1. Chapter 13
TRANSFER OF HEAT is minimized by multiple layers of beta cloth. These and other insulating materials protect spacecraft from hostile environmental conditions. (NASA)

2. 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.

3. 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.

4. 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
Convection is significantly affected by geometry of heated surfaces. (wall, ceiling, floor)

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

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

7. Water movement Cools at the surface Convection current Hot water rises
Cooler water sinks

8. Why is it windy at the seaside?

9. Where is the freezer compartment put in a fridge?
Cold air sinks
Where is the freezer compartment put in a fridge?
Freezer compartment
It is warmer at the bottom, so this warmer air rises and a convection current is set up.
It is put at the top, because cool air sinks, so it cools the food on the way down.

10. Heat Current
The heat current H is defined as the quantity of heat Q transferred per unit of time t in the direction from high temperature to low temperature.
Steam
Ice
Typical units are: J/s, cal/s, and Btu/h

11. Thermal Conductivity t1 t2
Dt = t2 - t1
The thermal conductivity k of a material is a measure of its ability to conduct heat.
H = Heat current (J/s)
A = Surface area (m2)
Dt = Temperature difference
L = Thickness of material

12. The SI Units for Conductivity
Hot
Cold
For Copper: k = 385 J/s m C0
Taken literally, this means that for a 1-m length of copper whose cross section is 1 m2 and whose end points differ in temperature by 1 C0, heat will be conducted at the rate of 1 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.

13. Thermal Conductivities
Examples of the two systems of units used for thermal conductivities of materials are given below:
Material
Copper:
385
2660
Concrete or Glass:
0.800
5.6
0.040
0.30
Corkboard:

14. Examples of Thermal Conductivity
Comparison of Heat Currents for Similar Conditions: L = 1 cm (0.39 in.); A = 1 m2 (10.8 ft2); Dt = 100 C0
2050 kJ/s
4980 Btu/h
Aluminum:
3850 kJ/s
9360 Btu/h
Copper:
Concrete or Glass:
8.00 kJ/s
19.4 Btu/h
0.400 kJ/s
9.72 Btu/h
Corkboard:

15. Example 1: A large glass window measures 2 m wide and 6 m high
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
200C
120C
Dt = t2 - t1 = 8 C0
0.015 m A
Q = ?
t = 1 h
Q = 18.4 MJ

16. 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 C and the outside surface is +250C. What is the interface temperature ti? ti
250C
-200C HA
8 cm 12 cm
Steady Flow
Note:

17. Rearranging factors gives:
Example 2 (Cont.): Finding the interface temperature for a composite wall. ti
250C
-200C HA
8 cm 12 cm
Steady Flow
Rearranging factors gives:

18. Example 2 (Cont.): Simplifying, we obtain:ti
250C
-200C HA
8 cm 12 cm
Steady Flow
0.075ti C = 250C - ti
From which:
ti = 21.90C
Knowing the interface temperature ti allows us to determine the rate of heat flow per unit of area, H/A.
The quantity H/A is same for cork or concrete:

19. Example 2 (Cont.): Constant steady state flow.
Over time H/A is constant so different k’s cause different Dt’s ti
250C
-200C HA
8 cm 12 cm
Steady Flow
Cork: Dt = 21.90C - (-200C) = 41.9 C0
Concrete: Dt = 250C C = 3.1 C0
Since H/A is the same, let’s just choose concrete alone:

20. Example 2 (Cont.): Constant steady state flow.ti
250C
-200C HA
8 cm 12 cm
Steady Flow
Cork: Dt = 21.90C - (-200C) = 41.9 C0
Concrete: Dt = 250C C = 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.
If A = 10 m2, the heat flow in 1 h would be ______?
745 kW

21. 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):
Emissivity, e : > e > 1
Stefan-Boltzman Constant : s = 5.67 x 10-8 W/m·K4

22. Power Radiated from Surface:
Find Power Radiated
Example 3: A spherical surface 12 cm in radius is heated to 6270C. The emissivity is What power is radiated?
A = m2
T = ; T = 900 K
Power Radiated from Surface:
P = 808 W

23. Definition of a black body
A black body is an ideal body which allows the whole of the incident radiation to pass into itself ( without reflecting the energy ) and absorbs within itself this whole incident radiation (without passing on the energy). This propety is valid for radiation corresponding to all wavelengths and to all angels of incidence. Therefore, the black body is an ideal absorber of incident radaition.
An oven with a small opening produces a very good blackbody spectrum.
Univ. of Oregon

24. A blackbody is a hypothetical object that is a perfect absorber of electromagnetic radiation at all wavelengths.
Stars closely approximate the behavior of blackbodies, as do other hot, dense objects.
The intensities of radiation emitted at various wavelengths by a blackbody at a given temperature are shown by a blackbody curve.

26. E = T4 Stefan-Boltzmann Law
The Stefan-Boltzmann law states that a blackbody radiates electromagnetic waves with a total energy flux E directly proportional to the fourth power of the Kelvin temperature T of the object:
E = T4

27. Conclusion
As the temperature increases, the peak wavelength emitted by the black body decreases.
As temperature increases, the total energy emitted increases, because the total area under the curve increases.
The curve gets infinitely close to the x-axis but never touches it.