler magnet

                          Fig(18) Principle of operation of Free-Electron laser

                          The free-electron laser is a highly tunable device which has been used to

                          generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

                          this range they are the highest power source Applications of free-electron

                          lasers are envisioned in isotope separation plasma heating for nuclear fusion

                          long-range high resolution radar and particle acceleration in accelerators

                          10

                          15 Pulsed operation

                          Pulsed operation of lasers refers to any laser not classified as continuous

                          wave so that the optical power appears in pulses of some duration at some

                          repetition rate This encompasses a wide range of technologies addressing a

                          number of different motivations Some lasers are pulsed simply because they

                          cannot be run in continuous mode

                          In other cases the application requires the production of pulses having as

                          large an energy as possible Since the pulse energy is equal to the average

                          power divided by the repitition rate this goal can sometimes be satisfied by

                          lowering the rate of pulses so that more energy can be built up in between

                          pulses In laser ablation for example a small volume of material at the surface

                          of a work piece can be evaporated if it is heated in a very short time whereas

                          supplying the energy gradually would allow for the heat to be absorbed into

                          the bulk of the piece never attaining a sufficiently high temperature at a

                          particular point

                          Other applications rely on the peak pulse power (rather than the energy in

                          the pulse) especially in order to obtain nonlinear optical effects For a given

                          pulse energy this requires creating pulses of the shortest possible duration

                          utilizing techniques such as Q-switching

                          16 Heat and heat capacity

                          When a sample is heated meaning it receives thermal energy from an

                          external source some of the introduced heat is converted into kinetic energy

                          the rest to other forms of internal energy specific to the material The amount

                          converted into kinetic energy causes the temperature of the material to rise

                          The amount of the temperature increase depends on how much heat was

                          added the size of the sample the original temperature of the sample and on

                          how the heat was added The two obvious choices on how to add the heat are

                          11

                          to add it holding volume constant or to add it holding pressure constant

                          (There may be other choices but they will not concern us)

                          Lets assume for the moment that we are going to add heat to our sample

                          holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                          subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                          be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                          amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                          (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                          lim119876119876119889119889rarr0

                          119876119876119889119889∆120549120549119889119889

                          = 120597120597119876119876120597120597120549120549119889119889

                          = 119862119862119889119889 (11)

                          We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                          capacity at constant volume Usually one quotes the molar heat capacity

                          119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                          (12)

                          We can rearrange Equation (11) as follows

                          119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                          Then we can integrate this equation to find the heat involved in a finite

                          change at constant volume

                          119876119876119889119889 = 119862119862119889119889

                          1205491205492

                          1205491205491

                          119889119889120549120549 (14)

                          If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                          out of the integral and the heat at constant volume becomes

                          119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                          Let us now go through the same sequence of steps except holding pressure

                          constant instead of volume Our initial definition of the heat capacity at

                          constant pressure 119862119862119875119875 R Rbecomes

                          lim119876119876119875119875rarr0

                          119876119876119875119875∆120549120549119875119875

                          = 120597120597119876119876120597120597120549120549119875119875

                          = 119862119862119875119875 (16)

                          The analogous molar heat capacity is

                          12

                          119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                          (17)

                          Equation (16) rearranges to

                          119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                          which integrates to give

                          119876119876119875119875 = 119862119862119875119875

                          1205491205492

                          1205491205491

                          119889119889120549120549 (19)

                          When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                          119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                          Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                          regarded as constant In these cases the heat capacity is fit to a polynomial (or

                          similar function) in 120549120549 For example some tables give the heat capacity as

                          119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                          where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                          dependent heat capacity the heat at constant pressure would integrate as

                          follows

                          119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                          1205491205492

                          1205491205491

                          119889119889120549120549 (112)

                          119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                          2 minus 12054912054912 + 119899119899

                          1205741205743

                          12054912054923 minus 1205491205491

                          3 (113)

                          Occasionally one finds a different form for the temperature dependent heat

                          capacity in the literature

                          119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                          When you do calculations with temperature dependent heat capacities you

                          must check to see which form is being used for 119862119862119875119875 We are using the

                          convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                          be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                          13

                          positive then heat was indeed absorbed by the system On the other hand if

                          119876119876 is negative it means that the system gave up heat to the surroundings

                          17 Thermal conductivity

                          In physics thermal conductivity 119896119896 is the property of a material that indicates

                          its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                          Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                          The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                          a piece of material The reciprocal of thermal conductivity is thermal

                          resistivity

                          18 Derivation in one dimension

                          The heat equation is derived from Fouriers law and conservation of energy

                          (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                          surface is proportional to the negative temperature gradient across the

                          surface

                          119902119902 = minus119896119896 120571120571120549120549 (115)

                          where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                          dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                          119902119902 = minus119896119896 120549120549119909119909 (116)

                          where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                          energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                          temperature 120549120549120549120549 That is

                          ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                          where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                          material Choosing zero energy at absolute zero temperature this can be

                          rewritten as

                          ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                          14

                          The increase in internal energy in a small spatial region of the material

                          (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                          given by

                          119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                          119909119909minus∆119909119909

                          = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                          119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                          119909119909minus∆119909119909

                          119905119905+∆119905119905

                          119905119905minus∆119905119905

                          (119)

                          Where the fundamental theorem of calculus was used Additionally with no

                          work done and absent any heat sources or sinks the change in internal energy

                          in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                          across the boundaries By Fouriers law this is

                          119896119896 120597120597120549120549120597120597119909119909

                          (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                          (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                          119905119905minus∆119905119905

                          = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                          119909119909+∆119909119909

                          119909119909minus∆119909119909

                          119905119905+∆119905119905

                          119905119905minus∆119905119905

                          (120)

                          again by the fundamental theorem of calculus By conservation of energy

                          119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                          119909119909minus∆119909119909

                          119905119905+∆119905119905

                          119905119905minus∆119905119905

                          = 0 (121)

                          This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                          Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                          Which can be rewritten as

                          120549120549119905119905 =119896119896119888119888119875119875 120588120588

                          120549120549119909119909119909119909 (122)

                          or

                          120597120597120549120549120597120597119905119905

                          =119896119896119888119888119875119875 120588120588

                          12059712059721205491205491205971205971199091199092 (123)

                          15

                          which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                          diffusivity and is often denoted 120572120572

                          19 Aim of present work

                          The goal of this study is to estimate the solution of partial differential

                          equation that governs the laser-solid interaction using numerical methods

                          The solution will been restricted into one dimensional situation in which we

                          assume that both the laser power density and thermal properties are

                          functions of time and temperature respectively In this project we attempt to

                          investigate the laser interaction with both lead and copper materials by

                          predicting the temperature gradient with the depth of the metals

                          16

                          Chapter Two

                          Theoretical Aspects

                          21 Introduction

                          When a laser interacts with a solid surface a variety of processes can

                          occur We are mainly interested in the interaction of pulsed lasers with a

                          solid surface in first instance a metal When such a laser interacts with a

                          copper surface the laser energy will be transformed into heat The

                          temperature of the solid material will increase leading to melting and

                          evaporation of the solid material

                          The evaporated material (vapour atoms) will expand Depending on the

                          applications this can happen in vacuum (or very low pressure) or in a

                          background gas (helium argon air)

                          22 One dimension laser heating equation

                          In general the one dimension laser heating processes of opaque solid slab is

                          represented as

                          120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                          ( 119870119870 119879119879120597120597 ) (21)

                          With boundary conditions and initial condition which represent the pre-

                          vaporization stage

                          minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                          minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                          17

                          119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                          where

                          119870119870 represents the thermal conductivity

                          120588120588 represents the density

                          119862119862 represents the specific heat

                          119879119879 represents the temperature

                          119879119879infin represents the ambient temperature

                          119879119879119907119907 represents the front surface vaporization

                          120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                          Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                          119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                          With the same boundary conditions as in equation (22)

                          where 119889119889119889119889 = 119870119870120588120588119862119862

                          which represents the thermal diffusion

                          But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                          equation (21) with this assuming implies

                          119879119879119905119905 = 1

                          120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                          With the same boundary and initial conditions in equation (22) Where 119870119870

                          represents the derivative of K with respect the temperature

                          23 Numerical solution of Initial value problems

                          An immense number of analytical solutions for conduction heat-transfer

                          problems have been accumulated in literature over the past 100 years Even so

                          in many practical situations the geometry or boundary conditions are such that an

                          analytical solution has not been obtained at all or if the solution has been

                          18

                          developed it involves such a complex series solution that numerical evaluation

                          becomes exceedingly difficult For such situation the most fruitful approach to

                          the problem is numerical techniques the basic principles of which we shall

                          outline in this section

                          One way to guarantee accuracy in the solution of an initial values problems

                          (IVP) is to solve the problem twice using step sizes h and h2 and compare

                          answers at the mesh points corresponding to the larger step size But this requires

                          a significant amount of computation for the smaller step size and must be

                          repeated if it is determined that the agreement is not good enough

                          24 Finite Difference Method

                          The finite difference method is one of several techniques for obtaining

                          numerical solutions to differential equations In all numerical solutions the

                          continuous partial differential equation (PDE) is replaced with a discrete

                          approximation In this context the word discrete means that the numerical

                          solution is known only at a finite number of points in the physical domain The

                          number of those points can be selected by the user of the numerical method In

                          general increasing the number of points not only increases the resolution but

                          also the accuracy of the numerical solution

                          The discrete approximation results in a set of algebraic equations that are

                          evaluated for the values of the discrete unknowns

                          The mesh is the set of locations where the discrete solution is computed

                          These points are called nodes and if one were to draw lines between adjacent

                          nodes in the domain the resulting image would resemble a net or mesh Two key

                          parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                          space and ∆119905119905 the local distance between adjacent time steps For the simple

                          examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                          19

                          The core idea of the finite-difference method is to replace continuous

                          derivatives with so-called difference formulas that involve only the discrete

                          values associated with positions on the mesh

                          Applying the finite-difference method to a differential equation involves

                          replacing all derivatives with difference formulas In the heat equation there are

                          derivatives with respect to time and derivatives with respect to space Using

                          different combinations of mesh points in the difference formulas results in

                          different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                          numerical solution obtained with any useful scheme will approach the true

                          solution to the original differential equation However the rate at which the

                          numerical solution approaches the true solution varies with the scheme

                          241 First Order Forward Difference

                          Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                          empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                          +1205751205751205971205972

                          2 1205971205972empty1205971205971205971205972

                          1205971205971

                          +1205751205751205971205973

                          3 1205971205973empty1205971205971205971205973

                          1205971205971

                          + ⋯ (25)

                          where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                          consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                          empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                          +∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          +∆1205971205973

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          + ⋯ (26)

                          Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                          120597120597empty120597120597120597120597120597120597119894119894

                          =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                          ∆120597120597minus∆1205971205972

                          1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          minus∆1205971205972

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          minus ⋯ (27)

                          Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                          hand side have been reduced by one

                          20

                          Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                          and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                          120597120597empty120597120597120597120597120597120597119894119894

                          =empty119894119894+1 minus empty119894119894

                          ∆120597120597minus∆1205971205972

                          1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          minus∆1205971205972

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          minus ⋯ (28)

                          The mean value theorem can be used to replace the higher order derivatives

                          ∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          +∆1205971205973

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          + ⋯ =∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120585120585 (29)

                          where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                          120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                          ∆120597120597+∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120585120585 (210)

                          120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                          ∆120597120597asymp∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120585120585 (211)

                          The term on the right hand side of previous equation is called the truncation

                          error of the finite difference approximation

                          In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                          unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                          truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                          analytical form) the big 119978119978 notation can be used to express the dependence of

                          the truncation error on the mesh spacing Note that the right hand side of last

                          equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                          the finite difference simulation Since this is the only parameter under the users

                          control that determines the error the truncation error is simply written

                          ∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120585120585= 119978119978(∆1205971205972) (212)

                          The equals sign in this expression is true in the order of magnitude sense In

                          other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                          21

                          equality Rather the expression means that the left hand side is a product of an

                          unknown constant and ∆1205971205972 Although the expression does not give us the exact

                          magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                          approaches zero as ∆120597120597 is reduced

                          Using big 119978119978 notation Equation (28) can be written

                          120597120597empty120597120597120597120597120597120597119894119894

                          =empty119894119894+1 minus empty119894119894

                          ∆120597120597+ 119978119978(∆120597120597) (213)

                          This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                          it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                          truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                          our control because we can choose the mesh size ∆120597120597 The part of the truncation

                          error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                          242 First Order Backward Difference

                          An alternative first order finite difference formula is obtained if the Taylor series

                          like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                          variables in place of all the unknowns one obtains

                          empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                          +∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          minus∆1205971205973

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          + ⋯ (214)

                          Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                          to get

                          120597120597empty120597120597120597120597120597120597119894119894

                          =empty119894119894+1 minus empty119894119894

                          ∆120597120597minus∆1205971205972

                          1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          minus∆1205971205972

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          minus ⋯ (215)

                          Or using big 119978119978 notation

                          120597120597empty120597120597120597120597120597120597119894119894

                          =empty119894119894 minus empty119894119894minus1

                          ∆120597120597+ 119978119978(∆120597120597) (216)

                          22

                          This is called the backward difference formula because it involves the values of

                          empty at 120597120597119894119894 and 120597120597119894119894minus1

                          The order of magnitude of the truncation error for the backward difference

                          approximation is the same as that of the forward difference approximation Can

                          we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                          truncation error The answer is yes

                          242 First Order Central Difference

                          Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                          empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                          +∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          +∆1205971205973

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          + ⋯ (217)

                          empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                          +∆1205971205972

                          2 1205971205972empty1205971205971205971205972

                          120597120597119894119894

                          minus∆1205971205973

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          + ⋯ (218)

                          Subtracting Equation (10) from Equation (9) yields

                          empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                          + 2∆1205971205973

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          + ⋯ (219)

                          Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                          120597120597empty120597120597120597120597120597120597119894119894

                          =empty119894119894+1 minus empty119894119894minus1

                          2∆120597120597minus∆1205971205972

                          3 1205971205973empty1205971205971205971205973

                          120597120597119894119894

                          minus ⋯ (220)

                          or

                          120597120597empty120597120597120597120597120597120597119894119894

                          =empty119894119894+1 minus empty119894119894minus1

                          2∆120597120597+ 119978119978(∆1205971205972) (221)

                          This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                          approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                          the truncation error for the central difference approximation goes to zero much

                          faster than the truncation error in forward and backward equations

                          23

                          25 Procedures

                          The simple case in this investigation was assuming the constant thermal

                          properties of the material First we assumed all the thermal properties of the

                          materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                          point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                          assume the constant energy after that the pulse of special shapes was selected

                          The numerical solution of equation (23) with boundary and initial conditions

                          in equation (22) was investigated using Matlab program as shown in Appendix

                          The equation of thermal conductivity and specific heat capacity of metal as a

                          function of temperature was obtained by best fitting of polynomials using

                          tabulated data in references

                          24

                          Chapter Three

                          Results and Discursion

                          31 Introduction

                          The development of laser has been an exciting chapter in the history of

                          science and engineering It has produced a new type of advice with potential for

                          application in an extremely wide variety of fields Mach basic development in

                          lasers were occurred during last 35 years The lasers interaction with metal and

                          vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                          The status of laser development and application were still rather rudimentary

                          The light emitted by laser is electro magnetic radiation this radiation has a wave

                          nature the waves consists of vibrating electric and magnetic fields many studies

                          have tried to find and solve models of laser interactions Some researchers

                          proposed the mathematical model related to the laser - plasma interaction and

                          the others have developed an analytical model to study the temperature

                          distribution in Infrared optical materials heated by laser pulses Also an attempt

                          have made to study the interaction of nanosecond pulsed lasers with material

                          from point of view using experimental technique and theoretical approach of

                          dimensional analysis

                          In this study we have evaluate the solution of partial difference equation

                          (PDE) that represent the laser interaction with solid situation in one dimension

                          assuming that the power density of laser and thermal properties are functions

                          with time and temperature respectively

                          25

                          32 Numerical solution with constant laser power density and constant

                          thermal properties

                          First we have taken the lead metal (Pb) with thermal properties

                          119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                          119862119862 = 014016119869119869119892119892119870119870

                          120588120588 = 10751 1198921198921198981198981198981198983

                          119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                          119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                          and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                          represent the area under laser influence

                          The numerical solution of equation (23) with boundary and initial conditions

                          in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                          of lead metal by explicit method using Matlab program give us the results as

                          shown in Fig (31)

                          Fig(31) Depth dependence of the temperature with the laser power density

                          1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                          26

                          33 Evaluation of function 119920119920(119957119957) of laser flux density

                          From following data that represent the energy (119869119869) with time (millie second)

                          Time 0 001 01 02 03 04 05 06 07 08

                          Energy 0 002 017 022 024 02 012 007 002 0

                          By using Matlab program the best polynomial with deduced from above data

                          was

                          119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                          minus 10069 1198981198985 (31)

                          As shown in Fig (32)

                          Fig(32) Laser energy as a function of time

                          Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                          normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                          maximum value (119864119864119898119898119907119907119898119898 )

                          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                          (119864119864119898119898119907119907119898119898 ) (32)

                          The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                          27

                          Fig(33) Normalized laser energy as a function of time

                          The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                          equal to 3 (total laser energy) ie

                          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                          08

                          00

                          119899119899119898119898 = 3 (33)

                          Therefore there exist a real number 119875119875 such that

                          119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                          08

                          00

                          119899119899119898119898 = 3 (34)

                          that implies 119875119875 = 68241 and

                          119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                          08

                          00

                          119899119899119898119898 = 3 (35)

                          The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                          08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                          119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                          00

                          (36)

                          28

                          Where 119863119863119898119898 put to balance the units of equation (36)

                          But integral

                          119868119868 = 119864119864119860119860

                          (37)

                          and from equations (35) (36) and (37) we have

                          119868119868 (119898119898)11989911989911989811989808

                          00

                          = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                          0800 119899119899119898119898

                          119860119860 119863119863119898119898 (38)

                          Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                          (38)

                          There fore

                          119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                          119860119860 119863119863119898119898 (39)

                          As shown in Fig(34) Matlab program was used to obtain the best polynomial

                          that agrees with result data

                          119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                          + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                          Fig(34) Time dependence of laser intensity

                          29

                          34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                          constant thermal properties

                          With all constant thermal properties of lead metal as in article (23) and

                          119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                          equation (23) with boundary and initial condition as in equation (22) and the

                          depth penetration is shown in Fig(35)

                          Fig(35) Depth dependence of the temperature when laser intensity function

                          of time and constant thermal properties of Lead

                          35 Evaluation the Thermal Conductivity as functions of temperature

                          The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                          temperature for Lead material was obtained by Matlab program using the

                          experimental data tabulated in researches

                          119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                          times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                          30

                          119879119879 ( 119870119870) 119870119870 119869119869

                          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                          300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                          The previous thermal conductivity data and the best fitting of the data are

                          shown in Fig (36)

                          Fig(36) The best fitting of thermal conductivity of Lead as a function of

                          temperature

                          31

                          36 Evaluation the Specific heat as functions of temperature

                          The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                          material was obtained from the following experimental data tacked from

                          literatures

                          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                          300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                          The best polynomial fitted for these data was

                          119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                          + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                          times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                          The specific heat capacity data and the best polynomial fitting of the data are

                          shown in Fig (37)

                          32

                          Fig(37) The best fitting of specific heat capacity of Lead as a function of

                          temperature

                          37 Evaluation the Density as functions of temperature

                          The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                          was used to find the best polynomial fitting

                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                          300 11330 400 11230 500 11130 600 11010 800 10430

                          1000 10190 1200 9940

                          The best polynomial of this data was

                          120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                          minus 45158 times 10minus12 1198791198794 (313)

                          33

                          The density of Lead as a function of temperature and the best polynomial fitting

                          are shown in Fig (38)

                          Fig(38) The best fitting of density of Lead as a function of temperature

                          38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                          variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                          We have deduced the solution of equation (24) with initial and boundary

                          condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                          and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                          respectively then by using Matlab program the depth penetration is shown in

                          Fig (39)

                          34

                          Fig(39) Depth dependence of the temperature for pulse laser on Lead

                          material

                          39 Laser interaction with copper material

                          The same time dependence of laser intensity as shown in Fig(34) with

                          thermal properties of copper was used to calculate the temperature distribution as

                          a function of depth penetration

                          The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                          copper material was obtained from the experimental data tabulated in literary

                          The Matlab program used to obtain the best polynomial equation that agrees

                          with the above data

                          119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                          minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                          times 10minus17 1198791198795 (314)

                          35

                          119879119879 ( 119870119870) 119870119870 119869119869

                          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                          100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                          The previous thermal conductivity data and the best fitting of the data are

                          shown in Fig (310)

                          Fig(310) The best fitting of thermal conductivity of Copper as a function of

                          temperature

                          36

                          The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                          material was obtained from the following experimental data tacked from

                          literatures

                          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                          100 0254

                          200 0357

                          273 0384

                          298 0387

                          400 0397

                          600 0416

                          800 0435

                          1000 0454

                          1100 0464

                          1200 0474

                          1300 0483

                          The best polynomial fitted for these data was

                          119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                          + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                          times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                          The specific heat capacity data and the best polynomial fitting of the data are

                          shown in Fig (311)

                          37

                          Fig(311) The best fitting of specific heat capacity of Copper as a function of

                          temperature

                          The density of copper 120588120588(119879119879) as a function of temperature tacked from

                          literature was used to find the best polynomial fitting

                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                          100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                          1000 8576 1100 8519 1200 8458 1300 8396

                          38

                          The best polynomial of this data was

                          120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                          minus 76765 times 10minus14 1198791198794 (316)

                          The density of copper as a function of temperature and the best polynomial

                          fitting are shown in Fig (312)

                          Fig(312) The best fitting of density of copper as a function of temperature

                          The depth penetration of laser energy for copper metal was calculated using

                          the polynomial equations of thermal conductivity specific heat capacity and

                          density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                          (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                          function of time the result was shown in Fig (313)

                          39

                          The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                          for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                          copper metal so the depth penetration of laser energy of lead metal was smaller

                          than that of copper metal this may be due to the high thermal conductivity and

                          high specific heat capacity of copper with that of lead metal

                          Fig(313) Depth dependence of the temperature for pulse laser on Copper

                          material

                          40

                          310 Conclusions

                          The Depth dependence of temperature for lead metal was investigated in two

                          case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                          thermal properties (thermal conductivity specific heat) and density of metal are

                          also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                          vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                          specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                          120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                          penetration depth in the first case is smaller than that of the second case about

                          (190) times

                          The temperature distribution as a function of depth dependence for copper

                          metal was also investigated in the case when the laser intensity vary with time

                          119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                          density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                          The depth penetration of laser energy of lead metal was found to be smaller

                          than that of copper metal this may be due to the high thermal conductivity and

                          high specific heat capacity of copper with that of lead metal

                          41

                          References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                          Sons Inc Hoboken New Jersey Canada (2003)

                          [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                          [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                          [9]

                          Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                          [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                          [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                          [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                          [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                          [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                          httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                          [10] httpwwwworldoflaserscomlaserprincipleshtm

                          [11] httpenwikipediaorgwikiLaserPulsed_operation

                          [12] httpenwikipediaorgwikiThermal_conductivity

                          [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                          [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                          [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                          42

                          Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                          This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                          This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                          43

                          E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                          This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                          This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                          This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                          44

                          title(Dencity as a function of temperature) hold off

                          This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                          45

                          elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                          46

                          r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                          47

                          for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                          48

                          for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                          49

                          6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                          This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                          50

                          u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                          51

                          alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                          52

                          715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                          • 01 Title
                            • 0TDr Omed Ghareb Abdullah
                              • Ch1-Laser
                                • 16 Heat and heat capacity
                                • 18 Derivation in one dimension
                                  • Ch2 laser2
                                  • Ch3 laser2
                                  • Appendix-Laser

                            9

                            pulses can be achieved because the stimulated emission cross-sections of the

                            laser transitions are relatively low allowing a large population inversion to

                            build up The power is also enhanced by the fact that the ground state of XeF

                            quickly dissociates so that there is little absorption to quench the laser pulse

                            action

                            144 Free-Electron Lasers

                            The radiation from a free-electron laser is produced from free electrons

                            which are forced to oscillate in a regular fashion by an applied field They are

                            therefore more like synchrotron light sources or microwave tubes than like

                            other lasers They are able to produce highly coherent collimated radiation

                            over a wide range of frequencies The magnetic field arrangement which

                            produces the alternating field is commonly called a wiggler magnet

                            Fig(18) Principle of operation of Free-Electron laser

                            The free-electron laser is a highly tunable device which has been used to

                            generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

                            this range they are the highest power source Applications of free-electron

                            lasers are envisioned in isotope separation plasma heating for nuclear fusion

                            long-range high resolution radar and particle acceleration in accelerators

                            10

                            15 Pulsed operation

                            Pulsed operation of lasers refers to any laser not classified as continuous

                            wave so that the optical power appears in pulses of some duration at some

                            repetition rate This encompasses a wide range of technologies addressing a

                            number of different motivations Some lasers are pulsed simply because they

                            cannot be run in continuous mode

                            In other cases the application requires the production of pulses having as

                            large an energy as possible Since the pulse energy is equal to the average

                            power divided by the repitition rate this goal can sometimes be satisfied by

                            lowering the rate of pulses so that more energy can be built up in between

                            pulses In laser ablation for example a small volume of material at the surface

                            of a work piece can be evaporated if it is heated in a very short time whereas

                            supplying the energy gradually would allow for the heat to be absorbed into

                            the bulk of the piece never attaining a sufficiently high temperature at a

                            particular point

                            Other applications rely on the peak pulse power (rather than the energy in

                            the pulse) especially in order to obtain nonlinear optical effects For a given

                            pulse energy this requires creating pulses of the shortest possible duration

                            utilizing techniques such as Q-switching

                            16 Heat and heat capacity

                            When a sample is heated meaning it receives thermal energy from an

                            external source some of the introduced heat is converted into kinetic energy

                            the rest to other forms of internal energy specific to the material The amount

                            converted into kinetic energy causes the temperature of the material to rise

                            The amount of the temperature increase depends on how much heat was

                            added the size of the sample the original temperature of the sample and on

                            how the heat was added The two obvious choices on how to add the heat are

                            11

                            to add it holding volume constant or to add it holding pressure constant

                            (There may be other choices but they will not concern us)

                            Lets assume for the moment that we are going to add heat to our sample

                            holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                            subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                            be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                            amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                            (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                            lim119876119876119889119889rarr0

                            119876119876119889119889∆120549120549119889119889

                            = 120597120597119876119876120597120597120549120549119889119889

                            = 119862119862119889119889 (11)

                            We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                            capacity at constant volume Usually one quotes the molar heat capacity

                            119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                            (12)

                            We can rearrange Equation (11) as follows

                            119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                            Then we can integrate this equation to find the heat involved in a finite

                            change at constant volume

                            119876119876119889119889 = 119862119862119889119889

                            1205491205492

                            1205491205491

                            119889119889120549120549 (14)

                            If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                            out of the integral and the heat at constant volume becomes

                            119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                            Let us now go through the same sequence of steps except holding pressure

                            constant instead of volume Our initial definition of the heat capacity at

                            constant pressure 119862119862119875119875 R Rbecomes

                            lim119876119876119875119875rarr0

                            119876119876119875119875∆120549120549119875119875

                            = 120597120597119876119876120597120597120549120549119875119875

                            = 119862119862119875119875 (16)

                            The analogous molar heat capacity is

                            12

                            119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                            (17)

                            Equation (16) rearranges to

                            119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                            which integrates to give

                            119876119876119875119875 = 119862119862119875119875

                            1205491205492

                            1205491205491

                            119889119889120549120549 (19)

                            When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                            119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                            Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                            regarded as constant In these cases the heat capacity is fit to a polynomial (or

                            similar function) in 120549120549 For example some tables give the heat capacity as

                            119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                            where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                            dependent heat capacity the heat at constant pressure would integrate as

                            follows

                            119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                            1205491205492

                            1205491205491

                            119889119889120549120549 (112)

                            119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                            2 minus 12054912054912 + 119899119899

                            1205741205743

                            12054912054923 minus 1205491205491

                            3 (113)

                            Occasionally one finds a different form for the temperature dependent heat

                            capacity in the literature

                            119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                            When you do calculations with temperature dependent heat capacities you

                            must check to see which form is being used for 119862119862119875119875 We are using the

                            convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                            be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                            13

                            positive then heat was indeed absorbed by the system On the other hand if

                            119876119876 is negative it means that the system gave up heat to the surroundings

                            17 Thermal conductivity

                            In physics thermal conductivity 119896119896 is the property of a material that indicates

                            its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                            Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                            The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                            a piece of material The reciprocal of thermal conductivity is thermal

                            resistivity

                            18 Derivation in one dimension

                            The heat equation is derived from Fouriers law and conservation of energy

                            (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                            surface is proportional to the negative temperature gradient across the

                            surface

                            119902119902 = minus119896119896 120571120571120549120549 (115)

                            where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                            dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                            119902119902 = minus119896119896 120549120549119909119909 (116)

                            where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                            energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                            temperature 120549120549120549120549 That is

                            ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                            where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                            material Choosing zero energy at absolute zero temperature this can be

                            rewritten as

                            ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                            14

                            The increase in internal energy in a small spatial region of the material

                            (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                            given by

                            119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                            119909119909minus∆119909119909

                            = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                            119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                            119909119909minus∆119909119909

                            119905119905+∆119905119905

                            119905119905minus∆119905119905

                            (119)

                            Where the fundamental theorem of calculus was used Additionally with no

                            work done and absent any heat sources or sinks the change in internal energy

                            in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                            across the boundaries By Fouriers law this is

                            119896119896 120597120597120549120549120597120597119909119909

                            (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                            (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                            119905119905minus∆119905119905

                            = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                            119909119909+∆119909119909

                            119909119909minus∆119909119909

                            119905119905+∆119905119905

                            119905119905minus∆119905119905

                            (120)

                            again by the fundamental theorem of calculus By conservation of energy

                            119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                            119909119909minus∆119909119909

                            119905119905+∆119905119905

                            119905119905minus∆119905119905

                            = 0 (121)

                            This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                            Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                            Which can be rewritten as

                            120549120549119905119905 =119896119896119888119888119875119875 120588120588

                            120549120549119909119909119909119909 (122)

                            or

                            120597120597120549120549120597120597119905119905

                            =119896119896119888119888119875119875 120588120588

                            12059712059721205491205491205971205971199091199092 (123)

                            15

                            which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                            diffusivity and is often denoted 120572120572

                            19 Aim of present work

                            The goal of this study is to estimate the solution of partial differential

                            equation that governs the laser-solid interaction using numerical methods

                            The solution will been restricted into one dimensional situation in which we

                            assume that both the laser power density and thermal properties are

                            functions of time and temperature respectively In this project we attempt to

                            investigate the laser interaction with both lead and copper materials by

                            predicting the temperature gradient with the depth of the metals

                            16

                            Chapter Two

                            Theoretical Aspects

                            21 Introduction

                            When a laser interacts with a solid surface a variety of processes can

                            occur We are mainly interested in the interaction of pulsed lasers with a

                            solid surface in first instance a metal When such a laser interacts with a

                            copper surface the laser energy will be transformed into heat The

                            temperature of the solid material will increase leading to melting and

                            evaporation of the solid material

                            The evaporated material (vapour atoms) will expand Depending on the

                            applications this can happen in vacuum (or very low pressure) or in a

                            background gas (helium argon air)

                            22 One dimension laser heating equation

                            In general the one dimension laser heating processes of opaque solid slab is

                            represented as

                            120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                            ( 119870119870 119879119879120597120597 ) (21)

                            With boundary conditions and initial condition which represent the pre-

                            vaporization stage

                            minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                            minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                            17

                            119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                            where

                            119870119870 represents the thermal conductivity

                            120588120588 represents the density

                            119862119862 represents the specific heat

                            119879119879 represents the temperature

                            119879119879infin represents the ambient temperature

                            119879119879119907119907 represents the front surface vaporization

                            120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                            Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                            119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                            With the same boundary conditions as in equation (22)

                            where 119889119889119889119889 = 119870119870120588120588119862119862

                            which represents the thermal diffusion

                            But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                            equation (21) with this assuming implies

                            119879119879119905119905 = 1

                            120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                            With the same boundary and initial conditions in equation (22) Where 119870119870

                            represents the derivative of K with respect the temperature

                            23 Numerical solution of Initial value problems

                            An immense number of analytical solutions for conduction heat-transfer

                            problems have been accumulated in literature over the past 100 years Even so

                            in many practical situations the geometry or boundary conditions are such that an

                            analytical solution has not been obtained at all or if the solution has been

                            18

                            developed it involves such a complex series solution that numerical evaluation

                            becomes exceedingly difficult For such situation the most fruitful approach to

                            the problem is numerical techniques the basic principles of which we shall

                            outline in this section

                            One way to guarantee accuracy in the solution of an initial values problems

                            (IVP) is to solve the problem twice using step sizes h and h2 and compare

                            answers at the mesh points corresponding to the larger step size But this requires

                            a significant amount of computation for the smaller step size and must be

                            repeated if it is determined that the agreement is not good enough

                            24 Finite Difference Method

                            The finite difference method is one of several techniques for obtaining

                            numerical solutions to differential equations In all numerical solutions the

                            continuous partial differential equation (PDE) is replaced with a discrete

                            approximation In this context the word discrete means that the numerical

                            solution is known only at a finite number of points in the physical domain The

                            number of those points can be selected by the user of the numerical method In

                            general increasing the number of points not only increases the resolution but

                            also the accuracy of the numerical solution

                            The discrete approximation results in a set of algebraic equations that are

                            evaluated for the values of the discrete unknowns

                            The mesh is the set of locations where the discrete solution is computed

                            These points are called nodes and if one were to draw lines between adjacent

                            nodes in the domain the resulting image would resemble a net or mesh Two key

                            parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                            space and ∆119905119905 the local distance between adjacent time steps For the simple

                            examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                            19

                            The core idea of the finite-difference method is to replace continuous

                            derivatives with so-called difference formulas that involve only the discrete

                            values associated with positions on the mesh

                            Applying the finite-difference method to a differential equation involves

                            replacing all derivatives with difference formulas In the heat equation there are

                            derivatives with respect to time and derivatives with respect to space Using

                            different combinations of mesh points in the difference formulas results in

                            different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                            numerical solution obtained with any useful scheme will approach the true

                            solution to the original differential equation However the rate at which the

                            numerical solution approaches the true solution varies with the scheme

                            241 First Order Forward Difference

                            Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                            empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                            +1205751205751205971205972

                            2 1205971205972empty1205971205971205971205972

                            1205971205971

                            +1205751205751205971205973

                            3 1205971205973empty1205971205971205971205973

                            1205971205971

                            + ⋯ (25)

                            where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                            consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                            empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                            +∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            +∆1205971205973

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            + ⋯ (26)

                            Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                            120597120597empty120597120597120597120597120597120597119894119894

                            =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                            ∆120597120597minus∆1205971205972

                            1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            minus∆1205971205972

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            minus ⋯ (27)

                            Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                            hand side have been reduced by one

                            20

                            Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                            and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                            120597120597empty120597120597120597120597120597120597119894119894

                            =empty119894119894+1 minus empty119894119894

                            ∆120597120597minus∆1205971205972

                            1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            minus∆1205971205972

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            minus ⋯ (28)

                            The mean value theorem can be used to replace the higher order derivatives

                            ∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            +∆1205971205973

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            + ⋯ =∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120585120585 (29)

                            where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                            120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                            ∆120597120597+∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120585120585 (210)

                            120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                            ∆120597120597asymp∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120585120585 (211)

                            The term on the right hand side of previous equation is called the truncation

                            error of the finite difference approximation

                            In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                            unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                            truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                            analytical form) the big 119978119978 notation can be used to express the dependence of

                            the truncation error on the mesh spacing Note that the right hand side of last

                            equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                            the finite difference simulation Since this is the only parameter under the users

                            control that determines the error the truncation error is simply written

                            ∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120585120585= 119978119978(∆1205971205972) (212)

                            The equals sign in this expression is true in the order of magnitude sense In

                            other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                            21

                            equality Rather the expression means that the left hand side is a product of an

                            unknown constant and ∆1205971205972 Although the expression does not give us the exact

                            magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                            approaches zero as ∆120597120597 is reduced

                            Using big 119978119978 notation Equation (28) can be written

                            120597120597empty120597120597120597120597120597120597119894119894

                            =empty119894119894+1 minus empty119894119894

                            ∆120597120597+ 119978119978(∆120597120597) (213)

                            This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                            it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                            truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                            our control because we can choose the mesh size ∆120597120597 The part of the truncation

                            error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                            242 First Order Backward Difference

                            An alternative first order finite difference formula is obtained if the Taylor series

                            like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                            variables in place of all the unknowns one obtains

                            empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                            +∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            minus∆1205971205973

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            + ⋯ (214)

                            Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                            to get

                            120597120597empty120597120597120597120597120597120597119894119894

                            =empty119894119894+1 minus empty119894119894

                            ∆120597120597minus∆1205971205972

                            1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            minus∆1205971205972

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            minus ⋯ (215)

                            Or using big 119978119978 notation

                            120597120597empty120597120597120597120597120597120597119894119894

                            =empty119894119894 minus empty119894119894minus1

                            ∆120597120597+ 119978119978(∆120597120597) (216)

                            22

                            This is called the backward difference formula because it involves the values of

                            empty at 120597120597119894119894 and 120597120597119894119894minus1

                            The order of magnitude of the truncation error for the backward difference

                            approximation is the same as that of the forward difference approximation Can

                            we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                            truncation error The answer is yes

                            242 First Order Central Difference

                            Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                            empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                            +∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            +∆1205971205973

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            + ⋯ (217)

                            empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                            +∆1205971205972

                            2 1205971205972empty1205971205971205971205972

                            120597120597119894119894

                            minus∆1205971205973

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            + ⋯ (218)

                            Subtracting Equation (10) from Equation (9) yields

                            empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                            + 2∆1205971205973

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            + ⋯ (219)

                            Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                            120597120597empty120597120597120597120597120597120597119894119894

                            =empty119894119894+1 minus empty119894119894minus1

                            2∆120597120597minus∆1205971205972

                            3 1205971205973empty1205971205971205971205973

                            120597120597119894119894

                            minus ⋯ (220)

                            or

                            120597120597empty120597120597120597120597120597120597119894119894

                            =empty119894119894+1 minus empty119894119894minus1

                            2∆120597120597+ 119978119978(∆1205971205972) (221)

                            This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                            approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                            the truncation error for the central difference approximation goes to zero much

                            faster than the truncation error in forward and backward equations

                            23

                            25 Procedures

                            The simple case in this investigation was assuming the constant thermal

                            properties of the material First we assumed all the thermal properties of the

                            materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                            point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                            assume the constant energy after that the pulse of special shapes was selected

                            The numerical solution of equation (23) with boundary and initial conditions

                            in equation (22) was investigated using Matlab program as shown in Appendix

                            The equation of thermal conductivity and specific heat capacity of metal as a

                            function of temperature was obtained by best fitting of polynomials using

                            tabulated data in references

                            24

                            Chapter Three

                            Results and Discursion

                            31 Introduction

                            The development of laser has been an exciting chapter in the history of

                            science and engineering It has produced a new type of advice with potential for

                            application in an extremely wide variety of fields Mach basic development in

                            lasers were occurred during last 35 years The lasers interaction with metal and

                            vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                            The status of laser development and application were still rather rudimentary

                            The light emitted by laser is electro magnetic radiation this radiation has a wave

                            nature the waves consists of vibrating electric and magnetic fields many studies

                            have tried to find and solve models of laser interactions Some researchers

                            proposed the mathematical model related to the laser - plasma interaction and

                            the others have developed an analytical model to study the temperature

                            distribution in Infrared optical materials heated by laser pulses Also an attempt

                            have made to study the interaction of nanosecond pulsed lasers with material

                            from point of view using experimental technique and theoretical approach of

                            dimensional analysis

                            In this study we have evaluate the solution of partial difference equation

                            (PDE) that represent the laser interaction with solid situation in one dimension

                            assuming that the power density of laser and thermal properties are functions

                            with time and temperature respectively

                            25

                            32 Numerical solution with constant laser power density and constant

                            thermal properties

                            First we have taken the lead metal (Pb) with thermal properties

                            119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                            119862119862 = 014016119869119869119892119892119870119870

                            120588120588 = 10751 1198921198921198981198981198981198983

                            119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                            119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                            and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                            represent the area under laser influence

                            The numerical solution of equation (23) with boundary and initial conditions

                            in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                            of lead metal by explicit method using Matlab program give us the results as

                            shown in Fig (31)

                            Fig(31) Depth dependence of the temperature with the laser power density

                            1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                            26

                            33 Evaluation of function 119920119920(119957119957) of laser flux density

                            From following data that represent the energy (119869119869) with time (millie second)

                            Time 0 001 01 02 03 04 05 06 07 08

                            Energy 0 002 017 022 024 02 012 007 002 0

                            By using Matlab program the best polynomial with deduced from above data

                            was

                            119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                            minus 10069 1198981198985 (31)

                            As shown in Fig (32)

                            Fig(32) Laser energy as a function of time

                            Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                            normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                            maximum value (119864119864119898119898119907119907119898119898 )

                            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                            (119864119864119898119898119907119907119898119898 ) (32)

                            The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                            27

                            Fig(33) Normalized laser energy as a function of time

                            The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                            equal to 3 (total laser energy) ie

                            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                            08

                            00

                            119899119899119898119898 = 3 (33)

                            Therefore there exist a real number 119875119875 such that

                            119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                            08

                            00

                            119899119899119898119898 = 3 (34)

                            that implies 119875119875 = 68241 and

                            119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                            08

                            00

                            119899119899119898119898 = 3 (35)

                            The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                            08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                            119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                            00

                            (36)

                            28

                            Where 119863119863119898119898 put to balance the units of equation (36)

                            But integral

                            119868119868 = 119864119864119860119860

                            (37)

                            and from equations (35) (36) and (37) we have

                            119868119868 (119898119898)11989911989911989811989808

                            00

                            = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                            0800 119899119899119898119898

                            119860119860 119863119863119898119898 (38)

                            Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                            (38)

                            There fore

                            119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                            119860119860 119863119863119898119898 (39)

                            As shown in Fig(34) Matlab program was used to obtain the best polynomial

                            that agrees with result data

                            119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                            + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                            Fig(34) Time dependence of laser intensity

                            29

                            34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                            constant thermal properties

                            With all constant thermal properties of lead metal as in article (23) and

                            119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                            equation (23) with boundary and initial condition as in equation (22) and the

                            depth penetration is shown in Fig(35)

                            Fig(35) Depth dependence of the temperature when laser intensity function

                            of time and constant thermal properties of Lead

                            35 Evaluation the Thermal Conductivity as functions of temperature

                            The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                            temperature for Lead material was obtained by Matlab program using the

                            experimental data tabulated in researches

                            119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                            times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                            30

                            119879119879 ( 119870119870) 119870119870 119869119869

                            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                            300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                            The previous thermal conductivity data and the best fitting of the data are

                            shown in Fig (36)

                            Fig(36) The best fitting of thermal conductivity of Lead as a function of

                            temperature

                            31

                            36 Evaluation the Specific heat as functions of temperature

                            The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                            material was obtained from the following experimental data tacked from

                            literatures

                            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                            300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                            The best polynomial fitted for these data was

                            119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                            + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                            times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                            The specific heat capacity data and the best polynomial fitting of the data are

                            shown in Fig (37)

                            32

                            Fig(37) The best fitting of specific heat capacity of Lead as a function of

                            temperature

                            37 Evaluation the Density as functions of temperature

                            The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                            was used to find the best polynomial fitting

                            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                            300 11330 400 11230 500 11130 600 11010 800 10430

                            1000 10190 1200 9940

                            The best polynomial of this data was

                            120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                            minus 45158 times 10minus12 1198791198794 (313)

                            33

                            The density of Lead as a function of temperature and the best polynomial fitting

                            are shown in Fig (38)

                            Fig(38) The best fitting of density of Lead as a function of temperature

                            38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                            variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                            We have deduced the solution of equation (24) with initial and boundary

                            condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                            and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                            respectively then by using Matlab program the depth penetration is shown in

                            Fig (39)

                            34

                            Fig(39) Depth dependence of the temperature for pulse laser on Lead

                            material

                            39 Laser interaction with copper material

                            The same time dependence of laser intensity as shown in Fig(34) with

                            thermal properties of copper was used to calculate the temperature distribution as

                            a function of depth penetration

                            The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                            copper material was obtained from the experimental data tabulated in literary

                            The Matlab program used to obtain the best polynomial equation that agrees

                            with the above data

                            119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                            minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                            times 10minus17 1198791198795 (314)

                            35

                            119879119879 ( 119870119870) 119870119870 119869119869

                            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                            100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                            The previous thermal conductivity data and the best fitting of the data are

                            shown in Fig (310)

                            Fig(310) The best fitting of thermal conductivity of Copper as a function of

                            temperature

                            36

                            The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                            material was obtained from the following experimental data tacked from

                            literatures

                            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                            100 0254

                            200 0357

                            273 0384

                            298 0387

                            400 0397

                            600 0416

                            800 0435

                            1000 0454

                            1100 0464

                            1200 0474

                            1300 0483

                            The best polynomial fitted for these data was

                            119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                            + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                            times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                            The specific heat capacity data and the best polynomial fitting of the data are

                            shown in Fig (311)

                            37

                            Fig(311) The best fitting of specific heat capacity of Copper as a function of

                            temperature

                            The density of copper 120588120588(119879119879) as a function of temperature tacked from

                            literature was used to find the best polynomial fitting

                            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                            100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                            1000 8576 1100 8519 1200 8458 1300 8396

                            38

                            The best polynomial of this data was

                            120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                            minus 76765 times 10minus14 1198791198794 (316)

                            The density of copper as a function of temperature and the best polynomial

                            fitting are shown in Fig (312)

                            Fig(312) The best fitting of density of copper as a function of temperature

                            The depth penetration of laser energy for copper metal was calculated using

                            the polynomial equations of thermal conductivity specific heat capacity and

                            density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                            (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                            function of time the result was shown in Fig (313)

                            39

                            The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                            for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                            copper metal so the depth penetration of laser energy of lead metal was smaller

                            than that of copper metal this may be due to the high thermal conductivity and

                            high specific heat capacity of copper with that of lead metal

                            Fig(313) Depth dependence of the temperature for pulse laser on Copper

                            material

                            40

                            310 Conclusions

                            The Depth dependence of temperature for lead metal was investigated in two

                            case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                            thermal properties (thermal conductivity specific heat) and density of metal are

                            also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                            vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                            specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                            120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                            penetration depth in the first case is smaller than that of the second case about

                            (190) times

                            The temperature distribution as a function of depth dependence for copper

                            metal was also investigated in the case when the laser intensity vary with time

                            119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                            density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                            The depth penetration of laser energy of lead metal was found to be smaller

                            than that of copper metal this may be due to the high thermal conductivity and

                            high specific heat capacity of copper with that of lead metal

                            41

                            References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                            Sons Inc Hoboken New Jersey Canada (2003)

                            [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                            [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                            [9]

                            Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                            [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                            [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                            [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                            [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                            [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                            httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                            [10] httpwwwworldoflaserscomlaserprincipleshtm

                            [11] httpenwikipediaorgwikiLaserPulsed_operation

                            [12] httpenwikipediaorgwikiThermal_conductivity

                            [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                            [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                            [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                            42

                            Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                            This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                            This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                            43

                            E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                            This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                            This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                            This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                            44

                            title(Dencity as a function of temperature) hold off

                            This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                            45

                            elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                            46

                            r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                            47

                            for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                            48

                            for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                            49

                            6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                            This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                            50

                            u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                            51

                            alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                            52

                            715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                            • 01 Title
                              • 0TDr Omed Ghareb Abdullah
                                • Ch1-Laser
                                  • 16 Heat and heat capacity
                                  • 18 Derivation in one dimension
                                    • Ch2 laser2
                                    • Ch3 laser2
                                    • Appendix-Laser

                              10

                              15 Pulsed operation

                              Pulsed operation of lasers refers to any laser not classified as continuous

                              wave so that the optical power appears in pulses of some duration at some

                              repetition rate This encompasses a wide range of technologies addressing a

                              number of different motivations Some lasers are pulsed simply because they

                              cannot be run in continuous mode

                              In other cases the application requires the production of pulses having as

                              large an energy as possible Since the pulse energy is equal to the average

                              power divided by the repitition rate this goal can sometimes be satisfied by

                              lowering the rate of pulses so that more energy can be built up in between

                              pulses In laser ablation for example a small volume of material at the surface

                              of a work piece can be evaporated if it is heated in a very short time whereas

                              supplying the energy gradually would allow for the heat to be absorbed into

                              the bulk of the piece never attaining a sufficiently high temperature at a

                              particular point

                              Other applications rely on the peak pulse power (rather than the energy in

                              the pulse) especially in order to obtain nonlinear optical effects For a given

                              pulse energy this requires creating pulses of the shortest possible duration

                              utilizing techniques such as Q-switching

                              16 Heat and heat capacity

                              When a sample is heated meaning it receives thermal energy from an

                              external source some of the introduced heat is converted into kinetic energy

                              the rest to other forms of internal energy specific to the material The amount

                              converted into kinetic energy causes the temperature of the material to rise

                              The amount of the temperature increase depends on how much heat was

                              added the size of the sample the original temperature of the sample and on

                              how the heat was added The two obvious choices on how to add the heat are

                              11

                              to add it holding volume constant or to add it holding pressure constant

                              (There may be other choices but they will not concern us)

                              Lets assume for the moment that we are going to add heat to our sample

                              holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                              subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                              be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                              amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                              (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                              lim119876119876119889119889rarr0

                              119876119876119889119889∆120549120549119889119889

                              = 120597120597119876119876120597120597120549120549119889119889

                              = 119862119862119889119889 (11)

                              We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                              capacity at constant volume Usually one quotes the molar heat capacity

                              119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                              (12)

                              We can rearrange Equation (11) as follows

                              119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                              Then we can integrate this equation to find the heat involved in a finite

                              change at constant volume

                              119876119876119889119889 = 119862119862119889119889

                              1205491205492

                              1205491205491

                              119889119889120549120549 (14)

                              If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                              out of the integral and the heat at constant volume becomes

                              119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                              Let us now go through the same sequence of steps except holding pressure

                              constant instead of volume Our initial definition of the heat capacity at

                              constant pressure 119862119862119875119875 R Rbecomes

                              lim119876119876119875119875rarr0

                              119876119876119875119875∆120549120549119875119875

                              = 120597120597119876119876120597120597120549120549119875119875

                              = 119862119862119875119875 (16)

                              The analogous molar heat capacity is

                              12

                              119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                              (17)

                              Equation (16) rearranges to

                              119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                              which integrates to give

                              119876119876119875119875 = 119862119862119875119875

                              1205491205492

                              1205491205491

                              119889119889120549120549 (19)

                              When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                              119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                              Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                              regarded as constant In these cases the heat capacity is fit to a polynomial (or

                              similar function) in 120549120549 For example some tables give the heat capacity as

                              119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                              where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                              dependent heat capacity the heat at constant pressure would integrate as

                              follows

                              119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                              1205491205492

                              1205491205491

                              119889119889120549120549 (112)

                              119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                              2 minus 12054912054912 + 119899119899

                              1205741205743

                              12054912054923 minus 1205491205491

                              3 (113)

                              Occasionally one finds a different form for the temperature dependent heat

                              capacity in the literature

                              119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                              When you do calculations with temperature dependent heat capacities you

                              must check to see which form is being used for 119862119862119875119875 We are using the

                              convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                              be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                              13

                              positive then heat was indeed absorbed by the system On the other hand if

                              119876119876 is negative it means that the system gave up heat to the surroundings

                              17 Thermal conductivity

                              In physics thermal conductivity 119896119896 is the property of a material that indicates

                              its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                              Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                              The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                              a piece of material The reciprocal of thermal conductivity is thermal

                              resistivity

                              18 Derivation in one dimension

                              The heat equation is derived from Fouriers law and conservation of energy

                              (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                              surface is proportional to the negative temperature gradient across the

                              surface

                              119902119902 = minus119896119896 120571120571120549120549 (115)

                              where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                              dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                              119902119902 = minus119896119896 120549120549119909119909 (116)

                              where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                              energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                              temperature 120549120549120549120549 That is

                              ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                              where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                              material Choosing zero energy at absolute zero temperature this can be

                              rewritten as

                              ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                              14

                              The increase in internal energy in a small spatial region of the material

                              (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                              given by

                              119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                              119909119909minus∆119909119909

                              = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                              119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                              119909119909minus∆119909119909

                              119905119905+∆119905119905

                              119905119905minus∆119905119905

                              (119)

                              Where the fundamental theorem of calculus was used Additionally with no

                              work done and absent any heat sources or sinks the change in internal energy

                              in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                              across the boundaries By Fouriers law this is

                              119896119896 120597120597120549120549120597120597119909119909

                              (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                              (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                              119905119905minus∆119905119905

                              = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                              119909119909+∆119909119909

                              119909119909minus∆119909119909

                              119905119905+∆119905119905

                              119905119905minus∆119905119905

                              (120)

                              again by the fundamental theorem of calculus By conservation of energy

                              119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                              119909119909minus∆119909119909

                              119905119905+∆119905119905

                              119905119905minus∆119905119905

                              = 0 (121)

                              This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                              Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                              Which can be rewritten as

                              120549120549119905119905 =119896119896119888119888119875119875 120588120588

                              120549120549119909119909119909119909 (122)

                              or

                              120597120597120549120549120597120597119905119905

                              =119896119896119888119888119875119875 120588120588

                              12059712059721205491205491205971205971199091199092 (123)

                              15

                              which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                              diffusivity and is often denoted 120572120572

                              19 Aim of present work

                              The goal of this study is to estimate the solution of partial differential

                              equation that governs the laser-solid interaction using numerical methods

                              The solution will been restricted into one dimensional situation in which we

                              assume that both the laser power density and thermal properties are

                              functions of time and temperature respectively In this project we attempt to

                              investigate the laser interaction with both lead and copper materials by

                              predicting the temperature gradient with the depth of the metals

                              16

                              Chapter Two

                              Theoretical Aspects

                              21 Introduction

                              When a laser interacts with a solid surface a variety of processes can

                              occur We are mainly interested in the interaction of pulsed lasers with a

                              solid surface in first instance a metal When such a laser interacts with a

                              copper surface the laser energy will be transformed into heat The

                              temperature of the solid material will increase leading to melting and

                              evaporation of the solid material

                              The evaporated material (vapour atoms) will expand Depending on the

                              applications this can happen in vacuum (or very low pressure) or in a

                              background gas (helium argon air)

                              22 One dimension laser heating equation

                              In general the one dimension laser heating processes of opaque solid slab is

                              represented as

                              120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                              ( 119870119870 119879119879120597120597 ) (21)

                              With boundary conditions and initial condition which represent the pre-

                              vaporization stage

                              minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                              minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                              17

                              119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                              where

                              119870119870 represents the thermal conductivity

                              120588120588 represents the density

                              119862119862 represents the specific heat

                              119879119879 represents the temperature

                              119879119879infin represents the ambient temperature

                              119879119879119907119907 represents the front surface vaporization

                              120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                              Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                              119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                              With the same boundary conditions as in equation (22)

                              where 119889119889119889119889 = 119870119870120588120588119862119862

                              which represents the thermal diffusion

                              But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                              equation (21) with this assuming implies

                              119879119879119905119905 = 1

                              120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                              With the same boundary and initial conditions in equation (22) Where 119870119870

                              represents the derivative of K with respect the temperature

                              23 Numerical solution of Initial value problems

                              An immense number of analytical solutions for conduction heat-transfer

                              problems have been accumulated in literature over the past 100 years Even so

                              in many practical situations the geometry or boundary conditions are such that an

                              analytical solution has not been obtained at all or if the solution has been

                              18

                              developed it involves such a complex series solution that numerical evaluation

                              becomes exceedingly difficult For such situation the most fruitful approach to

                              the problem is numerical techniques the basic principles of which we shall

                              outline in this section

                              One way to guarantee accuracy in the solution of an initial values problems

                              (IVP) is to solve the problem twice using step sizes h and h2 and compare

                              answers at the mesh points corresponding to the larger step size But this requires

                              a significant amount of computation for the smaller step size and must be

                              repeated if it is determined that the agreement is not good enough

                              24 Finite Difference Method

                              The finite difference method is one of several techniques for obtaining

                              numerical solutions to differential equations In all numerical solutions the

                              continuous partial differential equation (PDE) is replaced with a discrete

                              approximation In this context the word discrete means that the numerical

                              solution is known only at a finite number of points in the physical domain The

                              number of those points can be selected by the user of the numerical method In

                              general increasing the number of points not only increases the resolution but

                              also the accuracy of the numerical solution

                              The discrete approximation results in a set of algebraic equations that are

                              evaluated for the values of the discrete unknowns

                              The mesh is the set of locations where the discrete solution is computed

                              These points are called nodes and if one were to draw lines between adjacent

                              nodes in the domain the resulting image would resemble a net or mesh Two key

                              parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                              space and ∆119905119905 the local distance between adjacent time steps For the simple

                              examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                              19

                              The core idea of the finite-difference method is to replace continuous

                              derivatives with so-called difference formulas that involve only the discrete

                              values associated with positions on the mesh

                              Applying the finite-difference method to a differential equation involves

                              replacing all derivatives with difference formulas In the heat equation there are

                              derivatives with respect to time and derivatives with respect to space Using

                              different combinations of mesh points in the difference formulas results in

                              different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                              numerical solution obtained with any useful scheme will approach the true

                              solution to the original differential equation However the rate at which the

                              numerical solution approaches the true solution varies with the scheme

                              241 First Order Forward Difference

                              Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                              empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                              +1205751205751205971205972

                              2 1205971205972empty1205971205971205971205972

                              1205971205971

                              +1205751205751205971205973

                              3 1205971205973empty1205971205971205971205973

                              1205971205971

                              + ⋯ (25)

                              where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                              consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                              empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                              +∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              +∆1205971205973

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              + ⋯ (26)

                              Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                              120597120597empty120597120597120597120597120597120597119894119894

                              =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                              ∆120597120597minus∆1205971205972

                              1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              minus∆1205971205972

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              minus ⋯ (27)

                              Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                              hand side have been reduced by one

                              20

                              Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                              and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                              120597120597empty120597120597120597120597120597120597119894119894

                              =empty119894119894+1 minus empty119894119894

                              ∆120597120597minus∆1205971205972

                              1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              minus∆1205971205972

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              minus ⋯ (28)

                              The mean value theorem can be used to replace the higher order derivatives

                              ∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              +∆1205971205973

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              + ⋯ =∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120585120585 (29)

                              where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                              120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                              ∆120597120597+∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120585120585 (210)

                              120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                              ∆120597120597asymp∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120585120585 (211)

                              The term on the right hand side of previous equation is called the truncation

                              error of the finite difference approximation

                              In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                              unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                              truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                              analytical form) the big 119978119978 notation can be used to express the dependence of

                              the truncation error on the mesh spacing Note that the right hand side of last

                              equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                              the finite difference simulation Since this is the only parameter under the users

                              control that determines the error the truncation error is simply written

                              ∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120585120585= 119978119978(∆1205971205972) (212)

                              The equals sign in this expression is true in the order of magnitude sense In

                              other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                              21

                              equality Rather the expression means that the left hand side is a product of an

                              unknown constant and ∆1205971205972 Although the expression does not give us the exact

                              magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                              approaches zero as ∆120597120597 is reduced

                              Using big 119978119978 notation Equation (28) can be written

                              120597120597empty120597120597120597120597120597120597119894119894

                              =empty119894119894+1 minus empty119894119894

                              ∆120597120597+ 119978119978(∆120597120597) (213)

                              This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                              it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                              truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                              our control because we can choose the mesh size ∆120597120597 The part of the truncation

                              error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                              242 First Order Backward Difference

                              An alternative first order finite difference formula is obtained if the Taylor series

                              like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                              variables in place of all the unknowns one obtains

                              empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                              +∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              minus∆1205971205973

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              + ⋯ (214)

                              Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                              to get

                              120597120597empty120597120597120597120597120597120597119894119894

                              =empty119894119894+1 minus empty119894119894

                              ∆120597120597minus∆1205971205972

                              1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              minus∆1205971205972

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              minus ⋯ (215)

                              Or using big 119978119978 notation

                              120597120597empty120597120597120597120597120597120597119894119894

                              =empty119894119894 minus empty119894119894minus1

                              ∆120597120597+ 119978119978(∆120597120597) (216)

                              22

                              This is called the backward difference formula because it involves the values of

                              empty at 120597120597119894119894 and 120597120597119894119894minus1

                              The order of magnitude of the truncation error for the backward difference

                              approximation is the same as that of the forward difference approximation Can

                              we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                              truncation error The answer is yes

                              242 First Order Central Difference

                              Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                              empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                              +∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              +∆1205971205973

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              + ⋯ (217)

                              empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                              +∆1205971205972

                              2 1205971205972empty1205971205971205971205972

                              120597120597119894119894

                              minus∆1205971205973

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              + ⋯ (218)

                              Subtracting Equation (10) from Equation (9) yields

                              empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                              + 2∆1205971205973

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              + ⋯ (219)

                              Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                              120597120597empty120597120597120597120597120597120597119894119894

                              =empty119894119894+1 minus empty119894119894minus1

                              2∆120597120597minus∆1205971205972

                              3 1205971205973empty1205971205971205971205973

                              120597120597119894119894

                              minus ⋯ (220)

                              or

                              120597120597empty120597120597120597120597120597120597119894119894

                              =empty119894119894+1 minus empty119894119894minus1

                              2∆120597120597+ 119978119978(∆1205971205972) (221)

                              This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                              approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                              the truncation error for the central difference approximation goes to zero much

                              faster than the truncation error in forward and backward equations

                              23

                              25 Procedures

                              The simple case in this investigation was assuming the constant thermal

                              properties of the material First we assumed all the thermal properties of the

                              materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                              point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                              assume the constant energy after that the pulse of special shapes was selected

                              The numerical solution of equation (23) with boundary and initial conditions

                              in equation (22) was investigated using Matlab program as shown in Appendix

                              The equation of thermal conductivity and specific heat capacity of metal as a

                              function of temperature was obtained by best fitting of polynomials using

                              tabulated data in references

                              24

                              Chapter Three

                              Results and Discursion

                              31 Introduction

                              The development of laser has been an exciting chapter in the history of

                              science and engineering It has produced a new type of advice with potential for

                              application in an extremely wide variety of fields Mach basic development in

                              lasers were occurred during last 35 years The lasers interaction with metal and

                              vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                              The status of laser development and application were still rather rudimentary

                              The light emitted by laser is electro magnetic radiation this radiation has a wave

                              nature the waves consists of vibrating electric and magnetic fields many studies

                              have tried to find and solve models of laser interactions Some researchers

                              proposed the mathematical model related to the laser - plasma interaction and

                              the others have developed an analytical model to study the temperature

                              distribution in Infrared optical materials heated by laser pulses Also an attempt

                              have made to study the interaction of nanosecond pulsed lasers with material

                              from point of view using experimental technique and theoretical approach of

                              dimensional analysis

                              In this study we have evaluate the solution of partial difference equation

                              (PDE) that represent the laser interaction with solid situation in one dimension

                              assuming that the power density of laser and thermal properties are functions

                              with time and temperature respectively

                              25

                              32 Numerical solution with constant laser power density and constant

                              thermal properties

                              First we have taken the lead metal (Pb) with thermal properties

                              119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                              119862119862 = 014016119869119869119892119892119870119870

                              120588120588 = 10751 1198921198921198981198981198981198983

                              119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                              119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                              and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                              represent the area under laser influence

                              The numerical solution of equation (23) with boundary and initial conditions

                              in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                              of lead metal by explicit method using Matlab program give us the results as

                              shown in Fig (31)

                              Fig(31) Depth dependence of the temperature with the laser power density

                              1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                              26

                              33 Evaluation of function 119920119920(119957119957) of laser flux density

                              From following data that represent the energy (119869119869) with time (millie second)

                              Time 0 001 01 02 03 04 05 06 07 08

                              Energy 0 002 017 022 024 02 012 007 002 0

                              By using Matlab program the best polynomial with deduced from above data

                              was

                              119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                              minus 10069 1198981198985 (31)

                              As shown in Fig (32)

                              Fig(32) Laser energy as a function of time

                              Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                              normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                              maximum value (119864119864119898119898119907119907119898119898 )

                              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                              (119864119864119898119898119907119907119898119898 ) (32)

                              The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                              27

                              Fig(33) Normalized laser energy as a function of time

                              The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                              equal to 3 (total laser energy) ie

                              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                              08

                              00

                              119899119899119898119898 = 3 (33)

                              Therefore there exist a real number 119875119875 such that

                              119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                              08

                              00

                              119899119899119898119898 = 3 (34)

                              that implies 119875119875 = 68241 and

                              119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                              08

                              00

                              119899119899119898119898 = 3 (35)

                              The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                              08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                              119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                              00

                              (36)

                              28

                              Where 119863119863119898119898 put to balance the units of equation (36)

                              But integral

                              119868119868 = 119864119864119860119860

                              (37)

                              and from equations (35) (36) and (37) we have

                              119868119868 (119898119898)11989911989911989811989808

                              00

                              = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                              0800 119899119899119898119898

                              119860119860 119863119863119898119898 (38)

                              Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                              (38)

                              There fore

                              119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                              119860119860 119863119863119898119898 (39)

                              As shown in Fig(34) Matlab program was used to obtain the best polynomial

                              that agrees with result data

                              119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                              + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                              Fig(34) Time dependence of laser intensity

                              29

                              34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                              constant thermal properties

                              With all constant thermal properties of lead metal as in article (23) and

                              119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                              equation (23) with boundary and initial condition as in equation (22) and the

                              depth penetration is shown in Fig(35)

                              Fig(35) Depth dependence of the temperature when laser intensity function

                              of time and constant thermal properties of Lead

                              35 Evaluation the Thermal Conductivity as functions of temperature

                              The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                              temperature for Lead material was obtained by Matlab program using the

                              experimental data tabulated in researches

                              119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                              times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                              30

                              119879119879 ( 119870119870) 119870119870 119869119869

                              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                              300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                              The previous thermal conductivity data and the best fitting of the data are

                              shown in Fig (36)

                              Fig(36) The best fitting of thermal conductivity of Lead as a function of

                              temperature

                              31

                              36 Evaluation the Specific heat as functions of temperature

                              The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                              material was obtained from the following experimental data tacked from

                              literatures

                              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                              300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                              The best polynomial fitted for these data was

                              119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                              + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                              times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                              The specific heat capacity data and the best polynomial fitting of the data are

                              shown in Fig (37)

                              32

                              Fig(37) The best fitting of specific heat capacity of Lead as a function of

                              temperature

                              37 Evaluation the Density as functions of temperature

                              The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                              was used to find the best polynomial fitting

                              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                              300 11330 400 11230 500 11130 600 11010 800 10430

                              1000 10190 1200 9940

                              The best polynomial of this data was

                              120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                              minus 45158 times 10minus12 1198791198794 (313)

                              33

                              The density of Lead as a function of temperature and the best polynomial fitting

                              are shown in Fig (38)

                              Fig(38) The best fitting of density of Lead as a function of temperature

                              38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                              variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                              We have deduced the solution of equation (24) with initial and boundary

                              condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                              and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                              respectively then by using Matlab program the depth penetration is shown in

                              Fig (39)

                              34

                              Fig(39) Depth dependence of the temperature for pulse laser on Lead

                              material

                              39 Laser interaction with copper material

                              The same time dependence of laser intensity as shown in Fig(34) with

                              thermal properties of copper was used to calculate the temperature distribution as

                              a function of depth penetration

                              The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                              copper material was obtained from the experimental data tabulated in literary

                              The Matlab program used to obtain the best polynomial equation that agrees

                              with the above data

                              119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                              minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                              times 10minus17 1198791198795 (314)

                              35

                              119879119879 ( 119870119870) 119870119870 119869119869

                              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                              100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                              The previous thermal conductivity data and the best fitting of the data are

                              shown in Fig (310)

                              Fig(310) The best fitting of thermal conductivity of Copper as a function of

                              temperature

                              36

                              The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                              material was obtained from the following experimental data tacked from

                              literatures

                              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                              100 0254

                              200 0357

                              273 0384

                              298 0387

                              400 0397

                              600 0416

                              800 0435

                              1000 0454

                              1100 0464

                              1200 0474

                              1300 0483

                              The best polynomial fitted for these data was

                              119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                              + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                              times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                              The specific heat capacity data and the best polynomial fitting of the data are

                              shown in Fig (311)

                              37

                              Fig(311) The best fitting of specific heat capacity of Copper as a function of

                              temperature

                              The density of copper 120588120588(119879119879) as a function of temperature tacked from

                              literature was used to find the best polynomial fitting

                              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                              100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                              1000 8576 1100 8519 1200 8458 1300 8396

                              38

                              The best polynomial of this data was

                              120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                              minus 76765 times 10minus14 1198791198794 (316)

                              The density of copper as a function of temperature and the best polynomial

                              fitting are shown in Fig (312)

                              Fig(312) The best fitting of density of copper as a function of temperature

                              The depth penetration of laser energy for copper metal was calculated using

                              the polynomial equations of thermal conductivity specific heat capacity and

                              density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                              (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                              function of time the result was shown in Fig (313)

                              39

                              The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                              for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                              copper metal so the depth penetration of laser energy of lead metal was smaller

                              than that of copper metal this may be due to the high thermal conductivity and

                              high specific heat capacity of copper with that of lead metal

                              Fig(313) Depth dependence of the temperature for pulse laser on Copper

                              material

                              40

                              310 Conclusions

                              The Depth dependence of temperature for lead metal was investigated in two

                              case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                              thermal properties (thermal conductivity specific heat) and density of metal are

                              also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                              vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                              specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                              120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                              penetration depth in the first case is smaller than that of the second case about

                              (190) times

                              The temperature distribution as a function of depth dependence for copper

                              metal was also investigated in the case when the laser intensity vary with time

                              119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                              density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                              The depth penetration of laser energy of lead metal was found to be smaller

                              than that of copper metal this may be due to the high thermal conductivity and

                              high specific heat capacity of copper with that of lead metal

                              41

                              References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                              Sons Inc Hoboken New Jersey Canada (2003)

                              [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                              [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                              [9]

                              Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                              [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                              [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                              [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                              [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                              [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                              httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                              [10] httpwwwworldoflaserscomlaserprincipleshtm

                              [11] httpenwikipediaorgwikiLaserPulsed_operation

                              [12] httpenwikipediaorgwikiThermal_conductivity

                              [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                              [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                              [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                              42

                              Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                              This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                              This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                              43

                              E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                              This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                              This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                              This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                              44

                              title(Dencity as a function of temperature) hold off

                              This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                              45

                              elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                              46

                              r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                              47

                              for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                              48

                              for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                              49

                              6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                              This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                              50

                              u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                              51

                              alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                              52

                              715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                              • 01 Title
                                • 0TDr Omed Ghareb Abdullah
                                  • Ch1-Laser
                                    • 16 Heat and heat capacity
                                    • 18 Derivation in one dimension
                                      • Ch2 laser2
                                      • Ch3 laser2
                                      • Appendix-Laser

                                11

                                to add it holding volume constant or to add it holding pressure constant

                                (There may be other choices but they will not concern us)

                                Lets assume for the moment that we are going to add heat to our sample

                                holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

                                subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

                                be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

                                amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

                                (so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

                                lim119876119876119889119889rarr0

                                119876119876119889119889∆120549120549119889119889

                                = 120597120597119876119876120597120597120549120549119889119889

                                = 119862119862119889119889 (11)

                                We have given this derivative the symbol 119862119862119889119889 and we call it the heat

                                capacity at constant volume Usually one quotes the molar heat capacity

                                119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

                                (12)

                                We can rearrange Equation (11) as follows

                                119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

                                Then we can integrate this equation to find the heat involved in a finite

                                change at constant volume

                                119876119876119889119889 = 119862119862119889119889

                                1205491205492

                                1205491205491

                                119889119889120549120549 (14)

                                If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

                                out of the integral and the heat at constant volume becomes

                                119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

                                Let us now go through the same sequence of steps except holding pressure

                                constant instead of volume Our initial definition of the heat capacity at

                                constant pressure 119862119862119875119875 R Rbecomes

                                lim119876119876119875119875rarr0

                                119876119876119875119875∆120549120549119875119875

                                = 120597120597119876119876120597120597120549120549119875119875

                                = 119862119862119875119875 (16)

                                The analogous molar heat capacity is

                                12

                                119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                                (17)

                                Equation (16) rearranges to

                                119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                                which integrates to give

                                119876119876119875119875 = 119862119862119875119875

                                1205491205492

                                1205491205491

                                119889119889120549120549 (19)

                                When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                                119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                                Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                                regarded as constant In these cases the heat capacity is fit to a polynomial (or

                                similar function) in 120549120549 For example some tables give the heat capacity as

                                119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                                where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                                dependent heat capacity the heat at constant pressure would integrate as

                                follows

                                119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                                1205491205492

                                1205491205491

                                119889119889120549120549 (112)

                                119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                                2 minus 12054912054912 + 119899119899

                                1205741205743

                                12054912054923 minus 1205491205491

                                3 (113)

                                Occasionally one finds a different form for the temperature dependent heat

                                capacity in the literature

                                119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                                When you do calculations with temperature dependent heat capacities you

                                must check to see which form is being used for 119862119862119875119875 We are using the

                                convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                                be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                                13

                                positive then heat was indeed absorbed by the system On the other hand if

                                119876119876 is negative it means that the system gave up heat to the surroundings

                                17 Thermal conductivity

                                In physics thermal conductivity 119896119896 is the property of a material that indicates

                                its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                                Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                                The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                                a piece of material The reciprocal of thermal conductivity is thermal

                                resistivity

                                18 Derivation in one dimension

                                The heat equation is derived from Fouriers law and conservation of energy

                                (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                                surface is proportional to the negative temperature gradient across the

                                surface

                                119902119902 = minus119896119896 120571120571120549120549 (115)

                                where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                                dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                                119902119902 = minus119896119896 120549120549119909119909 (116)

                                where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                                energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                                temperature 120549120549120549120549 That is

                                ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                                where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                                material Choosing zero energy at absolute zero temperature this can be

                                rewritten as

                                ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                                14

                                The increase in internal energy in a small spatial region of the material

                                (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                                given by

                                119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                                119909119909minus∆119909119909

                                = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                                119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                119909119909minus∆119909119909

                                119905119905+∆119905119905

                                119905119905minus∆119905119905

                                (119)

                                Where the fundamental theorem of calculus was used Additionally with no

                                work done and absent any heat sources or sinks the change in internal energy

                                in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                                across the boundaries By Fouriers law this is

                                119896119896 120597120597120549120549120597120597119909119909

                                (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                                (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                                119905119905minus∆119905119905

                                = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                                119909119909+∆119909119909

                                119909119909minus∆119909119909

                                119905119905+∆119905119905

                                119905119905minus∆119905119905

                                (120)

                                again by the fundamental theorem of calculus By conservation of energy

                                119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                119909119909minus∆119909119909

                                119905119905+∆119905119905

                                119905119905minus∆119905119905

                                = 0 (121)

                                This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                                Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                                Which can be rewritten as

                                120549120549119905119905 =119896119896119888119888119875119875 120588120588

                                120549120549119909119909119909119909 (122)

                                or

                                120597120597120549120549120597120597119905119905

                                =119896119896119888119888119875119875 120588120588

                                12059712059721205491205491205971205971199091199092 (123)

                                15

                                which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                                diffusivity and is often denoted 120572120572

                                19 Aim of present work

                                The goal of this study is to estimate the solution of partial differential

                                equation that governs the laser-solid interaction using numerical methods

                                The solution will been restricted into one dimensional situation in which we

                                assume that both the laser power density and thermal properties are

                                functions of time and temperature respectively In this project we attempt to

                                investigate the laser interaction with both lead and copper materials by

                                predicting the temperature gradient with the depth of the metals

                                16

                                Chapter Two

                                Theoretical Aspects

                                21 Introduction

                                When a laser interacts with a solid surface a variety of processes can

                                occur We are mainly interested in the interaction of pulsed lasers with a

                                solid surface in first instance a metal When such a laser interacts with a

                                copper surface the laser energy will be transformed into heat The

                                temperature of the solid material will increase leading to melting and

                                evaporation of the solid material

                                The evaporated material (vapour atoms) will expand Depending on the

                                applications this can happen in vacuum (or very low pressure) or in a

                                background gas (helium argon air)

                                22 One dimension laser heating equation

                                In general the one dimension laser heating processes of opaque solid slab is

                                represented as

                                120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                                ( 119870119870 119879119879120597120597 ) (21)

                                With boundary conditions and initial condition which represent the pre-

                                vaporization stage

                                minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                                minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                                17

                                119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                                where

                                119870119870 represents the thermal conductivity

                                120588120588 represents the density

                                119862119862 represents the specific heat

                                119879119879 represents the temperature

                                119879119879infin represents the ambient temperature

                                119879119879119907119907 represents the front surface vaporization

                                120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                                Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                                119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                                With the same boundary conditions as in equation (22)

                                where 119889119889119889119889 = 119870119870120588120588119862119862

                                which represents the thermal diffusion

                                But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                                equation (21) with this assuming implies

                                119879119879119905119905 = 1

                                120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                                With the same boundary and initial conditions in equation (22) Where 119870119870

                                represents the derivative of K with respect the temperature

                                23 Numerical solution of Initial value problems

                                An immense number of analytical solutions for conduction heat-transfer

                                problems have been accumulated in literature over the past 100 years Even so

                                in many practical situations the geometry or boundary conditions are such that an

                                analytical solution has not been obtained at all or if the solution has been

                                18

                                developed it involves such a complex series solution that numerical evaluation

                                becomes exceedingly difficult For such situation the most fruitful approach to

                                the problem is numerical techniques the basic principles of which we shall

                                outline in this section

                                One way to guarantee accuracy in the solution of an initial values problems

                                (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                answers at the mesh points corresponding to the larger step size But this requires

                                a significant amount of computation for the smaller step size and must be

                                repeated if it is determined that the agreement is not good enough

                                24 Finite Difference Method

                                The finite difference method is one of several techniques for obtaining

                                numerical solutions to differential equations In all numerical solutions the

                                continuous partial differential equation (PDE) is replaced with a discrete

                                approximation In this context the word discrete means that the numerical

                                solution is known only at a finite number of points in the physical domain The

                                number of those points can be selected by the user of the numerical method In

                                general increasing the number of points not only increases the resolution but

                                also the accuracy of the numerical solution

                                The discrete approximation results in a set of algebraic equations that are

                                evaluated for the values of the discrete unknowns

                                The mesh is the set of locations where the discrete solution is computed

                                These points are called nodes and if one were to draw lines between adjacent

                                nodes in the domain the resulting image would resemble a net or mesh Two key

                                parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                space and ∆119905119905 the local distance between adjacent time steps For the simple

                                examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                19

                                The core idea of the finite-difference method is to replace continuous

                                derivatives with so-called difference formulas that involve only the discrete

                                values associated with positions on the mesh

                                Applying the finite-difference method to a differential equation involves

                                replacing all derivatives with difference formulas In the heat equation there are

                                derivatives with respect to time and derivatives with respect to space Using

                                different combinations of mesh points in the difference formulas results in

                                different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                numerical solution obtained with any useful scheme will approach the true

                                solution to the original differential equation However the rate at which the

                                numerical solution approaches the true solution varies with the scheme

                                241 First Order Forward Difference

                                Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                +1205751205751205971205972

                                2 1205971205972empty1205971205971205971205972

                                1205971205971

                                +1205751205751205971205973

                                3 1205971205973empty1205971205971205971205973

                                1205971205971

                                + ⋯ (25)

                                where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                +∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                +∆1205971205973

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                + ⋯ (26)

                                Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                120597120597empty120597120597120597120597120597120597119894119894

                                =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                ∆120597120597minus∆1205971205972

                                1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                minus∆1205971205972

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                minus ⋯ (27)

                                Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                hand side have been reduced by one

                                20

                                Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                120597120597empty120597120597120597120597120597120597119894119894

                                =empty119894119894+1 minus empty119894119894

                                ∆120597120597minus∆1205971205972

                                1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                minus∆1205971205972

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                minus ⋯ (28)

                                The mean value theorem can be used to replace the higher order derivatives

                                ∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                +∆1205971205973

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                + ⋯ =∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120585120585 (29)

                                where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                ∆120597120597+∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120585120585 (210)

                                120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                ∆120597120597asymp∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120585120585 (211)

                                The term on the right hand side of previous equation is called the truncation

                                error of the finite difference approximation

                                In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                analytical form) the big 119978119978 notation can be used to express the dependence of

                                the truncation error on the mesh spacing Note that the right hand side of last

                                equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                the finite difference simulation Since this is the only parameter under the users

                                control that determines the error the truncation error is simply written

                                ∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120585120585= 119978119978(∆1205971205972) (212)

                                The equals sign in this expression is true in the order of magnitude sense In

                                other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                21

                                equality Rather the expression means that the left hand side is a product of an

                                unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                approaches zero as ∆120597120597 is reduced

                                Using big 119978119978 notation Equation (28) can be written

                                120597120597empty120597120597120597120597120597120597119894119894

                                =empty119894119894+1 minus empty119894119894

                                ∆120597120597+ 119978119978(∆120597120597) (213)

                                This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                242 First Order Backward Difference

                                An alternative first order finite difference formula is obtained if the Taylor series

                                like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                variables in place of all the unknowns one obtains

                                empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                +∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                minus∆1205971205973

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                + ⋯ (214)

                                Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                to get

                                120597120597empty120597120597120597120597120597120597119894119894

                                =empty119894119894+1 minus empty119894119894

                                ∆120597120597minus∆1205971205972

                                1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                minus∆1205971205972

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                minus ⋯ (215)

                                Or using big 119978119978 notation

                                120597120597empty120597120597120597120597120597120597119894119894

                                =empty119894119894 minus empty119894119894minus1

                                ∆120597120597+ 119978119978(∆120597120597) (216)

                                22

                                This is called the backward difference formula because it involves the values of

                                empty at 120597120597119894119894 and 120597120597119894119894minus1

                                The order of magnitude of the truncation error for the backward difference

                                approximation is the same as that of the forward difference approximation Can

                                we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                truncation error The answer is yes

                                242 First Order Central Difference

                                Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                +∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                +∆1205971205973

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                + ⋯ (217)

                                empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                +∆1205971205972

                                2 1205971205972empty1205971205971205971205972

                                120597120597119894119894

                                minus∆1205971205973

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                + ⋯ (218)

                                Subtracting Equation (10) from Equation (9) yields

                                empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                + 2∆1205971205973

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                + ⋯ (219)

                                Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                120597120597empty120597120597120597120597120597120597119894119894

                                =empty119894119894+1 minus empty119894119894minus1

                                2∆120597120597minus∆1205971205972

                                3 1205971205973empty1205971205971205971205973

                                120597120597119894119894

                                minus ⋯ (220)

                                or

                                120597120597empty120597120597120597120597120597120597119894119894

                                =empty119894119894+1 minus empty119894119894minus1

                                2∆120597120597+ 119978119978(∆1205971205972) (221)

                                This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                the truncation error for the central difference approximation goes to zero much

                                faster than the truncation error in forward and backward equations

                                23

                                25 Procedures

                                The simple case in this investigation was assuming the constant thermal

                                properties of the material First we assumed all the thermal properties of the

                                materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                assume the constant energy after that the pulse of special shapes was selected

                                The numerical solution of equation (23) with boundary and initial conditions

                                in equation (22) was investigated using Matlab program as shown in Appendix

                                The equation of thermal conductivity and specific heat capacity of metal as a

                                function of temperature was obtained by best fitting of polynomials using

                                tabulated data in references

                                24

                                Chapter Three

                                Results and Discursion

                                31 Introduction

                                The development of laser has been an exciting chapter in the history of

                                science and engineering It has produced a new type of advice with potential for

                                application in an extremely wide variety of fields Mach basic development in

                                lasers were occurred during last 35 years The lasers interaction with metal and

                                vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                The status of laser development and application were still rather rudimentary

                                The light emitted by laser is electro magnetic radiation this radiation has a wave

                                nature the waves consists of vibrating electric and magnetic fields many studies

                                have tried to find and solve models of laser interactions Some researchers

                                proposed the mathematical model related to the laser - plasma interaction and

                                the others have developed an analytical model to study the temperature

                                distribution in Infrared optical materials heated by laser pulses Also an attempt

                                have made to study the interaction of nanosecond pulsed lasers with material

                                from point of view using experimental technique and theoretical approach of

                                dimensional analysis

                                In this study we have evaluate the solution of partial difference equation

                                (PDE) that represent the laser interaction with solid situation in one dimension

                                assuming that the power density of laser and thermal properties are functions

                                with time and temperature respectively

                                25

                                32 Numerical solution with constant laser power density and constant

                                thermal properties

                                First we have taken the lead metal (Pb) with thermal properties

                                119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                119862119862 = 014016119869119869119892119892119870119870

                                120588120588 = 10751 1198921198921198981198981198981198983

                                119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                represent the area under laser influence

                                The numerical solution of equation (23) with boundary and initial conditions

                                in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                of lead metal by explicit method using Matlab program give us the results as

                                shown in Fig (31)

                                Fig(31) Depth dependence of the temperature with the laser power density

                                1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                26

                                33 Evaluation of function 119920119920(119957119957) of laser flux density

                                From following data that represent the energy (119869119869) with time (millie second)

                                Time 0 001 01 02 03 04 05 06 07 08

                                Energy 0 002 017 022 024 02 012 007 002 0

                                By using Matlab program the best polynomial with deduced from above data

                                was

                                119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                minus 10069 1198981198985 (31)

                                As shown in Fig (32)

                                Fig(32) Laser energy as a function of time

                                Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                maximum value (119864119864119898119898119907119907119898119898 )

                                119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                (119864119864119898119898119907119907119898119898 ) (32)

                                The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                27

                                Fig(33) Normalized laser energy as a function of time

                                The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                equal to 3 (total laser energy) ie

                                119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                08

                                00

                                119899119899119898119898 = 3 (33)

                                Therefore there exist a real number 119875119875 such that

                                119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                08

                                00

                                119899119899119898119898 = 3 (34)

                                that implies 119875119875 = 68241 and

                                119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                08

                                00

                                119899119899119898119898 = 3 (35)

                                The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                00

                                (36)

                                28

                                Where 119863119863119898119898 put to balance the units of equation (36)

                                But integral

                                119868119868 = 119864119864119860119860

                                (37)

                                and from equations (35) (36) and (37) we have

                                119868119868 (119898119898)11989911989911989811989808

                                00

                                = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                0800 119899119899119898119898

                                119860119860 119863119863119898119898 (38)

                                Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                (38)

                                There fore

                                119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                119860119860 119863119863119898119898 (39)

                                As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                that agrees with result data

                                119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                Fig(34) Time dependence of laser intensity

                                29

                                34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                constant thermal properties

                                With all constant thermal properties of lead metal as in article (23) and

                                119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                equation (23) with boundary and initial condition as in equation (22) and the

                                depth penetration is shown in Fig(35)

                                Fig(35) Depth dependence of the temperature when laser intensity function

                                of time and constant thermal properties of Lead

                                35 Evaluation the Thermal Conductivity as functions of temperature

                                The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                temperature for Lead material was obtained by Matlab program using the

                                experimental data tabulated in researches

                                119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                30

                                119879119879 ( 119870119870) 119870119870 119869119869

                                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                The previous thermal conductivity data and the best fitting of the data are

                                shown in Fig (36)

                                Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                temperature

                                31

                                36 Evaluation the Specific heat as functions of temperature

                                The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                material was obtained from the following experimental data tacked from

                                literatures

                                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                The best polynomial fitted for these data was

                                119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                The specific heat capacity data and the best polynomial fitting of the data are

                                shown in Fig (37)

                                32

                                Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                temperature

                                37 Evaluation the Density as functions of temperature

                                The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                was used to find the best polynomial fitting

                                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                300 11330 400 11230 500 11130 600 11010 800 10430

                                1000 10190 1200 9940

                                The best polynomial of this data was

                                120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                minus 45158 times 10minus12 1198791198794 (313)

                                33

                                The density of Lead as a function of temperature and the best polynomial fitting

                                are shown in Fig (38)

                                Fig(38) The best fitting of density of Lead as a function of temperature

                                38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                We have deduced the solution of equation (24) with initial and boundary

                                condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                respectively then by using Matlab program the depth penetration is shown in

                                Fig (39)

                                34

                                Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                material

                                39 Laser interaction with copper material

                                The same time dependence of laser intensity as shown in Fig(34) with

                                thermal properties of copper was used to calculate the temperature distribution as

                                a function of depth penetration

                                The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                copper material was obtained from the experimental data tabulated in literary

                                The Matlab program used to obtain the best polynomial equation that agrees

                                with the above data

                                119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                times 10minus17 1198791198795 (314)

                                35

                                119879119879 ( 119870119870) 119870119870 119869119869

                                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                The previous thermal conductivity data and the best fitting of the data are

                                shown in Fig (310)

                                Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                temperature

                                36

                                The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                material was obtained from the following experimental data tacked from

                                literatures

                                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                100 0254

                                200 0357

                                273 0384

                                298 0387

                                400 0397

                                600 0416

                                800 0435

                                1000 0454

                                1100 0464

                                1200 0474

                                1300 0483

                                The best polynomial fitted for these data was

                                119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                The specific heat capacity data and the best polynomial fitting of the data are

                                shown in Fig (311)

                                37

                                Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                temperature

                                The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                literature was used to find the best polynomial fitting

                                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                1000 8576 1100 8519 1200 8458 1300 8396

                                38

                                The best polynomial of this data was

                                120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                minus 76765 times 10minus14 1198791198794 (316)

                                The density of copper as a function of temperature and the best polynomial

                                fitting are shown in Fig (312)

                                Fig(312) The best fitting of density of copper as a function of temperature

                                The depth penetration of laser energy for copper metal was calculated using

                                the polynomial equations of thermal conductivity specific heat capacity and

                                density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                function of time the result was shown in Fig (313)

                                39

                                The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                copper metal so the depth penetration of laser energy of lead metal was smaller

                                than that of copper metal this may be due to the high thermal conductivity and

                                high specific heat capacity of copper with that of lead metal

                                Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                material

                                40

                                310 Conclusions

                                The Depth dependence of temperature for lead metal was investigated in two

                                case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                thermal properties (thermal conductivity specific heat) and density of metal are

                                also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                penetration depth in the first case is smaller than that of the second case about

                                (190) times

                                The temperature distribution as a function of depth dependence for copper

                                metal was also investigated in the case when the laser intensity vary with time

                                119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                The depth penetration of laser energy of lead metal was found to be smaller

                                than that of copper metal this may be due to the high thermal conductivity and

                                high specific heat capacity of copper with that of lead metal

                                41

                                References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                Sons Inc Hoboken New Jersey Canada (2003)

                                [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                [9]

                                Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                [10] httpwwwworldoflaserscomlaserprincipleshtm

                                [11] httpenwikipediaorgwikiLaserPulsed_operation

                                [12] httpenwikipediaorgwikiThermal_conductivity

                                [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                42

                                Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                43

                                E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                44

                                title(Dencity as a function of temperature) hold off

                                This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                45

                                elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                46

                                r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                47

                                for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                48

                                for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                49

                                6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                50

                                u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                51

                                alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                52

                                715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                • 01 Title
                                  • 0TDr Omed Ghareb Abdullah
                                    • Ch1-Laser
                                      • 16 Heat and heat capacity
                                      • 18 Derivation in one dimension
                                        • Ch2 laser2
                                        • Ch3 laser2
                                        • Appendix-Laser

                                  12

                                  119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

                                  (17)

                                  Equation (16) rearranges to

                                  119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

                                  which integrates to give

                                  119876119876119875119875 = 119862119862119875119875

                                  1205491205492

                                  1205491205491

                                  119889119889120549120549 (19)

                                  When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

                                  119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

                                  Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

                                  regarded as constant In these cases the heat capacity is fit to a polynomial (or

                                  similar function) in 120549120549 For example some tables give the heat capacity as

                                  119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

                                  where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

                                  dependent heat capacity the heat at constant pressure would integrate as

                                  follows

                                  119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

                                  1205491205492

                                  1205491205491

                                  119889119889120549120549 (112)

                                  119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

                                  2 minus 12054912054912 + 119899119899

                                  1205741205743

                                  12054912054923 minus 1205491205491

                                  3 (113)

                                  Occasionally one finds a different form for the temperature dependent heat

                                  capacity in the literature

                                  119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

                                  When you do calculations with temperature dependent heat capacities you

                                  must check to see which form is being used for 119862119862119875119875 We are using the

                                  convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

                                  be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

                                  13

                                  positive then heat was indeed absorbed by the system On the other hand if

                                  119876119876 is negative it means that the system gave up heat to the surroundings

                                  17 Thermal conductivity

                                  In physics thermal conductivity 119896119896 is the property of a material that indicates

                                  its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                                  Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                                  The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                                  a piece of material The reciprocal of thermal conductivity is thermal

                                  resistivity

                                  18 Derivation in one dimension

                                  The heat equation is derived from Fouriers law and conservation of energy

                                  (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                                  surface is proportional to the negative temperature gradient across the

                                  surface

                                  119902119902 = minus119896119896 120571120571120549120549 (115)

                                  where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                                  dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                                  119902119902 = minus119896119896 120549120549119909119909 (116)

                                  where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                                  energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                                  temperature 120549120549120549120549 That is

                                  ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                                  where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                                  material Choosing zero energy at absolute zero temperature this can be

                                  rewritten as

                                  ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                                  14

                                  The increase in internal energy in a small spatial region of the material

                                  (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                                  given by

                                  119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                                  119909119909minus∆119909119909

                                  = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                                  119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                  119909119909minus∆119909119909

                                  119905119905+∆119905119905

                                  119905119905minus∆119905119905

                                  (119)

                                  Where the fundamental theorem of calculus was used Additionally with no

                                  work done and absent any heat sources or sinks the change in internal energy

                                  in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                                  across the boundaries By Fouriers law this is

                                  119896119896 120597120597120549120549120597120597119909119909

                                  (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                                  (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                                  119905119905minus∆119905119905

                                  = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                                  119909119909+∆119909119909

                                  119909119909minus∆119909119909

                                  119905119905+∆119905119905

                                  119905119905minus∆119905119905

                                  (120)

                                  again by the fundamental theorem of calculus By conservation of energy

                                  119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                  119909119909minus∆119909119909

                                  119905119905+∆119905119905

                                  119905119905minus∆119905119905

                                  = 0 (121)

                                  This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                                  Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                                  Which can be rewritten as

                                  120549120549119905119905 =119896119896119888119888119875119875 120588120588

                                  120549120549119909119909119909119909 (122)

                                  or

                                  120597120597120549120549120597120597119905119905

                                  =119896119896119888119888119875119875 120588120588

                                  12059712059721205491205491205971205971199091199092 (123)

                                  15

                                  which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                                  diffusivity and is often denoted 120572120572

                                  19 Aim of present work

                                  The goal of this study is to estimate the solution of partial differential

                                  equation that governs the laser-solid interaction using numerical methods

                                  The solution will been restricted into one dimensional situation in which we

                                  assume that both the laser power density and thermal properties are

                                  functions of time and temperature respectively In this project we attempt to

                                  investigate the laser interaction with both lead and copper materials by

                                  predicting the temperature gradient with the depth of the metals

                                  16

                                  Chapter Two

                                  Theoretical Aspects

                                  21 Introduction

                                  When a laser interacts with a solid surface a variety of processes can

                                  occur We are mainly interested in the interaction of pulsed lasers with a

                                  solid surface in first instance a metal When such a laser interacts with a

                                  copper surface the laser energy will be transformed into heat The

                                  temperature of the solid material will increase leading to melting and

                                  evaporation of the solid material

                                  The evaporated material (vapour atoms) will expand Depending on the

                                  applications this can happen in vacuum (or very low pressure) or in a

                                  background gas (helium argon air)

                                  22 One dimension laser heating equation

                                  In general the one dimension laser heating processes of opaque solid slab is

                                  represented as

                                  120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                                  ( 119870119870 119879119879120597120597 ) (21)

                                  With boundary conditions and initial condition which represent the pre-

                                  vaporization stage

                                  minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                                  minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                                  17

                                  119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                                  where

                                  119870119870 represents the thermal conductivity

                                  120588120588 represents the density

                                  119862119862 represents the specific heat

                                  119879119879 represents the temperature

                                  119879119879infin represents the ambient temperature

                                  119879119879119907119907 represents the front surface vaporization

                                  120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                                  Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                                  119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                                  With the same boundary conditions as in equation (22)

                                  where 119889119889119889119889 = 119870119870120588120588119862119862

                                  which represents the thermal diffusion

                                  But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                                  equation (21) with this assuming implies

                                  119879119879119905119905 = 1

                                  120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                                  With the same boundary and initial conditions in equation (22) Where 119870119870

                                  represents the derivative of K with respect the temperature

                                  23 Numerical solution of Initial value problems

                                  An immense number of analytical solutions for conduction heat-transfer

                                  problems have been accumulated in literature over the past 100 years Even so

                                  in many practical situations the geometry or boundary conditions are such that an

                                  analytical solution has not been obtained at all or if the solution has been

                                  18

                                  developed it involves such a complex series solution that numerical evaluation

                                  becomes exceedingly difficult For such situation the most fruitful approach to

                                  the problem is numerical techniques the basic principles of which we shall

                                  outline in this section

                                  One way to guarantee accuracy in the solution of an initial values problems

                                  (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                  answers at the mesh points corresponding to the larger step size But this requires

                                  a significant amount of computation for the smaller step size and must be

                                  repeated if it is determined that the agreement is not good enough

                                  24 Finite Difference Method

                                  The finite difference method is one of several techniques for obtaining

                                  numerical solutions to differential equations In all numerical solutions the

                                  continuous partial differential equation (PDE) is replaced with a discrete

                                  approximation In this context the word discrete means that the numerical

                                  solution is known only at a finite number of points in the physical domain The

                                  number of those points can be selected by the user of the numerical method In

                                  general increasing the number of points not only increases the resolution but

                                  also the accuracy of the numerical solution

                                  The discrete approximation results in a set of algebraic equations that are

                                  evaluated for the values of the discrete unknowns

                                  The mesh is the set of locations where the discrete solution is computed

                                  These points are called nodes and if one were to draw lines between adjacent

                                  nodes in the domain the resulting image would resemble a net or mesh Two key

                                  parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                  space and ∆119905119905 the local distance between adjacent time steps For the simple

                                  examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                  19

                                  The core idea of the finite-difference method is to replace continuous

                                  derivatives with so-called difference formulas that involve only the discrete

                                  values associated with positions on the mesh

                                  Applying the finite-difference method to a differential equation involves

                                  replacing all derivatives with difference formulas In the heat equation there are

                                  derivatives with respect to time and derivatives with respect to space Using

                                  different combinations of mesh points in the difference formulas results in

                                  different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                  numerical solution obtained with any useful scheme will approach the true

                                  solution to the original differential equation However the rate at which the

                                  numerical solution approaches the true solution varies with the scheme

                                  241 First Order Forward Difference

                                  Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                  empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                  +1205751205751205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  1205971205971

                                  +1205751205751205971205973

                                  3 1205971205973empty1205971205971205971205973

                                  1205971205971

                                  + ⋯ (25)

                                  where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                  consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                  empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                  +∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  +∆1205971205973

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  + ⋯ (26)

                                  Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                  120597120597empty120597120597120597120597120597120597119894119894

                                  =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                  ∆120597120597minus∆1205971205972

                                  1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  minus∆1205971205972

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  minus ⋯ (27)

                                  Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                  hand side have been reduced by one

                                  20

                                  Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                  and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                  120597120597empty120597120597120597120597120597120597119894119894

                                  =empty119894119894+1 minus empty119894119894

                                  ∆120597120597minus∆1205971205972

                                  1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  minus∆1205971205972

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  minus ⋯ (28)

                                  The mean value theorem can be used to replace the higher order derivatives

                                  ∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  +∆1205971205973

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  + ⋯ =∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120585120585 (29)

                                  where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                  120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                  ∆120597120597+∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120585120585 (210)

                                  120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                  ∆120597120597asymp∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120585120585 (211)

                                  The term on the right hand side of previous equation is called the truncation

                                  error of the finite difference approximation

                                  In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                  unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                  truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                  analytical form) the big 119978119978 notation can be used to express the dependence of

                                  the truncation error on the mesh spacing Note that the right hand side of last

                                  equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                  the finite difference simulation Since this is the only parameter under the users

                                  control that determines the error the truncation error is simply written

                                  ∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120585120585= 119978119978(∆1205971205972) (212)

                                  The equals sign in this expression is true in the order of magnitude sense In

                                  other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                  21

                                  equality Rather the expression means that the left hand side is a product of an

                                  unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                  magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                  approaches zero as ∆120597120597 is reduced

                                  Using big 119978119978 notation Equation (28) can be written

                                  120597120597empty120597120597120597120597120597120597119894119894

                                  =empty119894119894+1 minus empty119894119894

                                  ∆120597120597+ 119978119978(∆120597120597) (213)

                                  This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                  it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                  truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                  our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                  error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                  242 First Order Backward Difference

                                  An alternative first order finite difference formula is obtained if the Taylor series

                                  like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                  variables in place of all the unknowns one obtains

                                  empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                  +∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  minus∆1205971205973

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  + ⋯ (214)

                                  Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                  to get

                                  120597120597empty120597120597120597120597120597120597119894119894

                                  =empty119894119894+1 minus empty119894119894

                                  ∆120597120597minus∆1205971205972

                                  1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  minus∆1205971205972

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  minus ⋯ (215)

                                  Or using big 119978119978 notation

                                  120597120597empty120597120597120597120597120597120597119894119894

                                  =empty119894119894 minus empty119894119894minus1

                                  ∆120597120597+ 119978119978(∆120597120597) (216)

                                  22

                                  This is called the backward difference formula because it involves the values of

                                  empty at 120597120597119894119894 and 120597120597119894119894minus1

                                  The order of magnitude of the truncation error for the backward difference

                                  approximation is the same as that of the forward difference approximation Can

                                  we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                  truncation error The answer is yes

                                  242 First Order Central Difference

                                  Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                  empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                  +∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  +∆1205971205973

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  + ⋯ (217)

                                  empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                  +∆1205971205972

                                  2 1205971205972empty1205971205971205971205972

                                  120597120597119894119894

                                  minus∆1205971205973

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  + ⋯ (218)

                                  Subtracting Equation (10) from Equation (9) yields

                                  empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                  + 2∆1205971205973

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  + ⋯ (219)

                                  Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                  120597120597empty120597120597120597120597120597120597119894119894

                                  =empty119894119894+1 minus empty119894119894minus1

                                  2∆120597120597minus∆1205971205972

                                  3 1205971205973empty1205971205971205971205973

                                  120597120597119894119894

                                  minus ⋯ (220)

                                  or

                                  120597120597empty120597120597120597120597120597120597119894119894

                                  =empty119894119894+1 minus empty119894119894minus1

                                  2∆120597120597+ 119978119978(∆1205971205972) (221)

                                  This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                  approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                  the truncation error for the central difference approximation goes to zero much

                                  faster than the truncation error in forward and backward equations

                                  23

                                  25 Procedures

                                  The simple case in this investigation was assuming the constant thermal

                                  properties of the material First we assumed all the thermal properties of the

                                  materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                  point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                  assume the constant energy after that the pulse of special shapes was selected

                                  The numerical solution of equation (23) with boundary and initial conditions

                                  in equation (22) was investigated using Matlab program as shown in Appendix

                                  The equation of thermal conductivity and specific heat capacity of metal as a

                                  function of temperature was obtained by best fitting of polynomials using

                                  tabulated data in references

                                  24

                                  Chapter Three

                                  Results and Discursion

                                  31 Introduction

                                  The development of laser has been an exciting chapter in the history of

                                  science and engineering It has produced a new type of advice with potential for

                                  application in an extremely wide variety of fields Mach basic development in

                                  lasers were occurred during last 35 years The lasers interaction with metal and

                                  vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                  The status of laser development and application were still rather rudimentary

                                  The light emitted by laser is electro magnetic radiation this radiation has a wave

                                  nature the waves consists of vibrating electric and magnetic fields many studies

                                  have tried to find and solve models of laser interactions Some researchers

                                  proposed the mathematical model related to the laser - plasma interaction and

                                  the others have developed an analytical model to study the temperature

                                  distribution in Infrared optical materials heated by laser pulses Also an attempt

                                  have made to study the interaction of nanosecond pulsed lasers with material

                                  from point of view using experimental technique and theoretical approach of

                                  dimensional analysis

                                  In this study we have evaluate the solution of partial difference equation

                                  (PDE) that represent the laser interaction with solid situation in one dimension

                                  assuming that the power density of laser and thermal properties are functions

                                  with time and temperature respectively

                                  25

                                  32 Numerical solution with constant laser power density and constant

                                  thermal properties

                                  First we have taken the lead metal (Pb) with thermal properties

                                  119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                  119862119862 = 014016119869119869119892119892119870119870

                                  120588120588 = 10751 1198921198921198981198981198981198983

                                  119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                  119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                  and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                  represent the area under laser influence

                                  The numerical solution of equation (23) with boundary and initial conditions

                                  in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                  of lead metal by explicit method using Matlab program give us the results as

                                  shown in Fig (31)

                                  Fig(31) Depth dependence of the temperature with the laser power density

                                  1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                  26

                                  33 Evaluation of function 119920119920(119957119957) of laser flux density

                                  From following data that represent the energy (119869119869) with time (millie second)

                                  Time 0 001 01 02 03 04 05 06 07 08

                                  Energy 0 002 017 022 024 02 012 007 002 0

                                  By using Matlab program the best polynomial with deduced from above data

                                  was

                                  119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                  minus 10069 1198981198985 (31)

                                  As shown in Fig (32)

                                  Fig(32) Laser energy as a function of time

                                  Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                  normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                  maximum value (119864119864119898119898119907119907119898119898 )

                                  119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                  (119864119864119898119898119907119907119898119898 ) (32)

                                  The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                  27

                                  Fig(33) Normalized laser energy as a function of time

                                  The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                  equal to 3 (total laser energy) ie

                                  119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                  08

                                  00

                                  119899119899119898119898 = 3 (33)

                                  Therefore there exist a real number 119875119875 such that

                                  119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                  08

                                  00

                                  119899119899119898119898 = 3 (34)

                                  that implies 119875119875 = 68241 and

                                  119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                  08

                                  00

                                  119899119899119898119898 = 3 (35)

                                  The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                  08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                  119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                  00

                                  (36)

                                  28

                                  Where 119863119863119898119898 put to balance the units of equation (36)

                                  But integral

                                  119868119868 = 119864119864119860119860

                                  (37)

                                  and from equations (35) (36) and (37) we have

                                  119868119868 (119898119898)11989911989911989811989808

                                  00

                                  = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                  0800 119899119899119898119898

                                  119860119860 119863119863119898119898 (38)

                                  Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                  (38)

                                  There fore

                                  119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                  119860119860 119863119863119898119898 (39)

                                  As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                  that agrees with result data

                                  119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                  + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                  Fig(34) Time dependence of laser intensity

                                  29

                                  34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                  constant thermal properties

                                  With all constant thermal properties of lead metal as in article (23) and

                                  119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                  equation (23) with boundary and initial condition as in equation (22) and the

                                  depth penetration is shown in Fig(35)

                                  Fig(35) Depth dependence of the temperature when laser intensity function

                                  of time and constant thermal properties of Lead

                                  35 Evaluation the Thermal Conductivity as functions of temperature

                                  The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                  temperature for Lead material was obtained by Matlab program using the

                                  experimental data tabulated in researches

                                  119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                  times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                  30

                                  119879119879 ( 119870119870) 119870119870 119869119869

                                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                  300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                  The previous thermal conductivity data and the best fitting of the data are

                                  shown in Fig (36)

                                  Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                  temperature

                                  31

                                  36 Evaluation the Specific heat as functions of temperature

                                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                  material was obtained from the following experimental data tacked from

                                  literatures

                                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                  300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                  The best polynomial fitted for these data was

                                  119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                  + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                  times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                  The specific heat capacity data and the best polynomial fitting of the data are

                                  shown in Fig (37)

                                  32

                                  Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                  temperature

                                  37 Evaluation the Density as functions of temperature

                                  The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                  was used to find the best polynomial fitting

                                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                  300 11330 400 11230 500 11130 600 11010 800 10430

                                  1000 10190 1200 9940

                                  The best polynomial of this data was

                                  120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                  minus 45158 times 10minus12 1198791198794 (313)

                                  33

                                  The density of Lead as a function of temperature and the best polynomial fitting

                                  are shown in Fig (38)

                                  Fig(38) The best fitting of density of Lead as a function of temperature

                                  38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                  variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                  We have deduced the solution of equation (24) with initial and boundary

                                  condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                  and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                  respectively then by using Matlab program the depth penetration is shown in

                                  Fig (39)

                                  34

                                  Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                  material

                                  39 Laser interaction with copper material

                                  The same time dependence of laser intensity as shown in Fig(34) with

                                  thermal properties of copper was used to calculate the temperature distribution as

                                  a function of depth penetration

                                  The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                  copper material was obtained from the experimental data tabulated in literary

                                  The Matlab program used to obtain the best polynomial equation that agrees

                                  with the above data

                                  119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                  minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                  times 10minus17 1198791198795 (314)

                                  35

                                  119879119879 ( 119870119870) 119870119870 119869119869

                                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                  100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                  The previous thermal conductivity data and the best fitting of the data are

                                  shown in Fig (310)

                                  Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                  temperature

                                  36

                                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                  material was obtained from the following experimental data tacked from

                                  literatures

                                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                  100 0254

                                  200 0357

                                  273 0384

                                  298 0387

                                  400 0397

                                  600 0416

                                  800 0435

                                  1000 0454

                                  1100 0464

                                  1200 0474

                                  1300 0483

                                  The best polynomial fitted for these data was

                                  119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                  + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                  times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                  The specific heat capacity data and the best polynomial fitting of the data are

                                  shown in Fig (311)

                                  37

                                  Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                  temperature

                                  The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                  literature was used to find the best polynomial fitting

                                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                  100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                  1000 8576 1100 8519 1200 8458 1300 8396

                                  38

                                  The best polynomial of this data was

                                  120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                  minus 76765 times 10minus14 1198791198794 (316)

                                  The density of copper as a function of temperature and the best polynomial

                                  fitting are shown in Fig (312)

                                  Fig(312) The best fitting of density of copper as a function of temperature

                                  The depth penetration of laser energy for copper metal was calculated using

                                  the polynomial equations of thermal conductivity specific heat capacity and

                                  density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                  (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                  function of time the result was shown in Fig (313)

                                  39

                                  The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                  for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                  copper metal so the depth penetration of laser energy of lead metal was smaller

                                  than that of copper metal this may be due to the high thermal conductivity and

                                  high specific heat capacity of copper with that of lead metal

                                  Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                  material

                                  40

                                  310 Conclusions

                                  The Depth dependence of temperature for lead metal was investigated in two

                                  case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                  thermal properties (thermal conductivity specific heat) and density of metal are

                                  also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                  vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                  specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                  120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                  penetration depth in the first case is smaller than that of the second case about

                                  (190) times

                                  The temperature distribution as a function of depth dependence for copper

                                  metal was also investigated in the case when the laser intensity vary with time

                                  119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                  density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                  The depth penetration of laser energy of lead metal was found to be smaller

                                  than that of copper metal this may be due to the high thermal conductivity and

                                  high specific heat capacity of copper with that of lead metal

                                  41

                                  References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                  Sons Inc Hoboken New Jersey Canada (2003)

                                  [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                  [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                  [9]

                                  Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                  [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                  [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                  [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                  [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                  [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                  httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                  [10] httpwwwworldoflaserscomlaserprincipleshtm

                                  [11] httpenwikipediaorgwikiLaserPulsed_operation

                                  [12] httpenwikipediaorgwikiThermal_conductivity

                                  [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                  [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                  [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                  42

                                  Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                  This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                  This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                  43

                                  E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                  This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                  This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                  44

                                  title(Dencity as a function of temperature) hold off

                                  This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                  45

                                  elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                  46

                                  r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                  47

                                  for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                  48

                                  for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                  49

                                  6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                  50

                                  u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                  51

                                  alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                  52

                                  715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                  • 01 Title
                                    • 0TDr Omed Ghareb Abdullah
                                      • Ch1-Laser
                                        • 16 Heat and heat capacity
                                        • 18 Derivation in one dimension
                                          • Ch2 laser2
                                          • Ch3 laser2
                                          • Appendix-Laser

                                    13

                                    positive then heat was indeed absorbed by the system On the other hand if

                                    119876119876 is negative it means that the system gave up heat to the surroundings

                                    17 Thermal conductivity

                                    In physics thermal conductivity 119896119896 is the property of a material that indicates

                                    its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

                                    Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

                                    The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

                                    a piece of material The reciprocal of thermal conductivity is thermal

                                    resistivity

                                    18 Derivation in one dimension

                                    The heat equation is derived from Fouriers law and conservation of energy

                                    (Cannon 1984) By Fouriers law the flow rate of heat energy through a

                                    surface is proportional to the negative temperature gradient across the

                                    surface

                                    119902119902 = minus119896119896 120571120571120549120549 (115)

                                    where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

                                    dimension the gradient is an ordinary spatial derivative and so Fouriers law is

                                    119902119902 = minus119896119896 120549120549119909119909 (116)

                                    where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

                                    energy per unit volume in the material 120549120549119876119876 is proportional to the change in

                                    temperature 120549120549120549120549 That is

                                    ∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

                                    where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

                                    material Choosing zero energy at absolute zero temperature this can be

                                    rewritten as

                                    ∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

                                    14

                                    The increase in internal energy in a small spatial region of the material

                                    (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                                    given by

                                    119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                                    119909119909minus∆119909119909

                                    = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                                    119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                    119909119909minus∆119909119909

                                    119905119905+∆119905119905

                                    119905119905minus∆119905119905

                                    (119)

                                    Where the fundamental theorem of calculus was used Additionally with no

                                    work done and absent any heat sources or sinks the change in internal energy

                                    in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                                    across the boundaries By Fouriers law this is

                                    119896119896 120597120597120549120549120597120597119909119909

                                    (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                                    (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                                    119905119905minus∆119905119905

                                    = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                                    119909119909+∆119909119909

                                    119909119909minus∆119909119909

                                    119905119905+∆119905119905

                                    119905119905minus∆119905119905

                                    (120)

                                    again by the fundamental theorem of calculus By conservation of energy

                                    119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                    119909119909minus∆119909119909

                                    119905119905+∆119905119905

                                    119905119905minus∆119905119905

                                    = 0 (121)

                                    This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                                    Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                                    Which can be rewritten as

                                    120549120549119905119905 =119896119896119888119888119875119875 120588120588

                                    120549120549119909119909119909119909 (122)

                                    or

                                    120597120597120549120549120597120597119905119905

                                    =119896119896119888119888119875119875 120588120588

                                    12059712059721205491205491205971205971199091199092 (123)

                                    15

                                    which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                                    diffusivity and is often denoted 120572120572

                                    19 Aim of present work

                                    The goal of this study is to estimate the solution of partial differential

                                    equation that governs the laser-solid interaction using numerical methods

                                    The solution will been restricted into one dimensional situation in which we

                                    assume that both the laser power density and thermal properties are

                                    functions of time and temperature respectively In this project we attempt to

                                    investigate the laser interaction with both lead and copper materials by

                                    predicting the temperature gradient with the depth of the metals

                                    16

                                    Chapter Two

                                    Theoretical Aspects

                                    21 Introduction

                                    When a laser interacts with a solid surface a variety of processes can

                                    occur We are mainly interested in the interaction of pulsed lasers with a

                                    solid surface in first instance a metal When such a laser interacts with a

                                    copper surface the laser energy will be transformed into heat The

                                    temperature of the solid material will increase leading to melting and

                                    evaporation of the solid material

                                    The evaporated material (vapour atoms) will expand Depending on the

                                    applications this can happen in vacuum (or very low pressure) or in a

                                    background gas (helium argon air)

                                    22 One dimension laser heating equation

                                    In general the one dimension laser heating processes of opaque solid slab is

                                    represented as

                                    120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                                    ( 119870119870 119879119879120597120597 ) (21)

                                    With boundary conditions and initial condition which represent the pre-

                                    vaporization stage

                                    minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                                    minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                                    17

                                    119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                                    where

                                    119870119870 represents the thermal conductivity

                                    120588120588 represents the density

                                    119862119862 represents the specific heat

                                    119879119879 represents the temperature

                                    119879119879infin represents the ambient temperature

                                    119879119879119907119907 represents the front surface vaporization

                                    120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                                    Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                                    119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                                    With the same boundary conditions as in equation (22)

                                    where 119889119889119889119889 = 119870119870120588120588119862119862

                                    which represents the thermal diffusion

                                    But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                                    equation (21) with this assuming implies

                                    119879119879119905119905 = 1

                                    120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                                    With the same boundary and initial conditions in equation (22) Where 119870119870

                                    represents the derivative of K with respect the temperature

                                    23 Numerical solution of Initial value problems

                                    An immense number of analytical solutions for conduction heat-transfer

                                    problems have been accumulated in literature over the past 100 years Even so

                                    in many practical situations the geometry or boundary conditions are such that an

                                    analytical solution has not been obtained at all or if the solution has been

                                    18

                                    developed it involves such a complex series solution that numerical evaluation

                                    becomes exceedingly difficult For such situation the most fruitful approach to

                                    the problem is numerical techniques the basic principles of which we shall

                                    outline in this section

                                    One way to guarantee accuracy in the solution of an initial values problems

                                    (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                    answers at the mesh points corresponding to the larger step size But this requires

                                    a significant amount of computation for the smaller step size and must be

                                    repeated if it is determined that the agreement is not good enough

                                    24 Finite Difference Method

                                    The finite difference method is one of several techniques for obtaining

                                    numerical solutions to differential equations In all numerical solutions the

                                    continuous partial differential equation (PDE) is replaced with a discrete

                                    approximation In this context the word discrete means that the numerical

                                    solution is known only at a finite number of points in the physical domain The

                                    number of those points can be selected by the user of the numerical method In

                                    general increasing the number of points not only increases the resolution but

                                    also the accuracy of the numerical solution

                                    The discrete approximation results in a set of algebraic equations that are

                                    evaluated for the values of the discrete unknowns

                                    The mesh is the set of locations where the discrete solution is computed

                                    These points are called nodes and if one were to draw lines between adjacent

                                    nodes in the domain the resulting image would resemble a net or mesh Two key

                                    parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                    space and ∆119905119905 the local distance between adjacent time steps For the simple

                                    examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                    19

                                    The core idea of the finite-difference method is to replace continuous

                                    derivatives with so-called difference formulas that involve only the discrete

                                    values associated with positions on the mesh

                                    Applying the finite-difference method to a differential equation involves

                                    replacing all derivatives with difference formulas In the heat equation there are

                                    derivatives with respect to time and derivatives with respect to space Using

                                    different combinations of mesh points in the difference formulas results in

                                    different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                    numerical solution obtained with any useful scheme will approach the true

                                    solution to the original differential equation However the rate at which the

                                    numerical solution approaches the true solution varies with the scheme

                                    241 First Order Forward Difference

                                    Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                    empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                    +1205751205751205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    1205971205971

                                    +1205751205751205971205973

                                    3 1205971205973empty1205971205971205971205973

                                    1205971205971

                                    + ⋯ (25)

                                    where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                    consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                    empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                    +∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    +∆1205971205973

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    + ⋯ (26)

                                    Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                    120597120597empty120597120597120597120597120597120597119894119894

                                    =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                    ∆120597120597minus∆1205971205972

                                    1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    minus∆1205971205972

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    minus ⋯ (27)

                                    Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                    hand side have been reduced by one

                                    20

                                    Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                    and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                    120597120597empty120597120597120597120597120597120597119894119894

                                    =empty119894119894+1 minus empty119894119894

                                    ∆120597120597minus∆1205971205972

                                    1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    minus∆1205971205972

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    minus ⋯ (28)

                                    The mean value theorem can be used to replace the higher order derivatives

                                    ∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    +∆1205971205973

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    + ⋯ =∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120585120585 (29)

                                    where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                    120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                    ∆120597120597+∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120585120585 (210)

                                    120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                    ∆120597120597asymp∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120585120585 (211)

                                    The term on the right hand side of previous equation is called the truncation

                                    error of the finite difference approximation

                                    In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                    unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                    truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                    analytical form) the big 119978119978 notation can be used to express the dependence of

                                    the truncation error on the mesh spacing Note that the right hand side of last

                                    equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                    the finite difference simulation Since this is the only parameter under the users

                                    control that determines the error the truncation error is simply written

                                    ∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120585120585= 119978119978(∆1205971205972) (212)

                                    The equals sign in this expression is true in the order of magnitude sense In

                                    other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                    21

                                    equality Rather the expression means that the left hand side is a product of an

                                    unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                    magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                    approaches zero as ∆120597120597 is reduced

                                    Using big 119978119978 notation Equation (28) can be written

                                    120597120597empty120597120597120597120597120597120597119894119894

                                    =empty119894119894+1 minus empty119894119894

                                    ∆120597120597+ 119978119978(∆120597120597) (213)

                                    This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                    it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                    truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                    our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                    error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                    242 First Order Backward Difference

                                    An alternative first order finite difference formula is obtained if the Taylor series

                                    like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                    variables in place of all the unknowns one obtains

                                    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                    +∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    minus∆1205971205973

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    + ⋯ (214)

                                    Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                    to get

                                    120597120597empty120597120597120597120597120597120597119894119894

                                    =empty119894119894+1 minus empty119894119894

                                    ∆120597120597minus∆1205971205972

                                    1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    minus∆1205971205972

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    minus ⋯ (215)

                                    Or using big 119978119978 notation

                                    120597120597empty120597120597120597120597120597120597119894119894

                                    =empty119894119894 minus empty119894119894minus1

                                    ∆120597120597+ 119978119978(∆120597120597) (216)

                                    22

                                    This is called the backward difference formula because it involves the values of

                                    empty at 120597120597119894119894 and 120597120597119894119894minus1

                                    The order of magnitude of the truncation error for the backward difference

                                    approximation is the same as that of the forward difference approximation Can

                                    we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                    truncation error The answer is yes

                                    242 First Order Central Difference

                                    Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                    empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                    +∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    +∆1205971205973

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    + ⋯ (217)

                                    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                    +∆1205971205972

                                    2 1205971205972empty1205971205971205971205972

                                    120597120597119894119894

                                    minus∆1205971205973

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    + ⋯ (218)

                                    Subtracting Equation (10) from Equation (9) yields

                                    empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                    + 2∆1205971205973

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    + ⋯ (219)

                                    Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                    120597120597empty120597120597120597120597120597120597119894119894

                                    =empty119894119894+1 minus empty119894119894minus1

                                    2∆120597120597minus∆1205971205972

                                    3 1205971205973empty1205971205971205971205973

                                    120597120597119894119894

                                    minus ⋯ (220)

                                    or

                                    120597120597empty120597120597120597120597120597120597119894119894

                                    =empty119894119894+1 minus empty119894119894minus1

                                    2∆120597120597+ 119978119978(∆1205971205972) (221)

                                    This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                    approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                    the truncation error for the central difference approximation goes to zero much

                                    faster than the truncation error in forward and backward equations

                                    23

                                    25 Procedures

                                    The simple case in this investigation was assuming the constant thermal

                                    properties of the material First we assumed all the thermal properties of the

                                    materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                    point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                    assume the constant energy after that the pulse of special shapes was selected

                                    The numerical solution of equation (23) with boundary and initial conditions

                                    in equation (22) was investigated using Matlab program as shown in Appendix

                                    The equation of thermal conductivity and specific heat capacity of metal as a

                                    function of temperature was obtained by best fitting of polynomials using

                                    tabulated data in references

                                    24

                                    Chapter Three

                                    Results and Discursion

                                    31 Introduction

                                    The development of laser has been an exciting chapter in the history of

                                    science and engineering It has produced a new type of advice with potential for

                                    application in an extremely wide variety of fields Mach basic development in

                                    lasers were occurred during last 35 years The lasers interaction with metal and

                                    vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                    The status of laser development and application were still rather rudimentary

                                    The light emitted by laser is electro magnetic radiation this radiation has a wave

                                    nature the waves consists of vibrating electric and magnetic fields many studies

                                    have tried to find and solve models of laser interactions Some researchers

                                    proposed the mathematical model related to the laser - plasma interaction and

                                    the others have developed an analytical model to study the temperature

                                    distribution in Infrared optical materials heated by laser pulses Also an attempt

                                    have made to study the interaction of nanosecond pulsed lasers with material

                                    from point of view using experimental technique and theoretical approach of

                                    dimensional analysis

                                    In this study we have evaluate the solution of partial difference equation

                                    (PDE) that represent the laser interaction with solid situation in one dimension

                                    assuming that the power density of laser and thermal properties are functions

                                    with time and temperature respectively

                                    25

                                    32 Numerical solution with constant laser power density and constant

                                    thermal properties

                                    First we have taken the lead metal (Pb) with thermal properties

                                    119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                    119862119862 = 014016119869119869119892119892119870119870

                                    120588120588 = 10751 1198921198921198981198981198981198983

                                    119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                    119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                    and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                    represent the area under laser influence

                                    The numerical solution of equation (23) with boundary and initial conditions

                                    in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                    of lead metal by explicit method using Matlab program give us the results as

                                    shown in Fig (31)

                                    Fig(31) Depth dependence of the temperature with the laser power density

                                    1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                    26

                                    33 Evaluation of function 119920119920(119957119957) of laser flux density

                                    From following data that represent the energy (119869119869) with time (millie second)

                                    Time 0 001 01 02 03 04 05 06 07 08

                                    Energy 0 002 017 022 024 02 012 007 002 0

                                    By using Matlab program the best polynomial with deduced from above data

                                    was

                                    119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                    minus 10069 1198981198985 (31)

                                    As shown in Fig (32)

                                    Fig(32) Laser energy as a function of time

                                    Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                    normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                    maximum value (119864119864119898119898119907119907119898119898 )

                                    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                    (119864119864119898119898119907119907119898119898 ) (32)

                                    The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                    27

                                    Fig(33) Normalized laser energy as a function of time

                                    The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                    equal to 3 (total laser energy) ie

                                    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                    08

                                    00

                                    119899119899119898119898 = 3 (33)

                                    Therefore there exist a real number 119875119875 such that

                                    119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                    08

                                    00

                                    119899119899119898119898 = 3 (34)

                                    that implies 119875119875 = 68241 and

                                    119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                    08

                                    00

                                    119899119899119898119898 = 3 (35)

                                    The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                    08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                    119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                    00

                                    (36)

                                    28

                                    Where 119863119863119898119898 put to balance the units of equation (36)

                                    But integral

                                    119868119868 = 119864119864119860119860

                                    (37)

                                    and from equations (35) (36) and (37) we have

                                    119868119868 (119898119898)11989911989911989811989808

                                    00

                                    = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                    0800 119899119899119898119898

                                    119860119860 119863119863119898119898 (38)

                                    Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                    (38)

                                    There fore

                                    119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                    119860119860 119863119863119898119898 (39)

                                    As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                    that agrees with result data

                                    119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                    + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                    Fig(34) Time dependence of laser intensity

                                    29

                                    34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                    constant thermal properties

                                    With all constant thermal properties of lead metal as in article (23) and

                                    119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                    equation (23) with boundary and initial condition as in equation (22) and the

                                    depth penetration is shown in Fig(35)

                                    Fig(35) Depth dependence of the temperature when laser intensity function

                                    of time and constant thermal properties of Lead

                                    35 Evaluation the Thermal Conductivity as functions of temperature

                                    The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                    temperature for Lead material was obtained by Matlab program using the

                                    experimental data tabulated in researches

                                    119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                    times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                    30

                                    119879119879 ( 119870119870) 119870119870 119869119869

                                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                    300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                    The previous thermal conductivity data and the best fitting of the data are

                                    shown in Fig (36)

                                    Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                    temperature

                                    31

                                    36 Evaluation the Specific heat as functions of temperature

                                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                    material was obtained from the following experimental data tacked from

                                    literatures

                                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                    300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                    The best polynomial fitted for these data was

                                    119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                    + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                    times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                    The specific heat capacity data and the best polynomial fitting of the data are

                                    shown in Fig (37)

                                    32

                                    Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                    temperature

                                    37 Evaluation the Density as functions of temperature

                                    The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                    was used to find the best polynomial fitting

                                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                    300 11330 400 11230 500 11130 600 11010 800 10430

                                    1000 10190 1200 9940

                                    The best polynomial of this data was

                                    120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                    minus 45158 times 10minus12 1198791198794 (313)

                                    33

                                    The density of Lead as a function of temperature and the best polynomial fitting

                                    are shown in Fig (38)

                                    Fig(38) The best fitting of density of Lead as a function of temperature

                                    38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                    variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                    We have deduced the solution of equation (24) with initial and boundary

                                    condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                    and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                    respectively then by using Matlab program the depth penetration is shown in

                                    Fig (39)

                                    34

                                    Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                    material

                                    39 Laser interaction with copper material

                                    The same time dependence of laser intensity as shown in Fig(34) with

                                    thermal properties of copper was used to calculate the temperature distribution as

                                    a function of depth penetration

                                    The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                    copper material was obtained from the experimental data tabulated in literary

                                    The Matlab program used to obtain the best polynomial equation that agrees

                                    with the above data

                                    119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                    minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                    times 10minus17 1198791198795 (314)

                                    35

                                    119879119879 ( 119870119870) 119870119870 119869119869

                                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                    100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                    The previous thermal conductivity data and the best fitting of the data are

                                    shown in Fig (310)

                                    Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                    temperature

                                    36

                                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                    material was obtained from the following experimental data tacked from

                                    literatures

                                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                    100 0254

                                    200 0357

                                    273 0384

                                    298 0387

                                    400 0397

                                    600 0416

                                    800 0435

                                    1000 0454

                                    1100 0464

                                    1200 0474

                                    1300 0483

                                    The best polynomial fitted for these data was

                                    119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                    + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                    times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                    The specific heat capacity data and the best polynomial fitting of the data are

                                    shown in Fig (311)

                                    37

                                    Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                    temperature

                                    The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                    literature was used to find the best polynomial fitting

                                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                    100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                    1000 8576 1100 8519 1200 8458 1300 8396

                                    38

                                    The best polynomial of this data was

                                    120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                    minus 76765 times 10minus14 1198791198794 (316)

                                    The density of copper as a function of temperature and the best polynomial

                                    fitting are shown in Fig (312)

                                    Fig(312) The best fitting of density of copper as a function of temperature

                                    The depth penetration of laser energy for copper metal was calculated using

                                    the polynomial equations of thermal conductivity specific heat capacity and

                                    density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                    (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                    function of time the result was shown in Fig (313)

                                    39

                                    The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                    for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                    copper metal so the depth penetration of laser energy of lead metal was smaller

                                    than that of copper metal this may be due to the high thermal conductivity and

                                    high specific heat capacity of copper with that of lead metal

                                    Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                    material

                                    40

                                    310 Conclusions

                                    The Depth dependence of temperature for lead metal was investigated in two

                                    case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                    thermal properties (thermal conductivity specific heat) and density of metal are

                                    also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                    vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                    specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                    120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                    penetration depth in the first case is smaller than that of the second case about

                                    (190) times

                                    The temperature distribution as a function of depth dependence for copper

                                    metal was also investigated in the case when the laser intensity vary with time

                                    119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                    density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                    The depth penetration of laser energy of lead metal was found to be smaller

                                    than that of copper metal this may be due to the high thermal conductivity and

                                    high specific heat capacity of copper with that of lead metal

                                    41

                                    References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                    Sons Inc Hoboken New Jersey Canada (2003)

                                    [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                    [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                    [9]

                                    Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                    [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                    [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                    [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                    [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                    [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                    httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                    [10] httpwwwworldoflaserscomlaserprincipleshtm

                                    [11] httpenwikipediaorgwikiLaserPulsed_operation

                                    [12] httpenwikipediaorgwikiThermal_conductivity

                                    [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                    [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                    [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                    42

                                    Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                    This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                    This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                    43

                                    E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                    This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                    This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                    44

                                    title(Dencity as a function of temperature) hold off

                                    This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                    45

                                    elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                    46

                                    r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                    47

                                    for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                    48

                                    for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                    49

                                    6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                    50

                                    u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                    51

                                    alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                    52

                                    715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                    • 01 Title
                                      • 0TDr Omed Ghareb Abdullah
                                        • Ch1-Laser
                                          • 16 Heat and heat capacity
                                          • 18 Derivation in one dimension
                                            • Ch2 laser2
                                            • Ch3 laser2
                                            • Appendix-Laser

                                      14

                                      The increase in internal energy in a small spatial region of the material

                                      (119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

                                      given by

                                      119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

                                      119909119909minus∆119909119909

                                      = 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

                                      119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                      119909119909minus∆119909119909

                                      119905119905+∆119905119905

                                      119905119905minus∆119905119905

                                      (119)

                                      Where the fundamental theorem of calculus was used Additionally with no

                                      work done and absent any heat sources or sinks the change in internal energy

                                      in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

                                      across the boundaries By Fouriers law this is

                                      119896119896 120597120597120549120549120597120597119909119909

                                      (119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

                                      (119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

                                      119905119905minus∆119905119905

                                      = 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

                                      119909119909+∆119909119909

                                      119909119909minus∆119909119909

                                      119905119905+∆119905119905

                                      119905119905minus∆119905119905

                                      (120)

                                      again by the fundamental theorem of calculus By conservation of energy

                                      119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

                                      119909119909minus∆119909119909

                                      119905119905+∆119905119905

                                      119905119905minus∆119905119905

                                      = 0 (121)

                                      This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

                                      Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

                                      Which can be rewritten as

                                      120549120549119905119905 =119896119896119888119888119875119875 120588120588

                                      120549120549119909119909119909119909 (122)

                                      or

                                      120597120597120549120549120597120597119905119905

                                      =119896119896119888119888119875119875 120588120588

                                      12059712059721205491205491205971205971199091199092 (123)

                                      15

                                      which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                                      diffusivity and is often denoted 120572120572

                                      19 Aim of present work

                                      The goal of this study is to estimate the solution of partial differential

                                      equation that governs the laser-solid interaction using numerical methods

                                      The solution will been restricted into one dimensional situation in which we

                                      assume that both the laser power density and thermal properties are

                                      functions of time and temperature respectively In this project we attempt to

                                      investigate the laser interaction with both lead and copper materials by

                                      predicting the temperature gradient with the depth of the metals

                                      16

                                      Chapter Two

                                      Theoretical Aspects

                                      21 Introduction

                                      When a laser interacts with a solid surface a variety of processes can

                                      occur We are mainly interested in the interaction of pulsed lasers with a

                                      solid surface in first instance a metal When such a laser interacts with a

                                      copper surface the laser energy will be transformed into heat The

                                      temperature of the solid material will increase leading to melting and

                                      evaporation of the solid material

                                      The evaporated material (vapour atoms) will expand Depending on the

                                      applications this can happen in vacuum (or very low pressure) or in a

                                      background gas (helium argon air)

                                      22 One dimension laser heating equation

                                      In general the one dimension laser heating processes of opaque solid slab is

                                      represented as

                                      120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                                      ( 119870119870 119879119879120597120597 ) (21)

                                      With boundary conditions and initial condition which represent the pre-

                                      vaporization stage

                                      minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                                      minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                                      17

                                      119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                                      where

                                      119870119870 represents the thermal conductivity

                                      120588120588 represents the density

                                      119862119862 represents the specific heat

                                      119879119879 represents the temperature

                                      119879119879infin represents the ambient temperature

                                      119879119879119907119907 represents the front surface vaporization

                                      120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                                      Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                                      119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                                      With the same boundary conditions as in equation (22)

                                      where 119889119889119889119889 = 119870119870120588120588119862119862

                                      which represents the thermal diffusion

                                      But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                                      equation (21) with this assuming implies

                                      119879119879119905119905 = 1

                                      120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                                      With the same boundary and initial conditions in equation (22) Where 119870119870

                                      represents the derivative of K with respect the temperature

                                      23 Numerical solution of Initial value problems

                                      An immense number of analytical solutions for conduction heat-transfer

                                      problems have been accumulated in literature over the past 100 years Even so

                                      in many practical situations the geometry or boundary conditions are such that an

                                      analytical solution has not been obtained at all or if the solution has been

                                      18

                                      developed it involves such a complex series solution that numerical evaluation

                                      becomes exceedingly difficult For such situation the most fruitful approach to

                                      the problem is numerical techniques the basic principles of which we shall

                                      outline in this section

                                      One way to guarantee accuracy in the solution of an initial values problems

                                      (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                      answers at the mesh points corresponding to the larger step size But this requires

                                      a significant amount of computation for the smaller step size and must be

                                      repeated if it is determined that the agreement is not good enough

                                      24 Finite Difference Method

                                      The finite difference method is one of several techniques for obtaining

                                      numerical solutions to differential equations In all numerical solutions the

                                      continuous partial differential equation (PDE) is replaced with a discrete

                                      approximation In this context the word discrete means that the numerical

                                      solution is known only at a finite number of points in the physical domain The

                                      number of those points can be selected by the user of the numerical method In

                                      general increasing the number of points not only increases the resolution but

                                      also the accuracy of the numerical solution

                                      The discrete approximation results in a set of algebraic equations that are

                                      evaluated for the values of the discrete unknowns

                                      The mesh is the set of locations where the discrete solution is computed

                                      These points are called nodes and if one were to draw lines between adjacent

                                      nodes in the domain the resulting image would resemble a net or mesh Two key

                                      parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                      space and ∆119905119905 the local distance between adjacent time steps For the simple

                                      examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                      19

                                      The core idea of the finite-difference method is to replace continuous

                                      derivatives with so-called difference formulas that involve only the discrete

                                      values associated with positions on the mesh

                                      Applying the finite-difference method to a differential equation involves

                                      replacing all derivatives with difference formulas In the heat equation there are

                                      derivatives with respect to time and derivatives with respect to space Using

                                      different combinations of mesh points in the difference formulas results in

                                      different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                      numerical solution obtained with any useful scheme will approach the true

                                      solution to the original differential equation However the rate at which the

                                      numerical solution approaches the true solution varies with the scheme

                                      241 First Order Forward Difference

                                      Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                      empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                      +1205751205751205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      1205971205971

                                      +1205751205751205971205973

                                      3 1205971205973empty1205971205971205971205973

                                      1205971205971

                                      + ⋯ (25)

                                      where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                      consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                      empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                      +∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      +∆1205971205973

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      + ⋯ (26)

                                      Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                      120597120597empty120597120597120597120597120597120597119894119894

                                      =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                      ∆120597120597minus∆1205971205972

                                      1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      minus∆1205971205972

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      minus ⋯ (27)

                                      Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                      hand side have been reduced by one

                                      20

                                      Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                      and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                      120597120597empty120597120597120597120597120597120597119894119894

                                      =empty119894119894+1 minus empty119894119894

                                      ∆120597120597minus∆1205971205972

                                      1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      minus∆1205971205972

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      minus ⋯ (28)

                                      The mean value theorem can be used to replace the higher order derivatives

                                      ∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      +∆1205971205973

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      + ⋯ =∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120585120585 (29)

                                      where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                      120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                      ∆120597120597+∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120585120585 (210)

                                      120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                      ∆120597120597asymp∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120585120585 (211)

                                      The term on the right hand side of previous equation is called the truncation

                                      error of the finite difference approximation

                                      In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                      unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                      truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                      analytical form) the big 119978119978 notation can be used to express the dependence of

                                      the truncation error on the mesh spacing Note that the right hand side of last

                                      equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                      the finite difference simulation Since this is the only parameter under the users

                                      control that determines the error the truncation error is simply written

                                      ∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120585120585= 119978119978(∆1205971205972) (212)

                                      The equals sign in this expression is true in the order of magnitude sense In

                                      other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                      21

                                      equality Rather the expression means that the left hand side is a product of an

                                      unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                      magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                      approaches zero as ∆120597120597 is reduced

                                      Using big 119978119978 notation Equation (28) can be written

                                      120597120597empty120597120597120597120597120597120597119894119894

                                      =empty119894119894+1 minus empty119894119894

                                      ∆120597120597+ 119978119978(∆120597120597) (213)

                                      This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                      it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                      truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                      our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                      error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                      242 First Order Backward Difference

                                      An alternative first order finite difference formula is obtained if the Taylor series

                                      like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                      variables in place of all the unknowns one obtains

                                      empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                      +∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      minus∆1205971205973

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      + ⋯ (214)

                                      Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                      to get

                                      120597120597empty120597120597120597120597120597120597119894119894

                                      =empty119894119894+1 minus empty119894119894

                                      ∆120597120597minus∆1205971205972

                                      1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      minus∆1205971205972

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      minus ⋯ (215)

                                      Or using big 119978119978 notation

                                      120597120597empty120597120597120597120597120597120597119894119894

                                      =empty119894119894 minus empty119894119894minus1

                                      ∆120597120597+ 119978119978(∆120597120597) (216)

                                      22

                                      This is called the backward difference formula because it involves the values of

                                      empty at 120597120597119894119894 and 120597120597119894119894minus1

                                      The order of magnitude of the truncation error for the backward difference

                                      approximation is the same as that of the forward difference approximation Can

                                      we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                      truncation error The answer is yes

                                      242 First Order Central Difference

                                      Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                      empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                      +∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      +∆1205971205973

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      + ⋯ (217)

                                      empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                      +∆1205971205972

                                      2 1205971205972empty1205971205971205971205972

                                      120597120597119894119894

                                      minus∆1205971205973

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      + ⋯ (218)

                                      Subtracting Equation (10) from Equation (9) yields

                                      empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                      + 2∆1205971205973

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      + ⋯ (219)

                                      Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                      120597120597empty120597120597120597120597120597120597119894119894

                                      =empty119894119894+1 minus empty119894119894minus1

                                      2∆120597120597minus∆1205971205972

                                      3 1205971205973empty1205971205971205971205973

                                      120597120597119894119894

                                      minus ⋯ (220)

                                      or

                                      120597120597empty120597120597120597120597120597120597119894119894

                                      =empty119894119894+1 minus empty119894119894minus1

                                      2∆120597120597+ 119978119978(∆1205971205972) (221)

                                      This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                      approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                      the truncation error for the central difference approximation goes to zero much

                                      faster than the truncation error in forward and backward equations

                                      23

                                      25 Procedures

                                      The simple case in this investigation was assuming the constant thermal

                                      properties of the material First we assumed all the thermal properties of the

                                      materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                      point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                      assume the constant energy after that the pulse of special shapes was selected

                                      The numerical solution of equation (23) with boundary and initial conditions

                                      in equation (22) was investigated using Matlab program as shown in Appendix

                                      The equation of thermal conductivity and specific heat capacity of metal as a

                                      function of temperature was obtained by best fitting of polynomials using

                                      tabulated data in references

                                      24

                                      Chapter Three

                                      Results and Discursion

                                      31 Introduction

                                      The development of laser has been an exciting chapter in the history of

                                      science and engineering It has produced a new type of advice with potential for

                                      application in an extremely wide variety of fields Mach basic development in

                                      lasers were occurred during last 35 years The lasers interaction with metal and

                                      vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                      The status of laser development and application were still rather rudimentary

                                      The light emitted by laser is electro magnetic radiation this radiation has a wave

                                      nature the waves consists of vibrating electric and magnetic fields many studies

                                      have tried to find and solve models of laser interactions Some researchers

                                      proposed the mathematical model related to the laser - plasma interaction and

                                      the others have developed an analytical model to study the temperature

                                      distribution in Infrared optical materials heated by laser pulses Also an attempt

                                      have made to study the interaction of nanosecond pulsed lasers with material

                                      from point of view using experimental technique and theoretical approach of

                                      dimensional analysis

                                      In this study we have evaluate the solution of partial difference equation

                                      (PDE) that represent the laser interaction with solid situation in one dimension

                                      assuming that the power density of laser and thermal properties are functions

                                      with time and temperature respectively

                                      25

                                      32 Numerical solution with constant laser power density and constant

                                      thermal properties

                                      First we have taken the lead metal (Pb) with thermal properties

                                      119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                      119862119862 = 014016119869119869119892119892119870119870

                                      120588120588 = 10751 1198921198921198981198981198981198983

                                      119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                      119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                      and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                      represent the area under laser influence

                                      The numerical solution of equation (23) with boundary and initial conditions

                                      in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                      of lead metal by explicit method using Matlab program give us the results as

                                      shown in Fig (31)

                                      Fig(31) Depth dependence of the temperature with the laser power density

                                      1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                      26

                                      33 Evaluation of function 119920119920(119957119957) of laser flux density

                                      From following data that represent the energy (119869119869) with time (millie second)

                                      Time 0 001 01 02 03 04 05 06 07 08

                                      Energy 0 002 017 022 024 02 012 007 002 0

                                      By using Matlab program the best polynomial with deduced from above data

                                      was

                                      119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                      minus 10069 1198981198985 (31)

                                      As shown in Fig (32)

                                      Fig(32) Laser energy as a function of time

                                      Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                      normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                      maximum value (119864119864119898119898119907119907119898119898 )

                                      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                      (119864119864119898119898119907119907119898119898 ) (32)

                                      The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                      27

                                      Fig(33) Normalized laser energy as a function of time

                                      The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                      equal to 3 (total laser energy) ie

                                      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                      08

                                      00

                                      119899119899119898119898 = 3 (33)

                                      Therefore there exist a real number 119875119875 such that

                                      119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                      08

                                      00

                                      119899119899119898119898 = 3 (34)

                                      that implies 119875119875 = 68241 and

                                      119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                      08

                                      00

                                      119899119899119898119898 = 3 (35)

                                      The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                      08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                      119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                      00

                                      (36)

                                      28

                                      Where 119863119863119898119898 put to balance the units of equation (36)

                                      But integral

                                      119868119868 = 119864119864119860119860

                                      (37)

                                      and from equations (35) (36) and (37) we have

                                      119868119868 (119898119898)11989911989911989811989808

                                      00

                                      = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                      0800 119899119899119898119898

                                      119860119860 119863119863119898119898 (38)

                                      Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                      (38)

                                      There fore

                                      119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                      119860119860 119863119863119898119898 (39)

                                      As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                      that agrees with result data

                                      119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                      + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                      Fig(34) Time dependence of laser intensity

                                      29

                                      34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                      constant thermal properties

                                      With all constant thermal properties of lead metal as in article (23) and

                                      119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                      equation (23) with boundary and initial condition as in equation (22) and the

                                      depth penetration is shown in Fig(35)

                                      Fig(35) Depth dependence of the temperature when laser intensity function

                                      of time and constant thermal properties of Lead

                                      35 Evaluation the Thermal Conductivity as functions of temperature

                                      The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                      temperature for Lead material was obtained by Matlab program using the

                                      experimental data tabulated in researches

                                      119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                      times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                      30

                                      119879119879 ( 119870119870) 119870119870 119869119869

                                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                      300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                      The previous thermal conductivity data and the best fitting of the data are

                                      shown in Fig (36)

                                      Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                      temperature

                                      31

                                      36 Evaluation the Specific heat as functions of temperature

                                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                      material was obtained from the following experimental data tacked from

                                      literatures

                                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                      300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                      The best polynomial fitted for these data was

                                      119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                      + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                      times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                      The specific heat capacity data and the best polynomial fitting of the data are

                                      shown in Fig (37)

                                      32

                                      Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                      temperature

                                      37 Evaluation the Density as functions of temperature

                                      The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                      was used to find the best polynomial fitting

                                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                      300 11330 400 11230 500 11130 600 11010 800 10430

                                      1000 10190 1200 9940

                                      The best polynomial of this data was

                                      120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                      minus 45158 times 10minus12 1198791198794 (313)

                                      33

                                      The density of Lead as a function of temperature and the best polynomial fitting

                                      are shown in Fig (38)

                                      Fig(38) The best fitting of density of Lead as a function of temperature

                                      38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                      variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                      We have deduced the solution of equation (24) with initial and boundary

                                      condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                      and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                      respectively then by using Matlab program the depth penetration is shown in

                                      Fig (39)

                                      34

                                      Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                      material

                                      39 Laser interaction with copper material

                                      The same time dependence of laser intensity as shown in Fig(34) with

                                      thermal properties of copper was used to calculate the temperature distribution as

                                      a function of depth penetration

                                      The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                      copper material was obtained from the experimental data tabulated in literary

                                      The Matlab program used to obtain the best polynomial equation that agrees

                                      with the above data

                                      119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                      minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                      times 10minus17 1198791198795 (314)

                                      35

                                      119879119879 ( 119870119870) 119870119870 119869119869

                                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                      100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                      The previous thermal conductivity data and the best fitting of the data are

                                      shown in Fig (310)

                                      Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                      temperature

                                      36

                                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                      material was obtained from the following experimental data tacked from

                                      literatures

                                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                      100 0254

                                      200 0357

                                      273 0384

                                      298 0387

                                      400 0397

                                      600 0416

                                      800 0435

                                      1000 0454

                                      1100 0464

                                      1200 0474

                                      1300 0483

                                      The best polynomial fitted for these data was

                                      119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                      + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                      times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                      The specific heat capacity data and the best polynomial fitting of the data are

                                      shown in Fig (311)

                                      37

                                      Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                      temperature

                                      The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                      literature was used to find the best polynomial fitting

                                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                      100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                      1000 8576 1100 8519 1200 8458 1300 8396

                                      38

                                      The best polynomial of this data was

                                      120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                      minus 76765 times 10minus14 1198791198794 (316)

                                      The density of copper as a function of temperature and the best polynomial

                                      fitting are shown in Fig (312)

                                      Fig(312) The best fitting of density of copper as a function of temperature

                                      The depth penetration of laser energy for copper metal was calculated using

                                      the polynomial equations of thermal conductivity specific heat capacity and

                                      density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                      (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                      function of time the result was shown in Fig (313)

                                      39

                                      The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                      for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                      copper metal so the depth penetration of laser energy of lead metal was smaller

                                      than that of copper metal this may be due to the high thermal conductivity and

                                      high specific heat capacity of copper with that of lead metal

                                      Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                      material

                                      40

                                      310 Conclusions

                                      The Depth dependence of temperature for lead metal was investigated in two

                                      case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                      thermal properties (thermal conductivity specific heat) and density of metal are

                                      also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                      vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                      specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                      120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                      penetration depth in the first case is smaller than that of the second case about

                                      (190) times

                                      The temperature distribution as a function of depth dependence for copper

                                      metal was also investigated in the case when the laser intensity vary with time

                                      119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                      density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                      The depth penetration of laser energy of lead metal was found to be smaller

                                      than that of copper metal this may be due to the high thermal conductivity and

                                      high specific heat capacity of copper with that of lead metal

                                      41

                                      References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                      Sons Inc Hoboken New Jersey Canada (2003)

                                      [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                      [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                      [9]

                                      Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                      [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                      [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                      [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                      [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                      [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                      httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                      [10] httpwwwworldoflaserscomlaserprincipleshtm

                                      [11] httpenwikipediaorgwikiLaserPulsed_operation

                                      [12] httpenwikipediaorgwikiThermal_conductivity

                                      [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                      [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                      [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                      42

                                      Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                      This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                      This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                      43

                                      E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                      This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                      This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                      44

                                      title(Dencity as a function of temperature) hold off

                                      This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                      45

                                      elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                      46

                                      r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                      47

                                      for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                      48

                                      for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                      49

                                      6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                      50

                                      u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                      51

                                      alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                      52

                                      715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                      • 01 Title
                                        • 0TDr Omed Ghareb Abdullah
                                          • Ch1-Laser
                                            • 16 Heat and heat capacity
                                            • 18 Derivation in one dimension
                                              • Ch2 laser2
                                              • Ch3 laser2
                                              • Appendix-Laser

                                        15

                                        which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

                                        diffusivity and is often denoted 120572120572

                                        19 Aim of present work

                                        The goal of this study is to estimate the solution of partial differential

                                        equation that governs the laser-solid interaction using numerical methods

                                        The solution will been restricted into one dimensional situation in which we

                                        assume that both the laser power density and thermal properties are

                                        functions of time and temperature respectively In this project we attempt to

                                        investigate the laser interaction with both lead and copper materials by

                                        predicting the temperature gradient with the depth of the metals

                                        16

                                        Chapter Two

                                        Theoretical Aspects

                                        21 Introduction

                                        When a laser interacts with a solid surface a variety of processes can

                                        occur We are mainly interested in the interaction of pulsed lasers with a

                                        solid surface in first instance a metal When such a laser interacts with a

                                        copper surface the laser energy will be transformed into heat The

                                        temperature of the solid material will increase leading to melting and

                                        evaporation of the solid material

                                        The evaporated material (vapour atoms) will expand Depending on the

                                        applications this can happen in vacuum (or very low pressure) or in a

                                        background gas (helium argon air)

                                        22 One dimension laser heating equation

                                        In general the one dimension laser heating processes of opaque solid slab is

                                        represented as

                                        120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                                        ( 119870119870 119879119879120597120597 ) (21)

                                        With boundary conditions and initial condition which represent the pre-

                                        vaporization stage

                                        minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                                        minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                                        17

                                        119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                                        where

                                        119870119870 represents the thermal conductivity

                                        120588120588 represents the density

                                        119862119862 represents the specific heat

                                        119879119879 represents the temperature

                                        119879119879infin represents the ambient temperature

                                        119879119879119907119907 represents the front surface vaporization

                                        120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                                        Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                                        119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                                        With the same boundary conditions as in equation (22)

                                        where 119889119889119889119889 = 119870119870120588120588119862119862

                                        which represents the thermal diffusion

                                        But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                                        equation (21) with this assuming implies

                                        119879119879119905119905 = 1

                                        120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                                        With the same boundary and initial conditions in equation (22) Where 119870119870

                                        represents the derivative of K with respect the temperature

                                        23 Numerical solution of Initial value problems

                                        An immense number of analytical solutions for conduction heat-transfer

                                        problems have been accumulated in literature over the past 100 years Even so

                                        in many practical situations the geometry or boundary conditions are such that an

                                        analytical solution has not been obtained at all or if the solution has been

                                        18

                                        developed it involves such a complex series solution that numerical evaluation

                                        becomes exceedingly difficult For such situation the most fruitful approach to

                                        the problem is numerical techniques the basic principles of which we shall

                                        outline in this section

                                        One way to guarantee accuracy in the solution of an initial values problems

                                        (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                        answers at the mesh points corresponding to the larger step size But this requires

                                        a significant amount of computation for the smaller step size and must be

                                        repeated if it is determined that the agreement is not good enough

                                        24 Finite Difference Method

                                        The finite difference method is one of several techniques for obtaining

                                        numerical solutions to differential equations In all numerical solutions the

                                        continuous partial differential equation (PDE) is replaced with a discrete

                                        approximation In this context the word discrete means that the numerical

                                        solution is known only at a finite number of points in the physical domain The

                                        number of those points can be selected by the user of the numerical method In

                                        general increasing the number of points not only increases the resolution but

                                        also the accuracy of the numerical solution

                                        The discrete approximation results in a set of algebraic equations that are

                                        evaluated for the values of the discrete unknowns

                                        The mesh is the set of locations where the discrete solution is computed

                                        These points are called nodes and if one were to draw lines between adjacent

                                        nodes in the domain the resulting image would resemble a net or mesh Two key

                                        parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                        space and ∆119905119905 the local distance between adjacent time steps For the simple

                                        examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                        19

                                        The core idea of the finite-difference method is to replace continuous

                                        derivatives with so-called difference formulas that involve only the discrete

                                        values associated with positions on the mesh

                                        Applying the finite-difference method to a differential equation involves

                                        replacing all derivatives with difference formulas In the heat equation there are

                                        derivatives with respect to time and derivatives with respect to space Using

                                        different combinations of mesh points in the difference formulas results in

                                        different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                        numerical solution obtained with any useful scheme will approach the true

                                        solution to the original differential equation However the rate at which the

                                        numerical solution approaches the true solution varies with the scheme

                                        241 First Order Forward Difference

                                        Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                        empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                        +1205751205751205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        1205971205971

                                        +1205751205751205971205973

                                        3 1205971205973empty1205971205971205971205973

                                        1205971205971

                                        + ⋯ (25)

                                        where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                        consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                        empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                        +∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        +∆1205971205973

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        + ⋯ (26)

                                        Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                        120597120597empty120597120597120597120597120597120597119894119894

                                        =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                        ∆120597120597minus∆1205971205972

                                        1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        minus∆1205971205972

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        minus ⋯ (27)

                                        Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                        hand side have been reduced by one

                                        20

                                        Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                        and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                        120597120597empty120597120597120597120597120597120597119894119894

                                        =empty119894119894+1 minus empty119894119894

                                        ∆120597120597minus∆1205971205972

                                        1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        minus∆1205971205972

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        minus ⋯ (28)

                                        The mean value theorem can be used to replace the higher order derivatives

                                        ∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        +∆1205971205973

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        + ⋯ =∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120585120585 (29)

                                        where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                        120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                        ∆120597120597+∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120585120585 (210)

                                        120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                        ∆120597120597asymp∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120585120585 (211)

                                        The term on the right hand side of previous equation is called the truncation

                                        error of the finite difference approximation

                                        In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                        unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                        truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                        analytical form) the big 119978119978 notation can be used to express the dependence of

                                        the truncation error on the mesh spacing Note that the right hand side of last

                                        equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                        the finite difference simulation Since this is the only parameter under the users

                                        control that determines the error the truncation error is simply written

                                        ∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120585120585= 119978119978(∆1205971205972) (212)

                                        The equals sign in this expression is true in the order of magnitude sense In

                                        other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                        21

                                        equality Rather the expression means that the left hand side is a product of an

                                        unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                        magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                        approaches zero as ∆120597120597 is reduced

                                        Using big 119978119978 notation Equation (28) can be written

                                        120597120597empty120597120597120597120597120597120597119894119894

                                        =empty119894119894+1 minus empty119894119894

                                        ∆120597120597+ 119978119978(∆120597120597) (213)

                                        This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                        it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                        truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                        our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                        error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                        242 First Order Backward Difference

                                        An alternative first order finite difference formula is obtained if the Taylor series

                                        like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                        variables in place of all the unknowns one obtains

                                        empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                        +∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        minus∆1205971205973

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        + ⋯ (214)

                                        Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                        to get

                                        120597120597empty120597120597120597120597120597120597119894119894

                                        =empty119894119894+1 minus empty119894119894

                                        ∆120597120597minus∆1205971205972

                                        1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        minus∆1205971205972

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        minus ⋯ (215)

                                        Or using big 119978119978 notation

                                        120597120597empty120597120597120597120597120597120597119894119894

                                        =empty119894119894 minus empty119894119894minus1

                                        ∆120597120597+ 119978119978(∆120597120597) (216)

                                        22

                                        This is called the backward difference formula because it involves the values of

                                        empty at 120597120597119894119894 and 120597120597119894119894minus1

                                        The order of magnitude of the truncation error for the backward difference

                                        approximation is the same as that of the forward difference approximation Can

                                        we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                        truncation error The answer is yes

                                        242 First Order Central Difference

                                        Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                        empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                        +∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        +∆1205971205973

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        + ⋯ (217)

                                        empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                        +∆1205971205972

                                        2 1205971205972empty1205971205971205971205972

                                        120597120597119894119894

                                        minus∆1205971205973

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        + ⋯ (218)

                                        Subtracting Equation (10) from Equation (9) yields

                                        empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                        + 2∆1205971205973

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        + ⋯ (219)

                                        Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                        120597120597empty120597120597120597120597120597120597119894119894

                                        =empty119894119894+1 minus empty119894119894minus1

                                        2∆120597120597minus∆1205971205972

                                        3 1205971205973empty1205971205971205971205973

                                        120597120597119894119894

                                        minus ⋯ (220)

                                        or

                                        120597120597empty120597120597120597120597120597120597119894119894

                                        =empty119894119894+1 minus empty119894119894minus1

                                        2∆120597120597+ 119978119978(∆1205971205972) (221)

                                        This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                        approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                        the truncation error for the central difference approximation goes to zero much

                                        faster than the truncation error in forward and backward equations

                                        23

                                        25 Procedures

                                        The simple case in this investigation was assuming the constant thermal

                                        properties of the material First we assumed all the thermal properties of the

                                        materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                        point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                        assume the constant energy after that the pulse of special shapes was selected

                                        The numerical solution of equation (23) with boundary and initial conditions

                                        in equation (22) was investigated using Matlab program as shown in Appendix

                                        The equation of thermal conductivity and specific heat capacity of metal as a

                                        function of temperature was obtained by best fitting of polynomials using

                                        tabulated data in references

                                        24

                                        Chapter Three

                                        Results and Discursion

                                        31 Introduction

                                        The development of laser has been an exciting chapter in the history of

                                        science and engineering It has produced a new type of advice with potential for

                                        application in an extremely wide variety of fields Mach basic development in

                                        lasers were occurred during last 35 years The lasers interaction with metal and

                                        vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                        The status of laser development and application were still rather rudimentary

                                        The light emitted by laser is electro magnetic radiation this radiation has a wave

                                        nature the waves consists of vibrating electric and magnetic fields many studies

                                        have tried to find and solve models of laser interactions Some researchers

                                        proposed the mathematical model related to the laser - plasma interaction and

                                        the others have developed an analytical model to study the temperature

                                        distribution in Infrared optical materials heated by laser pulses Also an attempt

                                        have made to study the interaction of nanosecond pulsed lasers with material

                                        from point of view using experimental technique and theoretical approach of

                                        dimensional analysis

                                        In this study we have evaluate the solution of partial difference equation

                                        (PDE) that represent the laser interaction with solid situation in one dimension

                                        assuming that the power density of laser and thermal properties are functions

                                        with time and temperature respectively

                                        25

                                        32 Numerical solution with constant laser power density and constant

                                        thermal properties

                                        First we have taken the lead metal (Pb) with thermal properties

                                        119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                        119862119862 = 014016119869119869119892119892119870119870

                                        120588120588 = 10751 1198921198921198981198981198981198983

                                        119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                        119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                        and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                        represent the area under laser influence

                                        The numerical solution of equation (23) with boundary and initial conditions

                                        in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                        of lead metal by explicit method using Matlab program give us the results as

                                        shown in Fig (31)

                                        Fig(31) Depth dependence of the temperature with the laser power density

                                        1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                        26

                                        33 Evaluation of function 119920119920(119957119957) of laser flux density

                                        From following data that represent the energy (119869119869) with time (millie second)

                                        Time 0 001 01 02 03 04 05 06 07 08

                                        Energy 0 002 017 022 024 02 012 007 002 0

                                        By using Matlab program the best polynomial with deduced from above data

                                        was

                                        119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                        minus 10069 1198981198985 (31)

                                        As shown in Fig (32)

                                        Fig(32) Laser energy as a function of time

                                        Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                        normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                        maximum value (119864119864119898119898119907119907119898119898 )

                                        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                        (119864119864119898119898119907119907119898119898 ) (32)

                                        The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                        27

                                        Fig(33) Normalized laser energy as a function of time

                                        The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                        equal to 3 (total laser energy) ie

                                        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                        08

                                        00

                                        119899119899119898119898 = 3 (33)

                                        Therefore there exist a real number 119875119875 such that

                                        119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                        08

                                        00

                                        119899119899119898119898 = 3 (34)

                                        that implies 119875119875 = 68241 and

                                        119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                        08

                                        00

                                        119899119899119898119898 = 3 (35)

                                        The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                        08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                        119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                        00

                                        (36)

                                        28

                                        Where 119863119863119898119898 put to balance the units of equation (36)

                                        But integral

                                        119868119868 = 119864119864119860119860

                                        (37)

                                        and from equations (35) (36) and (37) we have

                                        119868119868 (119898119898)11989911989911989811989808

                                        00

                                        = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                        0800 119899119899119898119898

                                        119860119860 119863119863119898119898 (38)

                                        Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                        (38)

                                        There fore

                                        119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                        119860119860 119863119863119898119898 (39)

                                        As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                        that agrees with result data

                                        119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                        + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                        Fig(34) Time dependence of laser intensity

                                        29

                                        34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                        constant thermal properties

                                        With all constant thermal properties of lead metal as in article (23) and

                                        119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                        equation (23) with boundary and initial condition as in equation (22) and the

                                        depth penetration is shown in Fig(35)

                                        Fig(35) Depth dependence of the temperature when laser intensity function

                                        of time and constant thermal properties of Lead

                                        35 Evaluation the Thermal Conductivity as functions of temperature

                                        The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                        temperature for Lead material was obtained by Matlab program using the

                                        experimental data tabulated in researches

                                        119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                        times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                        30

                                        119879119879 ( 119870119870) 119870119870 119869119869

                                        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                        300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                        The previous thermal conductivity data and the best fitting of the data are

                                        shown in Fig (36)

                                        Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                        temperature

                                        31

                                        36 Evaluation the Specific heat as functions of temperature

                                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                        material was obtained from the following experimental data tacked from

                                        literatures

                                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                        300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                        The best polynomial fitted for these data was

                                        119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                        + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                        times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                        The specific heat capacity data and the best polynomial fitting of the data are

                                        shown in Fig (37)

                                        32

                                        Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                        temperature

                                        37 Evaluation the Density as functions of temperature

                                        The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                        was used to find the best polynomial fitting

                                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                        300 11330 400 11230 500 11130 600 11010 800 10430

                                        1000 10190 1200 9940

                                        The best polynomial of this data was

                                        120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                        minus 45158 times 10minus12 1198791198794 (313)

                                        33

                                        The density of Lead as a function of temperature and the best polynomial fitting

                                        are shown in Fig (38)

                                        Fig(38) The best fitting of density of Lead as a function of temperature

                                        38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                        variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                        We have deduced the solution of equation (24) with initial and boundary

                                        condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                        and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                        respectively then by using Matlab program the depth penetration is shown in

                                        Fig (39)

                                        34

                                        Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                        material

                                        39 Laser interaction with copper material

                                        The same time dependence of laser intensity as shown in Fig(34) with

                                        thermal properties of copper was used to calculate the temperature distribution as

                                        a function of depth penetration

                                        The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                        copper material was obtained from the experimental data tabulated in literary

                                        The Matlab program used to obtain the best polynomial equation that agrees

                                        with the above data

                                        119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                        minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                        times 10minus17 1198791198795 (314)

                                        35

                                        119879119879 ( 119870119870) 119870119870 119869119869

                                        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                        100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                        The previous thermal conductivity data and the best fitting of the data are

                                        shown in Fig (310)

                                        Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                        temperature

                                        36

                                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                        material was obtained from the following experimental data tacked from

                                        literatures

                                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                        100 0254

                                        200 0357

                                        273 0384

                                        298 0387

                                        400 0397

                                        600 0416

                                        800 0435

                                        1000 0454

                                        1100 0464

                                        1200 0474

                                        1300 0483

                                        The best polynomial fitted for these data was

                                        119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                        + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                        times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                        The specific heat capacity data and the best polynomial fitting of the data are

                                        shown in Fig (311)

                                        37

                                        Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                        temperature

                                        The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                        literature was used to find the best polynomial fitting

                                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                        100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                        1000 8576 1100 8519 1200 8458 1300 8396

                                        38

                                        The best polynomial of this data was

                                        120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                        minus 76765 times 10minus14 1198791198794 (316)

                                        The density of copper as a function of temperature and the best polynomial

                                        fitting are shown in Fig (312)

                                        Fig(312) The best fitting of density of copper as a function of temperature

                                        The depth penetration of laser energy for copper metal was calculated using

                                        the polynomial equations of thermal conductivity specific heat capacity and

                                        density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                        (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                        function of time the result was shown in Fig (313)

                                        39

                                        The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                        for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                        copper metal so the depth penetration of laser energy of lead metal was smaller

                                        than that of copper metal this may be due to the high thermal conductivity and

                                        high specific heat capacity of copper with that of lead metal

                                        Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                        material

                                        40

                                        310 Conclusions

                                        The Depth dependence of temperature for lead metal was investigated in two

                                        case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                        thermal properties (thermal conductivity specific heat) and density of metal are

                                        also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                        vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                        specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                        120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                        penetration depth in the first case is smaller than that of the second case about

                                        (190) times

                                        The temperature distribution as a function of depth dependence for copper

                                        metal was also investigated in the case when the laser intensity vary with time

                                        119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                        density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                        The depth penetration of laser energy of lead metal was found to be smaller

                                        than that of copper metal this may be due to the high thermal conductivity and

                                        high specific heat capacity of copper with that of lead metal

                                        41

                                        References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                        Sons Inc Hoboken New Jersey Canada (2003)

                                        [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                        [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                        [9]

                                        Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                        [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                        [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                        [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                        [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                        [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                        httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                        [10] httpwwwworldoflaserscomlaserprincipleshtm

                                        [11] httpenwikipediaorgwikiLaserPulsed_operation

                                        [12] httpenwikipediaorgwikiThermal_conductivity

                                        [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                        [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                        [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                        42

                                        Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                        This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                        This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                        43

                                        E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                        This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                        This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                        44

                                        title(Dencity as a function of temperature) hold off

                                        This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                        45

                                        elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                        46

                                        r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                        47

                                        for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                        48

                                        for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                        49

                                        6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                        50

                                        u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                        51

                                        alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                        52

                                        715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                        • 01 Title
                                          • 0TDr Omed Ghareb Abdullah
                                            • Ch1-Laser
                                              • 16 Heat and heat capacity
                                              • 18 Derivation in one dimension
                                                • Ch2 laser2
                                                • Ch3 laser2
                                                • Appendix-Laser

                                          16

                                          Chapter Two

                                          Theoretical Aspects

                                          21 Introduction

                                          When a laser interacts with a solid surface a variety of processes can

                                          occur We are mainly interested in the interaction of pulsed lasers with a

                                          solid surface in first instance a metal When such a laser interacts with a

                                          copper surface the laser energy will be transformed into heat The

                                          temperature of the solid material will increase leading to melting and

                                          evaporation of the solid material

                                          The evaporated material (vapour atoms) will expand Depending on the

                                          applications this can happen in vacuum (or very low pressure) or in a

                                          background gas (helium argon air)

                                          22 One dimension laser heating equation

                                          In general the one dimension laser heating processes of opaque solid slab is

                                          represented as

                                          120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

                                          ( 119870119870 119879119879120597120597 ) (21)

                                          With boundary conditions and initial condition which represent the pre-

                                          vaporization stage

                                          minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

                                          minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

                                          17

                                          119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                                          where

                                          119870119870 represents the thermal conductivity

                                          120588120588 represents the density

                                          119862119862 represents the specific heat

                                          119879119879 represents the temperature

                                          119879119879infin represents the ambient temperature

                                          119879119879119907119907 represents the front surface vaporization

                                          120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                                          Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                                          119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                                          With the same boundary conditions as in equation (22)

                                          where 119889119889119889119889 = 119870119870120588120588119862119862

                                          which represents the thermal diffusion

                                          But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                                          equation (21) with this assuming implies

                                          119879119879119905119905 = 1

                                          120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                                          With the same boundary and initial conditions in equation (22) Where 119870119870

                                          represents the derivative of K with respect the temperature

                                          23 Numerical solution of Initial value problems

                                          An immense number of analytical solutions for conduction heat-transfer

                                          problems have been accumulated in literature over the past 100 years Even so

                                          in many practical situations the geometry or boundary conditions are such that an

                                          analytical solution has not been obtained at all or if the solution has been

                                          18

                                          developed it involves such a complex series solution that numerical evaluation

                                          becomes exceedingly difficult For such situation the most fruitful approach to

                                          the problem is numerical techniques the basic principles of which we shall

                                          outline in this section

                                          One way to guarantee accuracy in the solution of an initial values problems

                                          (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                          answers at the mesh points corresponding to the larger step size But this requires

                                          a significant amount of computation for the smaller step size and must be

                                          repeated if it is determined that the agreement is not good enough

                                          24 Finite Difference Method

                                          The finite difference method is one of several techniques for obtaining

                                          numerical solutions to differential equations In all numerical solutions the

                                          continuous partial differential equation (PDE) is replaced with a discrete

                                          approximation In this context the word discrete means that the numerical

                                          solution is known only at a finite number of points in the physical domain The

                                          number of those points can be selected by the user of the numerical method In

                                          general increasing the number of points not only increases the resolution but

                                          also the accuracy of the numerical solution

                                          The discrete approximation results in a set of algebraic equations that are

                                          evaluated for the values of the discrete unknowns

                                          The mesh is the set of locations where the discrete solution is computed

                                          These points are called nodes and if one were to draw lines between adjacent

                                          nodes in the domain the resulting image would resemble a net or mesh Two key

                                          parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                          space and ∆119905119905 the local distance between adjacent time steps For the simple

                                          examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                          19

                                          The core idea of the finite-difference method is to replace continuous

                                          derivatives with so-called difference formulas that involve only the discrete

                                          values associated with positions on the mesh

                                          Applying the finite-difference method to a differential equation involves

                                          replacing all derivatives with difference formulas In the heat equation there are

                                          derivatives with respect to time and derivatives with respect to space Using

                                          different combinations of mesh points in the difference formulas results in

                                          different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                          numerical solution obtained with any useful scheme will approach the true

                                          solution to the original differential equation However the rate at which the

                                          numerical solution approaches the true solution varies with the scheme

                                          241 First Order Forward Difference

                                          Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                          empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                          +1205751205751205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          1205971205971

                                          +1205751205751205971205973

                                          3 1205971205973empty1205971205971205971205973

                                          1205971205971

                                          + ⋯ (25)

                                          where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                          consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                          empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                          +∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          +∆1205971205973

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          + ⋯ (26)

                                          Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                          120597120597empty120597120597120597120597120597120597119894119894

                                          =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                          ∆120597120597minus∆1205971205972

                                          1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          minus∆1205971205972

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          minus ⋯ (27)

                                          Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                          hand side have been reduced by one

                                          20

                                          Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                          and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                          120597120597empty120597120597120597120597120597120597119894119894

                                          =empty119894119894+1 minus empty119894119894

                                          ∆120597120597minus∆1205971205972

                                          1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          minus∆1205971205972

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          minus ⋯ (28)

                                          The mean value theorem can be used to replace the higher order derivatives

                                          ∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          +∆1205971205973

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          + ⋯ =∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120585120585 (29)

                                          where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                          120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                          ∆120597120597+∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120585120585 (210)

                                          120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                          ∆120597120597asymp∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120585120585 (211)

                                          The term on the right hand side of previous equation is called the truncation

                                          error of the finite difference approximation

                                          In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                          unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                          truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                          analytical form) the big 119978119978 notation can be used to express the dependence of

                                          the truncation error on the mesh spacing Note that the right hand side of last

                                          equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                          the finite difference simulation Since this is the only parameter under the users

                                          control that determines the error the truncation error is simply written

                                          ∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120585120585= 119978119978(∆1205971205972) (212)

                                          The equals sign in this expression is true in the order of magnitude sense In

                                          other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                          21

                                          equality Rather the expression means that the left hand side is a product of an

                                          unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                          magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                          approaches zero as ∆120597120597 is reduced

                                          Using big 119978119978 notation Equation (28) can be written

                                          120597120597empty120597120597120597120597120597120597119894119894

                                          =empty119894119894+1 minus empty119894119894

                                          ∆120597120597+ 119978119978(∆120597120597) (213)

                                          This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                          it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                          truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                          our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                          error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                          242 First Order Backward Difference

                                          An alternative first order finite difference formula is obtained if the Taylor series

                                          like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                          variables in place of all the unknowns one obtains

                                          empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                          +∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          minus∆1205971205973

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          + ⋯ (214)

                                          Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                          to get

                                          120597120597empty120597120597120597120597120597120597119894119894

                                          =empty119894119894+1 minus empty119894119894

                                          ∆120597120597minus∆1205971205972

                                          1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          minus∆1205971205972

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          minus ⋯ (215)

                                          Or using big 119978119978 notation

                                          120597120597empty120597120597120597120597120597120597119894119894

                                          =empty119894119894 minus empty119894119894minus1

                                          ∆120597120597+ 119978119978(∆120597120597) (216)

                                          22

                                          This is called the backward difference formula because it involves the values of

                                          empty at 120597120597119894119894 and 120597120597119894119894minus1

                                          The order of magnitude of the truncation error for the backward difference

                                          approximation is the same as that of the forward difference approximation Can

                                          we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                          truncation error The answer is yes

                                          242 First Order Central Difference

                                          Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                          empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                          +∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          +∆1205971205973

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          + ⋯ (217)

                                          empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                          +∆1205971205972

                                          2 1205971205972empty1205971205971205971205972

                                          120597120597119894119894

                                          minus∆1205971205973

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          + ⋯ (218)

                                          Subtracting Equation (10) from Equation (9) yields

                                          empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                          + 2∆1205971205973

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          + ⋯ (219)

                                          Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                          120597120597empty120597120597120597120597120597120597119894119894

                                          =empty119894119894+1 minus empty119894119894minus1

                                          2∆120597120597minus∆1205971205972

                                          3 1205971205973empty1205971205971205971205973

                                          120597120597119894119894

                                          minus ⋯ (220)

                                          or

                                          120597120597empty120597120597120597120597120597120597119894119894

                                          =empty119894119894+1 minus empty119894119894minus1

                                          2∆120597120597+ 119978119978(∆1205971205972) (221)

                                          This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                          approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                          the truncation error for the central difference approximation goes to zero much

                                          faster than the truncation error in forward and backward equations

                                          23

                                          25 Procedures

                                          The simple case in this investigation was assuming the constant thermal

                                          properties of the material First we assumed all the thermal properties of the

                                          materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                          point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                          assume the constant energy after that the pulse of special shapes was selected

                                          The numerical solution of equation (23) with boundary and initial conditions

                                          in equation (22) was investigated using Matlab program as shown in Appendix

                                          The equation of thermal conductivity and specific heat capacity of metal as a

                                          function of temperature was obtained by best fitting of polynomials using

                                          tabulated data in references

                                          24

                                          Chapter Three

                                          Results and Discursion

                                          31 Introduction

                                          The development of laser has been an exciting chapter in the history of

                                          science and engineering It has produced a new type of advice with potential for

                                          application in an extremely wide variety of fields Mach basic development in

                                          lasers were occurred during last 35 years The lasers interaction with metal and

                                          vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                          The status of laser development and application were still rather rudimentary

                                          The light emitted by laser is electro magnetic radiation this radiation has a wave

                                          nature the waves consists of vibrating electric and magnetic fields many studies

                                          have tried to find and solve models of laser interactions Some researchers

                                          proposed the mathematical model related to the laser - plasma interaction and

                                          the others have developed an analytical model to study the temperature

                                          distribution in Infrared optical materials heated by laser pulses Also an attempt

                                          have made to study the interaction of nanosecond pulsed lasers with material

                                          from point of view using experimental technique and theoretical approach of

                                          dimensional analysis

                                          In this study we have evaluate the solution of partial difference equation

                                          (PDE) that represent the laser interaction with solid situation in one dimension

                                          assuming that the power density of laser and thermal properties are functions

                                          with time and temperature respectively

                                          25

                                          32 Numerical solution with constant laser power density and constant

                                          thermal properties

                                          First we have taken the lead metal (Pb) with thermal properties

                                          119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                          119862119862 = 014016119869119869119892119892119870119870

                                          120588120588 = 10751 1198921198921198981198981198981198983

                                          119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                          119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                          and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                          represent the area under laser influence

                                          The numerical solution of equation (23) with boundary and initial conditions

                                          in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                          of lead metal by explicit method using Matlab program give us the results as

                                          shown in Fig (31)

                                          Fig(31) Depth dependence of the temperature with the laser power density

                                          1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                          26

                                          33 Evaluation of function 119920119920(119957119957) of laser flux density

                                          From following data that represent the energy (119869119869) with time (millie second)

                                          Time 0 001 01 02 03 04 05 06 07 08

                                          Energy 0 002 017 022 024 02 012 007 002 0

                                          By using Matlab program the best polynomial with deduced from above data

                                          was

                                          119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                          minus 10069 1198981198985 (31)

                                          As shown in Fig (32)

                                          Fig(32) Laser energy as a function of time

                                          Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                          normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                          maximum value (119864119864119898119898119907119907119898119898 )

                                          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                          (119864119864119898119898119907119907119898119898 ) (32)

                                          The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                          27

                                          Fig(33) Normalized laser energy as a function of time

                                          The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                          equal to 3 (total laser energy) ie

                                          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                          08

                                          00

                                          119899119899119898119898 = 3 (33)

                                          Therefore there exist a real number 119875119875 such that

                                          119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                          08

                                          00

                                          119899119899119898119898 = 3 (34)

                                          that implies 119875119875 = 68241 and

                                          119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                          08

                                          00

                                          119899119899119898119898 = 3 (35)

                                          The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                          08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                          119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                          00

                                          (36)

                                          28

                                          Where 119863119863119898119898 put to balance the units of equation (36)

                                          But integral

                                          119868119868 = 119864119864119860119860

                                          (37)

                                          and from equations (35) (36) and (37) we have

                                          119868119868 (119898119898)11989911989911989811989808

                                          00

                                          = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                          0800 119899119899119898119898

                                          119860119860 119863119863119898119898 (38)

                                          Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                          (38)

                                          There fore

                                          119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                          119860119860 119863119863119898119898 (39)

                                          As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                          that agrees with result data

                                          119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                          + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                          Fig(34) Time dependence of laser intensity

                                          29

                                          34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                          constant thermal properties

                                          With all constant thermal properties of lead metal as in article (23) and

                                          119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                          equation (23) with boundary and initial condition as in equation (22) and the

                                          depth penetration is shown in Fig(35)

                                          Fig(35) Depth dependence of the temperature when laser intensity function

                                          of time and constant thermal properties of Lead

                                          35 Evaluation the Thermal Conductivity as functions of temperature

                                          The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                          temperature for Lead material was obtained by Matlab program using the

                                          experimental data tabulated in researches

                                          119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                          times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                          30

                                          119879119879 ( 119870119870) 119870119870 119869119869

                                          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                          300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                          The previous thermal conductivity data and the best fitting of the data are

                                          shown in Fig (36)

                                          Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                          temperature

                                          31

                                          36 Evaluation the Specific heat as functions of temperature

                                          The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                          material was obtained from the following experimental data tacked from

                                          literatures

                                          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                          300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                          The best polynomial fitted for these data was

                                          119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                          + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                          times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                          The specific heat capacity data and the best polynomial fitting of the data are

                                          shown in Fig (37)

                                          32

                                          Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                          temperature

                                          37 Evaluation the Density as functions of temperature

                                          The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                          was used to find the best polynomial fitting

                                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                          300 11330 400 11230 500 11130 600 11010 800 10430

                                          1000 10190 1200 9940

                                          The best polynomial of this data was

                                          120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                          minus 45158 times 10minus12 1198791198794 (313)

                                          33

                                          The density of Lead as a function of temperature and the best polynomial fitting

                                          are shown in Fig (38)

                                          Fig(38) The best fitting of density of Lead as a function of temperature

                                          38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                          variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                          We have deduced the solution of equation (24) with initial and boundary

                                          condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                          and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                          respectively then by using Matlab program the depth penetration is shown in

                                          Fig (39)

                                          34

                                          Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                          material

                                          39 Laser interaction with copper material

                                          The same time dependence of laser intensity as shown in Fig(34) with

                                          thermal properties of copper was used to calculate the temperature distribution as

                                          a function of depth penetration

                                          The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                          copper material was obtained from the experimental data tabulated in literary

                                          The Matlab program used to obtain the best polynomial equation that agrees

                                          with the above data

                                          119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                          minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                          times 10minus17 1198791198795 (314)

                                          35

                                          119879119879 ( 119870119870) 119870119870 119869119869

                                          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                          100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                          The previous thermal conductivity data and the best fitting of the data are

                                          shown in Fig (310)

                                          Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                          temperature

                                          36

                                          The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                          material was obtained from the following experimental data tacked from

                                          literatures

                                          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                          100 0254

                                          200 0357

                                          273 0384

                                          298 0387

                                          400 0397

                                          600 0416

                                          800 0435

                                          1000 0454

                                          1100 0464

                                          1200 0474

                                          1300 0483

                                          The best polynomial fitted for these data was

                                          119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                          + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                          times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                          The specific heat capacity data and the best polynomial fitting of the data are

                                          shown in Fig (311)

                                          37

                                          Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                          temperature

                                          The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                          literature was used to find the best polynomial fitting

                                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                          100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                          1000 8576 1100 8519 1200 8458 1300 8396

                                          38

                                          The best polynomial of this data was

                                          120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                          minus 76765 times 10minus14 1198791198794 (316)

                                          The density of copper as a function of temperature and the best polynomial

                                          fitting are shown in Fig (312)

                                          Fig(312) The best fitting of density of copper as a function of temperature

                                          The depth penetration of laser energy for copper metal was calculated using

                                          the polynomial equations of thermal conductivity specific heat capacity and

                                          density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                          (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                          function of time the result was shown in Fig (313)

                                          39

                                          The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                          for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                          copper metal so the depth penetration of laser energy of lead metal was smaller

                                          than that of copper metal this may be due to the high thermal conductivity and

                                          high specific heat capacity of copper with that of lead metal

                                          Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                          material

                                          40

                                          310 Conclusions

                                          The Depth dependence of temperature for lead metal was investigated in two

                                          case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                          thermal properties (thermal conductivity specific heat) and density of metal are

                                          also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                          vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                          specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                          120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                          penetration depth in the first case is smaller than that of the second case about

                                          (190) times

                                          The temperature distribution as a function of depth dependence for copper

                                          metal was also investigated in the case when the laser intensity vary with time

                                          119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                          density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                          The depth penetration of laser energy of lead metal was found to be smaller

                                          than that of copper metal this may be due to the high thermal conductivity and

                                          high specific heat capacity of copper with that of lead metal

                                          41

                                          References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                          Sons Inc Hoboken New Jersey Canada (2003)

                                          [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                          [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                          [9]

                                          Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                          [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                          [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                          [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                          [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                          [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                          httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                          [10] httpwwwworldoflaserscomlaserprincipleshtm

                                          [11] httpenwikipediaorgwikiLaserPulsed_operation

                                          [12] httpenwikipediaorgwikiThermal_conductivity

                                          [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                          [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                          [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                          42

                                          Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                          This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                          This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                          43

                                          E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                          This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                          This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                          44

                                          title(Dencity as a function of temperature) hold off

                                          This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                          45

                                          elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                          46

                                          r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                          47

                                          for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                          48

                                          for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                          49

                                          6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                          50

                                          u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                          51

                                          alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                          52

                                          715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                          • 01 Title
                                            • 0TDr Omed Ghareb Abdullah
                                              • Ch1-Laser
                                                • 16 Heat and heat capacity
                                                • 18 Derivation in one dimension
                                                  • Ch2 laser2
                                                  • Ch3 laser2
                                                  • Appendix-Laser

                                            17

                                            119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

                                            where

                                            119870119870 represents the thermal conductivity

                                            120588120588 represents the density

                                            119862119862 represents the specific heat

                                            119879119879 represents the temperature

                                            119879119879infin represents the ambient temperature

                                            119879119879119907119907 represents the front surface vaporization

                                            120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

                                            Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

                                            119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

                                            With the same boundary conditions as in equation (22)

                                            where 119889119889119889119889 = 119870119870120588120588119862119862

                                            which represents the thermal diffusion

                                            But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

                                            equation (21) with this assuming implies

                                            119879119879119905119905 = 1

                                            120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

                                            With the same boundary and initial conditions in equation (22) Where 119870119870

                                            represents the derivative of K with respect the temperature

                                            23 Numerical solution of Initial value problems

                                            An immense number of analytical solutions for conduction heat-transfer

                                            problems have been accumulated in literature over the past 100 years Even so

                                            in many practical situations the geometry or boundary conditions are such that an

                                            analytical solution has not been obtained at all or if the solution has been

                                            18

                                            developed it involves such a complex series solution that numerical evaluation

                                            becomes exceedingly difficult For such situation the most fruitful approach to

                                            the problem is numerical techniques the basic principles of which we shall

                                            outline in this section

                                            One way to guarantee accuracy in the solution of an initial values problems

                                            (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                            answers at the mesh points corresponding to the larger step size But this requires

                                            a significant amount of computation for the smaller step size and must be

                                            repeated if it is determined that the agreement is not good enough

                                            24 Finite Difference Method

                                            The finite difference method is one of several techniques for obtaining

                                            numerical solutions to differential equations In all numerical solutions the

                                            continuous partial differential equation (PDE) is replaced with a discrete

                                            approximation In this context the word discrete means that the numerical

                                            solution is known only at a finite number of points in the physical domain The

                                            number of those points can be selected by the user of the numerical method In

                                            general increasing the number of points not only increases the resolution but

                                            also the accuracy of the numerical solution

                                            The discrete approximation results in a set of algebraic equations that are

                                            evaluated for the values of the discrete unknowns

                                            The mesh is the set of locations where the discrete solution is computed

                                            These points are called nodes and if one were to draw lines between adjacent

                                            nodes in the domain the resulting image would resemble a net or mesh Two key

                                            parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                            space and ∆119905119905 the local distance between adjacent time steps For the simple

                                            examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                            19

                                            The core idea of the finite-difference method is to replace continuous

                                            derivatives with so-called difference formulas that involve only the discrete

                                            values associated with positions on the mesh

                                            Applying the finite-difference method to a differential equation involves

                                            replacing all derivatives with difference formulas In the heat equation there are

                                            derivatives with respect to time and derivatives with respect to space Using

                                            different combinations of mesh points in the difference formulas results in

                                            different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                            numerical solution obtained with any useful scheme will approach the true

                                            solution to the original differential equation However the rate at which the

                                            numerical solution approaches the true solution varies with the scheme

                                            241 First Order Forward Difference

                                            Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                            empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                            +1205751205751205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            1205971205971

                                            +1205751205751205971205973

                                            3 1205971205973empty1205971205971205971205973

                                            1205971205971

                                            + ⋯ (25)

                                            where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                            consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                            empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                            +∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            +∆1205971205973

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            + ⋯ (26)

                                            Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                            120597120597empty120597120597120597120597120597120597119894119894

                                            =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                            ∆120597120597minus∆1205971205972

                                            1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            minus∆1205971205972

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            minus ⋯ (27)

                                            Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                            hand side have been reduced by one

                                            20

                                            Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                            and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                            120597120597empty120597120597120597120597120597120597119894119894

                                            =empty119894119894+1 minus empty119894119894

                                            ∆120597120597minus∆1205971205972

                                            1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            minus∆1205971205972

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            minus ⋯ (28)

                                            The mean value theorem can be used to replace the higher order derivatives

                                            ∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            +∆1205971205973

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            + ⋯ =∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120585120585 (29)

                                            where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                            120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                            ∆120597120597+∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120585120585 (210)

                                            120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                            ∆120597120597asymp∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120585120585 (211)

                                            The term on the right hand side of previous equation is called the truncation

                                            error of the finite difference approximation

                                            In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                            unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                            truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                            analytical form) the big 119978119978 notation can be used to express the dependence of

                                            the truncation error on the mesh spacing Note that the right hand side of last

                                            equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                            the finite difference simulation Since this is the only parameter under the users

                                            control that determines the error the truncation error is simply written

                                            ∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120585120585= 119978119978(∆1205971205972) (212)

                                            The equals sign in this expression is true in the order of magnitude sense In

                                            other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                            21

                                            equality Rather the expression means that the left hand side is a product of an

                                            unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                            magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                            approaches zero as ∆120597120597 is reduced

                                            Using big 119978119978 notation Equation (28) can be written

                                            120597120597empty120597120597120597120597120597120597119894119894

                                            =empty119894119894+1 minus empty119894119894

                                            ∆120597120597+ 119978119978(∆120597120597) (213)

                                            This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                            it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                            truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                            our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                            error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                            242 First Order Backward Difference

                                            An alternative first order finite difference formula is obtained if the Taylor series

                                            like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                            variables in place of all the unknowns one obtains

                                            empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                            +∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            minus∆1205971205973

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            + ⋯ (214)

                                            Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                            to get

                                            120597120597empty120597120597120597120597120597120597119894119894

                                            =empty119894119894+1 minus empty119894119894

                                            ∆120597120597minus∆1205971205972

                                            1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            minus∆1205971205972

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            minus ⋯ (215)

                                            Or using big 119978119978 notation

                                            120597120597empty120597120597120597120597120597120597119894119894

                                            =empty119894119894 minus empty119894119894minus1

                                            ∆120597120597+ 119978119978(∆120597120597) (216)

                                            22

                                            This is called the backward difference formula because it involves the values of

                                            empty at 120597120597119894119894 and 120597120597119894119894minus1

                                            The order of magnitude of the truncation error for the backward difference

                                            approximation is the same as that of the forward difference approximation Can

                                            we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                            truncation error The answer is yes

                                            242 First Order Central Difference

                                            Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                            empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                            +∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            +∆1205971205973

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            + ⋯ (217)

                                            empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                            +∆1205971205972

                                            2 1205971205972empty1205971205971205971205972

                                            120597120597119894119894

                                            minus∆1205971205973

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            + ⋯ (218)

                                            Subtracting Equation (10) from Equation (9) yields

                                            empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                            + 2∆1205971205973

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            + ⋯ (219)

                                            Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                            120597120597empty120597120597120597120597120597120597119894119894

                                            =empty119894119894+1 minus empty119894119894minus1

                                            2∆120597120597minus∆1205971205972

                                            3 1205971205973empty1205971205971205971205973

                                            120597120597119894119894

                                            minus ⋯ (220)

                                            or

                                            120597120597empty120597120597120597120597120597120597119894119894

                                            =empty119894119894+1 minus empty119894119894minus1

                                            2∆120597120597+ 119978119978(∆1205971205972) (221)

                                            This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                            approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                            the truncation error for the central difference approximation goes to zero much

                                            faster than the truncation error in forward and backward equations

                                            23

                                            25 Procedures

                                            The simple case in this investigation was assuming the constant thermal

                                            properties of the material First we assumed all the thermal properties of the

                                            materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                            point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                            assume the constant energy after that the pulse of special shapes was selected

                                            The numerical solution of equation (23) with boundary and initial conditions

                                            in equation (22) was investigated using Matlab program as shown in Appendix

                                            The equation of thermal conductivity and specific heat capacity of metal as a

                                            function of temperature was obtained by best fitting of polynomials using

                                            tabulated data in references

                                            24

                                            Chapter Three

                                            Results and Discursion

                                            31 Introduction

                                            The development of laser has been an exciting chapter in the history of

                                            science and engineering It has produced a new type of advice with potential for

                                            application in an extremely wide variety of fields Mach basic development in

                                            lasers were occurred during last 35 years The lasers interaction with metal and

                                            vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                            The status of laser development and application were still rather rudimentary

                                            The light emitted by laser is electro magnetic radiation this radiation has a wave

                                            nature the waves consists of vibrating electric and magnetic fields many studies

                                            have tried to find and solve models of laser interactions Some researchers

                                            proposed the mathematical model related to the laser - plasma interaction and

                                            the others have developed an analytical model to study the temperature

                                            distribution in Infrared optical materials heated by laser pulses Also an attempt

                                            have made to study the interaction of nanosecond pulsed lasers with material

                                            from point of view using experimental technique and theoretical approach of

                                            dimensional analysis

                                            In this study we have evaluate the solution of partial difference equation

                                            (PDE) that represent the laser interaction with solid situation in one dimension

                                            assuming that the power density of laser and thermal properties are functions

                                            with time and temperature respectively

                                            25

                                            32 Numerical solution with constant laser power density and constant

                                            thermal properties

                                            First we have taken the lead metal (Pb) with thermal properties

                                            119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                            119862119862 = 014016119869119869119892119892119870119870

                                            120588120588 = 10751 1198921198921198981198981198981198983

                                            119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                            119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                            and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                            represent the area under laser influence

                                            The numerical solution of equation (23) with boundary and initial conditions

                                            in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                            of lead metal by explicit method using Matlab program give us the results as

                                            shown in Fig (31)

                                            Fig(31) Depth dependence of the temperature with the laser power density

                                            1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                            26

                                            33 Evaluation of function 119920119920(119957119957) of laser flux density

                                            From following data that represent the energy (119869119869) with time (millie second)

                                            Time 0 001 01 02 03 04 05 06 07 08

                                            Energy 0 002 017 022 024 02 012 007 002 0

                                            By using Matlab program the best polynomial with deduced from above data

                                            was

                                            119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                            minus 10069 1198981198985 (31)

                                            As shown in Fig (32)

                                            Fig(32) Laser energy as a function of time

                                            Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                            normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                            maximum value (119864119864119898119898119907119907119898119898 )

                                            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                            (119864119864119898119898119907119907119898119898 ) (32)

                                            The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                            27

                                            Fig(33) Normalized laser energy as a function of time

                                            The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                            equal to 3 (total laser energy) ie

                                            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                            08

                                            00

                                            119899119899119898119898 = 3 (33)

                                            Therefore there exist a real number 119875119875 such that

                                            119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                            08

                                            00

                                            119899119899119898119898 = 3 (34)

                                            that implies 119875119875 = 68241 and

                                            119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                            08

                                            00

                                            119899119899119898119898 = 3 (35)

                                            The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                            08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                            119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                            00

                                            (36)

                                            28

                                            Where 119863119863119898119898 put to balance the units of equation (36)

                                            But integral

                                            119868119868 = 119864119864119860119860

                                            (37)

                                            and from equations (35) (36) and (37) we have

                                            119868119868 (119898119898)11989911989911989811989808

                                            00

                                            = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                            0800 119899119899119898119898

                                            119860119860 119863119863119898119898 (38)

                                            Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                            (38)

                                            There fore

                                            119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                            119860119860 119863119863119898119898 (39)

                                            As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                            that agrees with result data

                                            119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                            + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                            Fig(34) Time dependence of laser intensity

                                            29

                                            34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                            constant thermal properties

                                            With all constant thermal properties of lead metal as in article (23) and

                                            119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                            equation (23) with boundary and initial condition as in equation (22) and the

                                            depth penetration is shown in Fig(35)

                                            Fig(35) Depth dependence of the temperature when laser intensity function

                                            of time and constant thermal properties of Lead

                                            35 Evaluation the Thermal Conductivity as functions of temperature

                                            The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                            temperature for Lead material was obtained by Matlab program using the

                                            experimental data tabulated in researches

                                            119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                            times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                            30

                                            119879119879 ( 119870119870) 119870119870 119869119869

                                            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                            300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                            The previous thermal conductivity data and the best fitting of the data are

                                            shown in Fig (36)

                                            Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                            temperature

                                            31

                                            36 Evaluation the Specific heat as functions of temperature

                                            The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                            material was obtained from the following experimental data tacked from

                                            literatures

                                            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                            300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                            The best polynomial fitted for these data was

                                            119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                            + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                            times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                            The specific heat capacity data and the best polynomial fitting of the data are

                                            shown in Fig (37)

                                            32

                                            Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                            temperature

                                            37 Evaluation the Density as functions of temperature

                                            The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                            was used to find the best polynomial fitting

                                            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                            300 11330 400 11230 500 11130 600 11010 800 10430

                                            1000 10190 1200 9940

                                            The best polynomial of this data was

                                            120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                            minus 45158 times 10minus12 1198791198794 (313)

                                            33

                                            The density of Lead as a function of temperature and the best polynomial fitting

                                            are shown in Fig (38)

                                            Fig(38) The best fitting of density of Lead as a function of temperature

                                            38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                            variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                            We have deduced the solution of equation (24) with initial and boundary

                                            condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                            and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                            respectively then by using Matlab program the depth penetration is shown in

                                            Fig (39)

                                            34

                                            Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                            material

                                            39 Laser interaction with copper material

                                            The same time dependence of laser intensity as shown in Fig(34) with

                                            thermal properties of copper was used to calculate the temperature distribution as

                                            a function of depth penetration

                                            The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                            copper material was obtained from the experimental data tabulated in literary

                                            The Matlab program used to obtain the best polynomial equation that agrees

                                            with the above data

                                            119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                            minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                            times 10minus17 1198791198795 (314)

                                            35

                                            119879119879 ( 119870119870) 119870119870 119869119869

                                            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                            100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                            The previous thermal conductivity data and the best fitting of the data are

                                            shown in Fig (310)

                                            Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                            temperature

                                            36

                                            The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                            material was obtained from the following experimental data tacked from

                                            literatures

                                            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                            100 0254

                                            200 0357

                                            273 0384

                                            298 0387

                                            400 0397

                                            600 0416

                                            800 0435

                                            1000 0454

                                            1100 0464

                                            1200 0474

                                            1300 0483

                                            The best polynomial fitted for these data was

                                            119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                            + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                            times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                            The specific heat capacity data and the best polynomial fitting of the data are

                                            shown in Fig (311)

                                            37

                                            Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                            temperature

                                            The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                            literature was used to find the best polynomial fitting

                                            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                            100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                            1000 8576 1100 8519 1200 8458 1300 8396

                                            38

                                            The best polynomial of this data was

                                            120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                            minus 76765 times 10minus14 1198791198794 (316)

                                            The density of copper as a function of temperature and the best polynomial

                                            fitting are shown in Fig (312)

                                            Fig(312) The best fitting of density of copper as a function of temperature

                                            The depth penetration of laser energy for copper metal was calculated using

                                            the polynomial equations of thermal conductivity specific heat capacity and

                                            density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                            (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                            function of time the result was shown in Fig (313)

                                            39

                                            The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                            for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                            copper metal so the depth penetration of laser energy of lead metal was smaller

                                            than that of copper metal this may be due to the high thermal conductivity and

                                            high specific heat capacity of copper with that of lead metal

                                            Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                            material

                                            40

                                            310 Conclusions

                                            The Depth dependence of temperature for lead metal was investigated in two

                                            case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                            thermal properties (thermal conductivity specific heat) and density of metal are

                                            also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                            vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                            specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                            120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                            penetration depth in the first case is smaller than that of the second case about

                                            (190) times

                                            The temperature distribution as a function of depth dependence for copper

                                            metal was also investigated in the case when the laser intensity vary with time

                                            119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                            density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                            The depth penetration of laser energy of lead metal was found to be smaller

                                            than that of copper metal this may be due to the high thermal conductivity and

                                            high specific heat capacity of copper with that of lead metal

                                            41

                                            References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                            Sons Inc Hoboken New Jersey Canada (2003)

                                            [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                            [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                            [9]

                                            Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                            [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                            [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                            [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                            [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                            [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                            httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                            [10] httpwwwworldoflaserscomlaserprincipleshtm

                                            [11] httpenwikipediaorgwikiLaserPulsed_operation

                                            [12] httpenwikipediaorgwikiThermal_conductivity

                                            [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                            [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                            [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                            42

                                            Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                            This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                            This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                            43

                                            E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                            This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                            This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                            44

                                            title(Dencity as a function of temperature) hold off

                                            This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                            45

                                            elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                            46

                                            r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                            47

                                            for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                            48

                                            for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                            49

                                            6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                            50

                                            u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                            51

                                            alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                            52

                                            715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                            • 01 Title
                                              • 0TDr Omed Ghareb Abdullah
                                                • Ch1-Laser
                                                  • 16 Heat and heat capacity
                                                  • 18 Derivation in one dimension
                                                    • Ch2 laser2
                                                    • Ch3 laser2
                                                    • Appendix-Laser

                                              18

                                              developed it involves such a complex series solution that numerical evaluation

                                              becomes exceedingly difficult For such situation the most fruitful approach to

                                              the problem is numerical techniques the basic principles of which we shall

                                              outline in this section

                                              One way to guarantee accuracy in the solution of an initial values problems

                                              (IVP) is to solve the problem twice using step sizes h and h2 and compare

                                              answers at the mesh points corresponding to the larger step size But this requires

                                              a significant amount of computation for the smaller step size and must be

                                              repeated if it is determined that the agreement is not good enough

                                              24 Finite Difference Method

                                              The finite difference method is one of several techniques for obtaining

                                              numerical solutions to differential equations In all numerical solutions the

                                              continuous partial differential equation (PDE) is replaced with a discrete

                                              approximation In this context the word discrete means that the numerical

                                              solution is known only at a finite number of points in the physical domain The

                                              number of those points can be selected by the user of the numerical method In

                                              general increasing the number of points not only increases the resolution but

                                              also the accuracy of the numerical solution

                                              The discrete approximation results in a set of algebraic equations that are

                                              evaluated for the values of the discrete unknowns

                                              The mesh is the set of locations where the discrete solution is computed

                                              These points are called nodes and if one were to draw lines between adjacent

                                              nodes in the domain the resulting image would resemble a net or mesh Two key

                                              parameters of the mesh are ∆120597120597 the local distance between adjacent points in

                                              space and ∆119905119905 the local distance between adjacent time steps For the simple

                                              examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

                                              19

                                              The core idea of the finite-difference method is to replace continuous

                                              derivatives with so-called difference formulas that involve only the discrete

                                              values associated with positions on the mesh

                                              Applying the finite-difference method to a differential equation involves

                                              replacing all derivatives with difference formulas In the heat equation there are

                                              derivatives with respect to time and derivatives with respect to space Using

                                              different combinations of mesh points in the difference formulas results in

                                              different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                              numerical solution obtained with any useful scheme will approach the true

                                              solution to the original differential equation However the rate at which the

                                              numerical solution approaches the true solution varies with the scheme

                                              241 First Order Forward Difference

                                              Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                              empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                              +1205751205751205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              1205971205971

                                              +1205751205751205971205973

                                              3 1205971205973empty1205971205971205971205973

                                              1205971205971

                                              + ⋯ (25)

                                              where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                              consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                              empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                              +∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              +∆1205971205973

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              + ⋯ (26)

                                              Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                              120597120597empty120597120597120597120597120597120597119894119894

                                              =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                              ∆120597120597minus∆1205971205972

                                              1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              minus∆1205971205972

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              minus ⋯ (27)

                                              Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                              hand side have been reduced by one

                                              20

                                              Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                              and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                              120597120597empty120597120597120597120597120597120597119894119894

                                              =empty119894119894+1 minus empty119894119894

                                              ∆120597120597minus∆1205971205972

                                              1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              minus∆1205971205972

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              minus ⋯ (28)

                                              The mean value theorem can be used to replace the higher order derivatives

                                              ∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              +∆1205971205973

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              + ⋯ =∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120585120585 (29)

                                              where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                              120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                              ∆120597120597+∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120585120585 (210)

                                              120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                              ∆120597120597asymp∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120585120585 (211)

                                              The term on the right hand side of previous equation is called the truncation

                                              error of the finite difference approximation

                                              In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                              unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                              truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                              analytical form) the big 119978119978 notation can be used to express the dependence of

                                              the truncation error on the mesh spacing Note that the right hand side of last

                                              equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                              the finite difference simulation Since this is the only parameter under the users

                                              control that determines the error the truncation error is simply written

                                              ∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120585120585= 119978119978(∆1205971205972) (212)

                                              The equals sign in this expression is true in the order of magnitude sense In

                                              other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                              21

                                              equality Rather the expression means that the left hand side is a product of an

                                              unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                              magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                              approaches zero as ∆120597120597 is reduced

                                              Using big 119978119978 notation Equation (28) can be written

                                              120597120597empty120597120597120597120597120597120597119894119894

                                              =empty119894119894+1 minus empty119894119894

                                              ∆120597120597+ 119978119978(∆120597120597) (213)

                                              This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                              it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                              truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                              our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                              error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                              242 First Order Backward Difference

                                              An alternative first order finite difference formula is obtained if the Taylor series

                                              like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                              variables in place of all the unknowns one obtains

                                              empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                              +∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              minus∆1205971205973

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              + ⋯ (214)

                                              Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                              to get

                                              120597120597empty120597120597120597120597120597120597119894119894

                                              =empty119894119894+1 minus empty119894119894

                                              ∆120597120597minus∆1205971205972

                                              1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              minus∆1205971205972

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              minus ⋯ (215)

                                              Or using big 119978119978 notation

                                              120597120597empty120597120597120597120597120597120597119894119894

                                              =empty119894119894 minus empty119894119894minus1

                                              ∆120597120597+ 119978119978(∆120597120597) (216)

                                              22

                                              This is called the backward difference formula because it involves the values of

                                              empty at 120597120597119894119894 and 120597120597119894119894minus1

                                              The order of magnitude of the truncation error for the backward difference

                                              approximation is the same as that of the forward difference approximation Can

                                              we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                              truncation error The answer is yes

                                              242 First Order Central Difference

                                              Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                              empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                              +∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              +∆1205971205973

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              + ⋯ (217)

                                              empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                              +∆1205971205972

                                              2 1205971205972empty1205971205971205971205972

                                              120597120597119894119894

                                              minus∆1205971205973

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              + ⋯ (218)

                                              Subtracting Equation (10) from Equation (9) yields

                                              empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                              + 2∆1205971205973

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              + ⋯ (219)

                                              Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                              120597120597empty120597120597120597120597120597120597119894119894

                                              =empty119894119894+1 minus empty119894119894minus1

                                              2∆120597120597minus∆1205971205972

                                              3 1205971205973empty1205971205971205971205973

                                              120597120597119894119894

                                              minus ⋯ (220)

                                              or

                                              120597120597empty120597120597120597120597120597120597119894119894

                                              =empty119894119894+1 minus empty119894119894minus1

                                              2∆120597120597+ 119978119978(∆1205971205972) (221)

                                              This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                              approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                              the truncation error for the central difference approximation goes to zero much

                                              faster than the truncation error in forward and backward equations

                                              23

                                              25 Procedures

                                              The simple case in this investigation was assuming the constant thermal

                                              properties of the material First we assumed all the thermal properties of the

                                              materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                              point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                              assume the constant energy after that the pulse of special shapes was selected

                                              The numerical solution of equation (23) with boundary and initial conditions

                                              in equation (22) was investigated using Matlab program as shown in Appendix

                                              The equation of thermal conductivity and specific heat capacity of metal as a

                                              function of temperature was obtained by best fitting of polynomials using

                                              tabulated data in references

                                              24

                                              Chapter Three

                                              Results and Discursion

                                              31 Introduction

                                              The development of laser has been an exciting chapter in the history of

                                              science and engineering It has produced a new type of advice with potential for

                                              application in an extremely wide variety of fields Mach basic development in

                                              lasers were occurred during last 35 years The lasers interaction with metal and

                                              vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                              The status of laser development and application were still rather rudimentary

                                              The light emitted by laser is electro magnetic radiation this radiation has a wave

                                              nature the waves consists of vibrating electric and magnetic fields many studies

                                              have tried to find and solve models of laser interactions Some researchers

                                              proposed the mathematical model related to the laser - plasma interaction and

                                              the others have developed an analytical model to study the temperature

                                              distribution in Infrared optical materials heated by laser pulses Also an attempt

                                              have made to study the interaction of nanosecond pulsed lasers with material

                                              from point of view using experimental technique and theoretical approach of

                                              dimensional analysis

                                              In this study we have evaluate the solution of partial difference equation

                                              (PDE) that represent the laser interaction with solid situation in one dimension

                                              assuming that the power density of laser and thermal properties are functions

                                              with time and temperature respectively

                                              25

                                              32 Numerical solution with constant laser power density and constant

                                              thermal properties

                                              First we have taken the lead metal (Pb) with thermal properties

                                              119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                              119862119862 = 014016119869119869119892119892119870119870

                                              120588120588 = 10751 1198921198921198981198981198981198983

                                              119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                              119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                              and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                              represent the area under laser influence

                                              The numerical solution of equation (23) with boundary and initial conditions

                                              in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                              of lead metal by explicit method using Matlab program give us the results as

                                              shown in Fig (31)

                                              Fig(31) Depth dependence of the temperature with the laser power density

                                              1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                              26

                                              33 Evaluation of function 119920119920(119957119957) of laser flux density

                                              From following data that represent the energy (119869119869) with time (millie second)

                                              Time 0 001 01 02 03 04 05 06 07 08

                                              Energy 0 002 017 022 024 02 012 007 002 0

                                              By using Matlab program the best polynomial with deduced from above data

                                              was

                                              119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                              minus 10069 1198981198985 (31)

                                              As shown in Fig (32)

                                              Fig(32) Laser energy as a function of time

                                              Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                              normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                              maximum value (119864119864119898119898119907119907119898119898 )

                                              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                              (119864119864119898119898119907119907119898119898 ) (32)

                                              The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                              27

                                              Fig(33) Normalized laser energy as a function of time

                                              The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                              equal to 3 (total laser energy) ie

                                              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                              08

                                              00

                                              119899119899119898119898 = 3 (33)

                                              Therefore there exist a real number 119875119875 such that

                                              119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                              08

                                              00

                                              119899119899119898119898 = 3 (34)

                                              that implies 119875119875 = 68241 and

                                              119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                              08

                                              00

                                              119899119899119898119898 = 3 (35)

                                              The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                              08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                              119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                              00

                                              (36)

                                              28

                                              Where 119863119863119898119898 put to balance the units of equation (36)

                                              But integral

                                              119868119868 = 119864119864119860119860

                                              (37)

                                              and from equations (35) (36) and (37) we have

                                              119868119868 (119898119898)11989911989911989811989808

                                              00

                                              = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                              0800 119899119899119898119898

                                              119860119860 119863119863119898119898 (38)

                                              Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                              (38)

                                              There fore

                                              119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                              119860119860 119863119863119898119898 (39)

                                              As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                              that agrees with result data

                                              119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                              + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                              Fig(34) Time dependence of laser intensity

                                              29

                                              34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                              constant thermal properties

                                              With all constant thermal properties of lead metal as in article (23) and

                                              119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                              equation (23) with boundary and initial condition as in equation (22) and the

                                              depth penetration is shown in Fig(35)

                                              Fig(35) Depth dependence of the temperature when laser intensity function

                                              of time and constant thermal properties of Lead

                                              35 Evaluation the Thermal Conductivity as functions of temperature

                                              The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                              temperature for Lead material was obtained by Matlab program using the

                                              experimental data tabulated in researches

                                              119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                              times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                              30

                                              119879119879 ( 119870119870) 119870119870 119869119869

                                              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                              300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                              The previous thermal conductivity data and the best fitting of the data are

                                              shown in Fig (36)

                                              Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                              temperature

                                              31

                                              36 Evaluation the Specific heat as functions of temperature

                                              The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                              material was obtained from the following experimental data tacked from

                                              literatures

                                              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                              300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                              The best polynomial fitted for these data was

                                              119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                              + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                              times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                              The specific heat capacity data and the best polynomial fitting of the data are

                                              shown in Fig (37)

                                              32

                                              Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                              temperature

                                              37 Evaluation the Density as functions of temperature

                                              The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                              was used to find the best polynomial fitting

                                              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                              300 11330 400 11230 500 11130 600 11010 800 10430

                                              1000 10190 1200 9940

                                              The best polynomial of this data was

                                              120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                              minus 45158 times 10minus12 1198791198794 (313)

                                              33

                                              The density of Lead as a function of temperature and the best polynomial fitting

                                              are shown in Fig (38)

                                              Fig(38) The best fitting of density of Lead as a function of temperature

                                              38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                              variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                              We have deduced the solution of equation (24) with initial and boundary

                                              condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                              and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                              respectively then by using Matlab program the depth penetration is shown in

                                              Fig (39)

                                              34

                                              Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                              material

                                              39 Laser interaction with copper material

                                              The same time dependence of laser intensity as shown in Fig(34) with

                                              thermal properties of copper was used to calculate the temperature distribution as

                                              a function of depth penetration

                                              The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                              copper material was obtained from the experimental data tabulated in literary

                                              The Matlab program used to obtain the best polynomial equation that agrees

                                              with the above data

                                              119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                              minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                              times 10minus17 1198791198795 (314)

                                              35

                                              119879119879 ( 119870119870) 119870119870 119869119869

                                              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                              100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                              The previous thermal conductivity data and the best fitting of the data are

                                              shown in Fig (310)

                                              Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                              temperature

                                              36

                                              The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                              material was obtained from the following experimental data tacked from

                                              literatures

                                              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                              100 0254

                                              200 0357

                                              273 0384

                                              298 0387

                                              400 0397

                                              600 0416

                                              800 0435

                                              1000 0454

                                              1100 0464

                                              1200 0474

                                              1300 0483

                                              The best polynomial fitted for these data was

                                              119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                              + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                              times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                              The specific heat capacity data and the best polynomial fitting of the data are

                                              shown in Fig (311)

                                              37

                                              Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                              temperature

                                              The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                              literature was used to find the best polynomial fitting

                                              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                              100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                              1000 8576 1100 8519 1200 8458 1300 8396

                                              38

                                              The best polynomial of this data was

                                              120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                              minus 76765 times 10minus14 1198791198794 (316)

                                              The density of copper as a function of temperature and the best polynomial

                                              fitting are shown in Fig (312)

                                              Fig(312) The best fitting of density of copper as a function of temperature

                                              The depth penetration of laser energy for copper metal was calculated using

                                              the polynomial equations of thermal conductivity specific heat capacity and

                                              density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                              (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                              function of time the result was shown in Fig (313)

                                              39

                                              The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                              for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                              copper metal so the depth penetration of laser energy of lead metal was smaller

                                              than that of copper metal this may be due to the high thermal conductivity and

                                              high specific heat capacity of copper with that of lead metal

                                              Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                              material

                                              40

                                              310 Conclusions

                                              The Depth dependence of temperature for lead metal was investigated in two

                                              case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                              thermal properties (thermal conductivity specific heat) and density of metal are

                                              also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                              vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                              specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                              120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                              penetration depth in the first case is smaller than that of the second case about

                                              (190) times

                                              The temperature distribution as a function of depth dependence for copper

                                              metal was also investigated in the case when the laser intensity vary with time

                                              119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                              density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                              The depth penetration of laser energy of lead metal was found to be smaller

                                              than that of copper metal this may be due to the high thermal conductivity and

                                              high specific heat capacity of copper with that of lead metal

                                              41

                                              References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                              Sons Inc Hoboken New Jersey Canada (2003)

                                              [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                              [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                              [9]

                                              Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                              [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                              [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                              [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                              [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                              [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                              httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                              [10] httpwwwworldoflaserscomlaserprincipleshtm

                                              [11] httpenwikipediaorgwikiLaserPulsed_operation

                                              [12] httpenwikipediaorgwikiThermal_conductivity

                                              [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                              [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                              [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                              42

                                              Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                              This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                              This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                              43

                                              E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                              This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                              This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                              44

                                              title(Dencity as a function of temperature) hold off

                                              This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                              45

                                              elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                              46

                                              r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                              47

                                              for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                              48

                                              for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                              49

                                              6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                              50

                                              u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                              51

                                              alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                              52

                                              715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                              • 01 Title
                                                • 0TDr Omed Ghareb Abdullah
                                                  • Ch1-Laser
                                                    • 16 Heat and heat capacity
                                                    • 18 Derivation in one dimension
                                                      • Ch2 laser2
                                                      • Ch3 laser2
                                                      • Appendix-Laser

                                                19

                                                The core idea of the finite-difference method is to replace continuous

                                                derivatives with so-called difference formulas that involve only the discrete

                                                values associated with positions on the mesh

                                                Applying the finite-difference method to a differential equation involves

                                                replacing all derivatives with difference formulas In the heat equation there are

                                                derivatives with respect to time and derivatives with respect to space Using

                                                different combinations of mesh points in the difference formulas results in

                                                different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

                                                numerical solution obtained with any useful scheme will approach the true

                                                solution to the original differential equation However the rate at which the

                                                numerical solution approaches the true solution varies with the scheme

                                                241 First Order Forward Difference

                                                Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

                                                empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

                                                +1205751205751205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                1205971205971

                                                +1205751205751205971205973

                                                3 1205971205973empty1205971205971205971205973

                                                1205971205971

                                                + ⋯ (25)

                                                where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

                                                consider the value of empty at the location of the 120597120597119894119894+1 mesh line

                                                empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                +∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                +∆1205971205973

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                + ⋯ (26)

                                                Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                                120597120597empty120597120597120597120597120597120597119894119894

                                                =empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

                                                ∆120597120597minus∆1205971205972

                                                1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                minus∆1205971205972

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                minus ⋯ (27)

                                                Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

                                                hand side have been reduced by one

                                                20

                                                Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                                and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                                120597120597empty120597120597120597120597120597120597119894119894

                                                =empty119894119894+1 minus empty119894119894

                                                ∆120597120597minus∆1205971205972

                                                1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                minus∆1205971205972

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                minus ⋯ (28)

                                                The mean value theorem can be used to replace the higher order derivatives

                                                ∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                +∆1205971205973

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                + ⋯ =∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120585120585 (29)

                                                where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                                120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                                ∆120597120597+∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120585120585 (210)

                                                120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                                ∆120597120597asymp∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120585120585 (211)

                                                The term on the right hand side of previous equation is called the truncation

                                                error of the finite difference approximation

                                                In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                                unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                                truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                                analytical form) the big 119978119978 notation can be used to express the dependence of

                                                the truncation error on the mesh spacing Note that the right hand side of last

                                                equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                                the finite difference simulation Since this is the only parameter under the users

                                                control that determines the error the truncation error is simply written

                                                ∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120585120585= 119978119978(∆1205971205972) (212)

                                                The equals sign in this expression is true in the order of magnitude sense In

                                                other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                                21

                                                equality Rather the expression means that the left hand side is a product of an

                                                unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                                magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                                approaches zero as ∆120597120597 is reduced

                                                Using big 119978119978 notation Equation (28) can be written

                                                120597120597empty120597120597120597120597120597120597119894119894

                                                =empty119894119894+1 minus empty119894119894

                                                ∆120597120597+ 119978119978(∆120597120597) (213)

                                                This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                                it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                                truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                                our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                                error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                                242 First Order Backward Difference

                                                An alternative first order finite difference formula is obtained if the Taylor series

                                                like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                                variables in place of all the unknowns one obtains

                                                empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                +∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                minus∆1205971205973

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                + ⋯ (214)

                                                Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                                to get

                                                120597120597empty120597120597120597120597120597120597119894119894

                                                =empty119894119894+1 minus empty119894119894

                                                ∆120597120597minus∆1205971205972

                                                1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                minus∆1205971205972

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                minus ⋯ (215)

                                                Or using big 119978119978 notation

                                                120597120597empty120597120597120597120597120597120597119894119894

                                                =empty119894119894 minus empty119894119894minus1

                                                ∆120597120597+ 119978119978(∆120597120597) (216)

                                                22

                                                This is called the backward difference formula because it involves the values of

                                                empty at 120597120597119894119894 and 120597120597119894119894minus1

                                                The order of magnitude of the truncation error for the backward difference

                                                approximation is the same as that of the forward difference approximation Can

                                                we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                                truncation error The answer is yes

                                                242 First Order Central Difference

                                                Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                                empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                +∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                +∆1205971205973

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                + ⋯ (217)

                                                empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                +∆1205971205972

                                                2 1205971205972empty1205971205971205971205972

                                                120597120597119894119894

                                                minus∆1205971205973

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                + ⋯ (218)

                                                Subtracting Equation (10) from Equation (9) yields

                                                empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                + 2∆1205971205973

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                + ⋯ (219)

                                                Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                                120597120597empty120597120597120597120597120597120597119894119894

                                                =empty119894119894+1 minus empty119894119894minus1

                                                2∆120597120597minus∆1205971205972

                                                3 1205971205973empty1205971205971205971205973

                                                120597120597119894119894

                                                minus ⋯ (220)

                                                or

                                                120597120597empty120597120597120597120597120597120597119894119894

                                                =empty119894119894+1 minus empty119894119894minus1

                                                2∆120597120597+ 119978119978(∆1205971205972) (221)

                                                This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                                approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                                the truncation error for the central difference approximation goes to zero much

                                                faster than the truncation error in forward and backward equations

                                                23

                                                25 Procedures

                                                The simple case in this investigation was assuming the constant thermal

                                                properties of the material First we assumed all the thermal properties of the

                                                materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                                point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                                assume the constant energy after that the pulse of special shapes was selected

                                                The numerical solution of equation (23) with boundary and initial conditions

                                                in equation (22) was investigated using Matlab program as shown in Appendix

                                                The equation of thermal conductivity and specific heat capacity of metal as a

                                                function of temperature was obtained by best fitting of polynomials using

                                                tabulated data in references

                                                24

                                                Chapter Three

                                                Results and Discursion

                                                31 Introduction

                                                The development of laser has been an exciting chapter in the history of

                                                science and engineering It has produced a new type of advice with potential for

                                                application in an extremely wide variety of fields Mach basic development in

                                                lasers were occurred during last 35 years The lasers interaction with metal and

                                                vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                                The status of laser development and application were still rather rudimentary

                                                The light emitted by laser is electro magnetic radiation this radiation has a wave

                                                nature the waves consists of vibrating electric and magnetic fields many studies

                                                have tried to find and solve models of laser interactions Some researchers

                                                proposed the mathematical model related to the laser - plasma interaction and

                                                the others have developed an analytical model to study the temperature

                                                distribution in Infrared optical materials heated by laser pulses Also an attempt

                                                have made to study the interaction of nanosecond pulsed lasers with material

                                                from point of view using experimental technique and theoretical approach of

                                                dimensional analysis

                                                In this study we have evaluate the solution of partial difference equation

                                                (PDE) that represent the laser interaction with solid situation in one dimension

                                                assuming that the power density of laser and thermal properties are functions

                                                with time and temperature respectively

                                                25

                                                32 Numerical solution with constant laser power density and constant

                                                thermal properties

                                                First we have taken the lead metal (Pb) with thermal properties

                                                119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                                119862119862 = 014016119869119869119892119892119870119870

                                                120588120588 = 10751 1198921198921198981198981198981198983

                                                119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                                119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                                and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                                represent the area under laser influence

                                                The numerical solution of equation (23) with boundary and initial conditions

                                                in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                                of lead metal by explicit method using Matlab program give us the results as

                                                shown in Fig (31)

                                                Fig(31) Depth dependence of the temperature with the laser power density

                                                1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                                26

                                                33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                From following data that represent the energy (119869119869) with time (millie second)

                                                Time 0 001 01 02 03 04 05 06 07 08

                                                Energy 0 002 017 022 024 02 012 007 002 0

                                                By using Matlab program the best polynomial with deduced from above data

                                                was

                                                119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                minus 10069 1198981198985 (31)

                                                As shown in Fig (32)

                                                Fig(32) Laser energy as a function of time

                                                Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                maximum value (119864119864119898119898119907119907119898119898 )

                                                119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                (119864119864119898119898119907119907119898119898 ) (32)

                                                The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                27

                                                Fig(33) Normalized laser energy as a function of time

                                                The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                equal to 3 (total laser energy) ie

                                                119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                08

                                                00

                                                119899119899119898119898 = 3 (33)

                                                Therefore there exist a real number 119875119875 such that

                                                119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                08

                                                00

                                                119899119899119898119898 = 3 (34)

                                                that implies 119875119875 = 68241 and

                                                119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                08

                                                00

                                                119899119899119898119898 = 3 (35)

                                                The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                00

                                                (36)

                                                28

                                                Where 119863119863119898119898 put to balance the units of equation (36)

                                                But integral

                                                119868119868 = 119864119864119860119860

                                                (37)

                                                and from equations (35) (36) and (37) we have

                                                119868119868 (119898119898)11989911989911989811989808

                                                00

                                                = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                0800 119899119899119898119898

                                                119860119860 119863119863119898119898 (38)

                                                Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                (38)

                                                There fore

                                                119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                119860119860 119863119863119898119898 (39)

                                                As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                that agrees with result data

                                                119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                Fig(34) Time dependence of laser intensity

                                                29

                                                34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                constant thermal properties

                                                With all constant thermal properties of lead metal as in article (23) and

                                                119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                equation (23) with boundary and initial condition as in equation (22) and the

                                                depth penetration is shown in Fig(35)

                                                Fig(35) Depth dependence of the temperature when laser intensity function

                                                of time and constant thermal properties of Lead

                                                35 Evaluation the Thermal Conductivity as functions of temperature

                                                The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                temperature for Lead material was obtained by Matlab program using the

                                                experimental data tabulated in researches

                                                119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                30

                                                119879119879 ( 119870119870) 119870119870 119869119869

                                                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                The previous thermal conductivity data and the best fitting of the data are

                                                shown in Fig (36)

                                                Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                temperature

                                                31

                                                36 Evaluation the Specific heat as functions of temperature

                                                The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                material was obtained from the following experimental data tacked from

                                                literatures

                                                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                The best polynomial fitted for these data was

                                                119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                The specific heat capacity data and the best polynomial fitting of the data are

                                                shown in Fig (37)

                                                32

                                                Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                temperature

                                                37 Evaluation the Density as functions of temperature

                                                The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                was used to find the best polynomial fitting

                                                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                300 11330 400 11230 500 11130 600 11010 800 10430

                                                1000 10190 1200 9940

                                                The best polynomial of this data was

                                                120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                minus 45158 times 10minus12 1198791198794 (313)

                                                33

                                                The density of Lead as a function of temperature and the best polynomial fitting

                                                are shown in Fig (38)

                                                Fig(38) The best fitting of density of Lead as a function of temperature

                                                38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                We have deduced the solution of equation (24) with initial and boundary

                                                condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                respectively then by using Matlab program the depth penetration is shown in

                                                Fig (39)

                                                34

                                                Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                material

                                                39 Laser interaction with copper material

                                                The same time dependence of laser intensity as shown in Fig(34) with

                                                thermal properties of copper was used to calculate the temperature distribution as

                                                a function of depth penetration

                                                The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                copper material was obtained from the experimental data tabulated in literary

                                                The Matlab program used to obtain the best polynomial equation that agrees

                                                with the above data

                                                119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                times 10minus17 1198791198795 (314)

                                                35

                                                119879119879 ( 119870119870) 119870119870 119869119869

                                                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                The previous thermal conductivity data and the best fitting of the data are

                                                shown in Fig (310)

                                                Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                temperature

                                                36

                                                The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                material was obtained from the following experimental data tacked from

                                                literatures

                                                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                100 0254

                                                200 0357

                                                273 0384

                                                298 0387

                                                400 0397

                                                600 0416

                                                800 0435

                                                1000 0454

                                                1100 0464

                                                1200 0474

                                                1300 0483

                                                The best polynomial fitted for these data was

                                                119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                The specific heat capacity data and the best polynomial fitting of the data are

                                                shown in Fig (311)

                                                37

                                                Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                temperature

                                                The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                literature was used to find the best polynomial fitting

                                                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                1000 8576 1100 8519 1200 8458 1300 8396

                                                38

                                                The best polynomial of this data was

                                                120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                minus 76765 times 10minus14 1198791198794 (316)

                                                The density of copper as a function of temperature and the best polynomial

                                                fitting are shown in Fig (312)

                                                Fig(312) The best fitting of density of copper as a function of temperature

                                                The depth penetration of laser energy for copper metal was calculated using

                                                the polynomial equations of thermal conductivity specific heat capacity and

                                                density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                function of time the result was shown in Fig (313)

                                                39

                                                The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                copper metal so the depth penetration of laser energy of lead metal was smaller

                                                than that of copper metal this may be due to the high thermal conductivity and

                                                high specific heat capacity of copper with that of lead metal

                                                Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                material

                                                40

                                                310 Conclusions

                                                The Depth dependence of temperature for lead metal was investigated in two

                                                case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                thermal properties (thermal conductivity specific heat) and density of metal are

                                                also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                penetration depth in the first case is smaller than that of the second case about

                                                (190) times

                                                The temperature distribution as a function of depth dependence for copper

                                                metal was also investigated in the case when the laser intensity vary with time

                                                119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                The depth penetration of laser energy of lead metal was found to be smaller

                                                than that of copper metal this may be due to the high thermal conductivity and

                                                high specific heat capacity of copper with that of lead metal

                                                41

                                                References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                Sons Inc Hoboken New Jersey Canada (2003)

                                                [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                [9]

                                                Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                [12] httpenwikipediaorgwikiThermal_conductivity

                                                [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                42

                                                Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                43

                                                E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                44

                                                title(Dencity as a function of temperature) hold off

                                                This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                45

                                                elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                46

                                                r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                47

                                                for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                48

                                                for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                49

                                                6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                50

                                                u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                51

                                                alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                52

                                                715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                • 01 Title
                                                  • 0TDr Omed Ghareb Abdullah
                                                    • Ch1-Laser
                                                      • 16 Heat and heat capacity
                                                      • 18 Derivation in one dimension
                                                        • Ch2 laser2
                                                        • Ch3 laser2
                                                        • Appendix-Laser

                                                  20

                                                  Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

                                                  and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

                                                  120597120597empty120597120597120597120597120597120597119894119894

                                                  =empty119894119894+1 minus empty119894119894

                                                  ∆120597120597minus∆1205971205972

                                                  1205971205972empty1205971205971205971205972

                                                  120597120597119894119894

                                                  minus∆1205971205972

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  minus ⋯ (28)

                                                  The mean value theorem can be used to replace the higher order derivatives

                                                  ∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120597120597119894119894

                                                  +∆1205971205973

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  + ⋯ =∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120585120585 (29)

                                                  where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

                                                  120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

                                                  ∆120597120597+∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120585120585 (210)

                                                  120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

                                                  ∆120597120597asymp∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120585120585 (211)

                                                  The term on the right hand side of previous equation is called the truncation

                                                  error of the finite difference approximation

                                                  In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

                                                  unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

                                                  truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

                                                  analytical form) the big 119978119978 notation can be used to express the dependence of

                                                  the truncation error on the mesh spacing Note that the right hand side of last

                                                  equation contains the mesh parameter ∆120597120597 which is chosen by the person using

                                                  the finite difference simulation Since this is the only parameter under the users

                                                  control that determines the error the truncation error is simply written

                                                  ∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120585120585= 119978119978(∆1205971205972) (212)

                                                  The equals sign in this expression is true in the order of magnitude sense In

                                                  other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

                                                  21

                                                  equality Rather the expression means that the left hand side is a product of an

                                                  unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                                  magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                                  approaches zero as ∆120597120597 is reduced

                                                  Using big 119978119978 notation Equation (28) can be written

                                                  120597120597empty120597120597120597120597120597120597119894119894

                                                  =empty119894119894+1 minus empty119894119894

                                                  ∆120597120597+ 119978119978(∆120597120597) (213)

                                                  This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                                  it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                                  truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                                  our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                                  error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                                  242 First Order Backward Difference

                                                  An alternative first order finite difference formula is obtained if the Taylor series

                                                  like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                                  variables in place of all the unknowns one obtains

                                                  empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                  +∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120597120597119894119894

                                                  minus∆1205971205973

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  + ⋯ (214)

                                                  Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                                  to get

                                                  120597120597empty120597120597120597120597120597120597119894119894

                                                  =empty119894119894+1 minus empty119894119894

                                                  ∆120597120597minus∆1205971205972

                                                  1205971205972empty1205971205971205971205972

                                                  120597120597119894119894

                                                  minus∆1205971205972

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  minus ⋯ (215)

                                                  Or using big 119978119978 notation

                                                  120597120597empty120597120597120597120597120597120597119894119894

                                                  =empty119894119894 minus empty119894119894minus1

                                                  ∆120597120597+ 119978119978(∆120597120597) (216)

                                                  22

                                                  This is called the backward difference formula because it involves the values of

                                                  empty at 120597120597119894119894 and 120597120597119894119894minus1

                                                  The order of magnitude of the truncation error for the backward difference

                                                  approximation is the same as that of the forward difference approximation Can

                                                  we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                                  truncation error The answer is yes

                                                  242 First Order Central Difference

                                                  Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                                  empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                  +∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120597120597119894119894

                                                  +∆1205971205973

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  + ⋯ (217)

                                                  empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                  +∆1205971205972

                                                  2 1205971205972empty1205971205971205971205972

                                                  120597120597119894119894

                                                  minus∆1205971205973

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  + ⋯ (218)

                                                  Subtracting Equation (10) from Equation (9) yields

                                                  empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                  + 2∆1205971205973

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  + ⋯ (219)

                                                  Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                                  120597120597empty120597120597120597120597120597120597119894119894

                                                  =empty119894119894+1 minus empty119894119894minus1

                                                  2∆120597120597minus∆1205971205972

                                                  3 1205971205973empty1205971205971205971205973

                                                  120597120597119894119894

                                                  minus ⋯ (220)

                                                  or

                                                  120597120597empty120597120597120597120597120597120597119894119894

                                                  =empty119894119894+1 minus empty119894119894minus1

                                                  2∆120597120597+ 119978119978(∆1205971205972) (221)

                                                  This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                                  approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                                  the truncation error for the central difference approximation goes to zero much

                                                  faster than the truncation error in forward and backward equations

                                                  23

                                                  25 Procedures

                                                  The simple case in this investigation was assuming the constant thermal

                                                  properties of the material First we assumed all the thermal properties of the

                                                  materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                                  point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                                  assume the constant energy after that the pulse of special shapes was selected

                                                  The numerical solution of equation (23) with boundary and initial conditions

                                                  in equation (22) was investigated using Matlab program as shown in Appendix

                                                  The equation of thermal conductivity and specific heat capacity of metal as a

                                                  function of temperature was obtained by best fitting of polynomials using

                                                  tabulated data in references

                                                  24

                                                  Chapter Three

                                                  Results and Discursion

                                                  31 Introduction

                                                  The development of laser has been an exciting chapter in the history of

                                                  science and engineering It has produced a new type of advice with potential for

                                                  application in an extremely wide variety of fields Mach basic development in

                                                  lasers were occurred during last 35 years The lasers interaction with metal and

                                                  vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                                  The status of laser development and application were still rather rudimentary

                                                  The light emitted by laser is electro magnetic radiation this radiation has a wave

                                                  nature the waves consists of vibrating electric and magnetic fields many studies

                                                  have tried to find and solve models of laser interactions Some researchers

                                                  proposed the mathematical model related to the laser - plasma interaction and

                                                  the others have developed an analytical model to study the temperature

                                                  distribution in Infrared optical materials heated by laser pulses Also an attempt

                                                  have made to study the interaction of nanosecond pulsed lasers with material

                                                  from point of view using experimental technique and theoretical approach of

                                                  dimensional analysis

                                                  In this study we have evaluate the solution of partial difference equation

                                                  (PDE) that represent the laser interaction with solid situation in one dimension

                                                  assuming that the power density of laser and thermal properties are functions

                                                  with time and temperature respectively

                                                  25

                                                  32 Numerical solution with constant laser power density and constant

                                                  thermal properties

                                                  First we have taken the lead metal (Pb) with thermal properties

                                                  119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                                  119862119862 = 014016119869119869119892119892119870119870

                                                  120588120588 = 10751 1198921198921198981198981198981198983

                                                  119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                                  119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                                  and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                                  represent the area under laser influence

                                                  The numerical solution of equation (23) with boundary and initial conditions

                                                  in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                                  of lead metal by explicit method using Matlab program give us the results as

                                                  shown in Fig (31)

                                                  Fig(31) Depth dependence of the temperature with the laser power density

                                                  1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                                  26

                                                  33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                  From following data that represent the energy (119869119869) with time (millie second)

                                                  Time 0 001 01 02 03 04 05 06 07 08

                                                  Energy 0 002 017 022 024 02 012 007 002 0

                                                  By using Matlab program the best polynomial with deduced from above data

                                                  was

                                                  119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                  minus 10069 1198981198985 (31)

                                                  As shown in Fig (32)

                                                  Fig(32) Laser energy as a function of time

                                                  Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                  normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                  maximum value (119864119864119898119898119907119907119898119898 )

                                                  119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                  (119864119864119898119898119907119907119898119898 ) (32)

                                                  The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                  27

                                                  Fig(33) Normalized laser energy as a function of time

                                                  The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                  equal to 3 (total laser energy) ie

                                                  119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                  08

                                                  00

                                                  119899119899119898119898 = 3 (33)

                                                  Therefore there exist a real number 119875119875 such that

                                                  119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                  08

                                                  00

                                                  119899119899119898119898 = 3 (34)

                                                  that implies 119875119875 = 68241 and

                                                  119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                  08

                                                  00

                                                  119899119899119898119898 = 3 (35)

                                                  The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                  08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                  119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                  00

                                                  (36)

                                                  28

                                                  Where 119863119863119898119898 put to balance the units of equation (36)

                                                  But integral

                                                  119868119868 = 119864119864119860119860

                                                  (37)

                                                  and from equations (35) (36) and (37) we have

                                                  119868119868 (119898119898)11989911989911989811989808

                                                  00

                                                  = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                  0800 119899119899119898119898

                                                  119860119860 119863119863119898119898 (38)

                                                  Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                  (38)

                                                  There fore

                                                  119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                  119860119860 119863119863119898119898 (39)

                                                  As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                  that agrees with result data

                                                  119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                  + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                  Fig(34) Time dependence of laser intensity

                                                  29

                                                  34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                  constant thermal properties

                                                  With all constant thermal properties of lead metal as in article (23) and

                                                  119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                  equation (23) with boundary and initial condition as in equation (22) and the

                                                  depth penetration is shown in Fig(35)

                                                  Fig(35) Depth dependence of the temperature when laser intensity function

                                                  of time and constant thermal properties of Lead

                                                  35 Evaluation the Thermal Conductivity as functions of temperature

                                                  The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                  temperature for Lead material was obtained by Matlab program using the

                                                  experimental data tabulated in researches

                                                  119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                  times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                  30

                                                  119879119879 ( 119870119870) 119870119870 119869119869

                                                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                  300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                  The previous thermal conductivity data and the best fitting of the data are

                                                  shown in Fig (36)

                                                  Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                  temperature

                                                  31

                                                  36 Evaluation the Specific heat as functions of temperature

                                                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                  material was obtained from the following experimental data tacked from

                                                  literatures

                                                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                  300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                  The best polynomial fitted for these data was

                                                  119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                  + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                  times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                  The specific heat capacity data and the best polynomial fitting of the data are

                                                  shown in Fig (37)

                                                  32

                                                  Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                  temperature

                                                  37 Evaluation the Density as functions of temperature

                                                  The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                  was used to find the best polynomial fitting

                                                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                  300 11330 400 11230 500 11130 600 11010 800 10430

                                                  1000 10190 1200 9940

                                                  The best polynomial of this data was

                                                  120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                  minus 45158 times 10minus12 1198791198794 (313)

                                                  33

                                                  The density of Lead as a function of temperature and the best polynomial fitting

                                                  are shown in Fig (38)

                                                  Fig(38) The best fitting of density of Lead as a function of temperature

                                                  38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                  variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                  We have deduced the solution of equation (24) with initial and boundary

                                                  condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                  and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                  respectively then by using Matlab program the depth penetration is shown in

                                                  Fig (39)

                                                  34

                                                  Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                  material

                                                  39 Laser interaction with copper material

                                                  The same time dependence of laser intensity as shown in Fig(34) with

                                                  thermal properties of copper was used to calculate the temperature distribution as

                                                  a function of depth penetration

                                                  The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                  copper material was obtained from the experimental data tabulated in literary

                                                  The Matlab program used to obtain the best polynomial equation that agrees

                                                  with the above data

                                                  119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                  minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                  times 10minus17 1198791198795 (314)

                                                  35

                                                  119879119879 ( 119870119870) 119870119870 119869119869

                                                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                  100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                  The previous thermal conductivity data and the best fitting of the data are

                                                  shown in Fig (310)

                                                  Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                  temperature

                                                  36

                                                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                  material was obtained from the following experimental data tacked from

                                                  literatures

                                                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                  100 0254

                                                  200 0357

                                                  273 0384

                                                  298 0387

                                                  400 0397

                                                  600 0416

                                                  800 0435

                                                  1000 0454

                                                  1100 0464

                                                  1200 0474

                                                  1300 0483

                                                  The best polynomial fitted for these data was

                                                  119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                  + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                  times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                  The specific heat capacity data and the best polynomial fitting of the data are

                                                  shown in Fig (311)

                                                  37

                                                  Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                  temperature

                                                  The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                  literature was used to find the best polynomial fitting

                                                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                  100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                  1000 8576 1100 8519 1200 8458 1300 8396

                                                  38

                                                  The best polynomial of this data was

                                                  120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                  minus 76765 times 10minus14 1198791198794 (316)

                                                  The density of copper as a function of temperature and the best polynomial

                                                  fitting are shown in Fig (312)

                                                  Fig(312) The best fitting of density of copper as a function of temperature

                                                  The depth penetration of laser energy for copper metal was calculated using

                                                  the polynomial equations of thermal conductivity specific heat capacity and

                                                  density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                  (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                  function of time the result was shown in Fig (313)

                                                  39

                                                  The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                  for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                  copper metal so the depth penetration of laser energy of lead metal was smaller

                                                  than that of copper metal this may be due to the high thermal conductivity and

                                                  high specific heat capacity of copper with that of lead metal

                                                  Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                  material

                                                  40

                                                  310 Conclusions

                                                  The Depth dependence of temperature for lead metal was investigated in two

                                                  case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                  thermal properties (thermal conductivity specific heat) and density of metal are

                                                  also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                  vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                  specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                  120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                  penetration depth in the first case is smaller than that of the second case about

                                                  (190) times

                                                  The temperature distribution as a function of depth dependence for copper

                                                  metal was also investigated in the case when the laser intensity vary with time

                                                  119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                  density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                  The depth penetration of laser energy of lead metal was found to be smaller

                                                  than that of copper metal this may be due to the high thermal conductivity and

                                                  high specific heat capacity of copper with that of lead metal

                                                  41

                                                  References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                  Sons Inc Hoboken New Jersey Canada (2003)

                                                  [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                  [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                  [9]

                                                  Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                  [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                  [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                  [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                  [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                  [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                  httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                  [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                  [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                  [12] httpenwikipediaorgwikiThermal_conductivity

                                                  [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                  [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                  [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                  42

                                                  Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                  This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                  This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                  43

                                                  E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                  This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                  This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                  44

                                                  title(Dencity as a function of temperature) hold off

                                                  This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                  45

                                                  elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                  46

                                                  r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                  47

                                                  for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                  48

                                                  for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                  49

                                                  6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                  50

                                                  u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                  51

                                                  alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                  52

                                                  715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                  • 01 Title
                                                    • 0TDr Omed Ghareb Abdullah
                                                      • Ch1-Laser
                                                        • 16 Heat and heat capacity
                                                        • 18 Derivation in one dimension
                                                          • Ch2 laser2
                                                          • Ch3 laser2
                                                          • Appendix-Laser

                                                    21

                                                    equality Rather the expression means that the left hand side is a product of an

                                                    unknown constant and ∆1205971205972 Although the expression does not give us the exact

                                                    magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

                                                    approaches zero as ∆120597120597 is reduced

                                                    Using big 119978119978 notation Equation (28) can be written

                                                    120597120597empty120597120597120597120597120597120597119894119894

                                                    =empty119894119894+1 minus empty119894119894

                                                    ∆120597120597+ 119978119978(∆120597120597) (213)

                                                    This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

                                                    it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

                                                    truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

                                                    our control because we can choose the mesh size ∆120597120597 The part of the truncation

                                                    error that is not under our control is |120597120597empty120597120597120597120597|120585120585

                                                    242 First Order Backward Difference

                                                    An alternative first order finite difference formula is obtained if the Taylor series

                                                    like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

                                                    variables in place of all the unknowns one obtains

                                                    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                    +∆1205971205972

                                                    2 1205971205972empty1205971205971205971205972

                                                    120597120597119894119894

                                                    minus∆1205971205973

                                                    3 1205971205973empty1205971205971205971205973

                                                    120597120597119894119894

                                                    + ⋯ (214)

                                                    Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

                                                    to get

                                                    120597120597empty120597120597120597120597120597120597119894119894

                                                    =empty119894119894+1 minus empty119894119894

                                                    ∆120597120597minus∆1205971205972

                                                    1205971205972empty1205971205971205971205972

                                                    120597120597119894119894

                                                    minus∆1205971205972

                                                    3 1205971205973empty1205971205971205971205973

                                                    120597120597119894119894

                                                    minus ⋯ (215)

                                                    Or using big 119978119978 notation

                                                    120597120597empty120597120597120597120597120597120597119894119894

                                                    =empty119894119894 minus empty119894119894minus1

                                                    ∆120597120597+ 119978119978(∆120597120597) (216)

                                                    22

                                                    This is called the backward difference formula because it involves the values of

                                                    empty at 120597120597119894119894 and 120597120597119894119894minus1

                                                    The order of magnitude of the truncation error for the backward difference

                                                    approximation is the same as that of the forward difference approximation Can

                                                    we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                                    truncation error The answer is yes

                                                    242 First Order Central Difference

                                                    Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                                    empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                    +∆1205971205972

                                                    2 1205971205972empty1205971205971205971205972

                                                    120597120597119894119894

                                                    +∆1205971205973

                                                    3 1205971205973empty1205971205971205971205973

                                                    120597120597119894119894

                                                    + ⋯ (217)

                                                    empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                    +∆1205971205972

                                                    2 1205971205972empty1205971205971205971205972

                                                    120597120597119894119894

                                                    minus∆1205971205973

                                                    3 1205971205973empty1205971205971205971205973

                                                    120597120597119894119894

                                                    + ⋯ (218)

                                                    Subtracting Equation (10) from Equation (9) yields

                                                    empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                    + 2∆1205971205973

                                                    3 1205971205973empty1205971205971205971205973

                                                    120597120597119894119894

                                                    + ⋯ (219)

                                                    Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                                    120597120597empty120597120597120597120597120597120597119894119894

                                                    =empty119894119894+1 minus empty119894119894minus1

                                                    2∆120597120597minus∆1205971205972

                                                    3 1205971205973empty1205971205971205971205973

                                                    120597120597119894119894

                                                    minus ⋯ (220)

                                                    or

                                                    120597120597empty120597120597120597120597120597120597119894119894

                                                    =empty119894119894+1 minus empty119894119894minus1

                                                    2∆120597120597+ 119978119978(∆1205971205972) (221)

                                                    This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                                    approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                                    the truncation error for the central difference approximation goes to zero much

                                                    faster than the truncation error in forward and backward equations

                                                    23

                                                    25 Procedures

                                                    The simple case in this investigation was assuming the constant thermal

                                                    properties of the material First we assumed all the thermal properties of the

                                                    materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                                    point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                                    assume the constant energy after that the pulse of special shapes was selected

                                                    The numerical solution of equation (23) with boundary and initial conditions

                                                    in equation (22) was investigated using Matlab program as shown in Appendix

                                                    The equation of thermal conductivity and specific heat capacity of metal as a

                                                    function of temperature was obtained by best fitting of polynomials using

                                                    tabulated data in references

                                                    24

                                                    Chapter Three

                                                    Results and Discursion

                                                    31 Introduction

                                                    The development of laser has been an exciting chapter in the history of

                                                    science and engineering It has produced a new type of advice with potential for

                                                    application in an extremely wide variety of fields Mach basic development in

                                                    lasers were occurred during last 35 years The lasers interaction with metal and

                                                    vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                                    The status of laser development and application were still rather rudimentary

                                                    The light emitted by laser is electro magnetic radiation this radiation has a wave

                                                    nature the waves consists of vibrating electric and magnetic fields many studies

                                                    have tried to find and solve models of laser interactions Some researchers

                                                    proposed the mathematical model related to the laser - plasma interaction and

                                                    the others have developed an analytical model to study the temperature

                                                    distribution in Infrared optical materials heated by laser pulses Also an attempt

                                                    have made to study the interaction of nanosecond pulsed lasers with material

                                                    from point of view using experimental technique and theoretical approach of

                                                    dimensional analysis

                                                    In this study we have evaluate the solution of partial difference equation

                                                    (PDE) that represent the laser interaction with solid situation in one dimension

                                                    assuming that the power density of laser and thermal properties are functions

                                                    with time and temperature respectively

                                                    25

                                                    32 Numerical solution with constant laser power density and constant

                                                    thermal properties

                                                    First we have taken the lead metal (Pb) with thermal properties

                                                    119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                                    119862119862 = 014016119869119869119892119892119870119870

                                                    120588120588 = 10751 1198921198921198981198981198981198983

                                                    119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                                    119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                                    and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                                    represent the area under laser influence

                                                    The numerical solution of equation (23) with boundary and initial conditions

                                                    in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                                    of lead metal by explicit method using Matlab program give us the results as

                                                    shown in Fig (31)

                                                    Fig(31) Depth dependence of the temperature with the laser power density

                                                    1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                                    26

                                                    33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                    From following data that represent the energy (119869119869) with time (millie second)

                                                    Time 0 001 01 02 03 04 05 06 07 08

                                                    Energy 0 002 017 022 024 02 012 007 002 0

                                                    By using Matlab program the best polynomial with deduced from above data

                                                    was

                                                    119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                    minus 10069 1198981198985 (31)

                                                    As shown in Fig (32)

                                                    Fig(32) Laser energy as a function of time

                                                    Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                    normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                    maximum value (119864119864119898119898119907119907119898119898 )

                                                    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                    (119864119864119898119898119907119907119898119898 ) (32)

                                                    The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                    27

                                                    Fig(33) Normalized laser energy as a function of time

                                                    The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                    equal to 3 (total laser energy) ie

                                                    119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                    08

                                                    00

                                                    119899119899119898119898 = 3 (33)

                                                    Therefore there exist a real number 119875119875 such that

                                                    119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                    08

                                                    00

                                                    119899119899119898119898 = 3 (34)

                                                    that implies 119875119875 = 68241 and

                                                    119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                    08

                                                    00

                                                    119899119899119898119898 = 3 (35)

                                                    The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                    08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                    119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                    00

                                                    (36)

                                                    28

                                                    Where 119863119863119898119898 put to balance the units of equation (36)

                                                    But integral

                                                    119868119868 = 119864119864119860119860

                                                    (37)

                                                    and from equations (35) (36) and (37) we have

                                                    119868119868 (119898119898)11989911989911989811989808

                                                    00

                                                    = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                    0800 119899119899119898119898

                                                    119860119860 119863119863119898119898 (38)

                                                    Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                    (38)

                                                    There fore

                                                    119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                    119860119860 119863119863119898119898 (39)

                                                    As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                    that agrees with result data

                                                    119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                    + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                    Fig(34) Time dependence of laser intensity

                                                    29

                                                    34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                    constant thermal properties

                                                    With all constant thermal properties of lead metal as in article (23) and

                                                    119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                    equation (23) with boundary and initial condition as in equation (22) and the

                                                    depth penetration is shown in Fig(35)

                                                    Fig(35) Depth dependence of the temperature when laser intensity function

                                                    of time and constant thermal properties of Lead

                                                    35 Evaluation the Thermal Conductivity as functions of temperature

                                                    The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                    temperature for Lead material was obtained by Matlab program using the

                                                    experimental data tabulated in researches

                                                    119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                    times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                    30

                                                    119879119879 ( 119870119870) 119870119870 119869119869

                                                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                    300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                    The previous thermal conductivity data and the best fitting of the data are

                                                    shown in Fig (36)

                                                    Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                    temperature

                                                    31

                                                    36 Evaluation the Specific heat as functions of temperature

                                                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                    material was obtained from the following experimental data tacked from

                                                    literatures

                                                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                    300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                    The best polynomial fitted for these data was

                                                    119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                    + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                    times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                    The specific heat capacity data and the best polynomial fitting of the data are

                                                    shown in Fig (37)

                                                    32

                                                    Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                    temperature

                                                    37 Evaluation the Density as functions of temperature

                                                    The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                    was used to find the best polynomial fitting

                                                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                    300 11330 400 11230 500 11130 600 11010 800 10430

                                                    1000 10190 1200 9940

                                                    The best polynomial of this data was

                                                    120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                    minus 45158 times 10minus12 1198791198794 (313)

                                                    33

                                                    The density of Lead as a function of temperature and the best polynomial fitting

                                                    are shown in Fig (38)

                                                    Fig(38) The best fitting of density of Lead as a function of temperature

                                                    38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                    variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                    We have deduced the solution of equation (24) with initial and boundary

                                                    condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                    and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                    respectively then by using Matlab program the depth penetration is shown in

                                                    Fig (39)

                                                    34

                                                    Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                    material

                                                    39 Laser interaction with copper material

                                                    The same time dependence of laser intensity as shown in Fig(34) with

                                                    thermal properties of copper was used to calculate the temperature distribution as

                                                    a function of depth penetration

                                                    The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                    copper material was obtained from the experimental data tabulated in literary

                                                    The Matlab program used to obtain the best polynomial equation that agrees

                                                    with the above data

                                                    119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                    minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                    times 10minus17 1198791198795 (314)

                                                    35

                                                    119879119879 ( 119870119870) 119870119870 119869119869

                                                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                    100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                    The previous thermal conductivity data and the best fitting of the data are

                                                    shown in Fig (310)

                                                    Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                    temperature

                                                    36

                                                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                    material was obtained from the following experimental data tacked from

                                                    literatures

                                                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                    100 0254

                                                    200 0357

                                                    273 0384

                                                    298 0387

                                                    400 0397

                                                    600 0416

                                                    800 0435

                                                    1000 0454

                                                    1100 0464

                                                    1200 0474

                                                    1300 0483

                                                    The best polynomial fitted for these data was

                                                    119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                    + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                    times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                    The specific heat capacity data and the best polynomial fitting of the data are

                                                    shown in Fig (311)

                                                    37

                                                    Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                    temperature

                                                    The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                    literature was used to find the best polynomial fitting

                                                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                    100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                    1000 8576 1100 8519 1200 8458 1300 8396

                                                    38

                                                    The best polynomial of this data was

                                                    120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                    minus 76765 times 10minus14 1198791198794 (316)

                                                    The density of copper as a function of temperature and the best polynomial

                                                    fitting are shown in Fig (312)

                                                    Fig(312) The best fitting of density of copper as a function of temperature

                                                    The depth penetration of laser energy for copper metal was calculated using

                                                    the polynomial equations of thermal conductivity specific heat capacity and

                                                    density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                    (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                    function of time the result was shown in Fig (313)

                                                    39

                                                    The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                    for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                    copper metal so the depth penetration of laser energy of lead metal was smaller

                                                    than that of copper metal this may be due to the high thermal conductivity and

                                                    high specific heat capacity of copper with that of lead metal

                                                    Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                    material

                                                    40

                                                    310 Conclusions

                                                    The Depth dependence of temperature for lead metal was investigated in two

                                                    case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                    thermal properties (thermal conductivity specific heat) and density of metal are

                                                    also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                    vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                    specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                    120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                    penetration depth in the first case is smaller than that of the second case about

                                                    (190) times

                                                    The temperature distribution as a function of depth dependence for copper

                                                    metal was also investigated in the case when the laser intensity vary with time

                                                    119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                    density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                    The depth penetration of laser energy of lead metal was found to be smaller

                                                    than that of copper metal this may be due to the high thermal conductivity and

                                                    high specific heat capacity of copper with that of lead metal

                                                    41

                                                    References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                    Sons Inc Hoboken New Jersey Canada (2003)

                                                    [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                    [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                    [9]

                                                    Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                    [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                    [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                    [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                    [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                    [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                    httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                    [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                    [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                    [12] httpenwikipediaorgwikiThermal_conductivity

                                                    [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                    [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                    [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                    42

                                                    Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                    This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                    This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                    43

                                                    E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                    This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                    This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                    44

                                                    title(Dencity as a function of temperature) hold off

                                                    This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                    45

                                                    elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                    46

                                                    r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                    47

                                                    for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                    48

                                                    for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                    49

                                                    6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                    50

                                                    u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                    51

                                                    alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                    52

                                                    715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                    • 01 Title
                                                      • 0TDr Omed Ghareb Abdullah
                                                        • Ch1-Laser
                                                          • 16 Heat and heat capacity
                                                          • 18 Derivation in one dimension
                                                            • Ch2 laser2
                                                            • Ch3 laser2
                                                            • Appendix-Laser

                                                      22

                                                      This is called the backward difference formula because it involves the values of

                                                      empty at 120597120597119894119894 and 120597120597119894119894minus1

                                                      The order of magnitude of the truncation error for the backward difference

                                                      approximation is the same as that of the forward difference approximation Can

                                                      we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

                                                      truncation error The answer is yes

                                                      242 First Order Central Difference

                                                      Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

                                                      empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                      +∆1205971205972

                                                      2 1205971205972empty1205971205971205971205972

                                                      120597120597119894119894

                                                      +∆1205971205973

                                                      3 1205971205973empty1205971205971205971205973

                                                      120597120597119894119894

                                                      + ⋯ (217)

                                                      empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                      +∆1205971205972

                                                      2 1205971205972empty1205971205971205971205972

                                                      120597120597119894119894

                                                      minus∆1205971205973

                                                      3 1205971205973empty1205971205971205971205973

                                                      120597120597119894119894

                                                      + ⋯ (218)

                                                      Subtracting Equation (10) from Equation (9) yields

                                                      empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

                                                      + 2∆1205971205973

                                                      3 1205971205973empty1205971205971205971205973

                                                      120597120597119894119894

                                                      + ⋯ (219)

                                                      Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

                                                      120597120597empty120597120597120597120597120597120597119894119894

                                                      =empty119894119894+1 minus empty119894119894minus1

                                                      2∆120597120597minus∆1205971205972

                                                      3 1205971205973empty1205971205971205971205973

                                                      120597120597119894119894

                                                      minus ⋯ (220)

                                                      or

                                                      120597120597empty120597120597120597120597120597120597119894119894

                                                      =empty119894119894+1 minus empty119894119894minus1

                                                      2∆120597120597+ 119978119978(∆1205971205972) (221)

                                                      This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

                                                      approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

                                                      the truncation error for the central difference approximation goes to zero much

                                                      faster than the truncation error in forward and backward equations

                                                      23

                                                      25 Procedures

                                                      The simple case in this investigation was assuming the constant thermal

                                                      properties of the material First we assumed all the thermal properties of the

                                                      materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                                      point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                                      assume the constant energy after that the pulse of special shapes was selected

                                                      The numerical solution of equation (23) with boundary and initial conditions

                                                      in equation (22) was investigated using Matlab program as shown in Appendix

                                                      The equation of thermal conductivity and specific heat capacity of metal as a

                                                      function of temperature was obtained by best fitting of polynomials using

                                                      tabulated data in references

                                                      24

                                                      Chapter Three

                                                      Results and Discursion

                                                      31 Introduction

                                                      The development of laser has been an exciting chapter in the history of

                                                      science and engineering It has produced a new type of advice with potential for

                                                      application in an extremely wide variety of fields Mach basic development in

                                                      lasers were occurred during last 35 years The lasers interaction with metal and

                                                      vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                                      The status of laser development and application were still rather rudimentary

                                                      The light emitted by laser is electro magnetic radiation this radiation has a wave

                                                      nature the waves consists of vibrating electric and magnetic fields many studies

                                                      have tried to find and solve models of laser interactions Some researchers

                                                      proposed the mathematical model related to the laser - plasma interaction and

                                                      the others have developed an analytical model to study the temperature

                                                      distribution in Infrared optical materials heated by laser pulses Also an attempt

                                                      have made to study the interaction of nanosecond pulsed lasers with material

                                                      from point of view using experimental technique and theoretical approach of

                                                      dimensional analysis

                                                      In this study we have evaluate the solution of partial difference equation

                                                      (PDE) that represent the laser interaction with solid situation in one dimension

                                                      assuming that the power density of laser and thermal properties are functions

                                                      with time and temperature respectively

                                                      25

                                                      32 Numerical solution with constant laser power density and constant

                                                      thermal properties

                                                      First we have taken the lead metal (Pb) with thermal properties

                                                      119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                                      119862119862 = 014016119869119869119892119892119870119870

                                                      120588120588 = 10751 1198921198921198981198981198981198983

                                                      119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                                      119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                                      and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                                      represent the area under laser influence

                                                      The numerical solution of equation (23) with boundary and initial conditions

                                                      in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                                      of lead metal by explicit method using Matlab program give us the results as

                                                      shown in Fig (31)

                                                      Fig(31) Depth dependence of the temperature with the laser power density

                                                      1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                                      26

                                                      33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                      From following data that represent the energy (119869119869) with time (millie second)

                                                      Time 0 001 01 02 03 04 05 06 07 08

                                                      Energy 0 002 017 022 024 02 012 007 002 0

                                                      By using Matlab program the best polynomial with deduced from above data

                                                      was

                                                      119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                      minus 10069 1198981198985 (31)

                                                      As shown in Fig (32)

                                                      Fig(32) Laser energy as a function of time

                                                      Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                      normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                      maximum value (119864119864119898119898119907119907119898119898 )

                                                      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                      (119864119864119898119898119907119907119898119898 ) (32)

                                                      The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                      27

                                                      Fig(33) Normalized laser energy as a function of time

                                                      The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                      equal to 3 (total laser energy) ie

                                                      119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                      08

                                                      00

                                                      119899119899119898119898 = 3 (33)

                                                      Therefore there exist a real number 119875119875 such that

                                                      119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                      08

                                                      00

                                                      119899119899119898119898 = 3 (34)

                                                      that implies 119875119875 = 68241 and

                                                      119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                      08

                                                      00

                                                      119899119899119898119898 = 3 (35)

                                                      The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                      08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                      119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                      00

                                                      (36)

                                                      28

                                                      Where 119863119863119898119898 put to balance the units of equation (36)

                                                      But integral

                                                      119868119868 = 119864119864119860119860

                                                      (37)

                                                      and from equations (35) (36) and (37) we have

                                                      119868119868 (119898119898)11989911989911989811989808

                                                      00

                                                      = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                      0800 119899119899119898119898

                                                      119860119860 119863119863119898119898 (38)

                                                      Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                      (38)

                                                      There fore

                                                      119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                      119860119860 119863119863119898119898 (39)

                                                      As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                      that agrees with result data

                                                      119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                      + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                      Fig(34) Time dependence of laser intensity

                                                      29

                                                      34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                      constant thermal properties

                                                      With all constant thermal properties of lead metal as in article (23) and

                                                      119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                      equation (23) with boundary and initial condition as in equation (22) and the

                                                      depth penetration is shown in Fig(35)

                                                      Fig(35) Depth dependence of the temperature when laser intensity function

                                                      of time and constant thermal properties of Lead

                                                      35 Evaluation the Thermal Conductivity as functions of temperature

                                                      The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                      temperature for Lead material was obtained by Matlab program using the

                                                      experimental data tabulated in researches

                                                      119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                      times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                      30

                                                      119879119879 ( 119870119870) 119870119870 119869119869

                                                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                      300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                      The previous thermal conductivity data and the best fitting of the data are

                                                      shown in Fig (36)

                                                      Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                      temperature

                                                      31

                                                      36 Evaluation the Specific heat as functions of temperature

                                                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                      material was obtained from the following experimental data tacked from

                                                      literatures

                                                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                      300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                      The best polynomial fitted for these data was

                                                      119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                      + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                      times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                      The specific heat capacity data and the best polynomial fitting of the data are

                                                      shown in Fig (37)

                                                      32

                                                      Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                      temperature

                                                      37 Evaluation the Density as functions of temperature

                                                      The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                      was used to find the best polynomial fitting

                                                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                      300 11330 400 11230 500 11130 600 11010 800 10430

                                                      1000 10190 1200 9940

                                                      The best polynomial of this data was

                                                      120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                      minus 45158 times 10minus12 1198791198794 (313)

                                                      33

                                                      The density of Lead as a function of temperature and the best polynomial fitting

                                                      are shown in Fig (38)

                                                      Fig(38) The best fitting of density of Lead as a function of temperature

                                                      38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                      variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                      We have deduced the solution of equation (24) with initial and boundary

                                                      condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                      and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                      respectively then by using Matlab program the depth penetration is shown in

                                                      Fig (39)

                                                      34

                                                      Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                      material

                                                      39 Laser interaction with copper material

                                                      The same time dependence of laser intensity as shown in Fig(34) with

                                                      thermal properties of copper was used to calculate the temperature distribution as

                                                      a function of depth penetration

                                                      The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                      copper material was obtained from the experimental data tabulated in literary

                                                      The Matlab program used to obtain the best polynomial equation that agrees

                                                      with the above data

                                                      119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                      minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                      times 10minus17 1198791198795 (314)

                                                      35

                                                      119879119879 ( 119870119870) 119870119870 119869119869

                                                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                      100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                      The previous thermal conductivity data and the best fitting of the data are

                                                      shown in Fig (310)

                                                      Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                      temperature

                                                      36

                                                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                      material was obtained from the following experimental data tacked from

                                                      literatures

                                                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                      100 0254

                                                      200 0357

                                                      273 0384

                                                      298 0387

                                                      400 0397

                                                      600 0416

                                                      800 0435

                                                      1000 0454

                                                      1100 0464

                                                      1200 0474

                                                      1300 0483

                                                      The best polynomial fitted for these data was

                                                      119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                      + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                      times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                      The specific heat capacity data and the best polynomial fitting of the data are

                                                      shown in Fig (311)

                                                      37

                                                      Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                      temperature

                                                      The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                      literature was used to find the best polynomial fitting

                                                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                      100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                      1000 8576 1100 8519 1200 8458 1300 8396

                                                      38

                                                      The best polynomial of this data was

                                                      120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                      minus 76765 times 10minus14 1198791198794 (316)

                                                      The density of copper as a function of temperature and the best polynomial

                                                      fitting are shown in Fig (312)

                                                      Fig(312) The best fitting of density of copper as a function of temperature

                                                      The depth penetration of laser energy for copper metal was calculated using

                                                      the polynomial equations of thermal conductivity specific heat capacity and

                                                      density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                      (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                      function of time the result was shown in Fig (313)

                                                      39

                                                      The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                      for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                      copper metal so the depth penetration of laser energy of lead metal was smaller

                                                      than that of copper metal this may be due to the high thermal conductivity and

                                                      high specific heat capacity of copper with that of lead metal

                                                      Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                      material

                                                      40

                                                      310 Conclusions

                                                      The Depth dependence of temperature for lead metal was investigated in two

                                                      case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                      thermal properties (thermal conductivity specific heat) and density of metal are

                                                      also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                      vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                      specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                      120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                      penetration depth in the first case is smaller than that of the second case about

                                                      (190) times

                                                      The temperature distribution as a function of depth dependence for copper

                                                      metal was also investigated in the case when the laser intensity vary with time

                                                      119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                      density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                      The depth penetration of laser energy of lead metal was found to be smaller

                                                      than that of copper metal this may be due to the high thermal conductivity and

                                                      high specific heat capacity of copper with that of lead metal

                                                      41

                                                      References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                      Sons Inc Hoboken New Jersey Canada (2003)

                                                      [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                      [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                      [9]

                                                      Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                      [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                      [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                      [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                      [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                      [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                      httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                      [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                      [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                      [12] httpenwikipediaorgwikiThermal_conductivity

                                                      [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                      [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                      [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                      42

                                                      Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                      This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                      This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                      43

                                                      E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                      This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                      This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                      44

                                                      title(Dencity as a function of temperature) hold off

                                                      This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                      45

                                                      elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                      46

                                                      r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                      47

                                                      for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                      48

                                                      for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                      49

                                                      6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                      50

                                                      u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                      51

                                                      alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                      52

                                                      715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                      • 01 Title
                                                        • 0TDr Omed Ghareb Abdullah
                                                          • Ch1-Laser
                                                            • 16 Heat and heat capacity
                                                            • 18 Derivation in one dimension
                                                              • Ch2 laser2
                                                              • Ch3 laser2
                                                              • Appendix-Laser

                                                        23

                                                        25 Procedures

                                                        The simple case in this investigation was assuming the constant thermal

                                                        properties of the material First we assumed all the thermal properties of the

                                                        materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

                                                        point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

                                                        assume the constant energy after that the pulse of special shapes was selected

                                                        The numerical solution of equation (23) with boundary and initial conditions

                                                        in equation (22) was investigated using Matlab program as shown in Appendix

                                                        The equation of thermal conductivity and specific heat capacity of metal as a

                                                        function of temperature was obtained by best fitting of polynomials using

                                                        tabulated data in references

                                                        24

                                                        Chapter Three

                                                        Results and Discursion

                                                        31 Introduction

                                                        The development of laser has been an exciting chapter in the history of

                                                        science and engineering It has produced a new type of advice with potential for

                                                        application in an extremely wide variety of fields Mach basic development in

                                                        lasers were occurred during last 35 years The lasers interaction with metal and

                                                        vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                                        The status of laser development and application were still rather rudimentary

                                                        The light emitted by laser is electro magnetic radiation this radiation has a wave

                                                        nature the waves consists of vibrating electric and magnetic fields many studies

                                                        have tried to find and solve models of laser interactions Some researchers

                                                        proposed the mathematical model related to the laser - plasma interaction and

                                                        the others have developed an analytical model to study the temperature

                                                        distribution in Infrared optical materials heated by laser pulses Also an attempt

                                                        have made to study the interaction of nanosecond pulsed lasers with material

                                                        from point of view using experimental technique and theoretical approach of

                                                        dimensional analysis

                                                        In this study we have evaluate the solution of partial difference equation

                                                        (PDE) that represent the laser interaction with solid situation in one dimension

                                                        assuming that the power density of laser and thermal properties are functions

                                                        with time and temperature respectively

                                                        25

                                                        32 Numerical solution with constant laser power density and constant

                                                        thermal properties

                                                        First we have taken the lead metal (Pb) with thermal properties

                                                        119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                                        119862119862 = 014016119869119869119892119892119870119870

                                                        120588120588 = 10751 1198921198921198981198981198981198983

                                                        119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                                        119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                                        and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                                        represent the area under laser influence

                                                        The numerical solution of equation (23) with boundary and initial conditions

                                                        in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                                        of lead metal by explicit method using Matlab program give us the results as

                                                        shown in Fig (31)

                                                        Fig(31) Depth dependence of the temperature with the laser power density

                                                        1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                                        26

                                                        33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                        From following data that represent the energy (119869119869) with time (millie second)

                                                        Time 0 001 01 02 03 04 05 06 07 08

                                                        Energy 0 002 017 022 024 02 012 007 002 0

                                                        By using Matlab program the best polynomial with deduced from above data

                                                        was

                                                        119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                        minus 10069 1198981198985 (31)

                                                        As shown in Fig (32)

                                                        Fig(32) Laser energy as a function of time

                                                        Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                        normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                        maximum value (119864119864119898119898119907119907119898119898 )

                                                        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                        (119864119864119898119898119907119907119898119898 ) (32)

                                                        The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                        27

                                                        Fig(33) Normalized laser energy as a function of time

                                                        The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                        equal to 3 (total laser energy) ie

                                                        119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                        08

                                                        00

                                                        119899119899119898119898 = 3 (33)

                                                        Therefore there exist a real number 119875119875 such that

                                                        119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                        08

                                                        00

                                                        119899119899119898119898 = 3 (34)

                                                        that implies 119875119875 = 68241 and

                                                        119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                        08

                                                        00

                                                        119899119899119898119898 = 3 (35)

                                                        The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                        08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                        119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                        00

                                                        (36)

                                                        28

                                                        Where 119863119863119898119898 put to balance the units of equation (36)

                                                        But integral

                                                        119868119868 = 119864119864119860119860

                                                        (37)

                                                        and from equations (35) (36) and (37) we have

                                                        119868119868 (119898119898)11989911989911989811989808

                                                        00

                                                        = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                        0800 119899119899119898119898

                                                        119860119860 119863119863119898119898 (38)

                                                        Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                        (38)

                                                        There fore

                                                        119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                        119860119860 119863119863119898119898 (39)

                                                        As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                        that agrees with result data

                                                        119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                        + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                        Fig(34) Time dependence of laser intensity

                                                        29

                                                        34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                        constant thermal properties

                                                        With all constant thermal properties of lead metal as in article (23) and

                                                        119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                        equation (23) with boundary and initial condition as in equation (22) and the

                                                        depth penetration is shown in Fig(35)

                                                        Fig(35) Depth dependence of the temperature when laser intensity function

                                                        of time and constant thermal properties of Lead

                                                        35 Evaluation the Thermal Conductivity as functions of temperature

                                                        The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                        temperature for Lead material was obtained by Matlab program using the

                                                        experimental data tabulated in researches

                                                        119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                        times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                        30

                                                        119879119879 ( 119870119870) 119870119870 119869119869

                                                        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                        300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                        The previous thermal conductivity data and the best fitting of the data are

                                                        shown in Fig (36)

                                                        Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                        temperature

                                                        31

                                                        36 Evaluation the Specific heat as functions of temperature

                                                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                        material was obtained from the following experimental data tacked from

                                                        literatures

                                                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                        300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                        The best polynomial fitted for these data was

                                                        119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                        + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                        times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                        The specific heat capacity data and the best polynomial fitting of the data are

                                                        shown in Fig (37)

                                                        32

                                                        Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                        temperature

                                                        37 Evaluation the Density as functions of temperature

                                                        The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                        was used to find the best polynomial fitting

                                                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                        300 11330 400 11230 500 11130 600 11010 800 10430

                                                        1000 10190 1200 9940

                                                        The best polynomial of this data was

                                                        120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                        minus 45158 times 10minus12 1198791198794 (313)

                                                        33

                                                        The density of Lead as a function of temperature and the best polynomial fitting

                                                        are shown in Fig (38)

                                                        Fig(38) The best fitting of density of Lead as a function of temperature

                                                        38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                        variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                        We have deduced the solution of equation (24) with initial and boundary

                                                        condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                        and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                        respectively then by using Matlab program the depth penetration is shown in

                                                        Fig (39)

                                                        34

                                                        Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                        material

                                                        39 Laser interaction with copper material

                                                        The same time dependence of laser intensity as shown in Fig(34) with

                                                        thermal properties of copper was used to calculate the temperature distribution as

                                                        a function of depth penetration

                                                        The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                        copper material was obtained from the experimental data tabulated in literary

                                                        The Matlab program used to obtain the best polynomial equation that agrees

                                                        with the above data

                                                        119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                        minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                        times 10minus17 1198791198795 (314)

                                                        35

                                                        119879119879 ( 119870119870) 119870119870 119869119869

                                                        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                        100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                        The previous thermal conductivity data and the best fitting of the data are

                                                        shown in Fig (310)

                                                        Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                        temperature

                                                        36

                                                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                        material was obtained from the following experimental data tacked from

                                                        literatures

                                                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                        100 0254

                                                        200 0357

                                                        273 0384

                                                        298 0387

                                                        400 0397

                                                        600 0416

                                                        800 0435

                                                        1000 0454

                                                        1100 0464

                                                        1200 0474

                                                        1300 0483

                                                        The best polynomial fitted for these data was

                                                        119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                        + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                        times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                        The specific heat capacity data and the best polynomial fitting of the data are

                                                        shown in Fig (311)

                                                        37

                                                        Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                        temperature

                                                        The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                        literature was used to find the best polynomial fitting

                                                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                        100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                        1000 8576 1100 8519 1200 8458 1300 8396

                                                        38

                                                        The best polynomial of this data was

                                                        120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                        minus 76765 times 10minus14 1198791198794 (316)

                                                        The density of copper as a function of temperature and the best polynomial

                                                        fitting are shown in Fig (312)

                                                        Fig(312) The best fitting of density of copper as a function of temperature

                                                        The depth penetration of laser energy for copper metal was calculated using

                                                        the polynomial equations of thermal conductivity specific heat capacity and

                                                        density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                        (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                        function of time the result was shown in Fig (313)

                                                        39

                                                        The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                        for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                        copper metal so the depth penetration of laser energy of lead metal was smaller

                                                        than that of copper metal this may be due to the high thermal conductivity and

                                                        high specific heat capacity of copper with that of lead metal

                                                        Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                        material

                                                        40

                                                        310 Conclusions

                                                        The Depth dependence of temperature for lead metal was investigated in two

                                                        case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                        thermal properties (thermal conductivity specific heat) and density of metal are

                                                        also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                        vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                        specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                        120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                        penetration depth in the first case is smaller than that of the second case about

                                                        (190) times

                                                        The temperature distribution as a function of depth dependence for copper

                                                        metal was also investigated in the case when the laser intensity vary with time

                                                        119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                        density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                        The depth penetration of laser energy of lead metal was found to be smaller

                                                        than that of copper metal this may be due to the high thermal conductivity and

                                                        high specific heat capacity of copper with that of lead metal

                                                        41

                                                        References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                        Sons Inc Hoboken New Jersey Canada (2003)

                                                        [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                        [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                        [9]

                                                        Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                        [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                        [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                        [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                        [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                        [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                        httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                        [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                        [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                        [12] httpenwikipediaorgwikiThermal_conductivity

                                                        [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                        [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                        [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                        42

                                                        Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                        This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                        This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                        43

                                                        E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                        This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                        This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                        44

                                                        title(Dencity as a function of temperature) hold off

                                                        This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                        45

                                                        elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                        46

                                                        r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                        47

                                                        for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                        48

                                                        for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                        49

                                                        6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                        50

                                                        u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                        51

                                                        alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                        52

                                                        715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                        • 01 Title
                                                          • 0TDr Omed Ghareb Abdullah
                                                            • Ch1-Laser
                                                              • 16 Heat and heat capacity
                                                              • 18 Derivation in one dimension
                                                                • Ch2 laser2
                                                                • Ch3 laser2
                                                                • Appendix-Laser

                                                          24

                                                          Chapter Three

                                                          Results and Discursion

                                                          31 Introduction

                                                          The development of laser has been an exciting chapter in the history of

                                                          science and engineering It has produced a new type of advice with potential for

                                                          application in an extremely wide variety of fields Mach basic development in

                                                          lasers were occurred during last 35 years The lasers interaction with metal and

                                                          vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

                                                          The status of laser development and application were still rather rudimentary

                                                          The light emitted by laser is electro magnetic radiation this radiation has a wave

                                                          nature the waves consists of vibrating electric and magnetic fields many studies

                                                          have tried to find and solve models of laser interactions Some researchers

                                                          proposed the mathematical model related to the laser - plasma interaction and

                                                          the others have developed an analytical model to study the temperature

                                                          distribution in Infrared optical materials heated by laser pulses Also an attempt

                                                          have made to study the interaction of nanosecond pulsed lasers with material

                                                          from point of view using experimental technique and theoretical approach of

                                                          dimensional analysis

                                                          In this study we have evaluate the solution of partial difference equation

                                                          (PDE) that represent the laser interaction with solid situation in one dimension

                                                          assuming that the power density of laser and thermal properties are functions

                                                          with time and temperature respectively

                                                          25

                                                          32 Numerical solution with constant laser power density and constant

                                                          thermal properties

                                                          First we have taken the lead metal (Pb) with thermal properties

                                                          119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                                          119862119862 = 014016119869119869119892119892119870119870

                                                          120588120588 = 10751 1198921198921198981198981198981198983

                                                          119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                                          119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                                          and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                                          represent the area under laser influence

                                                          The numerical solution of equation (23) with boundary and initial conditions

                                                          in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                                          of lead metal by explicit method using Matlab program give us the results as

                                                          shown in Fig (31)

                                                          Fig(31) Depth dependence of the temperature with the laser power density

                                                          1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                                          26

                                                          33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                          From following data that represent the energy (119869119869) with time (millie second)

                                                          Time 0 001 01 02 03 04 05 06 07 08

                                                          Energy 0 002 017 022 024 02 012 007 002 0

                                                          By using Matlab program the best polynomial with deduced from above data

                                                          was

                                                          119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                          minus 10069 1198981198985 (31)

                                                          As shown in Fig (32)

                                                          Fig(32) Laser energy as a function of time

                                                          Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                          normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                          maximum value (119864119864119898119898119907119907119898119898 )

                                                          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                          (119864119864119898119898119907119907119898119898 ) (32)

                                                          The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                          27

                                                          Fig(33) Normalized laser energy as a function of time

                                                          The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                          equal to 3 (total laser energy) ie

                                                          119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                          08

                                                          00

                                                          119899119899119898119898 = 3 (33)

                                                          Therefore there exist a real number 119875119875 such that

                                                          119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                          08

                                                          00

                                                          119899119899119898119898 = 3 (34)

                                                          that implies 119875119875 = 68241 and

                                                          119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                          08

                                                          00

                                                          119899119899119898119898 = 3 (35)

                                                          The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                          08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                          119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                          00

                                                          (36)

                                                          28

                                                          Where 119863119863119898119898 put to balance the units of equation (36)

                                                          But integral

                                                          119868119868 = 119864119864119860119860

                                                          (37)

                                                          and from equations (35) (36) and (37) we have

                                                          119868119868 (119898119898)11989911989911989811989808

                                                          00

                                                          = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                          0800 119899119899119898119898

                                                          119860119860 119863119863119898119898 (38)

                                                          Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                          (38)

                                                          There fore

                                                          119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                          119860119860 119863119863119898119898 (39)

                                                          As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                          that agrees with result data

                                                          119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                          + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                          Fig(34) Time dependence of laser intensity

                                                          29

                                                          34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                          constant thermal properties

                                                          With all constant thermal properties of lead metal as in article (23) and

                                                          119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                          equation (23) with boundary and initial condition as in equation (22) and the

                                                          depth penetration is shown in Fig(35)

                                                          Fig(35) Depth dependence of the temperature when laser intensity function

                                                          of time and constant thermal properties of Lead

                                                          35 Evaluation the Thermal Conductivity as functions of temperature

                                                          The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                          temperature for Lead material was obtained by Matlab program using the

                                                          experimental data tabulated in researches

                                                          119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                          times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                          30

                                                          119879119879 ( 119870119870) 119870119870 119869119869

                                                          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                          300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                          The previous thermal conductivity data and the best fitting of the data are

                                                          shown in Fig (36)

                                                          Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                          temperature

                                                          31

                                                          36 Evaluation the Specific heat as functions of temperature

                                                          The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                          material was obtained from the following experimental data tacked from

                                                          literatures

                                                          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                          300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                          The best polynomial fitted for these data was

                                                          119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                          + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                          times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                          The specific heat capacity data and the best polynomial fitting of the data are

                                                          shown in Fig (37)

                                                          32

                                                          Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                          temperature

                                                          37 Evaluation the Density as functions of temperature

                                                          The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                          was used to find the best polynomial fitting

                                                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                          300 11330 400 11230 500 11130 600 11010 800 10430

                                                          1000 10190 1200 9940

                                                          The best polynomial of this data was

                                                          120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                          minus 45158 times 10minus12 1198791198794 (313)

                                                          33

                                                          The density of Lead as a function of temperature and the best polynomial fitting

                                                          are shown in Fig (38)

                                                          Fig(38) The best fitting of density of Lead as a function of temperature

                                                          38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                          variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                          We have deduced the solution of equation (24) with initial and boundary

                                                          condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                          and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                          respectively then by using Matlab program the depth penetration is shown in

                                                          Fig (39)

                                                          34

                                                          Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                          material

                                                          39 Laser interaction with copper material

                                                          The same time dependence of laser intensity as shown in Fig(34) with

                                                          thermal properties of copper was used to calculate the temperature distribution as

                                                          a function of depth penetration

                                                          The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                          copper material was obtained from the experimental data tabulated in literary

                                                          The Matlab program used to obtain the best polynomial equation that agrees

                                                          with the above data

                                                          119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                          minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                          times 10minus17 1198791198795 (314)

                                                          35

                                                          119879119879 ( 119870119870) 119870119870 119869119869

                                                          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                          100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                          The previous thermal conductivity data and the best fitting of the data are

                                                          shown in Fig (310)

                                                          Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                          temperature

                                                          36

                                                          The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                          material was obtained from the following experimental data tacked from

                                                          literatures

                                                          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                          100 0254

                                                          200 0357

                                                          273 0384

                                                          298 0387

                                                          400 0397

                                                          600 0416

                                                          800 0435

                                                          1000 0454

                                                          1100 0464

                                                          1200 0474

                                                          1300 0483

                                                          The best polynomial fitted for these data was

                                                          119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                          + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                          times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                          The specific heat capacity data and the best polynomial fitting of the data are

                                                          shown in Fig (311)

                                                          37

                                                          Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                          temperature

                                                          The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                          literature was used to find the best polynomial fitting

                                                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                          100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                          1000 8576 1100 8519 1200 8458 1300 8396

                                                          38

                                                          The best polynomial of this data was

                                                          120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                          minus 76765 times 10minus14 1198791198794 (316)

                                                          The density of copper as a function of temperature and the best polynomial

                                                          fitting are shown in Fig (312)

                                                          Fig(312) The best fitting of density of copper as a function of temperature

                                                          The depth penetration of laser energy for copper metal was calculated using

                                                          the polynomial equations of thermal conductivity specific heat capacity and

                                                          density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                          (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                          function of time the result was shown in Fig (313)

                                                          39

                                                          The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                          for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                          copper metal so the depth penetration of laser energy of lead metal was smaller

                                                          than that of copper metal this may be due to the high thermal conductivity and

                                                          high specific heat capacity of copper with that of lead metal

                                                          Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                          material

                                                          40

                                                          310 Conclusions

                                                          The Depth dependence of temperature for lead metal was investigated in two

                                                          case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                          thermal properties (thermal conductivity specific heat) and density of metal are

                                                          also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                          vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                          specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                          120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                          penetration depth in the first case is smaller than that of the second case about

                                                          (190) times

                                                          The temperature distribution as a function of depth dependence for copper

                                                          metal was also investigated in the case when the laser intensity vary with time

                                                          119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                          density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                          The depth penetration of laser energy of lead metal was found to be smaller

                                                          than that of copper metal this may be due to the high thermal conductivity and

                                                          high specific heat capacity of copper with that of lead metal

                                                          41

                                                          References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                          Sons Inc Hoboken New Jersey Canada (2003)

                                                          [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                          [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                          [9]

                                                          Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                          [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                          [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                          [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                          [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                          [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                          httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                          [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                          [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                          [12] httpenwikipediaorgwikiThermal_conductivity

                                                          [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                          [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                          [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                          42

                                                          Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                          This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                          This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                          43

                                                          E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                          This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                          This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                          44

                                                          title(Dencity as a function of temperature) hold off

                                                          This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                          45

                                                          elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                          46

                                                          r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                          47

                                                          for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                          48

                                                          for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                          49

                                                          6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                          50

                                                          u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                          51

                                                          alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                          52

                                                          715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                          • 01 Title
                                                            • 0TDr Omed Ghareb Abdullah
                                                              • Ch1-Laser
                                                                • 16 Heat and heat capacity
                                                                • 18 Derivation in one dimension
                                                                  • Ch2 laser2
                                                                  • Ch3 laser2
                                                                  • Appendix-Laser

                                                            25

                                                            32 Numerical solution with constant laser power density and constant

                                                            thermal properties

                                                            First we have taken the lead metal (Pb) with thermal properties

                                                            119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

                                                            119862119862 = 014016119869119869119892119892119870119870

                                                            120588120588 = 10751 1198921198921198981198981198981198983

                                                            119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

                                                            119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

                                                            and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

                                                            represent the area under laser influence

                                                            The numerical solution of equation (23) with boundary and initial conditions

                                                            in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

                                                            of lead metal by explicit method using Matlab program give us the results as

                                                            shown in Fig (31)

                                                            Fig(31) Depth dependence of the temperature with the laser power density

                                                            1198681198680 = 76 times 106 119882119882119898119898119898119898 2

                                                            26

                                                            33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                            From following data that represent the energy (119869119869) with time (millie second)

                                                            Time 0 001 01 02 03 04 05 06 07 08

                                                            Energy 0 002 017 022 024 02 012 007 002 0

                                                            By using Matlab program the best polynomial with deduced from above data

                                                            was

                                                            119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                            minus 10069 1198981198985 (31)

                                                            As shown in Fig (32)

                                                            Fig(32) Laser energy as a function of time

                                                            Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                            normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                            maximum value (119864119864119898119898119907119907119898119898 )

                                                            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                            (119864119864119898119898119907119907119898119898 ) (32)

                                                            The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                            27

                                                            Fig(33) Normalized laser energy as a function of time

                                                            The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                            equal to 3 (total laser energy) ie

                                                            119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                            08

                                                            00

                                                            119899119899119898119898 = 3 (33)

                                                            Therefore there exist a real number 119875119875 such that

                                                            119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                            08

                                                            00

                                                            119899119899119898119898 = 3 (34)

                                                            that implies 119875119875 = 68241 and

                                                            119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                            08

                                                            00

                                                            119899119899119898119898 = 3 (35)

                                                            The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                            08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                            119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                            00

                                                            (36)

                                                            28

                                                            Where 119863119863119898119898 put to balance the units of equation (36)

                                                            But integral

                                                            119868119868 = 119864119864119860119860

                                                            (37)

                                                            and from equations (35) (36) and (37) we have

                                                            119868119868 (119898119898)11989911989911989811989808

                                                            00

                                                            = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                            0800 119899119899119898119898

                                                            119860119860 119863119863119898119898 (38)

                                                            Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                            (38)

                                                            There fore

                                                            119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                            119860119860 119863119863119898119898 (39)

                                                            As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                            that agrees with result data

                                                            119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                            + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                            Fig(34) Time dependence of laser intensity

                                                            29

                                                            34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                            constant thermal properties

                                                            With all constant thermal properties of lead metal as in article (23) and

                                                            119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                            equation (23) with boundary and initial condition as in equation (22) and the

                                                            depth penetration is shown in Fig(35)

                                                            Fig(35) Depth dependence of the temperature when laser intensity function

                                                            of time and constant thermal properties of Lead

                                                            35 Evaluation the Thermal Conductivity as functions of temperature

                                                            The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                            temperature for Lead material was obtained by Matlab program using the

                                                            experimental data tabulated in researches

                                                            119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                            times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                            30

                                                            119879119879 ( 119870119870) 119870119870 119869119869

                                                            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                            300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                            The previous thermal conductivity data and the best fitting of the data are

                                                            shown in Fig (36)

                                                            Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                            temperature

                                                            31

                                                            36 Evaluation the Specific heat as functions of temperature

                                                            The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                            material was obtained from the following experimental data tacked from

                                                            literatures

                                                            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                            300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                            The best polynomial fitted for these data was

                                                            119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                            + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                            times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                            The specific heat capacity data and the best polynomial fitting of the data are

                                                            shown in Fig (37)

                                                            32

                                                            Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                            temperature

                                                            37 Evaluation the Density as functions of temperature

                                                            The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                            was used to find the best polynomial fitting

                                                            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                            300 11330 400 11230 500 11130 600 11010 800 10430

                                                            1000 10190 1200 9940

                                                            The best polynomial of this data was

                                                            120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                            minus 45158 times 10minus12 1198791198794 (313)

                                                            33

                                                            The density of Lead as a function of temperature and the best polynomial fitting

                                                            are shown in Fig (38)

                                                            Fig(38) The best fitting of density of Lead as a function of temperature

                                                            38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                            variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                            We have deduced the solution of equation (24) with initial and boundary

                                                            condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                            and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                            respectively then by using Matlab program the depth penetration is shown in

                                                            Fig (39)

                                                            34

                                                            Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                            material

                                                            39 Laser interaction with copper material

                                                            The same time dependence of laser intensity as shown in Fig(34) with

                                                            thermal properties of copper was used to calculate the temperature distribution as

                                                            a function of depth penetration

                                                            The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                            copper material was obtained from the experimental data tabulated in literary

                                                            The Matlab program used to obtain the best polynomial equation that agrees

                                                            with the above data

                                                            119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                            minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                            times 10minus17 1198791198795 (314)

                                                            35

                                                            119879119879 ( 119870119870) 119870119870 119869119869

                                                            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                            100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                            The previous thermal conductivity data and the best fitting of the data are

                                                            shown in Fig (310)

                                                            Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                            temperature

                                                            36

                                                            The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                            material was obtained from the following experimental data tacked from

                                                            literatures

                                                            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                            100 0254

                                                            200 0357

                                                            273 0384

                                                            298 0387

                                                            400 0397

                                                            600 0416

                                                            800 0435

                                                            1000 0454

                                                            1100 0464

                                                            1200 0474

                                                            1300 0483

                                                            The best polynomial fitted for these data was

                                                            119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                            + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                            times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                            The specific heat capacity data and the best polynomial fitting of the data are

                                                            shown in Fig (311)

                                                            37

                                                            Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                            temperature

                                                            The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                            literature was used to find the best polynomial fitting

                                                            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                            100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                            1000 8576 1100 8519 1200 8458 1300 8396

                                                            38

                                                            The best polynomial of this data was

                                                            120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                            minus 76765 times 10minus14 1198791198794 (316)

                                                            The density of copper as a function of temperature and the best polynomial

                                                            fitting are shown in Fig (312)

                                                            Fig(312) The best fitting of density of copper as a function of temperature

                                                            The depth penetration of laser energy for copper metal was calculated using

                                                            the polynomial equations of thermal conductivity specific heat capacity and

                                                            density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                            (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                            function of time the result was shown in Fig (313)

                                                            39

                                                            The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                            for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                            copper metal so the depth penetration of laser energy of lead metal was smaller

                                                            than that of copper metal this may be due to the high thermal conductivity and

                                                            high specific heat capacity of copper with that of lead metal

                                                            Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                            material

                                                            40

                                                            310 Conclusions

                                                            The Depth dependence of temperature for lead metal was investigated in two

                                                            case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                            thermal properties (thermal conductivity specific heat) and density of metal are

                                                            also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                            vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                            specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                            120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                            penetration depth in the first case is smaller than that of the second case about

                                                            (190) times

                                                            The temperature distribution as a function of depth dependence for copper

                                                            metal was also investigated in the case when the laser intensity vary with time

                                                            119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                            density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                            The depth penetration of laser energy of lead metal was found to be smaller

                                                            than that of copper metal this may be due to the high thermal conductivity and

                                                            high specific heat capacity of copper with that of lead metal

                                                            41

                                                            References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                            Sons Inc Hoboken New Jersey Canada (2003)

                                                            [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                            [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                            [9]

                                                            Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                            [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                            [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                            [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                            [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                            [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                            httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                            [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                            [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                            [12] httpenwikipediaorgwikiThermal_conductivity

                                                            [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                            [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                            [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                            42

                                                            Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                            This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                            This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                            43

                                                            E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                            This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                            This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                            44

                                                            title(Dencity as a function of temperature) hold off

                                                            This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                            45

                                                            elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                            46

                                                            r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                            47

                                                            for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                            48

                                                            for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                            49

                                                            6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                            50

                                                            u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                            51

                                                            alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                            52

                                                            715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                            • 01 Title
                                                              • 0TDr Omed Ghareb Abdullah
                                                                • Ch1-Laser
                                                                  • 16 Heat and heat capacity
                                                                  • 18 Derivation in one dimension
                                                                    • Ch2 laser2
                                                                    • Ch3 laser2
                                                                    • Appendix-Laser

                                                              26

                                                              33 Evaluation of function 119920119920(119957119957) of laser flux density

                                                              From following data that represent the energy (119869119869) with time (millie second)

                                                              Time 0 001 01 02 03 04 05 06 07 08

                                                              Energy 0 002 017 022 024 02 012 007 002 0

                                                              By using Matlab program the best polynomial with deduced from above data

                                                              was

                                                              119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

                                                              minus 10069 1198981198985 (31)

                                                              As shown in Fig (32)

                                                              Fig(32) Laser energy as a function of time

                                                              Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

                                                              normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

                                                              maximum value (119864119864119898119898119907119907119898119898 )

                                                              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

                                                              (119864119864119898119898119907119907119898119898 ) (32)

                                                              The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

                                                              27

                                                              Fig(33) Normalized laser energy as a function of time

                                                              The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                              equal to 3 (total laser energy) ie

                                                              119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                              08

                                                              00

                                                              119899119899119898119898 = 3 (33)

                                                              Therefore there exist a real number 119875119875 such that

                                                              119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                              08

                                                              00

                                                              119899119899119898119898 = 3 (34)

                                                              that implies 119875119875 = 68241 and

                                                              119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                              08

                                                              00

                                                              119899119899119898119898 = 3 (35)

                                                              The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                              08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                              119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                              00

                                                              (36)

                                                              28

                                                              Where 119863119863119898119898 put to balance the units of equation (36)

                                                              But integral

                                                              119868119868 = 119864119864119860119860

                                                              (37)

                                                              and from equations (35) (36) and (37) we have

                                                              119868119868 (119898119898)11989911989911989811989808

                                                              00

                                                              = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                              0800 119899119899119898119898

                                                              119860119860 119863119863119898119898 (38)

                                                              Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                              (38)

                                                              There fore

                                                              119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                              119860119860 119863119863119898119898 (39)

                                                              As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                              that agrees with result data

                                                              119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                              + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                              Fig(34) Time dependence of laser intensity

                                                              29

                                                              34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                              constant thermal properties

                                                              With all constant thermal properties of lead metal as in article (23) and

                                                              119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                              equation (23) with boundary and initial condition as in equation (22) and the

                                                              depth penetration is shown in Fig(35)

                                                              Fig(35) Depth dependence of the temperature when laser intensity function

                                                              of time and constant thermal properties of Lead

                                                              35 Evaluation the Thermal Conductivity as functions of temperature

                                                              The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                              temperature for Lead material was obtained by Matlab program using the

                                                              experimental data tabulated in researches

                                                              119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                              times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                              30

                                                              119879119879 ( 119870119870) 119870119870 119869119869

                                                              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                              300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                              The previous thermal conductivity data and the best fitting of the data are

                                                              shown in Fig (36)

                                                              Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                              temperature

                                                              31

                                                              36 Evaluation the Specific heat as functions of temperature

                                                              The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                              material was obtained from the following experimental data tacked from

                                                              literatures

                                                              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                              300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                              The best polynomial fitted for these data was

                                                              119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                              + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                              times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                              The specific heat capacity data and the best polynomial fitting of the data are

                                                              shown in Fig (37)

                                                              32

                                                              Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                              temperature

                                                              37 Evaluation the Density as functions of temperature

                                                              The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                              was used to find the best polynomial fitting

                                                              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                              300 11330 400 11230 500 11130 600 11010 800 10430

                                                              1000 10190 1200 9940

                                                              The best polynomial of this data was

                                                              120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                              minus 45158 times 10minus12 1198791198794 (313)

                                                              33

                                                              The density of Lead as a function of temperature and the best polynomial fitting

                                                              are shown in Fig (38)

                                                              Fig(38) The best fitting of density of Lead as a function of temperature

                                                              38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                              variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                              We have deduced the solution of equation (24) with initial and boundary

                                                              condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                              and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                              respectively then by using Matlab program the depth penetration is shown in

                                                              Fig (39)

                                                              34

                                                              Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                              material

                                                              39 Laser interaction with copper material

                                                              The same time dependence of laser intensity as shown in Fig(34) with

                                                              thermal properties of copper was used to calculate the temperature distribution as

                                                              a function of depth penetration

                                                              The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                              copper material was obtained from the experimental data tabulated in literary

                                                              The Matlab program used to obtain the best polynomial equation that agrees

                                                              with the above data

                                                              119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                              minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                              times 10minus17 1198791198795 (314)

                                                              35

                                                              119879119879 ( 119870119870) 119870119870 119869119869

                                                              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                              100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                              The previous thermal conductivity data and the best fitting of the data are

                                                              shown in Fig (310)

                                                              Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                              temperature

                                                              36

                                                              The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                              material was obtained from the following experimental data tacked from

                                                              literatures

                                                              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                              100 0254

                                                              200 0357

                                                              273 0384

                                                              298 0387

                                                              400 0397

                                                              600 0416

                                                              800 0435

                                                              1000 0454

                                                              1100 0464

                                                              1200 0474

                                                              1300 0483

                                                              The best polynomial fitted for these data was

                                                              119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                              + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                              times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                              The specific heat capacity data and the best polynomial fitting of the data are

                                                              shown in Fig (311)

                                                              37

                                                              Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                              temperature

                                                              The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                              literature was used to find the best polynomial fitting

                                                              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                              100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                              1000 8576 1100 8519 1200 8458 1300 8396

                                                              38

                                                              The best polynomial of this data was

                                                              120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                              minus 76765 times 10minus14 1198791198794 (316)

                                                              The density of copper as a function of temperature and the best polynomial

                                                              fitting are shown in Fig (312)

                                                              Fig(312) The best fitting of density of copper as a function of temperature

                                                              The depth penetration of laser energy for copper metal was calculated using

                                                              the polynomial equations of thermal conductivity specific heat capacity and

                                                              density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                              (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                              function of time the result was shown in Fig (313)

                                                              39

                                                              The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                              for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                              copper metal so the depth penetration of laser energy of lead metal was smaller

                                                              than that of copper metal this may be due to the high thermal conductivity and

                                                              high specific heat capacity of copper with that of lead metal

                                                              Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                              material

                                                              40

                                                              310 Conclusions

                                                              The Depth dependence of temperature for lead metal was investigated in two

                                                              case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                              thermal properties (thermal conductivity specific heat) and density of metal are

                                                              also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                              vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                              specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                              120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                              penetration depth in the first case is smaller than that of the second case about

                                                              (190) times

                                                              The temperature distribution as a function of depth dependence for copper

                                                              metal was also investigated in the case when the laser intensity vary with time

                                                              119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                              density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                              The depth penetration of laser energy of lead metal was found to be smaller

                                                              than that of copper metal this may be due to the high thermal conductivity and

                                                              high specific heat capacity of copper with that of lead metal

                                                              41

                                                              References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                              Sons Inc Hoboken New Jersey Canada (2003)

                                                              [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                              [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                              [9]

                                                              Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                              [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                              [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                              [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                              [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                              [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                              httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                              [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                              [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                              [12] httpenwikipediaorgwikiThermal_conductivity

                                                              [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                              [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                              [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                              42

                                                              Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                              This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                              This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                              43

                                                              E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                              This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                              This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                              44

                                                              title(Dencity as a function of temperature) hold off

                                                              This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                              45

                                                              elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                              46

                                                              r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                              47

                                                              for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                              48

                                                              for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                              49

                                                              6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                              50

                                                              u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                              51

                                                              alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                              52

                                                              715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                              • 01 Title
                                                                • 0TDr Omed Ghareb Abdullah
                                                                  • Ch1-Laser
                                                                    • 16 Heat and heat capacity
                                                                    • 18 Derivation in one dimension
                                                                      • Ch2 laser2
                                                                      • Ch3 laser2
                                                                      • Appendix-Laser

                                                                27

                                                                Fig(33) Normalized laser energy as a function of time

                                                                The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

                                                                equal to 3 (total laser energy) ie

                                                                119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                                08

                                                                00

                                                                119899119899119898119898 = 3 (33)

                                                                Therefore there exist a real number 119875119875 such that

                                                                119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                                08

                                                                00

                                                                119899119899119898119898 = 3 (34)

                                                                that implies 119875119875 = 68241 and

                                                                119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                                08

                                                                00

                                                                119899119899119898119898 = 3 (35)

                                                                The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

                                                                08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

                                                                119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

                                                                00

                                                                (36)

                                                                28

                                                                Where 119863119863119898119898 put to balance the units of equation (36)

                                                                But integral

                                                                119868119868 = 119864119864119860119860

                                                                (37)

                                                                and from equations (35) (36) and (37) we have

                                                                119868119868 (119898119898)11989911989911989811989808

                                                                00

                                                                = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                                0800 119899119899119898119898

                                                                119860119860 119863119863119898119898 (38)

                                                                Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                                (38)

                                                                There fore

                                                                119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                                119860119860 119863119863119898119898 (39)

                                                                As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                                that agrees with result data

                                                                119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                                + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                                Fig(34) Time dependence of laser intensity

                                                                29

                                                                34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                constant thermal properties

                                                                With all constant thermal properties of lead metal as in article (23) and

                                                                119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                                equation (23) with boundary and initial condition as in equation (22) and the

                                                                depth penetration is shown in Fig(35)

                                                                Fig(35) Depth dependence of the temperature when laser intensity function

                                                                of time and constant thermal properties of Lead

                                                                35 Evaluation the Thermal Conductivity as functions of temperature

                                                                The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                                temperature for Lead material was obtained by Matlab program using the

                                                                experimental data tabulated in researches

                                                                119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                                times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                                30

                                                                119879119879 ( 119870119870) 119870119870 119869119869

                                                                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                                The previous thermal conductivity data and the best fitting of the data are

                                                                shown in Fig (36)

                                                                Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                                temperature

                                                                31

                                                                36 Evaluation the Specific heat as functions of temperature

                                                                The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                                material was obtained from the following experimental data tacked from

                                                                literatures

                                                                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                                The best polynomial fitted for these data was

                                                                119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                                + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                                times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                                The specific heat capacity data and the best polynomial fitting of the data are

                                                                shown in Fig (37)

                                                                32

                                                                Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                                temperature

                                                                37 Evaluation the Density as functions of temperature

                                                                The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                                was used to find the best polynomial fitting

                                                                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                300 11330 400 11230 500 11130 600 11010 800 10430

                                                                1000 10190 1200 9940

                                                                The best polynomial of this data was

                                                                120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                                minus 45158 times 10minus12 1198791198794 (313)

                                                                33

                                                                The density of Lead as a function of temperature and the best polynomial fitting

                                                                are shown in Fig (38)

                                                                Fig(38) The best fitting of density of Lead as a function of temperature

                                                                38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                                We have deduced the solution of equation (24) with initial and boundary

                                                                condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                                and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                                respectively then by using Matlab program the depth penetration is shown in

                                                                Fig (39)

                                                                34

                                                                Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                material

                                                                39 Laser interaction with copper material

                                                                The same time dependence of laser intensity as shown in Fig(34) with

                                                                thermal properties of copper was used to calculate the temperature distribution as

                                                                a function of depth penetration

                                                                The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                copper material was obtained from the experimental data tabulated in literary

                                                                The Matlab program used to obtain the best polynomial equation that agrees

                                                                with the above data

                                                                119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                times 10minus17 1198791198795 (314)

                                                                35

                                                                119879119879 ( 119870119870) 119870119870 119869119869

                                                                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                The previous thermal conductivity data and the best fitting of the data are

                                                                shown in Fig (310)

                                                                Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                temperature

                                                                36

                                                                The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                material was obtained from the following experimental data tacked from

                                                                literatures

                                                                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                100 0254

                                                                200 0357

                                                                273 0384

                                                                298 0387

                                                                400 0397

                                                                600 0416

                                                                800 0435

                                                                1000 0454

                                                                1100 0464

                                                                1200 0474

                                                                1300 0483

                                                                The best polynomial fitted for these data was

                                                                119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                The specific heat capacity data and the best polynomial fitting of the data are

                                                                shown in Fig (311)

                                                                37

                                                                Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                temperature

                                                                The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                literature was used to find the best polynomial fitting

                                                                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                1000 8576 1100 8519 1200 8458 1300 8396

                                                                38

                                                                The best polynomial of this data was

                                                                120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                minus 76765 times 10minus14 1198791198794 (316)

                                                                The density of copper as a function of temperature and the best polynomial

                                                                fitting are shown in Fig (312)

                                                                Fig(312) The best fitting of density of copper as a function of temperature

                                                                The depth penetration of laser energy for copper metal was calculated using

                                                                the polynomial equations of thermal conductivity specific heat capacity and

                                                                density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                function of time the result was shown in Fig (313)

                                                                39

                                                                The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                than that of copper metal this may be due to the high thermal conductivity and

                                                                high specific heat capacity of copper with that of lead metal

                                                                Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                material

                                                                40

                                                                310 Conclusions

                                                                The Depth dependence of temperature for lead metal was investigated in two

                                                                case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                thermal properties (thermal conductivity specific heat) and density of metal are

                                                                also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                penetration depth in the first case is smaller than that of the second case about

                                                                (190) times

                                                                The temperature distribution as a function of depth dependence for copper

                                                                metal was also investigated in the case when the laser intensity vary with time

                                                                119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                The depth penetration of laser energy of lead metal was found to be smaller

                                                                than that of copper metal this may be due to the high thermal conductivity and

                                                                high specific heat capacity of copper with that of lead metal

                                                                41

                                                                References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                Sons Inc Hoboken New Jersey Canada (2003)

                                                                [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                [9]

                                                                Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                [12] httpenwikipediaorgwikiThermal_conductivity

                                                                [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                42

                                                                Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                43

                                                                E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                44

                                                                title(Dencity as a function of temperature) hold off

                                                                This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                45

                                                                elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                46

                                                                r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                47

                                                                for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                48

                                                                for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                49

                                                                6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                50

                                                                u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                51

                                                                alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                52

                                                                715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                • 01 Title
                                                                  • 0TDr Omed Ghareb Abdullah
                                                                    • Ch1-Laser
                                                                      • 16 Heat and heat capacity
                                                                      • 18 Derivation in one dimension
                                                                        • Ch2 laser2
                                                                        • Ch3 laser2
                                                                        • Appendix-Laser

                                                                  28

                                                                  Where 119863119863119898119898 put to balance the units of equation (36)

                                                                  But integral

                                                                  119868119868 = 119864119864119860119860

                                                                  (37)

                                                                  and from equations (35) (36) and (37) we have

                                                                  119868119868 (119898119898)11989911989911989811989808

                                                                  00

                                                                  = 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                                  0800 119899119899119898119898

                                                                  119860119860 119863119863119898119898 (38)

                                                                  Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

                                                                  (38)

                                                                  There fore

                                                                  119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

                                                                  119860119860 119863119863119898119898 (39)

                                                                  As shown in Fig(34) Matlab program was used to obtain the best polynomial

                                                                  that agrees with result data

                                                                  119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

                                                                  + 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

                                                                  Fig(34) Time dependence of laser intensity

                                                                  29

                                                                  34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                  constant thermal properties

                                                                  With all constant thermal properties of lead metal as in article (23) and

                                                                  119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                                  equation (23) with boundary and initial condition as in equation (22) and the

                                                                  depth penetration is shown in Fig(35)

                                                                  Fig(35) Depth dependence of the temperature when laser intensity function

                                                                  of time and constant thermal properties of Lead

                                                                  35 Evaluation the Thermal Conductivity as functions of temperature

                                                                  The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                                  temperature for Lead material was obtained by Matlab program using the

                                                                  experimental data tabulated in researches

                                                                  119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                                  times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                                  30

                                                                  119879119879 ( 119870119870) 119870119870 119869119869

                                                                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                  300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                                  The previous thermal conductivity data and the best fitting of the data are

                                                                  shown in Fig (36)

                                                                  Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                                  temperature

                                                                  31

                                                                  36 Evaluation the Specific heat as functions of temperature

                                                                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                                  material was obtained from the following experimental data tacked from

                                                                  literatures

                                                                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                  300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                                  The best polynomial fitted for these data was

                                                                  119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                                  + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                                  times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                                  The specific heat capacity data and the best polynomial fitting of the data are

                                                                  shown in Fig (37)

                                                                  32

                                                                  Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                                  temperature

                                                                  37 Evaluation the Density as functions of temperature

                                                                  The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                                  was used to find the best polynomial fitting

                                                                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                  300 11330 400 11230 500 11130 600 11010 800 10430

                                                                  1000 10190 1200 9940

                                                                  The best polynomial of this data was

                                                                  120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                                  minus 45158 times 10minus12 1198791198794 (313)

                                                                  33

                                                                  The density of Lead as a function of temperature and the best polynomial fitting

                                                                  are shown in Fig (38)

                                                                  Fig(38) The best fitting of density of Lead as a function of temperature

                                                                  38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                  variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                                  We have deduced the solution of equation (24) with initial and boundary

                                                                  condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                                  and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                                  respectively then by using Matlab program the depth penetration is shown in

                                                                  Fig (39)

                                                                  34

                                                                  Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                  material

                                                                  39 Laser interaction with copper material

                                                                  The same time dependence of laser intensity as shown in Fig(34) with

                                                                  thermal properties of copper was used to calculate the temperature distribution as

                                                                  a function of depth penetration

                                                                  The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                  copper material was obtained from the experimental data tabulated in literary

                                                                  The Matlab program used to obtain the best polynomial equation that agrees

                                                                  with the above data

                                                                  119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                  minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                  times 10minus17 1198791198795 (314)

                                                                  35

                                                                  119879119879 ( 119870119870) 119870119870 119869119869

                                                                  119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                  100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                  The previous thermal conductivity data and the best fitting of the data are

                                                                  shown in Fig (310)

                                                                  Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                  temperature

                                                                  36

                                                                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                  material was obtained from the following experimental data tacked from

                                                                  literatures

                                                                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                  100 0254

                                                                  200 0357

                                                                  273 0384

                                                                  298 0387

                                                                  400 0397

                                                                  600 0416

                                                                  800 0435

                                                                  1000 0454

                                                                  1100 0464

                                                                  1200 0474

                                                                  1300 0483

                                                                  The best polynomial fitted for these data was

                                                                  119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                  + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                  times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                  The specific heat capacity data and the best polynomial fitting of the data are

                                                                  shown in Fig (311)

                                                                  37

                                                                  Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                  temperature

                                                                  The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                  literature was used to find the best polynomial fitting

                                                                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                  100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                  1000 8576 1100 8519 1200 8458 1300 8396

                                                                  38

                                                                  The best polynomial of this data was

                                                                  120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                  minus 76765 times 10minus14 1198791198794 (316)

                                                                  The density of copper as a function of temperature and the best polynomial

                                                                  fitting are shown in Fig (312)

                                                                  Fig(312) The best fitting of density of copper as a function of temperature

                                                                  The depth penetration of laser energy for copper metal was calculated using

                                                                  the polynomial equations of thermal conductivity specific heat capacity and

                                                                  density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                  (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                  function of time the result was shown in Fig (313)

                                                                  39

                                                                  The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                  for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                  copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                  than that of copper metal this may be due to the high thermal conductivity and

                                                                  high specific heat capacity of copper with that of lead metal

                                                                  Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                  material

                                                                  40

                                                                  310 Conclusions

                                                                  The Depth dependence of temperature for lead metal was investigated in two

                                                                  case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                  thermal properties (thermal conductivity specific heat) and density of metal are

                                                                  also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                  vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                  specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                  120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                  penetration depth in the first case is smaller than that of the second case about

                                                                  (190) times

                                                                  The temperature distribution as a function of depth dependence for copper

                                                                  metal was also investigated in the case when the laser intensity vary with time

                                                                  119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                  density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                  The depth penetration of laser energy of lead metal was found to be smaller

                                                                  than that of copper metal this may be due to the high thermal conductivity and

                                                                  high specific heat capacity of copper with that of lead metal

                                                                  41

                                                                  References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                  Sons Inc Hoboken New Jersey Canada (2003)

                                                                  [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                  [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                  [9]

                                                                  Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                  [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                  [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                  [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                  [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                  [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                  httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                  [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                  [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                  [12] httpenwikipediaorgwikiThermal_conductivity

                                                                  [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                  [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                  [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                  42

                                                                  Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                  This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                  This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                  43

                                                                  E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                  This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                  This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                  44

                                                                  title(Dencity as a function of temperature) hold off

                                                                  This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                  45

                                                                  elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                  46

                                                                  r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                  47

                                                                  for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                  48

                                                                  for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                  49

                                                                  6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                  50

                                                                  u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                  51

                                                                  alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                  52

                                                                  715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                  • 01 Title
                                                                    • 0TDr Omed Ghareb Abdullah
                                                                      • Ch1-Laser
                                                                        • 16 Heat and heat capacity
                                                                        • 18 Derivation in one dimension
                                                                          • Ch2 laser2
                                                                          • Ch3 laser2
                                                                          • Appendix-Laser

                                                                    29

                                                                    34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                    constant thermal properties

                                                                    With all constant thermal properties of lead metal as in article (23) and

                                                                    119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

                                                                    equation (23) with boundary and initial condition as in equation (22) and the

                                                                    depth penetration is shown in Fig(35)

                                                                    Fig(35) Depth dependence of the temperature when laser intensity function

                                                                    of time and constant thermal properties of Lead

                                                                    35 Evaluation the Thermal Conductivity as functions of temperature

                                                                    The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

                                                                    temperature for Lead material was obtained by Matlab program using the

                                                                    experimental data tabulated in researches

                                                                    119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

                                                                    times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

                                                                    30

                                                                    119879119879 ( 119870119870) 119870119870 119869119869

                                                                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                    300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                                    The previous thermal conductivity data and the best fitting of the data are

                                                                    shown in Fig (36)

                                                                    Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                                    temperature

                                                                    31

                                                                    36 Evaluation the Specific heat as functions of temperature

                                                                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                                    material was obtained from the following experimental data tacked from

                                                                    literatures

                                                                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                    300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                                    The best polynomial fitted for these data was

                                                                    119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                                    + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                                    times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                                    The specific heat capacity data and the best polynomial fitting of the data are

                                                                    shown in Fig (37)

                                                                    32

                                                                    Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                                    temperature

                                                                    37 Evaluation the Density as functions of temperature

                                                                    The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                                    was used to find the best polynomial fitting

                                                                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                    300 11330 400 11230 500 11130 600 11010 800 10430

                                                                    1000 10190 1200 9940

                                                                    The best polynomial of this data was

                                                                    120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                                    minus 45158 times 10minus12 1198791198794 (313)

                                                                    33

                                                                    The density of Lead as a function of temperature and the best polynomial fitting

                                                                    are shown in Fig (38)

                                                                    Fig(38) The best fitting of density of Lead as a function of temperature

                                                                    38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                    variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                                    We have deduced the solution of equation (24) with initial and boundary

                                                                    condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                                    and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                                    respectively then by using Matlab program the depth penetration is shown in

                                                                    Fig (39)

                                                                    34

                                                                    Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                    material

                                                                    39 Laser interaction with copper material

                                                                    The same time dependence of laser intensity as shown in Fig(34) with

                                                                    thermal properties of copper was used to calculate the temperature distribution as

                                                                    a function of depth penetration

                                                                    The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                    copper material was obtained from the experimental data tabulated in literary

                                                                    The Matlab program used to obtain the best polynomial equation that agrees

                                                                    with the above data

                                                                    119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                    minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                    times 10minus17 1198791198795 (314)

                                                                    35

                                                                    119879119879 ( 119870119870) 119870119870 119869119869

                                                                    119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                    100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                    The previous thermal conductivity data and the best fitting of the data are

                                                                    shown in Fig (310)

                                                                    Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                    temperature

                                                                    36

                                                                    The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                    material was obtained from the following experimental data tacked from

                                                                    literatures

                                                                    119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                    100 0254

                                                                    200 0357

                                                                    273 0384

                                                                    298 0387

                                                                    400 0397

                                                                    600 0416

                                                                    800 0435

                                                                    1000 0454

                                                                    1100 0464

                                                                    1200 0474

                                                                    1300 0483

                                                                    The best polynomial fitted for these data was

                                                                    119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                    + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                    times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                    The specific heat capacity data and the best polynomial fitting of the data are

                                                                    shown in Fig (311)

                                                                    37

                                                                    Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                    temperature

                                                                    The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                    literature was used to find the best polynomial fitting

                                                                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                    100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                    1000 8576 1100 8519 1200 8458 1300 8396

                                                                    38

                                                                    The best polynomial of this data was

                                                                    120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                    minus 76765 times 10minus14 1198791198794 (316)

                                                                    The density of copper as a function of temperature and the best polynomial

                                                                    fitting are shown in Fig (312)

                                                                    Fig(312) The best fitting of density of copper as a function of temperature

                                                                    The depth penetration of laser energy for copper metal was calculated using

                                                                    the polynomial equations of thermal conductivity specific heat capacity and

                                                                    density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                    (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                    function of time the result was shown in Fig (313)

                                                                    39

                                                                    The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                    for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                    copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                    than that of copper metal this may be due to the high thermal conductivity and

                                                                    high specific heat capacity of copper with that of lead metal

                                                                    Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                    material

                                                                    40

                                                                    310 Conclusions

                                                                    The Depth dependence of temperature for lead metal was investigated in two

                                                                    case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                    thermal properties (thermal conductivity specific heat) and density of metal are

                                                                    also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                    vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                    specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                    120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                    penetration depth in the first case is smaller than that of the second case about

                                                                    (190) times

                                                                    The temperature distribution as a function of depth dependence for copper

                                                                    metal was also investigated in the case when the laser intensity vary with time

                                                                    119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                    density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                    The depth penetration of laser energy of lead metal was found to be smaller

                                                                    than that of copper metal this may be due to the high thermal conductivity and

                                                                    high specific heat capacity of copper with that of lead metal

                                                                    41

                                                                    References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                    Sons Inc Hoboken New Jersey Canada (2003)

                                                                    [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                    [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                    [9]

                                                                    Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                    [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                    [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                    [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                    [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                    [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                    httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                    [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                    [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                    [12] httpenwikipediaorgwikiThermal_conductivity

                                                                    [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                    [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                    [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                    42

                                                                    Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                    This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                    This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                    43

                                                                    E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                    This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                    This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                    44

                                                                    title(Dencity as a function of temperature) hold off

                                                                    This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                    45

                                                                    elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                    46

                                                                    r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                    47

                                                                    for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                    48

                                                                    for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                    49

                                                                    6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                    50

                                                                    u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                    51

                                                                    alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                    52

                                                                    715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                    • 01 Title
                                                                      • 0TDr Omed Ghareb Abdullah
                                                                        • Ch1-Laser
                                                                          • 16 Heat and heat capacity
                                                                          • 18 Derivation in one dimension
                                                                            • Ch2 laser2
                                                                            • Ch3 laser2
                                                                            • Appendix-Laser

                                                                      30

                                                                      119879119879 ( 119870119870) 119870119870 119869119869

                                                                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                      300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

                                                                      The previous thermal conductivity data and the best fitting of the data are

                                                                      shown in Fig (36)

                                                                      Fig(36) The best fitting of thermal conductivity of Lead as a function of

                                                                      temperature

                                                                      31

                                                                      36 Evaluation the Specific heat as functions of temperature

                                                                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                                      material was obtained from the following experimental data tacked from

                                                                      literatures

                                                                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                      300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                                      The best polynomial fitted for these data was

                                                                      119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                                      + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                                      times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                                      The specific heat capacity data and the best polynomial fitting of the data are

                                                                      shown in Fig (37)

                                                                      32

                                                                      Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                                      temperature

                                                                      37 Evaluation the Density as functions of temperature

                                                                      The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                                      was used to find the best polynomial fitting

                                                                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                      300 11330 400 11230 500 11130 600 11010 800 10430

                                                                      1000 10190 1200 9940

                                                                      The best polynomial of this data was

                                                                      120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                                      minus 45158 times 10minus12 1198791198794 (313)

                                                                      33

                                                                      The density of Lead as a function of temperature and the best polynomial fitting

                                                                      are shown in Fig (38)

                                                                      Fig(38) The best fitting of density of Lead as a function of temperature

                                                                      38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                      variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                                      We have deduced the solution of equation (24) with initial and boundary

                                                                      condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                                      and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                                      respectively then by using Matlab program the depth penetration is shown in

                                                                      Fig (39)

                                                                      34

                                                                      Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                      material

                                                                      39 Laser interaction with copper material

                                                                      The same time dependence of laser intensity as shown in Fig(34) with

                                                                      thermal properties of copper was used to calculate the temperature distribution as

                                                                      a function of depth penetration

                                                                      The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                      copper material was obtained from the experimental data tabulated in literary

                                                                      The Matlab program used to obtain the best polynomial equation that agrees

                                                                      with the above data

                                                                      119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                      minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                      times 10minus17 1198791198795 (314)

                                                                      35

                                                                      119879119879 ( 119870119870) 119870119870 119869119869

                                                                      119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                      100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                      The previous thermal conductivity data and the best fitting of the data are

                                                                      shown in Fig (310)

                                                                      Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                      temperature

                                                                      36

                                                                      The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                      material was obtained from the following experimental data tacked from

                                                                      literatures

                                                                      119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                      100 0254

                                                                      200 0357

                                                                      273 0384

                                                                      298 0387

                                                                      400 0397

                                                                      600 0416

                                                                      800 0435

                                                                      1000 0454

                                                                      1100 0464

                                                                      1200 0474

                                                                      1300 0483

                                                                      The best polynomial fitted for these data was

                                                                      119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                      + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                      times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                      The specific heat capacity data and the best polynomial fitting of the data are

                                                                      shown in Fig (311)

                                                                      37

                                                                      Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                      temperature

                                                                      The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                      literature was used to find the best polynomial fitting

                                                                      119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                      100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                      1000 8576 1100 8519 1200 8458 1300 8396

                                                                      38

                                                                      The best polynomial of this data was

                                                                      120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                      minus 76765 times 10minus14 1198791198794 (316)

                                                                      The density of copper as a function of temperature and the best polynomial

                                                                      fitting are shown in Fig (312)

                                                                      Fig(312) The best fitting of density of copper as a function of temperature

                                                                      The depth penetration of laser energy for copper metal was calculated using

                                                                      the polynomial equations of thermal conductivity specific heat capacity and

                                                                      density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                      (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                      function of time the result was shown in Fig (313)

                                                                      39

                                                                      The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                      for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                      copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                      than that of copper metal this may be due to the high thermal conductivity and

                                                                      high specific heat capacity of copper with that of lead metal

                                                                      Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                      material

                                                                      40

                                                                      310 Conclusions

                                                                      The Depth dependence of temperature for lead metal was investigated in two

                                                                      case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                      thermal properties (thermal conductivity specific heat) and density of metal are

                                                                      also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                      vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                      specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                      120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                      penetration depth in the first case is smaller than that of the second case about

                                                                      (190) times

                                                                      The temperature distribution as a function of depth dependence for copper

                                                                      metal was also investigated in the case when the laser intensity vary with time

                                                                      119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                      density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                      The depth penetration of laser energy of lead metal was found to be smaller

                                                                      than that of copper metal this may be due to the high thermal conductivity and

                                                                      high specific heat capacity of copper with that of lead metal

                                                                      41

                                                                      References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                      Sons Inc Hoboken New Jersey Canada (2003)

                                                                      [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                      [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                      [9]

                                                                      Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                      [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                      [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                      [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                      [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                      [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                      httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                      [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                      [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                      [12] httpenwikipediaorgwikiThermal_conductivity

                                                                      [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                      [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                      [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                      42

                                                                      Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                      This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                      This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                      43

                                                                      E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                      This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                      This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                      44

                                                                      title(Dencity as a function of temperature) hold off

                                                                      This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                      45

                                                                      elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                      46

                                                                      r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                      47

                                                                      for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                      48

                                                                      for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                      49

                                                                      6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                      50

                                                                      u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                      51

                                                                      alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                      52

                                                                      715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                      • 01 Title
                                                                        • 0TDr Omed Ghareb Abdullah
                                                                          • Ch1-Laser
                                                                            • 16 Heat and heat capacity
                                                                            • 18 Derivation in one dimension
                                                                              • Ch2 laser2
                                                                              • Ch3 laser2
                                                                              • Appendix-Laser

                                                                        31

                                                                        36 Evaluation the Specific heat as functions of temperature

                                                                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

                                                                        material was obtained from the following experimental data tacked from

                                                                        literatures

                                                                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                        300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

                                                                        The best polynomial fitted for these data was

                                                                        119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

                                                                        + 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

                                                                        times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

                                                                        The specific heat capacity data and the best polynomial fitting of the data are

                                                                        shown in Fig (37)

                                                                        32

                                                                        Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                                        temperature

                                                                        37 Evaluation the Density as functions of temperature

                                                                        The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                                        was used to find the best polynomial fitting

                                                                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                        300 11330 400 11230 500 11130 600 11010 800 10430

                                                                        1000 10190 1200 9940

                                                                        The best polynomial of this data was

                                                                        120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                                        minus 45158 times 10minus12 1198791198794 (313)

                                                                        33

                                                                        The density of Lead as a function of temperature and the best polynomial fitting

                                                                        are shown in Fig (38)

                                                                        Fig(38) The best fitting of density of Lead as a function of temperature

                                                                        38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                        variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                                        We have deduced the solution of equation (24) with initial and boundary

                                                                        condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                                        and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                                        respectively then by using Matlab program the depth penetration is shown in

                                                                        Fig (39)

                                                                        34

                                                                        Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                        material

                                                                        39 Laser interaction with copper material

                                                                        The same time dependence of laser intensity as shown in Fig(34) with

                                                                        thermal properties of copper was used to calculate the temperature distribution as

                                                                        a function of depth penetration

                                                                        The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                        copper material was obtained from the experimental data tabulated in literary

                                                                        The Matlab program used to obtain the best polynomial equation that agrees

                                                                        with the above data

                                                                        119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                        minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                        times 10minus17 1198791198795 (314)

                                                                        35

                                                                        119879119879 ( 119870119870) 119870119870 119869119869

                                                                        119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                        100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                        The previous thermal conductivity data and the best fitting of the data are

                                                                        shown in Fig (310)

                                                                        Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                        temperature

                                                                        36

                                                                        The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                        material was obtained from the following experimental data tacked from

                                                                        literatures

                                                                        119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                        100 0254

                                                                        200 0357

                                                                        273 0384

                                                                        298 0387

                                                                        400 0397

                                                                        600 0416

                                                                        800 0435

                                                                        1000 0454

                                                                        1100 0464

                                                                        1200 0474

                                                                        1300 0483

                                                                        The best polynomial fitted for these data was

                                                                        119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                        + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                        times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                        The specific heat capacity data and the best polynomial fitting of the data are

                                                                        shown in Fig (311)

                                                                        37

                                                                        Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                        temperature

                                                                        The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                        literature was used to find the best polynomial fitting

                                                                        119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                        100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                        1000 8576 1100 8519 1200 8458 1300 8396

                                                                        38

                                                                        The best polynomial of this data was

                                                                        120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                        minus 76765 times 10minus14 1198791198794 (316)

                                                                        The density of copper as a function of temperature and the best polynomial

                                                                        fitting are shown in Fig (312)

                                                                        Fig(312) The best fitting of density of copper as a function of temperature

                                                                        The depth penetration of laser energy for copper metal was calculated using

                                                                        the polynomial equations of thermal conductivity specific heat capacity and

                                                                        density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                        (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                        function of time the result was shown in Fig (313)

                                                                        39

                                                                        The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                        for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                        copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                        than that of copper metal this may be due to the high thermal conductivity and

                                                                        high specific heat capacity of copper with that of lead metal

                                                                        Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                        material

                                                                        40

                                                                        310 Conclusions

                                                                        The Depth dependence of temperature for lead metal was investigated in two

                                                                        case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                        thermal properties (thermal conductivity specific heat) and density of metal are

                                                                        also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                        vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                        specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                        120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                        penetration depth in the first case is smaller than that of the second case about

                                                                        (190) times

                                                                        The temperature distribution as a function of depth dependence for copper

                                                                        metal was also investigated in the case when the laser intensity vary with time

                                                                        119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                        density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                        The depth penetration of laser energy of lead metal was found to be smaller

                                                                        than that of copper metal this may be due to the high thermal conductivity and

                                                                        high specific heat capacity of copper with that of lead metal

                                                                        41

                                                                        References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                        Sons Inc Hoboken New Jersey Canada (2003)

                                                                        [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                        [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                        [9]

                                                                        Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                        [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                        [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                        [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                        [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                        [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                        httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                        [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                        [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                        [12] httpenwikipediaorgwikiThermal_conductivity

                                                                        [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                        [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                        [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                        42

                                                                        Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                        This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                        This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                        43

                                                                        E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                        This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                        This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                        44

                                                                        title(Dencity as a function of temperature) hold off

                                                                        This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                        45

                                                                        elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                        46

                                                                        r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                        47

                                                                        for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                        48

                                                                        for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                        49

                                                                        6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                        50

                                                                        u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                        51

                                                                        alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                        52

                                                                        715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                        • 01 Title
                                                                          • 0TDr Omed Ghareb Abdullah
                                                                            • Ch1-Laser
                                                                              • 16 Heat and heat capacity
                                                                              • 18 Derivation in one dimension
                                                                                • Ch2 laser2
                                                                                • Ch3 laser2
                                                                                • Appendix-Laser

                                                                          32

                                                                          Fig(37) The best fitting of specific heat capacity of Lead as a function of

                                                                          temperature

                                                                          37 Evaluation the Density as functions of temperature

                                                                          The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

                                                                          was used to find the best polynomial fitting

                                                                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                          300 11330 400 11230 500 11130 600 11010 800 10430

                                                                          1000 10190 1200 9940

                                                                          The best polynomial of this data was

                                                                          120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

                                                                          minus 45158 times 10minus12 1198791198794 (313)

                                                                          33

                                                                          The density of Lead as a function of temperature and the best polynomial fitting

                                                                          are shown in Fig (38)

                                                                          Fig(38) The best fitting of density of Lead as a function of temperature

                                                                          38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                          variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                                          We have deduced the solution of equation (24) with initial and boundary

                                                                          condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                                          and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                                          respectively then by using Matlab program the depth penetration is shown in

                                                                          Fig (39)

                                                                          34

                                                                          Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                          material

                                                                          39 Laser interaction with copper material

                                                                          The same time dependence of laser intensity as shown in Fig(34) with

                                                                          thermal properties of copper was used to calculate the temperature distribution as

                                                                          a function of depth penetration

                                                                          The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                          copper material was obtained from the experimental data tabulated in literary

                                                                          The Matlab program used to obtain the best polynomial equation that agrees

                                                                          with the above data

                                                                          119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                          minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                          times 10minus17 1198791198795 (314)

                                                                          35

                                                                          119879119879 ( 119870119870) 119870119870 119869119869

                                                                          119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                          100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                          The previous thermal conductivity data and the best fitting of the data are

                                                                          shown in Fig (310)

                                                                          Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                          temperature

                                                                          36

                                                                          The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                          material was obtained from the following experimental data tacked from

                                                                          literatures

                                                                          119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                          100 0254

                                                                          200 0357

                                                                          273 0384

                                                                          298 0387

                                                                          400 0397

                                                                          600 0416

                                                                          800 0435

                                                                          1000 0454

                                                                          1100 0464

                                                                          1200 0474

                                                                          1300 0483

                                                                          The best polynomial fitted for these data was

                                                                          119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                          + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                          times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                          The specific heat capacity data and the best polynomial fitting of the data are

                                                                          shown in Fig (311)

                                                                          37

                                                                          Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                          temperature

                                                                          The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                          literature was used to find the best polynomial fitting

                                                                          119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                          100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                          1000 8576 1100 8519 1200 8458 1300 8396

                                                                          38

                                                                          The best polynomial of this data was

                                                                          120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                          minus 76765 times 10minus14 1198791198794 (316)

                                                                          The density of copper as a function of temperature and the best polynomial

                                                                          fitting are shown in Fig (312)

                                                                          Fig(312) The best fitting of density of copper as a function of temperature

                                                                          The depth penetration of laser energy for copper metal was calculated using

                                                                          the polynomial equations of thermal conductivity specific heat capacity and

                                                                          density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                          (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                          function of time the result was shown in Fig (313)

                                                                          39

                                                                          The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                          for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                          copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                          than that of copper metal this may be due to the high thermal conductivity and

                                                                          high specific heat capacity of copper with that of lead metal

                                                                          Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                          material

                                                                          40

                                                                          310 Conclusions

                                                                          The Depth dependence of temperature for lead metal was investigated in two

                                                                          case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                          thermal properties (thermal conductivity specific heat) and density of metal are

                                                                          also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                          vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                          specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                          120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                          penetration depth in the first case is smaller than that of the second case about

                                                                          (190) times

                                                                          The temperature distribution as a function of depth dependence for copper

                                                                          metal was also investigated in the case when the laser intensity vary with time

                                                                          119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                          density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                          The depth penetration of laser energy of lead metal was found to be smaller

                                                                          than that of copper metal this may be due to the high thermal conductivity and

                                                                          high specific heat capacity of copper with that of lead metal

                                                                          41

                                                                          References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                          Sons Inc Hoboken New Jersey Canada (2003)

                                                                          [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                          [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                          [9]

                                                                          Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                          [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                          [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                          [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                          [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                          [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                          httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                          [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                          [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                          [12] httpenwikipediaorgwikiThermal_conductivity

                                                                          [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                          [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                          [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                          42

                                                                          Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                          This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                          This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                          43

                                                                          E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                          This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                          This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                          44

                                                                          title(Dencity as a function of temperature) hold off

                                                                          This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                          45

                                                                          elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                          46

                                                                          r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                          47

                                                                          for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                          48

                                                                          for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                          49

                                                                          6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                          50

                                                                          u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                          51

                                                                          alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                          52

                                                                          715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                          • 01 Title
                                                                            • 0TDr Omed Ghareb Abdullah
                                                                              • Ch1-Laser
                                                                                • 16 Heat and heat capacity
                                                                                • 18 Derivation in one dimension
                                                                                  • Ch2 laser2
                                                                                  • Ch3 laser2
                                                                                  • Appendix-Laser

                                                                            33

                                                                            The density of Lead as a function of temperature and the best polynomial fitting

                                                                            are shown in Fig (38)

                                                                            Fig(38) The best fitting of density of Lead as a function of temperature

                                                                            38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

                                                                            variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

                                                                            We have deduced the solution of equation (24) with initial and boundary

                                                                            condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

                                                                            and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

                                                                            respectively then by using Matlab program the depth penetration is shown in

                                                                            Fig (39)

                                                                            34

                                                                            Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                            material

                                                                            39 Laser interaction with copper material

                                                                            The same time dependence of laser intensity as shown in Fig(34) with

                                                                            thermal properties of copper was used to calculate the temperature distribution as

                                                                            a function of depth penetration

                                                                            The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                            copper material was obtained from the experimental data tabulated in literary

                                                                            The Matlab program used to obtain the best polynomial equation that agrees

                                                                            with the above data

                                                                            119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                            minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                            times 10minus17 1198791198795 (314)

                                                                            35

                                                                            119879119879 ( 119870119870) 119870119870 119869119869

                                                                            119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                            100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                            The previous thermal conductivity data and the best fitting of the data are

                                                                            shown in Fig (310)

                                                                            Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                            temperature

                                                                            36

                                                                            The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                            material was obtained from the following experimental data tacked from

                                                                            literatures

                                                                            119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                            100 0254

                                                                            200 0357

                                                                            273 0384

                                                                            298 0387

                                                                            400 0397

                                                                            600 0416

                                                                            800 0435

                                                                            1000 0454

                                                                            1100 0464

                                                                            1200 0474

                                                                            1300 0483

                                                                            The best polynomial fitted for these data was

                                                                            119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                            + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                            times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                            The specific heat capacity data and the best polynomial fitting of the data are

                                                                            shown in Fig (311)

                                                                            37

                                                                            Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                            temperature

                                                                            The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                            literature was used to find the best polynomial fitting

                                                                            119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                            100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                            1000 8576 1100 8519 1200 8458 1300 8396

                                                                            38

                                                                            The best polynomial of this data was

                                                                            120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                            minus 76765 times 10minus14 1198791198794 (316)

                                                                            The density of copper as a function of temperature and the best polynomial

                                                                            fitting are shown in Fig (312)

                                                                            Fig(312) The best fitting of density of copper as a function of temperature

                                                                            The depth penetration of laser energy for copper metal was calculated using

                                                                            the polynomial equations of thermal conductivity specific heat capacity and

                                                                            density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                            (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                            function of time the result was shown in Fig (313)

                                                                            39

                                                                            The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                            for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                            copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                            than that of copper metal this may be due to the high thermal conductivity and

                                                                            high specific heat capacity of copper with that of lead metal

                                                                            Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                            material

                                                                            40

                                                                            310 Conclusions

                                                                            The Depth dependence of temperature for lead metal was investigated in two

                                                                            case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                            thermal properties (thermal conductivity specific heat) and density of metal are

                                                                            also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                            vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                            specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                            120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                            penetration depth in the first case is smaller than that of the second case about

                                                                            (190) times

                                                                            The temperature distribution as a function of depth dependence for copper

                                                                            metal was also investigated in the case when the laser intensity vary with time

                                                                            119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                            density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                            The depth penetration of laser energy of lead metal was found to be smaller

                                                                            than that of copper metal this may be due to the high thermal conductivity and

                                                                            high specific heat capacity of copper with that of lead metal

                                                                            41

                                                                            References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                            Sons Inc Hoboken New Jersey Canada (2003)

                                                                            [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                            [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                            [9]

                                                                            Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                            [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                            [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                            [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                            [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                            [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                            httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                            [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                            [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                            [12] httpenwikipediaorgwikiThermal_conductivity

                                                                            [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                            [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                            [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                            42

                                                                            Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                            This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                            This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                            43

                                                                            E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                            This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                            This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                            44

                                                                            title(Dencity as a function of temperature) hold off

                                                                            This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                            45

                                                                            elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                            46

                                                                            r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                            47

                                                                            for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                            48

                                                                            for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                            49

                                                                            6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                            50

                                                                            u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                            51

                                                                            alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                            52

                                                                            715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                            • 01 Title
                                                                              • 0TDr Omed Ghareb Abdullah
                                                                                • Ch1-Laser
                                                                                  • 16 Heat and heat capacity
                                                                                  • 18 Derivation in one dimension
                                                                                    • Ch2 laser2
                                                                                    • Ch3 laser2
                                                                                    • Appendix-Laser

                                                                              34

                                                                              Fig(39) Depth dependence of the temperature for pulse laser on Lead

                                                                              material

                                                                              39 Laser interaction with copper material

                                                                              The same time dependence of laser intensity as shown in Fig(34) with

                                                                              thermal properties of copper was used to calculate the temperature distribution as

                                                                              a function of depth penetration

                                                                              The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

                                                                              copper material was obtained from the experimental data tabulated in literary

                                                                              The Matlab program used to obtain the best polynomial equation that agrees

                                                                              with the above data

                                                                              119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

                                                                              minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

                                                                              times 10minus17 1198791198795 (314)

                                                                              35

                                                                              119879119879 ( 119870119870) 119870119870 119869119869

                                                                              119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                              100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                              The previous thermal conductivity data and the best fitting of the data are

                                                                              shown in Fig (310)

                                                                              Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                              temperature

                                                                              36

                                                                              The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                              material was obtained from the following experimental data tacked from

                                                                              literatures

                                                                              119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                              100 0254

                                                                              200 0357

                                                                              273 0384

                                                                              298 0387

                                                                              400 0397

                                                                              600 0416

                                                                              800 0435

                                                                              1000 0454

                                                                              1100 0464

                                                                              1200 0474

                                                                              1300 0483

                                                                              The best polynomial fitted for these data was

                                                                              119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                              + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                              times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                              The specific heat capacity data and the best polynomial fitting of the data are

                                                                              shown in Fig (311)

                                                                              37

                                                                              Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                              temperature

                                                                              The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                              literature was used to find the best polynomial fitting

                                                                              119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                              100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                              1000 8576 1100 8519 1200 8458 1300 8396

                                                                              38

                                                                              The best polynomial of this data was

                                                                              120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                              minus 76765 times 10minus14 1198791198794 (316)

                                                                              The density of copper as a function of temperature and the best polynomial

                                                                              fitting are shown in Fig (312)

                                                                              Fig(312) The best fitting of density of copper as a function of temperature

                                                                              The depth penetration of laser energy for copper metal was calculated using

                                                                              the polynomial equations of thermal conductivity specific heat capacity and

                                                                              density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                              (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                              function of time the result was shown in Fig (313)

                                                                              39

                                                                              The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                              for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                              copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                              than that of copper metal this may be due to the high thermal conductivity and

                                                                              high specific heat capacity of copper with that of lead metal

                                                                              Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                              material

                                                                              40

                                                                              310 Conclusions

                                                                              The Depth dependence of temperature for lead metal was investigated in two

                                                                              case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                              thermal properties (thermal conductivity specific heat) and density of metal are

                                                                              also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                              vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                              specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                              120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                              penetration depth in the first case is smaller than that of the second case about

                                                                              (190) times

                                                                              The temperature distribution as a function of depth dependence for copper

                                                                              metal was also investigated in the case when the laser intensity vary with time

                                                                              119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                              density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                              The depth penetration of laser energy of lead metal was found to be smaller

                                                                              than that of copper metal this may be due to the high thermal conductivity and

                                                                              high specific heat capacity of copper with that of lead metal

                                                                              41

                                                                              References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                              Sons Inc Hoboken New Jersey Canada (2003)

                                                                              [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                              [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                              [9]

                                                                              Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                              [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                              [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                              [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                              [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                              [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                              httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                              [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                              [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                              [12] httpenwikipediaorgwikiThermal_conductivity

                                                                              [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                              [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                              [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                              42

                                                                              Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                              This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                              This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                              43

                                                                              E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                              This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                              This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                              44

                                                                              title(Dencity as a function of temperature) hold off

                                                                              This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                              45

                                                                              elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                              46

                                                                              r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                              47

                                                                              for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                              48

                                                                              for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                              49

                                                                              6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                              50

                                                                              u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                              51

                                                                              alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                              52

                                                                              715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                              • 01 Title
                                                                                • 0TDr Omed Ghareb Abdullah
                                                                                  • Ch1-Laser
                                                                                    • 16 Heat and heat capacity
                                                                                    • 18 Derivation in one dimension
                                                                                      • Ch2 laser2
                                                                                      • Ch3 laser2
                                                                                      • Appendix-Laser

                                                                                35

                                                                                119879119879 ( 119870119870) 119870119870 119869119869

                                                                                119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

                                                                                100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

                                                                                The previous thermal conductivity data and the best fitting of the data are

                                                                                shown in Fig (310)

                                                                                Fig(310) The best fitting of thermal conductivity of Copper as a function of

                                                                                temperature

                                                                                36

                                                                                The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                                material was obtained from the following experimental data tacked from

                                                                                literatures

                                                                                119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                                100 0254

                                                                                200 0357

                                                                                273 0384

                                                                                298 0387

                                                                                400 0397

                                                                                600 0416

                                                                                800 0435

                                                                                1000 0454

                                                                                1100 0464

                                                                                1200 0474

                                                                                1300 0483

                                                                                The best polynomial fitted for these data was

                                                                                119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                                + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                                times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                                The specific heat capacity data and the best polynomial fitting of the data are

                                                                                shown in Fig (311)

                                                                                37

                                                                                Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                                temperature

                                                                                The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                                literature was used to find the best polynomial fitting

                                                                                119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                                100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                                1000 8576 1100 8519 1200 8458 1300 8396

                                                                                38

                                                                                The best polynomial of this data was

                                                                                120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                                minus 76765 times 10minus14 1198791198794 (316)

                                                                                The density of copper as a function of temperature and the best polynomial

                                                                                fitting are shown in Fig (312)

                                                                                Fig(312) The best fitting of density of copper as a function of temperature

                                                                                The depth penetration of laser energy for copper metal was calculated using

                                                                                the polynomial equations of thermal conductivity specific heat capacity and

                                                                                density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                                (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                                function of time the result was shown in Fig (313)

                                                                                39

                                                                                The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                                for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                                copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                                than that of copper metal this may be due to the high thermal conductivity and

                                                                                high specific heat capacity of copper with that of lead metal

                                                                                Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                                material

                                                                                40

                                                                                310 Conclusions

                                                                                The Depth dependence of temperature for lead metal was investigated in two

                                                                                case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                                thermal properties (thermal conductivity specific heat) and density of metal are

                                                                                also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                                vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                                specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                                120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                                penetration depth in the first case is smaller than that of the second case about

                                                                                (190) times

                                                                                The temperature distribution as a function of depth dependence for copper

                                                                                metal was also investigated in the case when the laser intensity vary with time

                                                                                119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                                density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                                The depth penetration of laser energy of lead metal was found to be smaller

                                                                                than that of copper metal this may be due to the high thermal conductivity and

                                                                                high specific heat capacity of copper with that of lead metal

                                                                                41

                                                                                References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                                Sons Inc Hoboken New Jersey Canada (2003)

                                                                                [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                                [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                                [9]

                                                                                Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                                [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                                [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                                [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                                [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                                [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                                httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                                [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                                [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                                [12] httpenwikipediaorgwikiThermal_conductivity

                                                                                [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                                [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                                [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                                42

                                                                                Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                43

                                                                                E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                44

                                                                                title(Dencity as a function of temperature) hold off

                                                                                This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                45

                                                                                elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                46

                                                                                r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                47

                                                                                for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                48

                                                                                for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                49

                                                                                6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                50

                                                                                u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                51

                                                                                alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                52

                                                                                715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                • 01 Title
                                                                                  • 0TDr Omed Ghareb Abdullah
                                                                                    • Ch1-Laser
                                                                                      • 16 Heat and heat capacity
                                                                                      • 18 Derivation in one dimension
                                                                                        • Ch2 laser2
                                                                                        • Ch3 laser2
                                                                                        • Appendix-Laser

                                                                                  36

                                                                                  The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

                                                                                  material was obtained from the following experimental data tacked from

                                                                                  literatures

                                                                                  119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

                                                                                  100 0254

                                                                                  200 0357

                                                                                  273 0384

                                                                                  298 0387

                                                                                  400 0397

                                                                                  600 0416

                                                                                  800 0435

                                                                                  1000 0454

                                                                                  1100 0464

                                                                                  1200 0474

                                                                                  1300 0483

                                                                                  The best polynomial fitted for these data was

                                                                                  119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

                                                                                  + 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

                                                                                  times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

                                                                                  The specific heat capacity data and the best polynomial fitting of the data are

                                                                                  shown in Fig (311)

                                                                                  37

                                                                                  Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                                  temperature

                                                                                  The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                                  literature was used to find the best polynomial fitting

                                                                                  119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                                  100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                                  1000 8576 1100 8519 1200 8458 1300 8396

                                                                                  38

                                                                                  The best polynomial of this data was

                                                                                  120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                                  minus 76765 times 10minus14 1198791198794 (316)

                                                                                  The density of copper as a function of temperature and the best polynomial

                                                                                  fitting are shown in Fig (312)

                                                                                  Fig(312) The best fitting of density of copper as a function of temperature

                                                                                  The depth penetration of laser energy for copper metal was calculated using

                                                                                  the polynomial equations of thermal conductivity specific heat capacity and

                                                                                  density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                                  (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                                  function of time the result was shown in Fig (313)

                                                                                  39

                                                                                  The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                                  for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                                  copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                                  than that of copper metal this may be due to the high thermal conductivity and

                                                                                  high specific heat capacity of copper with that of lead metal

                                                                                  Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                                  material

                                                                                  40

                                                                                  310 Conclusions

                                                                                  The Depth dependence of temperature for lead metal was investigated in two

                                                                                  case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                                  thermal properties (thermal conductivity specific heat) and density of metal are

                                                                                  also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                                  vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                                  specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                                  120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                                  penetration depth in the first case is smaller than that of the second case about

                                                                                  (190) times

                                                                                  The temperature distribution as a function of depth dependence for copper

                                                                                  metal was also investigated in the case when the laser intensity vary with time

                                                                                  119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                                  density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                                  The depth penetration of laser energy of lead metal was found to be smaller

                                                                                  than that of copper metal this may be due to the high thermal conductivity and

                                                                                  high specific heat capacity of copper with that of lead metal

                                                                                  41

                                                                                  References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                                  Sons Inc Hoboken New Jersey Canada (2003)

                                                                                  [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                                  [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                                  [9]

                                                                                  Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                                  [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                                  [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                                  [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                                  [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                                  [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                                  httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                                  [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                                  [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                                  [12] httpenwikipediaorgwikiThermal_conductivity

                                                                                  [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                                  [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                                  [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                                  42

                                                                                  Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                  This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                  This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                  43

                                                                                  E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                  This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                  This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                  44

                                                                                  title(Dencity as a function of temperature) hold off

                                                                                  This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                  45

                                                                                  elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                  46

                                                                                  r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                  47

                                                                                  for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                  48

                                                                                  for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                  49

                                                                                  6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                  50

                                                                                  u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                  51

                                                                                  alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                  52

                                                                                  715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                  • 01 Title
                                                                                    • 0TDr Omed Ghareb Abdullah
                                                                                      • Ch1-Laser
                                                                                        • 16 Heat and heat capacity
                                                                                        • 18 Derivation in one dimension
                                                                                          • Ch2 laser2
                                                                                          • Ch3 laser2
                                                                                          • Appendix-Laser

                                                                                    37

                                                                                    Fig(311) The best fitting of specific heat capacity of Copper as a function of

                                                                                    temperature

                                                                                    The density of copper 120588120588(119879119879) as a function of temperature tacked from

                                                                                    literature was used to find the best polynomial fitting

                                                                                    119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

                                                                                    100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

                                                                                    1000 8576 1100 8519 1200 8458 1300 8396

                                                                                    38

                                                                                    The best polynomial of this data was

                                                                                    120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                                    minus 76765 times 10minus14 1198791198794 (316)

                                                                                    The density of copper as a function of temperature and the best polynomial

                                                                                    fitting are shown in Fig (312)

                                                                                    Fig(312) The best fitting of density of copper as a function of temperature

                                                                                    The depth penetration of laser energy for copper metal was calculated using

                                                                                    the polynomial equations of thermal conductivity specific heat capacity and

                                                                                    density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                                    (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                                    function of time the result was shown in Fig (313)

                                                                                    39

                                                                                    The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                                    for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                                    copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                                    than that of copper metal this may be due to the high thermal conductivity and

                                                                                    high specific heat capacity of copper with that of lead metal

                                                                                    Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                                    material

                                                                                    40

                                                                                    310 Conclusions

                                                                                    The Depth dependence of temperature for lead metal was investigated in two

                                                                                    case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                                    thermal properties (thermal conductivity specific heat) and density of metal are

                                                                                    also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                                    vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                                    specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                                    120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                                    penetration depth in the first case is smaller than that of the second case about

                                                                                    (190) times

                                                                                    The temperature distribution as a function of depth dependence for copper

                                                                                    metal was also investigated in the case when the laser intensity vary with time

                                                                                    119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                                    density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                                    The depth penetration of laser energy of lead metal was found to be smaller

                                                                                    than that of copper metal this may be due to the high thermal conductivity and

                                                                                    high specific heat capacity of copper with that of lead metal

                                                                                    41

                                                                                    References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                                    Sons Inc Hoboken New Jersey Canada (2003)

                                                                                    [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                                    [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                                    [9]

                                                                                    Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                                    [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                                    [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                                    [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                                    [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                                    [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                                    httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                                    [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                                    [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                                    [12] httpenwikipediaorgwikiThermal_conductivity

                                                                                    [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                                    [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                                    [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                                    42

                                                                                    Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                    This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                    This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                    43

                                                                                    E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                    This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                    This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                    44

                                                                                    title(Dencity as a function of temperature) hold off

                                                                                    This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                    45

                                                                                    elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                    46

                                                                                    r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                    47

                                                                                    for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                    48

                                                                                    for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                    49

                                                                                    6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                    50

                                                                                    u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                    51

                                                                                    alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                    52

                                                                                    715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                    • 01 Title
                                                                                      • 0TDr Omed Ghareb Abdullah
                                                                                        • Ch1-Laser
                                                                                          • 16 Heat and heat capacity
                                                                                          • 18 Derivation in one dimension
                                                                                            • Ch2 laser2
                                                                                            • Ch3 laser2
                                                                                            • Appendix-Laser

                                                                                      38

                                                                                      The best polynomial of this data was

                                                                                      120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

                                                                                      minus 76765 times 10minus14 1198791198794 (316)

                                                                                      The density of copper as a function of temperature and the best polynomial

                                                                                      fitting are shown in Fig (312)

                                                                                      Fig(312) The best fitting of density of copper as a function of temperature

                                                                                      The depth penetration of laser energy for copper metal was calculated using

                                                                                      the polynomial equations of thermal conductivity specific heat capacity and

                                                                                      density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

                                                                                      (equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

                                                                                      function of time the result was shown in Fig (313)

                                                                                      39

                                                                                      The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                                      for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                                      copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                                      than that of copper metal this may be due to the high thermal conductivity and

                                                                                      high specific heat capacity of copper with that of lead metal

                                                                                      Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                                      material

                                                                                      40

                                                                                      310 Conclusions

                                                                                      The Depth dependence of temperature for lead metal was investigated in two

                                                                                      case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                                      thermal properties (thermal conductivity specific heat) and density of metal are

                                                                                      also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                                      vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                                      specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                                      120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                                      penetration depth in the first case is smaller than that of the second case about

                                                                                      (190) times

                                                                                      The temperature distribution as a function of depth dependence for copper

                                                                                      metal was also investigated in the case when the laser intensity vary with time

                                                                                      119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                                      density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                                      The depth penetration of laser energy of lead metal was found to be smaller

                                                                                      than that of copper metal this may be due to the high thermal conductivity and

                                                                                      high specific heat capacity of copper with that of lead metal

                                                                                      41

                                                                                      References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                                      Sons Inc Hoboken New Jersey Canada (2003)

                                                                                      [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                                      [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                                      [9]

                                                                                      Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                                      [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                                      [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                                      [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                                      [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                                      [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                                      httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                                      [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                                      [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                                      [12] httpenwikipediaorgwikiThermal_conductivity

                                                                                      [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                                      [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                                      [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                                      42

                                                                                      Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                      This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                      This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                      43

                                                                                      E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                      This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                      This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                      44

                                                                                      title(Dencity as a function of temperature) hold off

                                                                                      This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                      45

                                                                                      elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                      46

                                                                                      r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                      47

                                                                                      for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                      48

                                                                                      for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                      49

                                                                                      6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                      50

                                                                                      u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                      51

                                                                                      alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                      52

                                                                                      715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                      • 01 Title
                                                                                        • 0TDr Omed Ghareb Abdullah
                                                                                          • Ch1-Laser
                                                                                            • 16 Heat and heat capacity
                                                                                            • 18 Derivation in one dimension
                                                                                              • Ch2 laser2
                                                                                              • Ch3 laser2
                                                                                              • Appendix-Laser

                                                                                        39

                                                                                        The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

                                                                                        for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

                                                                                        copper metal so the depth penetration of laser energy of lead metal was smaller

                                                                                        than that of copper metal this may be due to the high thermal conductivity and

                                                                                        high specific heat capacity of copper with that of lead metal

                                                                                        Fig(313) Depth dependence of the temperature for pulse laser on Copper

                                                                                        material

                                                                                        40

                                                                                        310 Conclusions

                                                                                        The Depth dependence of temperature for lead metal was investigated in two

                                                                                        case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                                        thermal properties (thermal conductivity specific heat) and density of metal are

                                                                                        also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                                        vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                                        specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                                        120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                                        penetration depth in the first case is smaller than that of the second case about

                                                                                        (190) times

                                                                                        The temperature distribution as a function of depth dependence for copper

                                                                                        metal was also investigated in the case when the laser intensity vary with time

                                                                                        119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                                        density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                                        The depth penetration of laser energy of lead metal was found to be smaller

                                                                                        than that of copper metal this may be due to the high thermal conductivity and

                                                                                        high specific heat capacity of copper with that of lead metal

                                                                                        41

                                                                                        References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                                        Sons Inc Hoboken New Jersey Canada (2003)

                                                                                        [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                                        [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                                        [9]

                                                                                        Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                                        [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                                        [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                                        [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                                        [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                                        [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                                        httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                                        [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                                        [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                                        [12] httpenwikipediaorgwikiThermal_conductivity

                                                                                        [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                                        [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                                        [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                                        42

                                                                                        Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                        This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                        This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                        43

                                                                                        E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                        This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                        This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                        44

                                                                                        title(Dencity as a function of temperature) hold off

                                                                                        This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                        45

                                                                                        elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                        46

                                                                                        r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                        47

                                                                                        for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                        48

                                                                                        for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                        49

                                                                                        6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                        50

                                                                                        u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                        51

                                                                                        alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                        52

                                                                                        715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                        • 01 Title
                                                                                          • 0TDr Omed Ghareb Abdullah
                                                                                            • Ch1-Laser
                                                                                              • 16 Heat and heat capacity
                                                                                              • 18 Derivation in one dimension
                                                                                                • Ch2 laser2
                                                                                                • Ch3 laser2
                                                                                                • Appendix-Laser

                                                                                          40

                                                                                          310 Conclusions

                                                                                          The Depth dependence of temperature for lead metal was investigated in two

                                                                                          case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

                                                                                          thermal properties (thermal conductivity specific heat) and density of metal are

                                                                                          also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

                                                                                          vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

                                                                                          specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

                                                                                          120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

                                                                                          penetration depth in the first case is smaller than that of the second case about

                                                                                          (190) times

                                                                                          The temperature distribution as a function of depth dependence for copper

                                                                                          metal was also investigated in the case when the laser intensity vary with time

                                                                                          119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

                                                                                          density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

                                                                                          The depth penetration of laser energy of lead metal was found to be smaller

                                                                                          than that of copper metal this may be due to the high thermal conductivity and

                                                                                          high specific heat capacity of copper with that of lead metal

                                                                                          41

                                                                                          References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                                          Sons Inc Hoboken New Jersey Canada (2003)

                                                                                          [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                                          [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                                          [9]

                                                                                          Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                                          [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                                          [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                                          [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                                          [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                                          [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                                          httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                                          [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                                          [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                                          [12] httpenwikipediaorgwikiThermal_conductivity

                                                                                          [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                                          [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                                          [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                                          42

                                                                                          Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                          This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                          This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                          43

                                                                                          E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                          This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                          This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                          44

                                                                                          title(Dencity as a function of temperature) hold off

                                                                                          This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                          45

                                                                                          elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                          46

                                                                                          r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                          47

                                                                                          for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                          This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                          48

                                                                                          for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                          49

                                                                                          6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                          50

                                                                                          u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                          51

                                                                                          alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                          52

                                                                                          715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                          • 01 Title
                                                                                            • 0TDr Omed Ghareb Abdullah
                                                                                              • Ch1-Laser
                                                                                                • 16 Heat and heat capacity
                                                                                                • 18 Derivation in one dimension
                                                                                                  • Ch2 laser2
                                                                                                  • Ch3 laser2
                                                                                                  • Appendix-Laser

                                                                                            41

                                                                                            References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

                                                                                            Sons Inc Hoboken New Jersey Canada (2003)

                                                                                            [2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

                                                                                            [3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

                                                                                            [9]

                                                                                            Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

                                                                                            [4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

                                                                                            [5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

                                                                                            [6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

                                                                                            [7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

                                                                                            [8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

                                                                                            httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

                                                                                            [10] httpwwwworldoflaserscomlaserprincipleshtm

                                                                                            [11] httpenwikipediaorgwikiLaserPulsed_operation

                                                                                            [12] httpenwikipediaorgwikiThermal_conductivity

                                                                                            [13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

                                                                                            [14] httpwebh01uaacbeplasmapageslaser-ablationhtml

                                                                                            [15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

                                                                                            42

                                                                                            Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                            This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                            This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                            43

                                                                                            E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                            This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                            This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                            44

                                                                                            title(Dencity as a function of temperature) hold off

                                                                                            This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                            45

                                                                                            elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                            46

                                                                                            r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                            47

                                                                                            for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                            This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                            48

                                                                                            for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                            49

                                                                                            6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                            50

                                                                                            u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                            51

                                                                                            alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                            52

                                                                                            715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                            • 01 Title
                                                                                              • 0TDr Omed Ghareb Abdullah
                                                                                                • Ch1-Laser
                                                                                                  • 16 Heat and heat capacity
                                                                                                  • 18 Derivation in one dimension
                                                                                                    • Ch2 laser2
                                                                                                    • Ch3 laser2
                                                                                                    • Appendix-Laser

                                                                                              42

                                                                                              Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                              This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                              This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

                                                                                              43

                                                                                              E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                              This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                              This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                              44

                                                                                              title(Dencity as a function of temperature) hold off

                                                                                              This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                              45

                                                                                              elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                              46

                                                                                              r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                              47

                                                                                              for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                              This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                              48

                                                                                              for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                              49

                                                                                              6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                              This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                              50

                                                                                              u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                              51

                                                                                              alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                              52

                                                                                              715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                              • 01 Title
                                                                                                • 0TDr Omed Ghareb Abdullah
                                                                                                  • Ch1-Laser
                                                                                                    • 16 Heat and heat capacity
                                                                                                    • 18 Derivation in one dimension
                                                                                                      • Ch2 laser2
                                                                                                      • Ch3 laser2
                                                                                                      • Appendix-Laser

                                                                                                43

                                                                                                E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

                                                                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

                                                                                                This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

                                                                                                This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

                                                                                                44

                                                                                                title(Dencity as a function of temperature) hold off

                                                                                                This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                                45

                                                                                                elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                                46

                                                                                                r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                                47

                                                                                                for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                                48

                                                                                                for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                                49

                                                                                                6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                                50

                                                                                                u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                51

                                                                                                alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                52

                                                                                                715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                • 01 Title
                                                                                                  • 0TDr Omed Ghareb Abdullah
                                                                                                    • Ch1-Laser
                                                                                                      • 16 Heat and heat capacity
                                                                                                      • 18 Derivation in one dimension
                                                                                                        • Ch2 laser2
                                                                                                        • Ch3 laser2
                                                                                                        • Appendix-Laser

                                                                                                  44

                                                                                                  title(Dencity as a function of temperature) hold off

                                                                                                  This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

                                                                                                  45

                                                                                                  elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                                  46

                                                                                                  r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                                  47

                                                                                                  for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                  This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                                  48

                                                                                                  for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                                  49

                                                                                                  6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                  This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                                  50

                                                                                                  u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                  51

                                                                                                  alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                  52

                                                                                                  715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                  • 01 Title
                                                                                                    • 0TDr Omed Ghareb Abdullah
                                                                                                      • Ch1-Laser
                                                                                                        • 16 Heat and heat capacity
                                                                                                        • 18 Derivation in one dimension
                                                                                                          • Ch2 laser2
                                                                                                          • Ch3 laser2
                                                                                                          • Appendix-Laser

                                                                                                    45

                                                                                                    elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

                                                                                                    46

                                                                                                    r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                                    47

                                                                                                    for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                    This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                                    48

                                                                                                    for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                                    49

                                                                                                    6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                    This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                                    50

                                                                                                    u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                    51

                                                                                                    alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                    52

                                                                                                    715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                    • 01 Title
                                                                                                      • 0TDr Omed Ghareb Abdullah
                                                                                                        • Ch1-Laser
                                                                                                          • 16 Heat and heat capacity
                                                                                                          • 18 Derivation in one dimension
                                                                                                            • Ch2 laser2
                                                                                                            • Ch3 laser2
                                                                                                            • Appendix-Laser

                                                                                                      46

                                                                                                      r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

                                                                                                      47

                                                                                                      for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                      This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                                      48

                                                                                                      for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                                      49

                                                                                                      6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                      This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                                      50

                                                                                                      u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                      51

                                                                                                      alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                      52

                                                                                                      715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                      • 01 Title
                                                                                                        • 0TDr Omed Ghareb Abdullah
                                                                                                          • Ch1-Laser
                                                                                                            • 16 Heat and heat capacity
                                                                                                            • 18 Derivation in one dimension
                                                                                                              • Ch2 laser2
                                                                                                              • Ch3 laser2
                                                                                                              • Appendix-Laser

                                                                                                        47

                                                                                                        for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                        This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

                                                                                                        48

                                                                                                        for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                                        49

                                                                                                        6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                        This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                                        50

                                                                                                        u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                        51

                                                                                                        alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                        52

                                                                                                        715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                        • 01 Title
                                                                                                          • 0TDr Omed Ghareb Abdullah
                                                                                                            • Ch1-Laser
                                                                                                              • 16 Heat and heat capacity
                                                                                                              • 18 Derivation in one dimension
                                                                                                                • Ch2 laser2
                                                                                                                • Ch3 laser2
                                                                                                                • Appendix-Laser

                                                                                                          48

                                                                                                          for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

                                                                                                          49

                                                                                                          6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                          This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                                          50

                                                                                                          u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                          51

                                                                                                          alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                          52

                                                                                                          715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                          • 01 Title
                                                                                                            • 0TDr Omed Ghareb Abdullah
                                                                                                              • Ch1-Laser
                                                                                                                • 16 Heat and heat capacity
                                                                                                                • 18 Derivation in one dimension
                                                                                                                  • Ch2 laser2
                                                                                                                  • Ch3 laser2
                                                                                                                  • Appendix-Laser

                                                                                                            49

                                                                                                            6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                            This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

                                                                                                            50

                                                                                                            u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                            51

                                                                                                            alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                            52

                                                                                                            715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                            • 01 Title
                                                                                                              • 0TDr Omed Ghareb Abdullah
                                                                                                                • Ch1-Laser
                                                                                                                  • 16 Heat and heat capacity
                                                                                                                  • 18 Derivation in one dimension
                                                                                                                    • Ch2 laser2
                                                                                                                    • Ch3 laser2
                                                                                                                    • Appendix-Laser

                                                                                                              50

                                                                                                              u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

                                                                                                              51

                                                                                                              alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                              52

                                                                                                              715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                              • 01 Title
                                                                                                                • 0TDr Omed Ghareb Abdullah
                                                                                                                  • Ch1-Laser
                                                                                                                    • 16 Heat and heat capacity
                                                                                                                    • 18 Derivation in one dimension
                                                                                                                      • Ch2 laser2
                                                                                                                      • Ch3 laser2
                                                                                                                      • Appendix-Laser

                                                                                                                51

                                                                                                                alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

                                                                                                                52

                                                                                                                715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                                • 01 Title
                                                                                                                  • 0TDr Omed Ghareb Abdullah
                                                                                                                    • Ch1-Laser
                                                                                                                      • 16 Heat and heat capacity
                                                                                                                      • 18 Derivation in one dimension
                                                                                                                        • Ch2 laser2
                                                                                                                        • Ch3 laser2
                                                                                                                        • Appendix-Laser

                                                                                                                  52

                                                                                                                  715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

                                                                                                                  • 01 Title
                                                                                                                    • 0TDr Omed Ghareb Abdullah
                                                                                                                      • Ch1-Laser
                                                                                                                        • 16 Heat and heat capacity
                                                                                                                        • 18 Derivation in one dimension
                                                                                                                          • Ch2 laser2
                                                                                                                          • Ch3 laser2
                                                                                                                          • Appendix-Laser

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