Transcript Reading Materials: Chapter 9 LAST LECTURE 1

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.
6
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
7
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
8
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
9
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
10
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)
11
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


12
Determining the size (H/T area) of the
exchanger
13
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
14
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
15
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
16
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)
17
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.
18
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
19
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


20
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

21
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


22
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

23
Summary
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