The document discusses heat transfer and related concepts including: 1) Heat is energy in transit and its units like joules and calories 2) Sensible heat associated with temperature change and specific heat capacity 3) Latent heat associated with phase changes and calculation of heat involved 4) Properties of steam and using steam tables to find properties at given conditions 5) Concepts and equations related to heat exchangers like calculating heat transfer between hot and cold fluids

1. Heat transfer
- Source of heat
- Heat transfer
- Steam and electricity as heating media
- Determination of requirement of amount of
steam/electrical energy
- Steam pressure
- Mathematical problems on heat transfer
1

2. What is Heat?
2

3. What is Heat?
Heat is energy in transit.
3

4. Units of Heat
• The SI unit is the joule (J),
which is equal to Newton-metre (Nm).
• Historically, heat was measured in terms of the ability
to raise the temperature of water.
• The calorie (cal): amount of heat needed to raise the
temperature of 1 gramme of water by 1 C0 (from
14.50C to 15.50C)
• In industry, the British thermal unit (Btu) is still used:
amount of heat needed to raise the temperature of 1 lb
of water by 1 F0 (from 630F to 640F)
4

5. Conversion between different
units of heat:
1 J = 0.2388 cal = 0.239x10-3 kcal = 60.189 Btu
1 cal = 4.186 J = 3.969 x 10-3 Btu
5

6. Sensible Heat
• What is 'sensible heat‘?
Sensible heat is associated
with a temperature change
6

7. Specific Heat Capacity
• To raise the temperature by 1 K, different
substances need different amount of energy
because substances have different molecular
configurations and bonding (eg: copper, water,
wood)
• The amount of energy needed to raise the
temperature of 1 kg of a substance by 1 K is
known as the specific heat capacity
• Specific heat capacity is denoted by c
7

8. Calculation of Sensible Heat
Q is the heat lost or gained by a substance
m is the mass of substance
c is the specific heat of substance which changes with temperature
T is the temperature
When temperature changes causes negligible changes in c,
where ΔT is the temperature change in the substance
8

9. Calculation of Sensible Heat
When temperature changes causes significant changes in c,
Q = m c ∆T cannot be used.
Instead, we use the following equation:
Q = ∆H = m ∆h
where ΔH is the enthalpy change in the substance
and ∆h is the specific enthalpy change in the substance.
To apply the above equation, the system should
remain at constant pressure and the associated
volume change must be negligibly small.
9

10. Calculation of Sensible Heat
Calculate the amount of heat required to raise the temperature
of 300 g Al from 25oC to 70oC.
Data: c = 0.896 J/g oC for Al
Q = m c ΔT (since c is taken as a constant)
= (300 g) (0.896 J/g oC)(70 - 25)oC
= 12,096 J
= 13.1 kJ
10

11. Exchange of Heat
Calculate the final temperature (tf), when 100 g iron at 80oC is
tossed into 53.5g of water at 25oC.
Data: c = 0.452 J/g oC for iron and 4.186 J/g oC for water
Heat lost by iron = Heat gained by water
(m c ΔT)iron = (m c ΔT)water
(100 g) (0.452 J/g oC)(80 - tf)oC
= (53.5 g) (4.186 J/g oC)(tf - 25)oC
80 - tf = 4.955 (tf -25)
tf = 34.2oC 11

12. Latent Heat
• What is ‘latent heat‘?
Latent heat is associated with
phase change of matter
12

13. Phases of Matter
13

14. Phase Change
• Heat required for phase changes:
» Melting: solid  liquid
» Vaporization: liquid  vapour
» Sublimation: solid  vapour
• Heat released by phase changes:
» Condensation: vapour  liquid
» Fusion: liquid  solid
» Deposition: vapour  solid
Prof. R. Shanthini 14
5&

15. Phase Diagram: Water
15

16. Phase Diagram: Water
Compressed liquid
Saturated liquid
Superheated
steam
Saturated steam
16

17. Phase Diagram: Water
Explain why water is at liquid
state at atm pressure
17

18. Phase Diagram: Carbon Dioxide
Explain why CO2 is at gas state
at atm pressure
Explain why CO2
cannot be made a
liquid at atm
pressure
18

19. Latent Heat
Latent heat is the amount of heat added per unit mass of
substance during a phase change
Latent heat of fusion is the amount of heat added to melt
a unit mass of ice OR it is the amount of heat removed
to freeze a unit mass of water.
Latent heat of vapourization is the amount of heat added
to vaporize a unit mass of water OR it is the amount of
heat removed to condense a unit mass of steam.
19

20. Water:
Specific Heat Capacities and Latent Heats
Specific heat of ice ≈ 2.06 J/g K (assumed constant)
Heat of fusion for ice/water ≈ 334 J/g (assumed constant)
Specific heat of water ≈ 4.18 J/g K (assumed constant)
Latent heat of vaporization cannot be assumed a
constant since it changes significantly with the pressure,
and could be found from the Steam Table
How to evaluate the sensible heat gained (or lost) by
superheated steam?
20

21. Water:
Specific Heat Capacities and Latent Heats
How to evaluate the sensible heat gained (or lost) by
superheated steam?
Q = m c ∆T
cannot be used since changes in c with changing
temperature is NOT negligible.
Instead, we use the following equation:
Q = ∆H = m ∆h
provided the system is at constant pressure and the
associated volume change is negligible.
Enthalpies could be referred from the Steam 21
Table

22. Properties of Steam
Learnt to refer to Steam Table to find properties of
steam such as saturated (or boiling point) temperature
and latent heat of vapourization at give pressures, and
enthalpies of superheated steam at various pressures and
temperatures.
Reference:
Chapter 6 of “Thermodynamics for Beginners with worked
examples” by R. Shanthini
(published by Science Education Unit, Faculty of Science,
University of Peradeniya)
(also uploaded at http://www.rshanthini.com/PM3125.htm)
22

23. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
-20oC ice
23

24. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
0oC melting point of ice
-20oC ice
24

25. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
120.2oC boiling point of water at 2 bar
Boiling point of water at 1 atm pressure is
100oC.
Boiling point of water at 2 bar is 120.2oC.
[Refer the Steam Table.]
0oC melting point of ice
-20oC ice
25

26. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
150oC superheated steam
Specific heat
120.2oC boiling point of water at 2 bar Latent heat
Specific heat
0oC melting point of ice Latent heat
Specific heat
-20oC ice
26

27. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
Specific heat required to raise the temperature of ice from -20 oCto 0oC
= (2 kg) (2.06 kJ/kg oC) [0 - (-20)]oC = 82.4 kJ
Latent heat required to turn ice into water at 0oC
= (2 kg) (334 kJ/kg) = 668 kJ
Specific heat required to raise the temperature of water from 0oC to
120.2oC
= (2 kg) (4.18 kJ/kg oC) [120.2 - 0)]oC = 1004.9 kJ
27

28. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
Latent heat required to turn water into steam at 120.2oC and at 2 bar
= (2 kg) (2202 kJ/kg) = 4404 kJ
[Latent heat of vapourization at 2 bar is 2202 kJ/kg as could be
referred to from the Steam Table]
Specific heat required to raise the temperature of steam from 120.2oC
to 150oC
= (2 kg) (2770 – 2707) kJ/kg = 126 kJ
[Enthalpy at 120.2oC and 2 bar is the saturated steam enthalpy of
2707 kJ/kg and the enthalpy at 150oC and 2 bar is 2770 kJ/kg as
could be referred to from the Steam Table]
28

29. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
Total amount of heat required
= 82.4 kJ + 668 kJ + 1004.9 kJ + 4404 kJ + 126 kJ
= 6285.3 kJ
29

30. Application: Heat Exchanger
It is an industrial equipment in which heat is transferred from a hot
fluid (a liquid or a gas) to a cold fluid (another liquid or gas) without
the two fluids having to mix together or come into direct contact.
Cold fluid Cold fluid
at TC,out at TC,in
Hot fluid
at TH,in Heat lost by the hot fluid Hot fluid
= Heat gained by the cold fluid at TH,out
30

31. Application: Heat Exchanger
31

32. Heat Exchanger
Heat lost by the hot fluid = Heat gained by the cold fluid
.m hot chot (TH,in – TH,out
.
)=m cold ccold (TC,out – TC,in)
mass flow rate mass flow rate
of hot fluid of cold fluid
Specific heat Specific heat
of hot fluid of cold fluid
Temperature Temperature
decrease in the increase in the
hot fluid cold fluid
32

33. Heat Exchanger
Heat lost by the hot fluid = Heat gained by the cold fluid
.m hot chot (TH,in – TH,out
.
)=m
cold ccold (TC,out – TC,in)
The above is true only under the following conditions:
(1) Heat exchanger is well insulated so that no heat is lost to the
environment
(2) There are no phase changes occurring within the heat
exchanger.
33

34. Heat Exchanger
If the heat exchanger is NOT well insulated, then
Heat lost by the hot fluid = Heat gained by the cold fluid
+ Heat lost to the environment
34

35. Worked Example 1 in Heat Exchanger
High pressure liquid water at 10 MPa (100 bar) and
30oC enters a series of heating tubes. Superheated
steam at 1.5 MPa (15 bar) and 200oC is sprayed over
the tubes and allowed to condense. The condensed
steam turns into saturated water which leaves the
heat exchanger. The high pressure water is to be
heated up to 170oC. What is the mass of steam
required per unit mass of incoming liquid water?
The heat exchanger is assumed to be well insulated
(adiabatic).
35

36. Solution to Worked Example 1 in Heat Exchanger
36

37. Solution to Worked Example 1 in Heat Exchanger contd.
High pressure (100 bar) water enters at 30oC and leaves at 198.3oC.
Boiling point of water at 100 bar is 311.0oC. Therefore, no phase
changes in the high pressure water that is getting heated up in the
heater.
Heat gained by high pressure water
= ccold (TC,out – TC,in)
= (4.18 kJ/kg oC) x (170-30)oC
= 585.2 kJ/kg
[You could calculate the above by taking the difference in enthalpies at
the 2 given states from tables available.]
37

38. Solution to Worked Example 1 in Heat Exchanger contd.
Superheated steam at 1.5 MPa (15 bar) and 200oC is sprayed over
the tubes and allowed to condense. The condensed steam turns into
saturated water which leaves the heat exchanger.
Heat lost by steam
= heat lost by superheated steam to become saturated steam
+ latent heat of steam lost for saturated steam to turn into
saturated water
= Enthalpy of superheated steam at 15 bar and 200oC
– Enthalpy of saturated steam at 15 bar
+ Latent heat of vapourization at 15 bar
= (2796 kJ/kg – 2792 kJ/kg) + 1947 kJ/kg = 1951 kJ/kg
38

39. Solution to Worked Example 1 in Heat Exchanger contd.
Since there is no heat loss from the heater,
Heat lost by steam = Heat gained by high pressure water
Mass flow rate of steam x 1951 kJ/kg
= Mass flow rate of water x 585.2 kJ/kg
Mass flow rate of steam / Mass flow rate of water
= 585.2 / 1951
= 0.30 kg stream / kg of water
39

40. Assignment
Give the design of a heat exchanger
which has the most effective heat
transfer properties.
Learning objectives:
1) To be able to appreciate heat transfer applications in pharmaceutical
industry
2) To become familiar with the working principles of various heat
exchangers
3) To get a mental picture of different heat exchangers so that solving
heat transfer problems in class becomes more interesting
40

41. Worked Example 2 in Heat Exchanger
Steam enters a heat exchanger at 10 bar and 200oC and
leaves it as saturated water at the same pressure. Feed-
water enters the heat exchanger at 25 bar and 80oC and
leaves at the same pressure and at a temperature 20oC
less than the exit temperature of the steam. Determine the
ratio of the mass flow rate of the steam to that of the
feed-water, neglecting heat losses from the heat
exchanger.
If the feed-water leaving the heat exchanger is fed
directly to a boiler to be converted to steam at 25 bar and
300oC, find the heat required by the boiler per kg of feed-
water.
41

42. Solution to Worked Example 2 in Heat Exchanger
- Steam enters at 10 bar and 200oC and leaves it as saturated water at
the same pressure.
- Saturation temperature of water at 10 bar is 179.9oC.
- Feed-water enters the heat exchanger at 25 bar and 80oC and leaves
at the same pressure and at a temperature 20oC less than the exit
temperature of the steam, which is 179.9oC.
- Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC.
- Therefore, no phase changes in the feed-water that is being heated.
Heat lost by steam = Heat gained by feed-water (with no heat losses)
Mass flow rate of steam x [2829 – 2778 + 2015] kJ/kg
= Mass flow rate of feed-water x [4.18 x (179.9-20-80) ] kJ/kg
Mass flow of steam / Mass flow of feed-water
= 333.98 / 2066 = 0.1617 kg stream / kg of water
42

43. Solution to Worked Example 1 in Heat Exchanger contd.
If the feed-water leaving the heat exchanger is fed directly to a
boiler to be converted to steam at 25 bar and 300oC, find the heat
required by the boiler per kg of feed-water.
- Temperature of feed-water leaving the heat exchanger is 159.9oC
- Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC
- The feed-water is converted to superheated steam at 300oC
Heat required by the boiler per kg of feed-water
= {4.18 x (223.9-159.9) + (1850+1831)/2
+ [(3138+3117)/2 – (2802+2803)/2]} kJ/kg
= {267.52 + 1840.5 + [3127.5 – 2802.5]} kJ/kg
= 2433 kJ/kg of feed-water
43

44. Heat Transfer
is the means by which
energy moves from
a hotter object to
a colder object
44

45. Mechanisms of Heat Transfer
Conduction
is the flow of heat by direct contact between a
warmer and a cooler body.
Convection
is the flow of heat carried by moving gas or liquid.
(warm air rises, gives up heat, cools, then falls)
Radiation
is the flow of heat without need of an intervening
medium.
(by infrared radiation, or light)
45

46. Mechanisms of Heat Transfer
Latent heat
Conduction
Convection
Radiation
Prof. R. Shanthini 46
5&

47. Conduction
HOT COLD
(lots of vibration) (not much vibration)
Heat travels
along the rod
47

48. Conduction
Conduction is the process whereby heat is transferred
directly through a material, any bulk motion of the
material playing no role in the transfer.
Those materials that conduct heat well are called
thermal conductors, while those that conduct heat poorly
are known as thermal insulators.
Most metals are excellent thermal conductors, while
wood, glass, and most plastics are common thermal
insulators.
The free electrons in metals are responsible for the
excellent thermal conductivity of metals.
48

49. Conduction: Fourier’s Law
Cross-sectional area A
L
Q = heat transferred
Q =k A ( )t
ΔT
L
k = thermal conductivity
A = cross sectional area
∆T = temperature difference
between two ends
What is the unit of k? L = length
t = duration of heat transfer
49

50. Thermal Conductivities
Substance Thermal Substance Thermal
Conductivity Conductivity
k [W/m.K] k [W/m.K]
Syrofoam 0.010 Glass 0.80
Air 0.026 Concrete 1.1
Wool 0.040 Iron 79
Wood 0.15 Aluminum 240
Body fat 0.20 Silver 420
Water 0.60 Diamond 2450
50

51. Conduction through Single Wall
Use Fourier’s Law:
T1
Q =k A ( )t
ΔT
L
. .
Q Q
T2 < T1 . k A (T1 – T2)
Q =
x Δx
Δx
51

52. Conduction through Single Wall
T1
. k A (T1 – T2)
Q =
Δx
. .
Q Q
T1 – T 2
=
T2 < T1 Δx/(kA)
x
Δx
Thermal resistance (in K/W)
(opposing heat flow) 52 52

53. Conduction through Composite Wall
T1 A B C
. T2 .
Q T3 Q
kA kB kC T4
x
ΔxA ΔxB ΔxC
. T1 – T 2 T2 – T 3 T3 – T 4
Q = = =
(Δx/kA)A (Δx/kA)B (Δx/kA)C
53
53

54. Conduction through Composite Wall
. T1 – T 2 T2 – T 3 T3 – T 4
Q = = =
(Δx/kA)A (Δx/kA)B (Δx/kA)C
.
[
Q (Δx/kA)A + (Δx/kA)B + (Δx/kA)C ]
= T 1 – T 2 + T2 – T 3 + T3 – T 4
. T1 – T 4
Q=
(Δx/kA)A + (Δx/kA)B + (Δx/kA)C
54
54

55. Example 1
An industrial furnace wall is constructed of 21 cm thick
fireclay brick having k = 1.04 W/m.K. This is covered on
the outer surface with 3 cm layer of insulating material
having k = 0.07 W/m.K. The innermost surface is at 1000oC
and the outermost surface is at 40oC. Calculate the steady
state heat transfer per area.
Solution: We start with the equation
. Tin – Tout
Q=
(Δx/kA)fireclay + (Δx/kA)insulation
55

56. Example 1 continued
. (1273.15 – 313.15) A
Q =
(0.21/1.04) + (0.03/0.07)
.
Q
= 1522.6 W/m2
A
56

57. Example 2
We want to reduce the heat loss in Example 1 to 960 W/m2.
What should be the insulation thickness?
Solution: We start with the equation
. Tin – Tout
Q=
(Δx/kA)fireclay + (Δx/kA)insulation
.
Q (1000 – 40)
= = 960 W/m2
A (0.21/1.04) + (Δx)insulation /0.07)
(Δx)insulation = 5.6 cm
57

58. Conduction through hollow-cylinder
ro
Ti ri
To
L
. Ti – T o
Q =
[ln(ro/ri)] / 2πkL
58

59. Conduction through the composite
r3
wall in a hollow-cylinder
r2 To
Material A
Ti r1
Material B
. Ti – T o
Q =
[ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL
59

60. Example 3
A thick walled tube of stainless steel ( k = 19 W/m.K) with
2-cm inner diameter and 4-cm outer diameter is covered
with a 3-cm layer of asbestos insulation (k = 0.2 W/m.K).
If the inside-wall temperature of the pipe is maintained at
600oC and the outside of the insulation at 100oC, calculate
the heat loss per meter of length.
Solution: We start with the equation
. Ti – T o
Q =
[ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL
60

61. Example 3 continued
. 2 π L ( 600 – 100)
Q =
[ln(2/1)] / 19 + [ln(5/2)] / 0.2
.
Q = 680 W/m
L
61

62. Mechanisms of Heat Transfer
Conduction
√
is the flow of heat by direct contact between a
warmer and a cooler body.
Convection
is the flow of heat carried by moving gas or liquid.
(warm air rises, gives up heat, cools, then falls)
Radiation
is the flow of heat without need of an intervening
medium.
(by infrared radiation, or light)
62

63. Convection
Convection is the process in which heat is carried from
place to place by the bulk movement of a fluid (gas or
liquid).
Convection currents are set up when
a pan of water is heated.
63

64. Convection
It explains why breezes come from the ocean in the day
and from the land at night
64

65. Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid
Heated surface at Tsurface
.
Qconv. = h A (Tsurface – Tfluid)
Area exposed
Heat transfer coefficient (in W/m2.K)
65

66. Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid
Heated surface at Tsurface
. Tsurface – Tfluid
Qconv. =
1/(hA)
Convective heat resistance (in K/W) 66

67. Example 4
The convection heat transfer coefficient between a surface at 50oC and
ambient air at 30oC is 20 W/m2.K. Calculate the heat flux leaving the
surface by convection.
Solution: Use Newton’s Law of cooling :
.
Flowing fluid at Tfluid = 30oC Q = h A (Tsurface – Tfluid)
conv.
= (20 W/m2.K) x A x (50-30)oC
Heated surface at Tsurface = 50oC
Heat flux leaving the surface:
.
Q
conv.
= 20 x 20 = 400 W/m2
h = 20 W/m2.K A
67

68. Example 5
Air at 300°C flows over a flat plate of dimensions 0.50 m by 0.25 m.
If the convection heat transfer coefficient is 250 W/m2.K, determine
the heat transfer rate from the air to one side of the plate when the
plate is maintained at 40°C.
Solution: Use Newton’s Law of cooling :
Flowing fluid at Tfluid = 300oC .
Q = h A (Tsurface – Tfluid)
conv.
Heated surface at Tsurface = 40oC = 250 W/m2.K x 0.125 m2
x (40 - 300)oC
= - 8125 W/m2
h = 250 W/m2.K
Heat is transferred from
A = 0.50x0.25 m2 the air to the plate.
68

69. Forced Convection
In forced convection, a fluid is forced by external forces
such as fans.
In forced convection over external surface:
Tfluid = the free stream temperature (T∞), or a
temperature far removed from the surface
In forced convection through a tube or channel:
Tfluid = the bulk temperature
69

70. Free Convection
In free convection, a fluid is circulated due to buoyancy
effects, in which less dense fluid near the heated surface rises
and thereby setting up convection.
In free (or partially forced) convection over
external surface:
Tfluid = (Tsurface + Tfree stream) / 2
In free or forced convection through a tube or
channel:
Tfluid = (Tinlet + Toutlet) / 2
70

71. Change of Phase Convection
Change-of-phase convection is observed with
boiling or condensation
.
It is a very complicated mechanism and
therefore will not be covered in this course.
71

72. Overall Heat Transfer through a Plane Wall
Fluid A
at TA > T1 T1
. .
Q Q
T2 Fluid B
at TB < T2
x
Δx
. T A – T1 T 1 – T2 T2– TB
Q = = =
1/(hAA) Δx/(kA) 1/(hBA)
72

73. Overall Heat Transfer through a Plane Wall
. T A – T1 T 1 – T2 T2– TB
Q = = =
1/(hAA) Δx/(kA) 1/(hBA)
. TA – TB
Q =
1/(hAA) + Δx/(kA) + 1/(hBA)
.
Q = U A (TA – TB)
where U is the overall heat transfer coefficient given by
1/U = 1/hA + Δx/k + 1/hB 73

74. Overall heat transfer through hollow-cylinder
Fluid A is inside the pipe ro
Fluid B is outside the pipe
TA > TB Ti r
i
To
L
.
Q = U A (TA – TB)
where
1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + 1/(hBAo)
74

75. Example 6
Steam at 120oC flows in an insulated pipe. The pipe is
mild steel (k = 45 W/m K) and has an inside radius of 5
cm and an outside radius of 5.5 cm. The pipe is covered
with a 2.5 cm layer of 85% magnesia (k = 0.07 W/m K).
The inside heat transfer coefficient (hi) is 85 W/m2 K, and
the outside coefficient (ho) is 12.5 W/m2 K. Determine the
heat transfer rate from the steam per m of pipe length, if
the surrounding air is at 35oC.
Solution: Start with
.
Q = U A (TA – TB) = U A (120 – 35)
What is UA?
75

76. Example 6 continued
1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + … + 1/(hBAo)
1/UA = 1/(85Ain) + ln(5.5/5) / 2π(45)L
+ ln(8/5.5) / 2π(0.07)L + 1/(12.5Aout)
Ain = 2π(0.05)L and Aout = 2π(0.08)L
1/UA = (0.235 + 0.0021 +5.35 + 1) / 2πL
76

77. Example 6 continued
UA = 2πL / (0.235 + 0.0021 +5.35 + 1)
.
Q = U A (120 – 35)
steel air
= 2πL (120 – 35) / (0.235 + 0.0021 +5.35 + 1)
steam insulation
= 81 L
.
Q / L = 81 W/m
77

78. Mechanisms of Heat Transfer
Conduction
√
is the flow of heat by direct contact between a
warmer and a cooler body.
Convection √
is the flow of heat carried by moving gas or liquid.
(warm air rises, gives up heat, cools, then falls)
Radiation
is the flow of heat without need of an intervening
medium.
(by infrared radiation, or light)
78

79. Radiation
Radiation is the process in which energy is transferred by
means of electromagnetic waves of wavelength band between
0.1 and 100 micrometers solely as a result of the temperature
of a surface.
Heat transfer by radiation
can take place through
vacuum. This is because
electromagnetic waves
can propagate through
empty space.
79

80. The Stefan–Boltzmann Law of Radiation
Q
= ε σ A T4
t
ε = emissivity, which takes a value between 0 (for
an ideal reflector) and 1 (for a black body).
σ = 5.668 x 10-8 W/m2.K4 is the Stefan-Boltzmann
constant
A = surface area of the radiator
T = temperature of the radiator in Kelvin.
80

81. Why is the
mother shielding
her cub?
Ratio of the surface
area of a cub to its
volume is much larger
than for its mother.
81

82. What is the Sun’s surface temperature?
The sun provides about 1000 W/m2 at the Earth's surface.
Assume the Sun's emissivity ε = 1
Distance from Sun to Earth = R = 1.5 x 1011 m
Radius of the Sun = r = 6.9 x 108 m
82

83. What is the Sun’s surface temperature?
Q
= ε σ A T4
t
(4 π 6.92 x 1016 m2)
(4 π 1.52 x 1022 m2)(1000 W/m2)
= 5.98 x 1018 m2
= 2.83 x 1026 W
2.83 x 1026 W
T4 =
(1) (5.67 x 10-8 W/m2.K4) (5.98 x 1018 m2)
ε
σ T = 5375 K
83

84. If object at temperature T is surrounded by
an environment at temperature T 0, the net
radioactive heat flow is:
Q
= ε σ A (T4 - To4 )
t
Temperature of the radiating surface
Temperature of the environment
84

85. Example 7
What is the rate at which radiation is emitted by a surface
of area 0.5 m2, emissivity 0.8, and temperature 150°C?
Solution:
Q [(273+150) K]4
= ε σ A T4
t
0.5 m2
0.8 5.67 x 10-8 W/m2.K4
Q
= (0.8) (5.67 x 10-8 W/m2.K4) (0.5 m2) (423 K)4
t
= 726 W
85

86. Example 8
If the surface of Example 7 is placed in a large, evacuated chamber
whose walls are maintained at 25°C, what is the net rate at which
radiation is exchanged between the surface and the chamber walls?
Solution: Q
= ε σ A (T4 - To4 )
t
[(273+25) K]4
[(273+150) K]4
Q
= (0.8) x (5.67 x 10-8 W/m2.K4) x (0.5 m2)
t
x [(423 K)4 -(298 K)4 ]
= 547 W
86

87. Example 8 continued
Note that 547 W of heat loss from the surface occurs
at the instant the surface is placed in the chamber. That
is, when the surface is at 150oC and the chamber wall
is at 25oC.
With increasing time, the surface would cool due to
the heat loss. Therefore its temperature, as well as the
heat loss, would decrease with increasing time.
Steady-state conditions would eventually be achieved
when the temperature of the surface reached that of the
surroundings.
87

88. Example 9
Under steady state operation, a 50 W incandescent light bulb has a
surface temperature of 135°C when the room air is at a temperature
of 25°C. If the bulb may be approximated as a 60 mm diameter
sphere with a diffuse, gray surface of emissivity 0.8, what is the
radiant heat transfer from the bulb surface to its surroundings?
Solution: Q
= ε σ A (T4 - To4 )
t
[(273+25) K]4
[(273+135) K]4
Q
= (0.8) x (5.67 x 10-8 J/s.m2.K4) x [π x (0.06) m2]
t
x [(408 K)4 -(298 K)4 ]
= 10.2 W (about 20% of the power is dissipated by radiation)
88