Solved HT problems.ppt

1PHYS1001Physics 1 REGULARModule 2 Thermal Physics

METHODS OF HEAT TRANSFER

CONDUCTION

CONVECTION

RADIATION

ptD_transfer.ppt

2

Overview of Thermal Physics Module:1. Thermodynamic Systems:

Work, Heat, Internal Energy0th, 1st and 2nd Law of Thermodynamics

2. Thermal Expansion3. Heat Capacity, Latent Heat4. Methods of Heat Transfer:

Conduction, Convection, Radiation5. Ideal Gases, Kinetic Theory Model6. Second Law of Thermodynamics

Entropy and Disorder7. Heat Engines, Refrigerators

3 METHODS OF HEAT TRANSFER energy transfer (heat, Q) due to a temperature difference, T

CONDUCTION CONVECTION RADIATION

§17.7 p591

system, T

Qnet

Environment, TE

European heat wave, 2003 ~35 000 deaths in France

References: University Physics 12th ed Young & Freedman

4

Live sheep tradeSunday, October 26, 2003

Sheep to shore ... finally

The Labor Opposition will pursue the Government over the cost of the "ship of death" saga, which ended on Friday when 50,000 Australian sheep at sea for three months began being unloading in Eritrea.

5Heat Conduction

Conduction is heat transfer by means of molecular agitation within a material without any motion of the material as a whole.

If one end of a metal rod is at a higher temperature, then energy will be transferred down the rod toward the colder end because the higher speed particles will collide with the slower ones with a net transfer of energy to the slower ones. TH

TC

conduction though glass

Q

Q

6

For conduction between two plane surfaces (eg heat loss through the wall of a house) the rate of heat transfer is

energy transferred

through slabQ

THTC

L

H CQ T Tk At L

dQ dTk Adt dx

Thermal conductivity k (W.m-1.K-1)

steady-state

heat current H = dQ/dt

Q QA

7

H CQ T Tk At L

dQ dTk Adt dx

steady-state

Thermal Conduction through a uniform slab

0 xLTC

TH

temperature gradient dTdx

8Material Thermal conductivity k (W.m-1.K-1)

diamond 2450

Cu 385

Al 205

Brick 0.2

Glass 0.8

Body fat 0.2

Water 0.6

Wood 0.2

Styrofoam 0.01

Air 0.024

Thermal conductivity, k property of the material

kdiamond very high: perfect heat sink, e.g. for high power laser diodes

khuman low: core temp relatively constant (37 oC)

kair very low: good insulator * home insulation * woolen clothing * windows double glazing Metals – good conductors: electrons transfer energy from hot to cold

9Heat Convection

Convection is heat transfer by mass motion of a fluid such as air or water when the heated fluid is caused to move away from the source of heat, carrying energy with it.

Convection above a hot surface occurs because hot air expands, becomes less dense and rises (natural or free). Convection assisted by breeze, pump or fan – forced convection.

Hot water is likewise less dense than cold water and rises, causing convection currents which transport energy.

Convection coefficient, hT between surface and air way from surface2 1~ ( )dQ h A T T

dt

10

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html

11Sea & Land Breezes, Monsoons

http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

35 oC 20 oC 17 oC11 oC

What is the role of heat capacity, c of water and soil?

12RADIATION Energy transferred by electromagnetic waves

All materials radiate thermal energy in amounts determined by their temperature, where the energy is carried by photons of light in the infrared and visible portions of the electromagnetic spectrum.

Thermal radiation wavelength ranges: IR ~ 100 - 0. 8 mVisible ~ 0.8 - 0.4 m 800 – 400 nmUV ~0.4 - 0.1 m

For exam: more detail than in the textbook

13Ludwig Boltzmann (1844-1906)

All objects above absolute zero emit radiant energy and the rate of emission increases and the peak wavelength decreases as the temperature of object increases

14

Thermography

15

Power absorbed by surface of an object

A, a

Qabs4abs

abs sdQP Aa T

dt

Ts

Absorption & Stefan-Boltzmann Law

Incident radiation (INTENSITY I - energy passing through a square metre every second Iinc = P / A Iabs = a Iinc

• Surface Area, A• Absorption coefficient, a = 0 to 1• Stefan-Boltzmann constant

σ = 5.67 x 10-8 W.m-2.K-4

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Power radiated from the surface of an object

A, e, T Qrad4radrad

dQP Ae Tdt

net rad absPP P

Emission & Stefan-Boltzmann Law

Pnet > 0 net heat transfer out of system

• Surface Area, A• Emissivity, e = 0 to 1• Stefan-Boltzmann constant

σ = 5.67 x 10-8 W.m-2.K-4

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A blackbody absorbs all the radiation incident upon it and emits the max possible radiation at all wavelengths (e = a = 1)

A graybody is a surface that absorbs a certain proportion of the energy of a blackbody, the constant being constant over the entire band of wavelengths(0 e = a < 1)

emissivity e

absorption coefficient (absorptivity) a

Stefan-Boltzmann constant = 5.6710-8 W.m-2.K-4

18Stefan-Boltzmann constant, = 5.67 x 10-8 W.m-2.K-4

* emissivity, e = 0 to 1 Blackbody, e = 1* Absorption coefficient, a = 0 to 1 * At a temperature T a = e all wavelengths* T > 700 oC visible radiation (dull red ~ 800 oC white ~ 2000 oC)* Black surface (e ~ 1) – good emitter / absorber* Polished surface (e ~ 0.01) – poor emitter / absorber, good reflector * Hot stars – blue* Cool stars - red

Water (e ~ 0.96) Earth (e ~ 0.3)

19

Problem D.4Semester 1, 2007 Examination

An igloo is a hemispherical enclosure built of ice. Elmo’s igloo has an inner radius of 2.55 mm and the thickness of the ice is 0.30 m. This thickness can be considered small compared to the radius. Heat leaks out of the igloo at a rate determined by the thermal conductivity of ice, kice = 1.67 W.m-1.K-1.

At what rate must thermal energy be generated inside the igloo to maintain a steady air temperature inside the igloo at 6.5 oC when the outside temperture is -40 oC?

Ignore all thermal energy losses by conduction through the ground, or any heat transfer by radiation or convection or leaks.

20SolutionIdentify / Setup

thickness t = 0.30 m

radius r = 2.55 mkice = 1.67 W.m-1.K-1

dQ dTk Adt dx

The rate of energy production must be equal to the rate of loss of thermal energy by conduction through the hemispherical ice wall.

Rate of energy transfer by conduction

6.5 oC

-40 oC

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dQ/dt = ? W

k = 1.67 W.m-1.K-1

dT = {6.5 – (– 40)} oC = 46.5 oC

thickness of ice, dx = 0.30 m

area, A = surface area of hemisphere = (4 R2) / 2 = 2 R2

Because the thickness of the ice is much smaller than either the inside or outside radius, it does not matter which radius is used – taking the average radius

R = (2.55 + 0.15) m = 2.70 m

22Execute

dQ/dt = – (1.67)(2)(2.70)2(46.5)/0.30 W

dQ/dt = – 1.2 104 W

dQ dTk Adt dx

Evaluatesensible valueunitssignificant figuresdid I answer the question ?

23

Problem D.5

Suppose a human could live for 120 min unclothed in air at 8 oC.

How long could they live in water at 8 oC?

24Problem D.5Suppose a human could live for 120 min unclothed in air at 8 oC. How long could they live in water at 8 oC?

Solution

Identify / SetupThermal conductivities kair = 0.024 W.m-1.K-1 kwater = 0.6 W.m-1.K-1

Execute

Evaluate

&

0.024(120) min 4.8 min0.6

W AW A

W A AW A

A W W

Q T Q T Q Tk A k A k At x t x t x

t k kt tt k k

25

Why do droplets of water dance over the very hot pan ?


Problem D.6

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Why do droplets of water dance over the very hot pan ?


Water at the bottom of the drops is evaporated and provides insulation against further evaporation.

27

Problem D.7

An aluminium pot contains water that is kept steadily boiling (100 ºC). The

bottom surface of the pot, which is 12 mm thick and 1.5104 mm2 in area,

is maintained at a temperature of 102 °C by an electric heating unit. Find

the rate at which heat is transferred through the bottom surface. Compare

this with a copper based pot. The thermal conductivities for aluminium and

copper are

kAl = 235 W.m-1.K-1 and kCu = 401 W.m-1.K-1

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SolutionIdentify / Setup

TH = 102 oC

TC = 100 oC

Base area A = 1.5x104 mm2 = 1.5x10-2 m2

Base thickness L = 12 mm = 12x10-3 m

kAl = 235 W.m-1.K-1

kCu = 401 W.m-1.K-1

dT/dx = (TH – TC) / L

dQ/dt = ? W

dQ dTk Adt dx

Execute

Al, dQ/dt = 5.9x102 W

Cu, dQ/dt = 1.0x103 W

Cu pots ~ 2 times more efficient

29“Body Heat”Heat lost by convection is very important for humansFor a naked person, h ~ 7.1 W.m-2.K-1

Assume the person’s surface area is 1.5 m2, skin temperature of 33 oC and surrounding temperature 29 oC.

A = 1.5 m2 T = 33 oC TE = 29 oC

dQ/dtconvection = h A T = (7.1)(1.5)(4) W = 43 W

If there is a breeze, convection losses are greater – wind chill factorViscosity of fluids slows natural convection near a stationary surface by producing a boundary layer which has about the same insulating values of 10 mm plywood.

30For the naked person, estimate the net rate of energy radiated. = 1.5 m2 T = 33 oC = (33 + 273) K = 306 KTE = (29 + 273) K = 302 K assume e = 1 = 5.67x10-8 W.m-2.K-4

Pradiation = e A (T 4 – TE4) = 39 W

dQ/dtloss = dQ/dtradiation + dQ/dtconvection = 39 W + 43 W = 82 W

Ans. similar to the rate at which heat is generated by the

body when resting

31

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html

Why not heat the water at the top?

CONVECTION

32


Warm air rises above the groundForced convection

CONVECTION

33

Why do we get this pollution haze?

Temperature inversion prevents air rising and the dispersing the pollution


CONVECTION

34

Why are the cooling coils at the top ?

In bathrooms, the heater is often near the ceiling. Problem ?


CONVECTION

Solved HT problems.ppt

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Solved HT problems.ppt