## Chapter 18. Heat Transfer

Chapter 18. Heat Transfer. A PowerPoint Presentation by. Paul E. Tippens, Professor of Physics. Southern Polytechnic State University. © 2007 ...

Chapter 18. Heat Transfer A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern 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 unit, you should be able to: • Demonstrate your understanding of conduction, convection, and radiation, and give examples. •

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

Solve problems involving the rate of radiation and emissivity of surfaces.

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

Heated fluid rises and is then replaced by cooler fluid, producing convection currents. Convection is significantly affected by geometry of heated surfaces. (wall, ceiling, floor)

Convection

Heat Transfer by Radiation Radiation is the process by which heat energy is transferred by electromagnetic waves. Radiation Atomic No medium is required !

Sun

Kinds of Heat Transfer Consider the operation of a typical coffee maker: Think about how heat is transferred by: Conduction? Convection? Radiation?

Heat Current The heat current H is defined as the quantity of heat Q transferred per unit of time  in the direction from high temperature to low temperature. Steam

Ice

H

Q

( J / s)

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

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

t1

t2

H = Heat current (J/s) A = Surface area (m2) t = Temperature difference

t = t2 - t1

L = Thickness of material

kAt H   L Q

QL k A t

J Units  s  m  C0

The SI Units for Conductivity Hot

Cold

QL k A t

For For Copper: Copper: kk = = 385 385 J/s J/s m m CC00 Taken literally, this means that for a 1-m length In SI units , typically small measures for length LL In SI units, typically small measures for length 2 of copper whose cross section is 1 m and and area AA must be converted to meters and and area must be converted to meters and whose end points differ in temperature by 1 C0, square meters, respectively, before substitution square meters, respectively, before heat will be conducted at the rate ofsubstitution 1 J/s. into into formulas. formulas.

Older Units for Conductivity t = 1 F0

h A=1 ft2 Q=1 Btu L = 1 in.

Older units, still active, use common measurements for area in ft2 time in hours, length in seconds, and quantity of heat in Btu’s. Glass k = 5.6 Btu in./ft2h F0

Taken literally, this means that for a 1-in. thick plate of glass whose area is 1 ft2 and whose sides differ in temperature by 1 F0, heat will be conducted at the rate of 5.6 Btu/h.

Thermal Conductivities Examples of the two systems of units used for thermal conductivities of materials are given below: Material

J/s  m  C

o

Btu  in/ft 2  h  F0

Copper:

385

2660

Concrete or Glass:

0.800

5.6

Corkboard:

0.040

0.30

Examples of Thermal Conductivity Comparison of Heat Currents for Similar Conditions: L = 1 cm (0.39 in.); A = 1 m2 (10.8 ft2); t = 100 C0 2050 kJ/s

4980 Btu/h

3850 kJ/s

9360 Btu/h

Concrete or Glass:

8.00 kJ/s

19.4 Btu/h

Corkboard:

0.400 kJ/s

9.72 Btu/h

Aluminum: Copper:

Example 1: A large glass window measures 2 m wide and 6 m high. The inside surface is at 200C and the outside surface is at 120C. How many joules of heat pass through this window in one hour? Assume L = 1.5 cm and that k = 0.8 J/s m C0. A = (2 m)(6 m) = 12

m2

kAt kAt H  ; Q L L  Q

(0.8 J/m  s  C0 )(12 m 2 )(8 C0 )(3600 s) Q 0.0150 m

Q Q == 18.4 18.4 MJ MJ

200C

120C

A = 1 h Q=? t = t2 - t1 = 8 C0 0.015 m

Example 2: The wall of a freezing plant is composed of 8 cm of corkboard and 12 cm of solid concrete. The inside surface is at -200C and the outside surface is +250C. What is the interface temperature ti? H H  ti   0 0C Note:   20 C 25  A Cork  A Concrete

k1 ti  (200 C)  L1

k2  250 C - ti  L2

k1 (ti  20 C) k2 (25 C - ti )  L1 L2 0

0

H A Steady Flow 8 cm 12 cm

Example 2 (Cont.): Finding the interface temperature for a composite wall. k1 (ti  200 C) k2 (250 C - ti ) -200C ti 250C

L1

L2 H A

Rearranging factors gives:

k1L 2 (ti  200 C)  (250 C - ti ) k2 L1

Steady Flow 8 cm 12 cm

k1L 2 (0.04 W/m  C0 )(0.12 m)   0.075 0 k2 L1 (0.8 W/m  C )(0.08 m)

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

(0.075)(ti  20 C)  (25 C - ti ) 0

0

-200C

ti

250C

0.075ti + 1.50C = 250C - ti

H A

tti i == 21.9 21.900CC

From which:

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

Steady Flow 8 cm 12 cm

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

kAt H k t H  ;   L A L Q

Example 2 (Cont.): Constant steady state flow. Over time H/A is constant so -200C different k’s cause different t’s Cork: t = 21.90C - (-200C) = 41.9 C0 Concrete: t = 250C - 21.90C = 3.1 C0

kAt H k t H  ;   L A L Q

ti

250C H A

Steady Flow 8 cm 12 cm

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

H k t (0.8 W/mC0 )(3.1 C0 )   A L 0.12 m

H  20.7 W/m 2 A

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

H 2  20.7 W/m A Cork: t =

21.90C

-

(-200C)

= 41.9

-200C C0

Concrete: t = 250C - 21.90C = 3.1 C0

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

ti

250C H A

Steady Flow 8 cm 12 cm 22, the IfIf A = 10 m A = 10 m , the heat heat flow flow in in 11 hh 745 kW would would be be ______? ______?

Radiation The rate of radiation R is the energy emitted per unit area per unit time (power per unit area). Rate of Radiation (W/m2):

Q P  R A A

Emissivity, Emissivity, ee ::

P R   e T 4 A

00 >> ee >> 11

Stefan-Boltzman Stefan-Boltzman Constant Constant ::  == 5.67 5.67 xx 10 10-8-8 W/m·K W/m·K44

Example 3: A spherical surface 12 cm in radius is heated to 6270C. The emissivity is 0.12. What power is radiated?

A  4 R  4 (0.12 m) 2

2

A

A = 0.181 m2 T = 627 + 273; T = 900 K

P  e AT

6270C

4

P  (0.12)(5.67 x 10 W/mK )(0.181 m )(900 K) -8

4

2

PP == 808 808 W W

4

Summary: Heat Transfer Conduction: Heat energy is transferred by adjacent molecular collisions inside a material. The medium itself does not move. Convection is the process by which heat energy is transferred by the actual mass motion of a heated fluid. Radiation is the process by which heat energy is transferred by electromagnetic waves.

Summary of Thermal Conductivity The thermal conductivity k of a material is a measure of its ability to conduct heat.

t1

t2

H = Heat current (J/s) A = Surface area (m2) t = Temperature difference

t = t2 - t1

L = Thickness of material

kAt H   L Q

QL k A t

J Units  s  m  C0

Summary of Radiation The rate of radiation R is the energy emitted per unit area per unit time (power per unit area). Rate of Radiation R  Q  P (W/m R 2): A A

P 4 R   e T A

Emissivity, Emissivity, ee :: Stefan-Boltzman Stefan-Boltzman Constant Constant :: -8 44 -8  == 5.67 x 10 W/m·K 5.67 x 10 W/m·K

00 >> ee >> 11

Summary of Formulas kAt H   L Q

QL k A t

kAt H k t H  ;   L A L Q

Q P  R A A

J Units  s  m  C0

P  e AT

P 4 R   e T A

4

CONCLUSION: Chapter 18 Transfer of Heat