Transcript CONVECTION HEAT TRANSFER Figure

Reading Materials: Chapter 9
Heat transfer results from a temperature difference.
LAST LECTURE
1
CONVECTION HEAT TRANSFER
Modes

forced
flow induced by external agency e.g. pump
eg. forced-draught air cooler, evaporators

natural
fluid motion caused by temperature-induced
density gradients within fluid
Examples
air flow over hot steam pipe, fireplace
circulation, cooling electronic devices
2
CONVECTION HEAT TRANSFER
Warm (lighter) air rises
Figure: Natural
convection flow
over a heated
steam pipe
Cool (more dense) air
falls to replace warm rising air
3
Modelling Convection
Forced convection generally most-effective transport of
energy from solid to fluid.
.Q
Engineer's prime
concern

Solid
Ts
Flowing
fluid
Tb
rate of convection

enables sizing of
equipment
4
Modelling Convection
Experimentally found that:
Q  A  Ts  Tb 
Q  hA  Ts  Tb 
h - convective heat transfer
coefficient.
Main problem
predict h value for:
• variety fluids & flow rates
• range of shapes
5
Resistance Concept
Rate equation
•
Q
Written in same form as Ohm’s Law:
surface
Ts
fluid
Tb
1
hA
Potential Difference  V 
Current flow I =
Resistance R 
Q  hA  Ts  Tb 
Ts  Tb  Ts  Tb



1/ hA
Ts
•
Q
Tb
R
(Ts-Tb)= driving force
(1/hA) – thermal resistance (R) for convection heat transfer.
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TYPICAL UNITS FOR h
-2
-1
-1
-2
-1
S.I.:
W m K or J s m K
British:
Btu hr ft (F deg)
Conversion:
1 W m K = 0.176 Btu hr ft (F deg)
-1 -2
-2
-1
-1
-1 -2
-1
Typical Values
free convection (air)
5 - 60
forced convection (air)
25 - 300
forced convection (water)
200 - 10,000
boiling water
2,000 - 25,000
condensing steam
4,000 - 110,000
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Illustration 27.1
Air at 20°C is blown over an electrical resistor to keep it cool.
The resistor is rated at 40,000 ohm and has a potential
difference of 200 volts applied across it.
The expected mean heat transfer coefficient between the
resistor surface and the air is 50 W m-2K-1.
What will be the surface temperature of the resistor, which
has a surface area of 2 cm2?
Air
20°C

h = 50 W m K

R = 40,000 
Potential = 200 V
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SOLUTION
Energy Balance
Generation = heat loss by convection
Rate of heat generation
PQ
2
V V
P  VI  V   
R R
V2
2002
Q

 1 watt
4
R
4x10
Convective loss
Q  hA(Ts  Tb )
1  (50)(2x10 4 )(Ts  20)
Ts  120o C
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Determining the size (H/T area) of the
exchanger
Q  h1A  T1  T2 
T2  T3 

 kA
h
x
2A
 T3  T4 
(10.25)
Q 
 T1  T4 
1
x
1


h1A kA h2 A

overall driving force
 resis tances
(10.26)
Figure 1. Heat transfer between two flowing fluids
separated by a rectangular
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Determining the size (H/T area) of the
exchanger
Q 
Q 
Figure 2: Heat transfer between
two flowing fluids separated by a
cylindrical wall
 Ti  To 
r
ln  o 
ri 
1
1



hi A i
2kL
ho A o
 Ti  To 
R1  R2  R3
(10.27)
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Overall heat-transfer coefficient
As a short-hand method of describing heatexchanger performance, we use the overall heattransfer coefficient,
Q  hA  Ts  Tb 
Qduty  Uo AT
Qduty
T
T


1
R
Uo A
(10.28)
where
Uo  overall heat-transfer coefficient W / m 2K


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Determining the size (H/T area) of the
exchanger
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Illustration 27.2
Consider the kettle below. For the conditions given, find the flame
temperature for the following values of the heat transfer coefficients:
hi (boiling) = 4000 W m-2K-1 ho (gas flame) = 40 W m-2K-1
•
Q = 1.88 kW
Water
hi
Flame
ho
FLAME
1
ho
T3
T2
T1
T0
?
.Q
T2
x
k
T1
To
1
hi
T3 = 100°C
BOILING
WATER
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Solution
Plane slab- area constant, eliminate A:
1
1
x
1



Uo A ho A kA h1A
1
1 1.2x10 3
1



Uo 40
204
4000
Uo  39.6 W m-2K -1
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Solution
Q  1883 W (as before)
A = 3.14x10-2 m2 (as before)
Qduty  Uo A  To  T3 
Qduty
 To  T3  
AUo
1883 

o

1514K

1514
C
 To  T3  
2
3.14x10  39.6 
Water
hi
Flame
ho
.Q
T3
T2
T1
To


To  1514  100  1614o C
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Determining the size (H/T area) of the
exchanger
(10.28)
Qduty  Uo ATavg
• The term Tavg in equation 10.28 represents the temperature
difference between the hot and cold streams averaged.
• For single-pass exchangers, the appropriate form of Tavg is
the log-mean temperature difference, Tlog mean (often
abbreviated LMTD), defined as
Log mean temperature difference  Tlog mean
Tlog mean
T1  T2


ln  T1


T
2

(10.29)
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Determining the size (H/T area) of the
exchanger
Qduty  Uo ATavg
A
Qduty
Uo Tlog mean
(10. 30)
For shell-and-tube exchangers, the
inside area (Ai) of the tubes is smaller
than the outside area (Ao). However,
the differences between Ai and Ao will
be neglected.
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Example
Saturated Steam,
Hot
280oF, mstream
Cold
Oil, 110oF, 960 lbm/min
Saturated water,
280oF, mstream
Oil, 35oF, 960 lbm/min
Balance on cold stream:
mCp  Tout  Tin  

 cold  Qduty
10.24b 

lbm  
Btu 
o 
 960
 0.74
110  35  F   Qduty

o 

min  
lbm F 

 cold
Btu
Qduty  53,280
min
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Example
How much area is required for the counter-current heat
exchanger in Example 10.5?
1
Saturated Steam,
280oF, mstream
2
Saturated water,
Hot
280oF, mstream
Cold
Oil, 35oF, 960 lbm/min
Oil, 110oF, 960 lbm/min
T1   280  35   245o F
T2   280  110   170o F
Tlog mean
245  170

 205o F
ln 245
170


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Example
From table 10.5

o
Uo  150 Btu / hr ft 2 F
A

53,280 Btu/min
150
Btu
hr ft 2 oF


60min
2
 104 ft
o
1hr
205 F

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Example
How much area is required for the co-current heat exchanger
in Example 10.5?
1
Saturated Steam,
280oF, mstream
Oil,
35oF,
2
Saturated water,
Hot
280oF, mstream
Cold
960 lbm/min
Oil, 110oF, 960 lbm/min
T1   280  110   170o F
T2   280  35   245o F
Tlog mean
170  245

 205o F
ln 170
245


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Example
From table 10.5

o
Uo  150 Btu / hr ft 2 F
A

53,280 Btu/min
150
Btu
hr ft 2 oF


60min
2
 104 ft
o
1hr
205 F

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Summary
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