1. PM3125: Lectures 6 to 9 Content of Lectures 6 to 12: 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
Prof. R. Shanthini & 12 March 2012

2. What is Heat?
Prof. R. Shanthini & 12 March 2012

3. Heat is energy in transit.
What is Heat?
Heat is energy in transit.
Prof. R. Shanthini & 12 March 2012

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)
Prof. R. Shanthini & 12 March 2012

5. Conversion between different
units of heat:
1 J = cal = 0.239x10-3 kcal = Btu
1 cal = J = x 10-3 Btu
Prof. R. Shanthini & 12 March 2012

6. Sensible heat is associated with a temperature change
What is 'sensible heat‘?
Sensible heat is associated with a temperature change
Prof. R. Shanthini & 12 March 2012

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
Prof. R. Shanthini & 12 March 2012

8. Calculation of Sensible Heat
Q = m c dT ∫
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,
Q = m c dT ∫
= m c ∆T
where ΔT is the temperature change in the substance
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

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 = J/g oC for Al
Q = m c ΔT (since c is taken as a constant)
= (300 g) (0.896 J/g oC)( )oC
= 12,096 J
= 13.1 kJ
Prof. R. Shanthini & 12 March 2012

11. Data: c = 0.452 J/g oC for iron and 4.186 J/g oC for water
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 = J/g oC for iron and 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 = (tf -25)
tf = 34.2oC
Prof. R. Shanthini & 12 March 2012

12. Latent heat is associated with phase change of matter
What is ‘latent heat‘?
Latent heat is associated with phase change of matter
Prof. R. Shanthini & 12 March 2012

13. Phases of Matter
Prof. R. Shanthini & 12 March 2012

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 & 12 March 2012

15. Phase Diagram: Water
Prof. R. Shanthini & 12 March 2012

16. Phase Diagram: Water Compressed liquid Saturated liquid Superheated
steam
Saturated steam
Prof. R. Shanthini & 12 March 2012

17. Phase Diagram: Water
Explain why water is at liquid state at atm pressure
Prof. R. Shanthini & 12 March 2012

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
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

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?
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012
Enthalpies could be referred from the Steam 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
Prof. R. Shanthini & 12 March 2012

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
Prof. R. Shanthini & 12 March 2012

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?
-20oC ice
0oC melting point of ice
Prof. R. Shanthini & 12 March 2012

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
Prof. R. Shanthini & 12 March 2012

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
Prof. R. Shanthini & 12 March 2012

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 -20oCto 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) [ )]oC = kJ
Prof. R. Shanthini & 12 March 2012

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) = 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]
Prof. R. Shanthini & 12 March 2012

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 kJ kJ kJ kJ
= kJ
Prof. R. Shanthini & 12 March 2012

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
at TC,in
Cold fluid
at TC,out
Hot fluid
at TH,in
Heat lost by the hot fluid
= Heat gained by the cold fluid
Hot fluid
at TH,out
Prof. R. Shanthini & 12 March 2012

31. Application: Heat Exchanger
Prof. R. Shanthini & 12 March 2012

32. mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in)
Heat Exchanger
mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in) .
Heat lost by the hot fluid = Heat gained by the cold fluid
mass flow rate of hot fluid
mass flow rate of cold fluid
Specific heat of hot fluid
Specific heat of cold fluid
Temperature increase in the cold fluid
Temperature decrease in the hot fluid
Prof. R. Shanthini & 12 March 2012

33. mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in)
Heat Exchanger
mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in) .
Heat lost by the hot fluid = Heat gained by the cold fluid
The above is true only under the following conditions:
Heat exchanger is well insulated so that no heat is lost to the environment
There are no phase changes occurring within the heat exchanger.
Prof. R. Shanthini & 12 March 2012

34. 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
Prof. R. Shanthini & 12 March 2012

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).
Prof. R. Shanthini & 12 March 2012

36. Solution to Worked Example 1 in Heat Exchanger
Prof. R. Shanthini & 12 March 2012

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
= kJ/kg
[You could calculate the above by taking the difference in enthalpies at the 2 given states from tables available.]
Prof. R. Shanthini & 12 March 2012

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) kJ/kg = 1951 kJ/kg
Prof. R. Shanthini & 12 March 2012

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 kJ/kg
Mass flow rate of steam / Mass flow rate of water
= / 1951
= 0.30 kg stream / kg of water
Prof. R. Shanthini & 12 March 2012

40. Assignment
Give the design of a heat exchanger which has the most effective heat transfer properties.
Learning objectives:
To be able to appreciate heat transfer applications in pharmaceutical industry
To become familiar with the working principles of various heat exchangers
To get a mental picture of different heat exchangers so that solving heat transfer problems in class becomes more interesting
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

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 ( )/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 – ] kJ/kg
= Mass flow rate of feed-water x [4.18 x ( ) ] kJ/kg
Mass flow of steam / Mass flow of feed-water
= / 2066 = kg stream / kg of water
Prof. R. Shanthini & 12 March 2012

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 ( )/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 ( ) + ( )/2
+ [( )/2 – ( )/2]} kJ/kg
= { [ – ]} kJ/kg
= 2433 kJ/kg of feed-water
Prof. R. Shanthini & 12 March 2012

44. Heat Transfer is the means by which energy moves from
a hotter object to
a colder object
Prof. R. Shanthini & 12 March 2012

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)
Prof. R. Shanthini & 12 March 2012

46. Mechanisms of Heat Transfer
Latent heat
Conduction
Convection
Radiation
Prof. R. Shanthini & 12 March 2012

47. Heat travels along the rod
Conduction
HOT
(lots of vibration)
COLD
(not much vibration)
Heat travels along the rod
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

49. Conduction: Fourier’s Law
Cross-sectional area A L
Q = heat transferred
k = thermal conductivity
A = cross sectional area
DT = temperature difference
between two ends
L = length
t = duration of heat transfer ΔT Q = L
k A t
( )
What is the unit of k?
Prof. R. Shanthini & 12 March 2012

50. Thermal Conductivities
Substance
Thermal
Conductivity
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
Prof. R. Shanthini & 12 March 2012

51. Conduction through Single Wall
Use Fourier’s Law: T1 ΔT Q = L
k A t
( ) Q . Q . Q .
k A (T1 – T2)
T2 T1 = x Δx Δx
Prof. R. Shanthini & 12 March 2012

52. Conduction through Single WallQ .
k A (T1 – T2) T1 = Δx Q . Q .
T1 – T2 =
Δx/(kA)
T2 T1 x Δx
Thermal resistance (in K/W)
(opposing heat flow)
Prof. R. Shanthini & 12 March 2012 52

53. Conduction through Composite WallB C T1 Q . Q . T2 T3 T4 kA kB kC x
ΔxA
ΔxB
ΔxC Q .
T1 – T2
T2 – T3
T3 – T4 = = =
(Δx/kA)A
(Δx/kA)B
(Δx/kA)C
Prof. R. Shanthini & 12 March 2012 53

54. Conduction through Composite Wall
T1 – T2 =
(Δx/kA)A Q .
T2 – T3
T3 – T4
(Δx/kA)C
(Δx/kA)B
+ (Δx/kA)B
(Δx/kA)A
+ (Δx/kA)C [ ] Q .
= T1 – T2 + T2 – T3 + T3 – T4 Q .
T1 – T4 =
+ (Δx/kA)B
(Δx/kA)A
+ (Δx/kA)C 54
Prof. R. Shanthini & 12 March 2012

55. Example 1 . Tin – Tout Q = + (Δx/kA)insulation (Δx/kA)fireclay
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 Q .
Tin – Tout =
(Δx/kA)fireclay
+ (Δx/kA)insulation
Prof. R. Shanthini & 12 March 2012

56. Example 1 continued . . Q (1000 – 40) A = (0.21/1.04) + (0.03/0.07) Q=
W/m2 A
Prof. R. Shanthini & 12 March 2012

57. Example 2 . . Q Tin – Tout = (Δx/kA)fireclay + (Δx/kA)insulation Q
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 Q .
Tin – Tout =
(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
Prof. R. Shanthini & 12 March 2012

58. Conduction through hollow-cylinderri To L Q .
Ti – To =
[ln(ro/ri)] / 2πkL
Prof. R. Shanthini & 12 March 2012

59. Conduction through the composite wall in a hollow-cylinderTo
Material A Ti r1
Material B Q .
Ti – To =
[ln(r2/r1)] / 2πkAL
+ [ln(r3/r2)] / 2πkBL
Prof. R. Shanthini & 12 March 2012

60. Example 3 . Ti – To Q = [ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL
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 Q .
Ti – To =
[ln(r2/r1)] / 2πkAL
+ [ln(r3/r2)] / 2πkBL
Prof. R. Shanthini & 12 March 2012

61. Example 3 continued . . 2 π L ( 600 – 100) Q = + [ln(5/2)] / 0.2
= 680 W/m L
Prof. R. Shanthini & 12 March 2012

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)
Prof. R. Shanthini & 12 March 2012

63. Convection currents are set up when a pan of water is heated.
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.
Prof. R. Shanthini & 12 March 2012

64. Convection
It explains why breezes come from the ocean in the day and from the land at night
Prof. R. Shanthini & 12 March 2012

65. Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid
Heated surface at Tsurface Q .
conv.
= h A (Tsurface – Tfluid)
Area exposed
Heat transfer coefficient (in W/m2.K)
Prof. R. Shanthini & 12 March 2012

66. Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid
Heated surface at Tsurface Q .
conv.
Tsurface – Tfluid =
1/(hA)
Convective heat resistance (in K/W)
Prof. R. Shanthini & 12 March 2012

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 : Q .
conv.
= h A (Tsurface – Tfluid)
Flowing fluid at Tfluid = 30oC
= (20 W/m2.K) x A x (50-30)oC
Heated surface at Tsurface = 50oC
Heat flux leaving the surface: . Q
conv. A
= 20 x 20
= 400 W/m2
h = 20 W/m2.K
Prof. R. Shanthini & 12 March 2012

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 : Q .
conv.
= h A (Tsurface – Tfluid)
Flowing fluid at Tfluid = 300oC
Heated surface at Tsurface = 40oC
= 250 W/m2.K x m2
x ( )oC
= W/m2
h = 250 W/m2.K
A = 0.50x0.25 m2
Heat is transferred from the air to the plate.
Prof. R. Shanthini & 12 March 2012

69. Forced Convection In forced convection over external surface:
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
Prof. R. Shanthini & 12 March 2012

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
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

72. Overall Heat Transfer through a Plane Wall
Fluid A
at TA > T1 T1 Q . Q . T2
Fluid B
at TB < T2 x Δx Q .
TA – T1 =
1/(hAA)
T1 – T2 =
Δx/(kA)
T2– TB =
1/(hBA)
Prof. R. Shanthini & 12 March 2012

73. Overall Heat Transfer through a Plane WallQ .
TA – T1 =
1/(hAA)
T1 – T2 =
Δx/(kA)
T2– TB =
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
Prof. R. Shanthini & 12 March 2012

74. Overall heat transfer through hollow-cylinderri ro Ti To
Fluid A is inside the pipe
Fluid B is outside the pipe
TA > TB L Q .
= U A
(TA – TB)
where
1/UA = 1/(hAAi)
+ ln(ro/ri) / 2πkL
+ 1/(hBAo)
Prof. R. Shanthini & 12 March 2012

75. Example 6 . Q = U A (TA – TB) = U A (120 – 35) What is UA?
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?
Prof. R. Shanthini & 12 March 2012

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 = ( ) / 2πL
Prof. R. Shanthini & 12 March 2012

77. Example 6 continued . . UA = 2πL / (0.235 + 0.0021 +5.35 + 1) Q = U A
(120 – 35)
steel
air
= 2πL
(120 – 35) / ( )
steam
insulation
= 81 L Q .
/ L
= 81 W/m
Prof. R. Shanthini & 12 March 2012

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) 
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

80. The Stefan–Boltzmann Law of RadiationQ t
= ε σ A T4
ε = emissivity, which takes a value between 0 (for
an ideal reflector) and 1 (for a black body).
σ = x 10-8 W/m2.K4 is the Stefan-Boltzmann
constant
A = surface area of the radiator
T = temperature of the radiator in Kelvin.
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

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    
Prof. R. Shanthini & 12 March 2012

83. What is the Sun’s surface temperature?Q t
= ε σ A T4
(4 π 6.92 x 1016 m2)
= x 1018 m2
(4 π 1.52 x 1022 m2)(1000 W/m2)
= 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
Prof. R. Shanthini & 12 March 2012

84. If object at temperature T is surrounded by an environment at temperature T0, the net radioactive heat flow is: Q t
= ε σ A (T4 - To4 )
Temperature of the radiating surface
Temperature of the environment
Prof. R. Shanthini & 12 March 2012

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:
[( ) K]4 Q t
= ε σ A T4
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
Prof. R. Shanthini & 12 March 2012

86. Example 8 Q = ε σ A (T4 - To4 ) t Q t Solution: [(273+25) K]4
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 t
= ε σ A (T4 - To4 )
[(273+25) K]4
[( ) K]4 Q =
(0.8) x (5.67 x 10-8 W/m2.K4) x (0.5 m2)
x [(423 K)4 -(298 K)4 ] t
= 547 W
Prof. R. Shanthini & 12 March 2012

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.
Prof. R. Shanthini & 12 March 2012

88. Example 9 Q = ε σ A (T4 - To4 ) t Q t Solution: [(273+25) K]4
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? Q t
= ε σ A (T4 - To4 )
Solution:
[(273+25) K]4
[( ) K]4 Q =
(0.8) x (5.67 x 10-8 J/s.m2.K4) x [π x (0.06) m2]
x [(408 K)4 -(298 K)4 ] t
= 10.2 W (about 20% of the power is dissipated by radiation)
Prof. R. Shanthini & 12 March 2012