1. PM3125: Lectures 4 to 6 Content of Lectures 1 to 6: 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
R. Shanthini May 2010

2. is the means by which energy moves from
Heat Transfer
is the means by which energy moves from
a hotter object to
a colder object
R. Shanthini May 2010

3. 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)
R. Shanthini May 2010

4. Mechanisms of Heat Transfer
Latent heat
Conduction
Convection
Radiation
R. Shanthini May 2010

5. Heat travels along the rod
Conduction
HOT
(lots of vibration)
COLD
(not much vibration)
Heat travels along the rod
R. Shanthini May 2010

6. 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.
R. Shanthini May 2010

7. 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?
R. Shanthini May 2010

8. 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
R. Shanthini May 2010

9. 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
R. Shanthini May 2010

10. 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)
R. Shanthini May 2010 10

11. 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
R. Shanthini May 2010 11

12. 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 12
R. Shanthini May 2010

13. 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
R. Shanthini May 2010

14. Example 1 continued . . Q (1000 – 40) A = (0.21/1.04) + (0.03/0.07) Q=
W/m2 A
R. Shanthini May 2010

15. 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
R. Shanthini May 2010

16. Conduction through hollow-cylinderri To L Q .
Ti – To =
[ln(ro/ri)] / 2πkL
R. Shanthini May 2010

17. 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
R. Shanthini May 2010

18. 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
R. Shanthini May 2010

19. Example 3 continued . . 2 π L ( 600 – 100) Q = + [ln(5/2)] / 0.2
= 680 W/m L
R. Shanthini May 2010

20. 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)
R. Shanthini May 2010

21. 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.
R. Shanthini May 2010

22. Convection
It explains why breezes come from the ocean in the day and from the land at night
R. Shanthini May 2010

23. 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)
R. Shanthini May 2010

24. 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)
R. Shanthini May 2010

25. 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
R. Shanthini May 2010

26. 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.
R. Shanthini May 2010

27. 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
R. Shanthini May 2010

28. 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
R. Shanthini May 2010

29. 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.
R. Shanthini May 2010

30. 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)
R. Shanthini May 2010

31. 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
R. Shanthini May 2010

32. 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)
R. Shanthini May 2010

33. 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?
R. Shanthini May 2010

34. 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
R. Shanthini May 2010

35. 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
R. Shanthini May 2010

36. 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) 
R. Shanthini May 2010

37. 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.
R. Shanthini May 2010

38. 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.
R. Shanthini May 2010

39. 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.
R. Shanthini May 2010

40. 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    
R. Shanthini May 2010

41. 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
R. Shanthini May 2010

42. 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
R. Shanthini May 2010

43. 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
R. Shanthini May 2010

44. 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
R. Shanthini May 2010

45. 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.
R. Shanthini May 2010

46. 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)
R. Shanthini May 2010

47. Mathematical Problems on Heat Exchanger
Tc,in
Th,out
Th,in
Tc,out . . .
Q = mc cc (Tc,out – Tc,in)
= mh ch (Th,in – Th,out)
R. Shanthini May 2010

48. Mathematical Problems on Heat Exchanger
Tc,in
Parallel-flow heat exchanger
Th,out
Th,in
Tc,in
Th,out
Th,in
Tc,out
Tc,out
high heat transfer
low heat transfer
R. Shanthini May 2010

49. Mathematical Problems on Heat Exchanger
Parallel-flow heat exchanger
Tc,in
Th,out
Th,in
Tc,out
ΔTa
ΔTb a b .
Q = U A ΔT
ΔTa - ΔTb
is the log mean temperature difference (LMTD)
where ΔT =
ln(ΔTa / ΔTb)
R. Shanthini May 2010

50. Mathematical Problems on Heat Exchanger
Tc,out
Counter-flow heat exchanger
Th,out
Th,in
Tc,in . . .
Q = mc cc (Tc,out – Tc,in)
= mh ch (Th,in – Th,out)
R. Shanthini May 2010

51. Mathematical Problems on Heat Exchanger
Tc,out
Counter-flow heat exchanger
Th,out
Th,in
Th,in
Tc,in
Th,out
Tc,out
Tc,in
R. Shanthini May 2010

52. Mathematical Problems on Heat Exchanger
Counter-flow heat exchanger
Tc,in
Tc,out
Th,in
Th,out
ΔTa
ΔTb a b .
Q = U A ΔT
ΔTa - ΔTb
is the log mean temperature difference (LMTD)
where ΔT =
ln(ΔTa / ΔTb)
R. Shanthini May 2010

53. Example in heat Exchanger Design
An exhaust pipe, 75 mm outside diameter, is cooled by surrounding it by an annular space containing water.
The hot gases enters the exhaust pipe at 350oC, gas flow rate being 200 kg/h, mean specific heat capacity at constant pressure 1.13 kJ/kg K, and comes out at 100oC.
Water enters from the mains at 25oC, flow rate 1400 kg/h, mean specific heat capacity 4.19 kJ/kg K.
The heat transfer coefficient for gases and water may be taken as 0.3 and 1.5 kW/m2 K and pipe thickness may be taken as negligible.
Calculate the required pipe length for (i) parallel flow, and for (ii) counter flow.
R. Shanthini May 2010

54. Example in heat Exchanger Design
Solution: . . .
Q = mc cc (Tc,out – Tc,in)
= mh ch (Th,in – Th,out)
(1400 kg/hr) (4.19 kJ/kg K) (Tc,out – 25)oC
= (200 kg/hr) (1.13 kJ/kg K) (350 – 100)oC
The temperature of water at the outlet = Tc,out = 34.63oC.
R. Shanthini May 2010

55. Example in heat Exchanger Design
Solution continued:
Parallel flow:
ΔTa = 350 – 25 = 325oC
ΔTb = 100 – = 65.37oC
ΔTa - ΔTb
325 – 65.37
ΔT = =
= 162oC
ln(ΔTa / ΔTb)
ln(325 / 65.37) .
Q = U A ΔT
= (UA) 162oC
What is UA?
R. Shanthini May 2010

56. Example in heat Exchanger Design
Solution continued:
1/U = 1/hwater + 1/hgases
= 1/ /0.3 = 4 (kW/m2 K)-1
Therefore, U = 0.25 kW/m2 K
A = π (outer diameter) (L) = π (0.075 m) (L m) . Q
= (UA) 162oC
= (0.25) π (0.075) L (162) kW .
What is Q?
R. Shanthini May 2010

57. Example in heat Exchanger Design
Solution continued: . . .
Q = mc cc (Tc,out – Tc,in)
= mh ch (Th,in – Th,out)
= (200 kg/h) (1.13 kJ/kg K) (350 – 100)oC
= kW
Substituting the above in . Q
= (UA) 162oC
= (0.25) π (0.075) L (162) kW
we get
L = 1.64 m
R. Shanthini May 2010

58. Example in heat Exchanger Design
Solution continued:
(ii) Counter flow:
ΔTa = 350 – = oC
ΔTb = 100 – 25 = 75oC
ΔTa - ΔTb
– 75
ΔT = =
= oC
ln(ΔTa / ΔTb)
ln( / 75) .
Q = U A ΔT
= (UA) oC .
Q = kW; U = 0.25 kW/m2 K ;
A = π (0.075) L m2
R. Shanthini May 2010
Therefore, L = 1.59 m

59. Other Heat Exchanger Types
Cross-flow heat exchanger with both fluids unmixed
The direction of fluids are perpendicular to each other.
The required surface area for this heat exchanger is usually calculated by using tables.
It is between the required surface area for counter-flow and parallel-flow heat exchangers.
R. Shanthini May 2010

60. Other Heat Exchanger Types
One shell pass and
two tube passes
Th,in
Tc,in
Tc,out
Th,out
The required surface area for this heat exchanger is calculated using tables.
R. Shanthini May 2010

61. Other Heat Exchanger Types
Two shell passes and
two tube passes
Th,in
Tc,in
Tc,out
Th,out
The required surface area for this heat exchanger is calculated using tables.
R. Shanthini May 2010

62. Batch Sterilization (method of heating):
Electrical
heating
Direct steam
sparging
Steam
heating
R. Shanthini May 2010

63. For batch heating with constant rate heat flow:
Total heat lost by the coil to the medium
= heat gained by the medium
M - mass of the medium
T0 - initial temperature of the medium
T - final temperature of the medium
c - specific heat of the medium
q - rate of heat transfer from the
electrical coil to the medium
t - duration of electrical heating .
Electrical
heating .
q t =
M c (T - T0)
R. Shanthini May 2010

64. For batch heating by direct steam sparging:
M - initial mass of the raw medium
T0 - initial temperature of the raw medium
ms - steam mass flow rate
t duration of steam sparging
H - enthalpy of steam relative to the
enthalpy at the initial temperature
of the raw medium (T0)
T - final temperature of the mixture
c specific heat of medium and water . . .
(ms t) (H + cT0)
+ M c T0
= (M + mst) c T
Direct steam
sparging . .
ms t H
= (M + ms t) c (T – T0)
R. Shanthini May 2010

65. ( ) For batch heating with isothermal heat source:
M - mass of the medium
T0 - initial temperature of the medium
TH - temperature of heat source (steam)
T - final temperature of the medium
c - specific heat of the medium
t - duration of steam heating
U - overall heat transfer coefficient
A - heat transfer area
( )
T0 - TH
Steam
heating
U A t = M c ln
T - TH
Could you prove the above?
R. Shanthini May 2010

66. ( ) ( ) For batch heating with isothermal heat source: T0 - TH
( )
T0 - TH
U A t = M c ln
T - TH
( )
U A t
T = TH + (T0 - TH) exp -
c M
Steam
heating
R. Shanthini May 2010

67. Example of batch heating by direct steam sparging:
A fermentor containing 40 m3 medium at 25oC is going to be sterilized by direct injection of saturated steam. The steam at 350 kPa absolute pressure is injected with a flow rate of 5000 kg/hr, which will be stopped when the medium temperature reaches 122oC. Determine the time taken to heat the medium.
Additional data required:
Enthalpy of saturated steam at 350 kPa = ??
Enthalpy of water at 25oC = ??
The heat capacity of the medium kJ/kg.K
The density of the medium are kJ/kg.K and 1000 kg/m3, respectively.)
R. Shanthini May 2010

68. . . Example of batch heating by direct steam sparging: ms t H
A fermentor containing 40 m3 medium at 25oC is going to be sterilized by direct injection of saturated steam. The steam at 350 kPa absolute pressure is injected with a flow rate of 5000 kg/hr, which will be stopped when the medium temperature reaches 122oC. Determine the time taken to heat the medium.
Additional data: The enthalpy of saturated steam at 350 kPa and water at 25oC are 2732 and 105 kJ/kg, respectively. The heat capacity and density of the medium are kJ/kg.K and 1000 kg/m3, respectively.
Solution:
Use the equation below: . .
ms t H
= (M + ms t) c (T – T0)
R. Shanthini May 2010

69. . . ms t H = (M + ms t) c (T – T0) (5000 kg/hr) (th) (2732-105) kJ/kg
= [(40 m3)(1000 kg/m3) + (5000 kg/hr)(th)](4.187 kJ/kg.K)(122-25)K
Taking the heating time (th) to be in hr, we get
(5000 th) (2627) kJ = [ t](4.187)(97)kJ
(5000 th) [2627 – x 97] = x x 97
th = hr
Therefore, the time taken to heat the medium is hours.
R. Shanthini May 2010

70. ( ) Example of batch heating with isothermal heat source: T0 - TH
A fermentor containing 40 m3 medium at 25oC is going to be sterilized by an isothermal heat source, which is saturated steam at 350 kPa absolute pressure. Heating will be stopped when the medium temperature reaches 122oC. Determine the time taken to heat the medium.
Additional data: The saturated temperature of steam at 350 kPa is 138.9oC. The heat capacity and density of the medium are kJ/kg.K and 1000 kg/m3, respectively.
Solution:
Use the equation below:
( )
T0 - TH
U A t = M c ln
T - TH
R. Shanthini May 2010

71. ( ) T0 - TH U A t = M c ln T - TH (2500 kJ/hr.m2.K) (40 m2) (tc)
( )
T0 - TH
U A t = M c ln
T - TH
(2500 kJ/hr.m2.K) (40 m2) (tc)
= (40 m3) (1000 kg/m3) (4.187 kJ/kg.K) ln[( )/( )]
Taking the heating time (th) to be in hr, we get
(2500 kJ/K) (40) (th) = (40) (1000) (4.187 kJ/K) ln[113.9/16.9]
(2500 kJ/K) (40) (th) = (40) (1000) (4.187 kJ/K) (1.908)
th = hr
Therefore, the time taken to heat the medium is hours.
R. Shanthini May 2010

72. Explain why heating with isothermal heat source takes twice the time taken by heating with steam sparging, even though we used the same steam.
R. Shanthini May 2010

73. Question from PM3125 / Jan 2010 past paper
A steel pipeline (inside diameter = mm; outside diameter = mm) contains saturated steam at 121.1oC. The line is insulated with 25.4 mm of asbestos. Assume that the inside surface temperature of the metal wall is at 121.1oC and the outer surface of the insulation is at 26.7oC. Taking the average value of ksteel as 45 W/m.K and that of kasbestos as W/m.K, calculate the following:
(a) Heat loss for 30.5 m of pipe length [10 marks]
(b) Mass (in kg) of steam condensed per hour in the pipe due to the heat loss [10 marks]
Additional data given on the next slide:
R. Shanthini May 2010

74. Question from PM3125 / Jan 2010 past paper
Additional Data:
i) Heat transfer rate through the pipe wall is given by,
where L is the length of pipe, T1 and T2 are the respective temperatures at the inner and outer surfaces of the insulated pipe, r1 and r2 are the respective inner and outer radius of the steel pipe, and r3 is the outer radius of the insulated pipe.
ii) Latent heat of vapourization of steam could be taken as 2200 kJ/kg.
R. Shanthini May 2010

75. Group Assignment will be uploaded at
(keep track of the site)
R. Shanthini May 2010

76. End of slides for the heat transfer lecture
R. Shanthini May 2010

77. Additional material not used in the lectures.
R. Shanthini May 2010

78. Critical Radius of InsulationTo r
Pipe
Insulation ro Ti ri Q .
Ti – To =
[ln(ro/ri)] /2πkPL
+ [ln(r/ro)] /2πkIL + 1/hairA
Pipe resistance could be neglected
A = 2 π r L
R. Shanthini May 2010

79. Critical Radius of InsulationQ .
Ti – To =
[ln(r/ro)] /2πkIL + 1/(hair 2πrL)
2π L ( Ti – To) =
[ln(r/ro)] /kI + 1/(hair r)
Convective resistance
Insulation
resistance
Increasing r increases insulation resistance and decreases heat transfer.
Increasing r decreases convective resistance and increases heat transfer.
R. Shanthini May 2010

80. Critical Radius of InsulationQ . d
/dr
= 0 at the critical radius of insulation,
which leads to rcr = kI / hair
If the outer radius of the pipe (ro) < rcr and if insulation is added to the pipe, heat losses will first increase and go through a maximum at the insulation radius of rcr and then decrease.
If the outer radius of the pipe (ro) > rcr and if insulation is added to the pipe, heat losses will continue to decrease.
R. Shanthini May 2010