L1condCartesian2015-2

Heat Transfer and Fluid FlowMMB801

Joanna Szmelter - Heat transfer

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CLASS TEST ARRANGEMENT 2014-2015The 45 minutes, open book, Class Test will take place on Friday in week6 13th March 2015 at 09:00 in rooms T003 (surnames beginning with A-J) and SMB014(surnames beginning with K-Z). 

Students with alternative arrangements will be individually informed by e-mail about their room allocation.

 The test cannot be moved.  If for exceptional reasons you are unable to come to the test, please follow the impaired performance procedure.

Heat TransferRecommended texts

Recommended reading.

C.A. Long “Essential Heat Transfer”, Longman 1999.

(Chapters 1,2 & 9)Y. Bayazitoglu, M. Nevati Ozisik, “Elements of Heat

Transfer”, McGraw-Hill Book Company, 1998. Y. A. Cengel “Heat Transfer. A Practical Approach”,

“Elements of Heat Transfer”, McGraw-Hill Book Company, 2003. (Chapters 1,2,3 &4)

Heat Transfer in Design and Manufacturing

Maximize desired heat transfer:- heating of materials for processing - household applications (kettle, iron, etc.)- cooling of electrical components

http://www.appliancist.comhttp://www.imaging1.com/thermal/images/Nikon-P4-chip-thermal.jpg

http://www.gspsteelprofiles.com

Heat Transfer in Design and Manufacturing

Minimize undesired heat transfer:- Maintaining a temperature difference

Thermos; Refrigerator; Windows

http://www.bbc.co.uk/schools/gcsebitesize/science/aqa_pre_2011/energy/heatrev3.shtml

http://www.appliancist.com

Everyday Example: A thermos Flask

Energy transfer (Q)from teato surroundings

T(tea)= 80oCT(surroundings) = 20oC

Poorly insulated; cools rapidly

Tea(150 ml,

80oC)

Insulation reduces conduction (often a vacuum)

Stopper; prevents convection

Reflective surface reduces radiation(often stainless steel)

Vtea = 150 ml

How much energy is transferred?How long does it take?

ThermodynamicsHeat Transfer

Why this design? How does it work?

What is Energy? Energy is a property of a substance

ALL substances have energy Various forms: kinetic (motion); potential; chemical

INTERNAL energy (U (J) or u (J/kg) where mass specific): Energy associated with substance’s constituent

molecules Increases with temperature

Energy is never created or destroyed, it is just changed from one form to another Or it is TRANSFERRED between substances

What is Heat Transfer?

Definition: heat transfer is the energy transfer due to a temperature difference

Study the ability of materials and components to either transmit, absorb, or prevent the transfer of heat

A RATE of energy transfer (units: J/s = W)

Related to thermodynamics:1st law: Energy conservation Law: The energy transferred out of one object = energy transferred into another object2nd law: Entropy: heat is transferred from the hot object to the cold object

Material Properties: Heat Capacity

Specific Heat (Cp): Amount of energy to increase the temperature of a given

mass of a substance (normally increasing 1 kg by 1 K)

Heat Capacity (C): Amount of energy to increase the temperature of a given

volume of a substance

Cp = Q/(mT) [J/kgK]

C= r*Cp [J/m3K]

Represents a substance’s capacity to store energy(Change in Specific Internal Energy =Du = CpDT)

A Heat Capacity Example: The Mug v The Thermos

V = 150 mlTinitial = 90°CTsurr. = 20°CCp(tea) = 4.2 kJ/kgKrtea = 980 kg/m3

Find the change in internal energy of the tea after 1 hour if:

a) The tea in the mug has cooled to 25°C

b) The tea in the thermos has cooled to 85°C

Answer: Part a): For the mug

Change in Specific Internal Energy =Du = CpDT Du = 4.2kJ/kgK * (90°C – 25°C) Du = 273kJ/kg

Change in internal energy of the tea in the mug = DU DU = m*Du = (r*V)*Du DU = 980 kg/m3 * 1.5 x 10-4 m3 * 273 kJ/kg DU = 40.1kJ

Part b): For the Thermos As above: Du = 21kJ/kg so DU = 3.1 kJ

The change in Total internal energy (DU) is the amount of heat transferred from the drink to the surroundings

What is heat transfer?

Study of the movement of heat energy from one place to another by:– Conduction – Convection – Radiation

Heat transfer occurs in many engineering applications and usually is a combination of the three heat transfer methods is present e.g. Engines Refrigeration and heating devicesManufacturing processesMechanical processesAnd from a small to a large scale – from solar systems to microelectronics

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Examples: Finite element solution of conduction equations

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ExamplesUrban boundary layers

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numerical models: applications to atmospheric flows

across a range of scales and physics

ΔX O(10-2) m O(102) m O(104) m O(107) m

Cloud turbulence Gravity waves Global flows Solar convection

Heat transfer by CONDUCTION

The transfer of heat energy, within a stationary medium

FROM a high temperature region (high molecular energy)

TO a low temperature region (low molecular energy)

Heat transfer has a direction as well as magnitude

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Mechanisms of Conduction

In solids – due to collision (vibrations) and motion of free electrons. This is why good conductors of electricity are good conductors of heat

In liquids and gases – due to molecular collision and molecular diffusion

Material Properties: Thermal Conductivity

Ability of a material to conduct heat Represented by k (units are W / m*K)

Definition: The rate at which heat is transferred through a unit thickness of a material per unit area per unit temperature difference

Related to collisions between molecules (gases, liquids) and molecular vibration (solids)

k varies between 0.026 W/mK for air and 2300 W/mK for diamond (graphite is less).

Liquids are typically in between: water ~0.61 W/mK Depends mainly on temperature but also pressure

Thermal Conductivity in Solids

Electron flow – dominant for metals (free electrons) Lattice vibration – depends on lattice structure

TLTH Tequal

Free electrons transfer energy(move from TH to TL)

Material k (W/mK)Air 0.024Water 0.58Copper 401Nickel 91Copper/nickel alloy (55/45) 23Silver 429Diamond(depends on orientation)

900-2300

Heat transfer by CONDUCTION

Q

Governed by FOURIER’S LAW, based on experimental observation

Where: is the rate of heat energy transfer (rate of heat flow) [W] or [J/s ] k is thermal conductivity [W/(K·m)], x is a linear distance [m]A is an area perpendicular to x [m²] T is temperature [oC or K].

Heat transfer by CONDUCTION per unit area

q

Where: is the heat flux i.e. the rate of heat energy transfer per unit area (W/m2),

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Materials k, W/(m C) at 300 K

Copper 401

Nickel 91

Aluminium 237

Bronze 52

Steel (mild) 70

Insulation materials 0.05 – 1.0

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Conduction The transfer of energy within a substance More energetic ‘particles’ pass energy to adjacent lower-

energy ‘particles’ Applies to solids, liquids, and vapours

Rate depends on: Material (thermal conductivity, k), thickness, orientation,

temperature difference, surface area

Thigh

Tlow

xTkAQconduction D

DhighQ lowQ

lowhigh QQ lowhigh TT

xD

xTkAQconduction D

D

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Rate of Change of Internal

Energy due to Conduction

+ Rate of Energy Generation in the element

= Rate of increase of internal energy over

time dt

tTmc

tE

tEQQQ gxxx

D

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Rate of increase of internal energy over time dt

tTcxA

tTcV

tTmc

tE

D

rr

Where m ≡ mass (kg)r ≡ density of the material (kg/m3)V ≡ volume of the element (m3)c ≡ specific heat (J/kg K)T ≡ temperature (K)t ≡ time (s)

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Derivation of 1D heat conduction equation

Apply an energy balance for the rate of

energy transfer

Taylor expansion gives

.

xQx

QQ

xQx

QQ

xQx

QQ

xxxx

xxxx

xxxx

D

D

D

D

D

D

)(

)(

)(

tTcxA

tEQQQ gxxx

D

D r

...!3

)(!2

)()()()( 3

3

2

2

D

D

D

D

xx

xfxx

xfxxxfxfxxf

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Derivation of 1D heat conduction equation

xAxTk

xQQ xxx D

D )(

tTcxAQQQ gxxx

D D r

From Taylor expansion

From Fourier law

Note also that

Returning to an energy balance for the rate of energy

transfer

1D heat conduction equation

. tTcg

xTk

x

r)(

Volumetric heat generation term W/m³gg

xTkAQx

xQx

QQ xxxx D

D )(

xAgQg D

xAD

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Heat conduction equation in Cartesian coordinates

1D conduction equation

3D conduction equation

tTcg

xTk

x

r)(

tTcg

zTk

zyTk

yxTk

x

r)()()(

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1D Heat conduction equation in Cartesian coordinates – SPECIAL CASES

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Kelvin temperature conversion formulae

from Kelvin to Kelvin Celsius

[ ] = [℃ K] − 273.15

[K] = [ ] + 273.15℃

Problems - Cartesian heat conductionExample 1.

The ends of a steel bar with thermal conductivity 60 W/K m are maintained at 100 oC and 500 oC. The length of the bar is 1 m. Find the temperature distribution.

T1 = 100 oC

T2 = 500 oC tTcg

xTk

x

r)(

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Subject to boundary conditions:

400

xT

02

2

xT

1CxT

21 CxCT

and temperature distribution:

4001001500

1000100

112

2211

CCT

CCCT

mxatCTT

mxatCTT

1500

0100

2

1

T1 = 100 oC, x=0

T2 = 500 oC

100400 xT

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x

T

0 1

100

500

slope=400

100400 xT400

xT

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Subject to boundary conditions

400

xT

02

2

xT

1CxT

21 CxCT

Temperature distribution:

4005001100

5000500

1211

2212

CCT

CCCT

mxatCTT

mxatCTT

0500

1100

2

1

T1 = 100 oC,

T2 = 500 oC x=0

Reverse the direction of the coordinate system

500400 xT