Chapter 18. Heat TransferChapter 18. Heat TransferA PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

TRANSFER OF HEAT is minimized by multiple layers of beta cloth. These and other insulating materials protect spacecraft from hostile environmental conditions. (NASA)

Objectives: After finishing this Objectives: After finishing this unit, you should be able to:unit, you should be able to:

•• Demonstrate your understanding of Demonstrate your understanding of conductionconduction, , convectionconvection, and, and radiationradiation, , and give examples.and give examples.

•• Solve Solve thermal conductivitythermal conductivity problems problems based on quantity of heat, length of based on quantity of heat, length of path, temperature, area, and time.path, temperature, area, and time.

•• Solve problems involving the Solve problems involving the rate of rate of radiationradiation and and emissivityemissivity of surfaces.of surfaces.

Heat Transfer by ConductionHeat Transfer by ConductionConduction is the process by which heat energy is transferred by adjacent molecular collisions inside a material. The medium itself does not move.

Conduction Direction

From hot to cold.

Heat Transfer by ConvectionHeat Transfer by Convection

Convection is the process by which heat energy is transferred by the actual mass motion of a heated fluid.

ConvectionHeated fluid rises and is then Heated fluid rises and is then replaced by cooler fluid, producing replaced by cooler fluid, producing convection currentsconvection currents..

Convection is significantly affected Convection is significantly affected by by geometry geometry of heated surfaces. of heated surfaces. (wall, ceiling, floor)(wall, ceiling, floor)

Heat Transfer by RadiationHeat Transfer by Radiation

Radiation

Sun

Radiation is the process by which heat energy is transferred by electromagnetic waves.

Atomic

No medium is required !No medium is required !

Kinds of Heat TransferKinds of Heat TransferConsider the operation of a typical coffee maker:Consider the operation of a typical coffee maker:

Think about how heat is Think about how heat is transferred by:transferred by:

Conduction?Conduction?

Convection?Convection?

Radiation?Radiation?

Heat CurrentHeat Current

SteamSteam IceIce

( / )QH J s

The The heat currentheat current HH is defined as the quantity is defined as the quantity of heat of heat QQ transferred per unit of time transferred per unit of time in the in the direction from high temperature to low direction from high temperature to low temperature.temperature.

Typical units are: Typical units are: J/s, cal/s, and Btu/hJ/s, cal/s, and Btu/h

H = Heat current (J/s)

A = Surface area (m2)

t = Temperature difference

L = Thickness of material

Thermal ConductivityThermal Conductivity

t1 t2

t = t2 - t1

The The thermal conductivity kthermal conductivity k of of a material is a measure of its a material is a measure of its ability to conduct heat.ability to conduct heat.

QLkA t

Q kA tHL

0

Js m C

Units

The SI Units for ConductivityThe SI Units for Conductivity

HotHot ColdCold QLkA t

For Copper: k = 385 J/s m C0For Copper: k = 385 J/s m C0

Taken literally, this means that for a Taken literally, this means that for a 11--mm length length of copper whose cross section is of copper whose cross section is 1 m1 m22 and and whose end points differ in temperature by whose end points differ in temperature by 1 C1 C00, , heat will be conducted at the rate of heat will be conducted at the rate of 1 J/s1 J/s..

In SI units, typically small measures for length L and area A must be converted to meters and square meters, respectively, before substitution into formulas.

In In SI unitsSI units, typically small measures for , typically small measures for lengthlength LL and and area Aarea A must be converted to meters and must be converted to meters and square meters, respectively, before substitution square meters, respectively, before substitution into formulas.into formulas.

Older Units for ConductivityOlder Units for Conductivity

Taken literally, this means that for a Taken literally, this means that for a 11--in.in. thick thick plate of glass whose area is plate of glass whose area is 1 ft1 ft22 and whose and whose sides differ in temperature by sides differ in temperature by 1 F1 F00, heat will be , heat will be conducted at the rate of conducted at the rate of 5.6 Btu/h5.6 Btu/h..

t = 1 F0

L = 1 in.

A=1 ft2

Q=1 Btu

h

Older units, still active, use Older units, still active, use common measurements for common measurements for area in area in ftft22 time in time in hourshours, , length in length in secondsseconds, and , and quantity of heat in quantity of heat in BtuBtu’’ss..

Glass k = 5.6 Btu in./ftGlass k = 5.6 Btu in./ft22h Fh F00

Thermal ConductivitiesThermal ConductivitiesExamples of the two systems of units used for

thermal conductivities of materials are given below:

Copper:Copper:

Concrete or Concrete or Glass:Glass:

Corkboard:Corkboard:

385385 26602660

0.8000.800 5.65.6

0.0400.040 0.300.30

MaterialMaterial oJ/s m C 2 0Btu in/ft h F

Examples of Thermal ConductivityExamples of Thermal Conductivity

Aluminum:Aluminum:

Comparison of Heat Currents for Similar Conditions: L = 1 cm (0.39 in.); A = 1 m2 (10.8 ft2); t = 100 C0

Copper:Copper:

Concrete or Concrete or Glass:Glass:

Corkboard:Corkboard:

2050 kJ/s2050 kJ/s 4980 Btu/h4980 Btu/h

3850 kJ/s3850 kJ/s 9360 Btu/h9360 Btu/h

8.00 kJ/s8.00 kJ/s 19.4 Btu/h19.4 Btu/h

0.400 kJ/s0.400 kJ/s 9.72 Btu/h9.72 Btu/h

Example 1:Example 1: A large glass window measures A large glass window measures 2 m2 m wide and wide and 6 m6 m high. The inside surface is high. The inside surface is at at 202000CC and the outside surface is at and the outside surface is at 121200CC. . How many joules of heat pass through this How many joules of heat pass through this window in window in one hourone hour? Assume ? Assume L =L = 1.5 cm1.5 cm and that and that k = 0.8 J/s m Ck = 0.8 J/s m C00. .

200C 120C

t = t2 - t1 = 8 C0

0.015 m

AQ = ?= 1 h

A = (2 m)(6 m) = 12 mA = (2 m)(6 m) = 12 m22

; Q kA t kA tH QL L

0 2 0(0.8 J/m s C )(12 m )(8 C )(3600 s)0.0150 m

Q

Q = 18.4 MJQ = 18.4 MJ

Example 2:Example 2: The wall of a freezing plant is The wall of a freezing plant is composed of composed of 8 cm8 cm of corkboard and of corkboard and 12 cm12 cm of solid concrete. The inside surface is at of solid concrete. The inside surface is at --202000CC and the outside surface is and the outside surface is +25+2500CC. . What is the interface temperature What is the interface temperature ttii ??

ttii 252500CC--202000CC

HH AA

8 cm 12 cm8 cm 12 cm

Steady Steady FlowFlow

Note:Note:Cork Concrete

H HA A

0 01 2

1 2

( 20 C) 25 C - L L

i ik t k t

0 01 2

1 2

( 20 C) (25 C - )L L

i ik t k t

Example 2 (Cont.):Example 2 (Cont.): Finding the interface Finding the interface temperature for a composite wall.temperature for a composite wall.

ttii 252500CC--202000CC

HH AA

8 cm 12 cm8 cm 12 cm

Steady Steady FlowFlow

0 01 2

1 2

( 20 C) (25 C - )L L

i ik t k t

Rearranging factors gives:Rearranging factors gives:

0 01 2

2 1

L ( 20 C) (25 C - )L i i

k t tk

01 2

02 1

L (0.04 W/m C )(0.12 m) 0.075L (0.8 W/m C )(0.08 m)

kk

Example 2 (Cont.):Example 2 (Cont.): Simplifying, we obtain:Simplifying, we obtain:

ttii 252500CC--202000CC

HH AA

8 cm 12 cm8 cm 12 cm

Steady Steady FlowFlow

0 0(0.075)( 20 C) (25 C - )i it t

0.0750.075ttii + + 1.51.500C = 25C = 2500C C -- ttii

From which:From which: ti = 21.90Cti = 21.90C

Knowing the interface temperature Knowing the interface temperature ttii allows us to determine the allows us to determine the rate of rate of heat flowheat flow per unit of area, H/Aper unit of area, H/A..

The quantity The quantity H/AH/A is same for cork or concrete:is same for cork or concrete:

H; A

Q kA t k tHL L

Example 2 (Cont.):Example 2 (Cont.): Constant steady state flow.Constant steady state flow.

ttii 252500CC--202000CC

HH AA

8 cm 12 cm8 cm 12 cm

Steady Steady FlowFlow

H; A

Q kA t k tHL L

Over time Over time H/AH/A is constant so is constant so different different kk’’ss cause different cause different tt’’ss

Cork:Cork: tt = 21.9= 21.900C C -- ((--202000C) = C) = 41.9 C41.9 C00

Concrete:Concrete: tt = 25= 2500C C -- 21.921.900C = C = 3.1 C3.1 C00

Since H/A is the same, letSince H/A is the same, let’’s just choose concrete alone:s just choose concrete alone:

0 0H (0.8 W/mC )(3.1 C )A 0.12 m

k tL

220.7 W/mHA

Example 2 (Cont.):Example 2 (Cont.): Constant steady state flow.Constant steady state flow.

ttii 252500CC--202000CC

HH AA

8 cm 12 cm8 cm 12 cm

Steady Steady FlowFlow

Cork:Cork: tt = 21.9= 21.900C C -- ((--202000C) = C) = 41.9 C41.9 C00

Concrete:Concrete: tt = 25= 2500C C -- 21.921.900C = C = 3.1 C3.1 C00

2 20.7 W/mHA

Note that Note that 20.7 Joules20.7 Joules of heat per of heat per secondsecond pass through the composite pass through the composite wall. However, the temperature wall. However, the temperature interval between the faces of the interval between the faces of the cork is cork is 13.5 times13.5 times as large as for the as large as for the concrete faces.concrete faces.

If A = 10 m2, the heat flow in 1 h would be ______?

If A = 10 m2, the heat flow in 1 h would be ______?745 kW745 kW

RadiationRadiationThe The rate of radiationrate of radiation R R is the energy emitted per is the energy emitted per unit area per unit time (power per unit area).unit area per unit time (power per unit area).

Q PRA A

Rate of RadiationRate of Radiation (W/m(W/m22):):

Emissivity, e : 0 > e > 1Emissivity, e : 0 > e > 1

Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4

Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4

4PR e TA

Example 3:Example 3: A spherical surface A spherical surface 12 cm12 cm in radius is heated to in radius is heated to 62762700CC. The emissivity is . The emissivity is 0.120.12. . What power is radiated?What power is radiated?

2 24 4 (0.12 m)A R

A = 0.181 mA = 0.181 m22

T = 627 + 273; T = 627 + 273; T = T = 900 K900 K

4P e AT-8 4 2 4(0.12)(5.67 x 10 W/mK )(0.181 m )(900 K)P

P = 808 WP = 808 WPower Radiated from Surface:Power Radiated from Surface:

A

6270C

Find Power Radiated

Summary: Heat TransferSummary: Heat Transfer

ConvectionConvection is the process by is the process by which heat energy is which heat energy is transferred by the actual transferred by the actual mass motion of a heated mass motion of a heated fluid.fluid.

Conduction: HeatConduction: Heat energy is energy is transferred by adjacent molecular transferred by adjacent molecular collisions inside a material. The collisions inside a material. The medium itself does not move.medium itself does not move.

Radiation is the process by which heat energy is transferred by electromagnetic waves.

Summary of Thermal ConductivitySummary of Thermal Conductivity

H = Heat current (J/s)

A = Surface area (m2)

t = Temperature difference

L = Thickness of material

t1 t2

t = t2 - t1

The The thermal conductivity kthermal conductivity k of of a material is a measure of its a material is a measure of its ability to conduct heat.ability to conduct heat.

QLkA t

Q kA tHL

0

Js m C

Units

Summary of RadiationSummary of Radiation

Rate of RadiationRate of Radiation (W/m(W/m22):):

The The rate of radiationrate of radiation R R is the energy emitted per is the energy emitted per unit area per unit time (power per unit area).unit area per unit time (power per unit area).

Q PRA A

Emissivity, e : 0 > e > 1Emissivity, e : 0 > e > 1

Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4

Stefan-Boltzman Constant : = 5.67 x 10-8 W/m·K4

4PR e TA

R

Summary of FormulasSummary of FormulasQLk

A t

Q kA tH

L

0

Js m C

Units

H; A

Q kA t k tHL L

Q PRA A

4PR e TA

4P e AT

CONCLUSION: Chapter 18CONCLUSION: Chapter 18 Transfer of HeatTransfer of Heat