Heat transfer / Heat exchanger Chapter6.ppt

6.1 INTRODUCTION
 A heat exchanger can be defined as any device
that transfers heat from one fluid to another that
are at different temperatures.
 Some of the specific applications of heat
exchangers are
 in space heating and air-conditioning,
 power production,
 waste heat recovery, and
 chemical processing.
2

6.2 HEAT EXCHANGER TYPES
Heat exchangers may be classified according to:
 Flow arrangement,
 Parallel flow,
 Counter flow,
 Cross flow.
 Construction type
 Tubular,
 Plate,
 Extended,
Regenerative.
3

6.2 HEAT EXCHANGER TYPES…
 Number of fluid passes,
 Two pass,
 Three pass, …
 Compactness
 Gas to liquid,
 Liquid to liquid,
 Phase change.
 Transfer processes
 Direct,
Indirect.
4

 Double pipe, cross-flow and shell-and-tube heat
exchangers are the common types of tubular heat
exchangers.
a) b)
Fig. 6.1 Double pipe heat exchangers a) Parallel flow
b) Counter flow
5

6
Fig. 6.2 Cross flow heat exchangers a) Both fluids unmixed
b) One fluid mixed and the other unmixed

7
Fig. 6.3 Shell and Tube heat exchanger with one shell pass
and one tube pass

8
Fig. 6.4 Shell and tube heat exchangers with more
fluid passes

9
Fig. 6.5 Compact heat exchangers

6.3 HEAT EXCHANGER ANALYSIS
10
There are different types of heat exchanger analysis
techniques, the most common ones being:
 The log mean temperature difference (LMTD) and
 The effectiveness-number of transfer units ( -NTU)
ε
methods.
The LMTD method is appropriate when we need to
select a heat exchanger where the temperature
changes and the mass flow rate are known.
The -NTU method is used when we want to predict
ε
the outlet temperatures of both fluids.

6.4 LOG MEAN TEMPERATURE
DIFFERENCE METHOD
11
Parallel flow heat exchangers
In heat exchanger analysis, the energy balance is
used along with the following assumptions:
 The heat exchanger is insulated from its surroundings,
and the heat exchange is only between the hot and
cold fluids,
 Axial conduction along the tubes is negligible,
 Potential and kinetic energy changes are negligible,
 The fluid specific heats are constant, and
 The overall heat transfer coefficient is constant.

DIFFERENCE METHOD…
12
Fig. 6.6 Temperature distribution in parallel flow arrangement

13
 According to the above assumptions, heat lost by
hot fluid is the same as heat gained by cold fluid.
 Heat lost by hot fluid
(6.1)
 Heat gained by cold fluid
(6.2)
 We can also express the heat transfer rate across
area dA as,
(6.3)
   
i
h
o
h
h
i
h
o
h
h
h T
T
C
T
T
c
m
Q ,
,
,
,
.
.






   
i
c
o
c
c
i
c
o
c
c
c T
T
C
T
T
c
m
Q ,
,
,
,
.
.




TdA
U
Q
d 

.

14
 T=Th-Tc is the local temperature difference
between the hot and cold fluids.
 To integrate equation (6.3), let’s make use of the
following relation
  c
h dT
dT
T
d 


  














c
h
c
h C
C
Q
d
C
Q
d
C
Q
d
T
d
1
1
.
.
.
  












c
h C
C
TdA
U
T
d
1
1
  dA
C
C
U
T
T
d
c
h












 1
1

15
 Integration of this equation gives
(6.4)
 

 












2
1
2
1
1
1
dA
C
C
U
T
T
d
c
h
       
 
 
1
2
.
,
,
,
,
.
.
,
,
.
,
,
1
2 1
1
ln
T
T
Q
UA
T
T
T
T
Q
UA
Q
T
T
Q
T
T
UA
C
C
UA
T
T
i
c
i
h
o
c
o
h
i
c
o
c
i
h
o
h
c
h

















 






















































1
2
1
2
.
ln
T
T
T
T
UA
Q

16
(6.5)
Tlm is the logmean temperature difference
   
lm
i
c
i
h
o
c
o
h
i
c
i
h
o
c
o
h
T
UA
T
T
T
T
T
T
T
T
UA
Q






























,
,
,
,
,
,
,
,
.
ln

17
Counter flow heat exchangers
Fig. 6.7 Temperature distribution in counter flow arrangement

18
 Equations (6.1) through (6.5) apply for counter flow
heat exchanger.
 However, for the counter flow exchanger the
endpoint temperature differences must now be
defined as
T1=Th,i-Tc,o (6.6)
T2=Th,o-Tc,I (6.7)

19
 And the log mean temperature difference will be,
(6.8)
 Then the heat transfer rate is
(6.9)
   

























































o
c
i
h
i
c
o
h
o
c
i
h
i
c
o
h
lm
T
T
T
T
T
T
T
T
T
T
T
T
T
,
,
,
,
,
,
,
,
1
2
1
2
ln
ln
lm
T
UA
Q 

.

20
Multi-pass and cross flow heat exchangers
 The log mean temperature difference method
developed earlier is limited to parallel and counter
flow heat exchangers.
 Similar relations can also be developed for cross flow
and multi-pass shell-and-tube heat exchangers but the
resulting relations are so complicated because of the
complex flow conditions.
 The easiest way is to relate the log mean temperature
difference of such heat exchangers with that of counter
flow heat exchanger using some correction factor F.

21
Tlm=FTlm,CF (6.10)
Where Tlm,CF is given by
(6.11)
The correction factor F for some arrangements can be
obtained from figures.
   

























































o
c
i
h
i
c
o
h
o
c
i
h
i
c
o
h
CF
lm
T
T
T
T
T
T
T
T
T
T
T
T
T
,
,
,
,
,
,
,
,
1
2
1
2
,
ln
ln

22
a) One-shell pass and multiple of two (2, 4, 6, etc) tube passes
1
1
1
2
t
T
t
t
P



1
2
2
1
t
t
T
T
R




23
b) Two-shell pass and multiple of four (4, 8, 12, etc) tube passes

24
c) Single-pass cross-flow with one fluid mixed and the other unmixed

25
Fig. 6.8 correction factor F for common shell and tube and
cross flow heat exchangers
d) Single-pass cross-flow with both fluids unmixed

26
Example 6.1
A counter flow, concentric tube heat exchanger is used to
cool the lubricating oil for a large industrial gas turbine
engine. The flow rate of cooling water through the inner
tube (Di=25mm) is 0.2kg/s, while the flow rate of oil
through the annulus (Do=45mm) is 0.1kg.s. The oil and
water enter at temperatures of 100 and 300
C,
respectively. If the outlet temperature of the oil is to be
600
C and the heat exchanger is 60m long, determine
the outlet temperature of the water, the LMTD and the
overall heat transfer coefficient.

27
Solution

28
 Properties of the fluids from tables: Unused engine
oil (Th = 800
C = 353 K): cp=2131J/kgK,
=3.25x10-2
Ns/m2
, k=0.138W/mK. Water
(Tc=350
C): cp=4178 J/kgK, =725x10-6
Ns/m2
,
k=0.625W/mK, Pr = 4.85.
 The required heat transfer rate may be obtained
from the overall energy balance for the hot fluid
  W
T
T
c
m
Q o
h
i
h
h
h 8524
)
60
100
(
2131
1
.
0
,
,
.
.







29
 The water outlet temperature is
 The LMTD for counter flow arrangement is
C
c
m
Q
T
T
c
c
i
c
o
c
0
.
.
,
, 2
.
40
4178
2
.
0
8524
30 





        2
.
43
2
.
40
100
30
60
ln
2
.
40
100
30
60
ln
ln
,
,
,
,
,
,
,
,
1
2
1
2






































































o
c
i
h
i
c
o
h
o
c
i
h
i
c
o
h
lm
T
T
T
T
T
T
T
T
T
T
T
T
T

30
 The overall heat transfer coefficient can be
obtained as,
K
m
W
T
L
D
Q
T
A
Q
U
lm
i
lm
2
.
.
/
87
.
41
2
.
43
60
025
.
0
8524












6.5 THE EFFECTIVENESS-NUMBER OF
TRANSFER UNITS (-NTU) METHOD
 The parameter P in the LMTD correction factor method
requires three temperatures for its computation.
 The inlet temperature of both the hot and cold streams
is usually a given, but when the cold-side outlet
temperature is not known, a trial-and-error method is
required to evaluate P.
 The trial-and-error procedure may be avoided in the
-
ε NTU method, and because of its suitability for
computer-aided design, the -
ε NTU method is gaining
more popularity.
31

TRANSFER UNITS (e-NTU) METHOD…
Kays and London (1984) have shown that the heat
exchanger transfer equations may be written in
dimensionless form, which results in three
dimensionless groups:
 Capacity rate ratio:
(6.14)
 Exchanger heat transfer effectiveness:
(6.15)
32
 
1
*
0
*
max
min


 C
C
C
C
 
1
0
max
.
.


 

Q
Q

This is the ratio of the actual heat transferred to the
maximum heat that could be transferred if the
exchanger were a counter-flow exchanger.
 Number of transfer units:
(6.16)
 The number of transfer units is a measure of the size
of the exchanger.
 Observe that if Ch>Cc then (T1-T2)<(t2-t1) and if
Cc>Ch then (T1-T2)>(t2-t1)
33
min
C
UA
NTU 

 The fluid that “might” experience the maximum
temperature change,T1-t1, is the fluid that has the
minimum capacity rate. Thus, the maximum possible
heat transfer can be expressed as
(6.17a)
Or
(6.17b)
34
   
h
c
c C
C
t
T
C
Q 

 1
1
max
.
   
c
h
h C
C
t
T
C
Q 

 1
1
max
.

 And either of these can be obtained with the
counter-flow exchanger. Therefore, the exchanger
effectiveness can be written as
(6.18)
 Observe that the value of will range between
ε
zero and unity and that for a given and , the
ε
actual heat transfer in the exchanger will be
(6.19)
35
 
 
 
 
1
1
min
1
2
1
1
min
2
1
max
.
.
t
T
C
t
t
C
t
T
C
T
T
C
Q
Q c
h








 
1
1
min
.
t
T
C
Q 

max
.
Q

36

Example 6.2
Hot oil (Cp=2200J/kgK) is to be cooled by water
(Cp=4180J/kgK) in a 1-shell-pass and 6-tube-pass heat
exchanger. The tubes are thin walled and are made of copper
with an internal diameter of 1.8cm. The length of each tube
pass in the heat exchanger is 3m, and the overall heat transfer
coefficient is 340W/m2
K. Water flows through the tubes at a
rate of 0.1kg/s, and the oil through shell at a rate of 0.2kg/s.
The water and oil enter at temperatures 180
C and 1600
C,
respectively. Determine the rate of heat transfer and the outlet
temperatures of the water and the oil.
37

Solution
 The heat capacity rate of oil
is:
 And the heat capacity rate of
water is:
 The minimum heat capacity
rate is:
38
K
W
c
m
C
oil
h /
440
2200
2
.
0
.










K
W
c
m
C
water
c /
418
4180
1
.
0
.










K
W
C
C c /
418
min 


 The maximum heat transfer rate is:
 The actual heat transfer rate is:
 Where the effectiveness is obtained from table
ε
6.1 as
 The capacity rate ratio is
39
    W
t
T
C
Q 59356
18
160
418
1
1
min
max
.





 
1
1
min
.
t
T
C
Q 

 
 
 
 
1
2
2
2
*
1
exp
*
1
*
1
exp
1
*
1
*
1
2





















C
NTU
C
C
NTU
C
C

95
.
0
440
418
*
max
min



C
C
C

 The number of transfer unit is
 The effectiveness will be
ε
 Then the actual heat transfer rate is:
40
    83
.
0
418
3
018
.
0
6
340
6
min
min










C
dL
U
C
UA
NTU
 
 
 
  44
.
0
95
.
0
1
83
.
0
exp
95
.
0
1
95
.
0
1
83
.
0
exp
1
95
.
0
1
95
.
0
1
2
1
2
2
2























W
Q
Q 91
.
26053
59356
44
.
0
max
.
.





 The outlet temperature of the water, t2, can be
obtained from the relation
 The outlet temperature of the oil, T2, can be
obtained from the relation
41
   
1
2
,
,
.
t
t
C
T
T
C
Q c
i
c
o
c
c 



C
C
Q
t
t
c
0
.
1
2 33
.
80
418
91
.
26053
18 





   
2
1
,
,
.
T
T
C
T
T
C
Q h
o
h
i
h
h 



C
C
Q
T
T
h
0
.
1
2 8
.
100
440
91
.
26053
160 






Example 6.3
In a textile manufacturing plant, the waste dyeing water
(Cp=4925J/kgK) at 750
C is to be used to preheat fresh
water (Cp=4180J/kgK) at 150
C at the same flow rate
in a double-pipe counter-flow heat exchanger. The
heat transfer surface area of the heat exchanger is
1.65m2
and the overall heat transfer coefficient is
625W/m2
K. If the rate of heat transfer in the heat
exchanger is 35kW, determine the outlet temperature
and the mass flow rate of each fluid stream.
42

43
Fresh
water
15C
Dyeing
water
75C
Tc,out
Th,out

Solution
 The temperature differences at the two ends of the
heat exchanger are
and
44
15
75
out
h,
in
c,
out
h,
2
out
c,
out
c,
in
h,
1










T
T
T
T
T
T
T
T
   

























15
75
ln
15
75
ln
,
,
,
,
2
1
2
1
out
h
out
c
out
h
out
c
lm
T
T
T
T
T
T
T
T
T

 Then the rate of heat transfer can be expressed as
(1)
 The rate of heat transfer can also be expressed as
(2)
(3)
Tc,out=41.40
C, Th,out=49.30
C,
45
   
















15
75
ln
15
75
65
.
1
625
.
0
35000
,
,
,
,
.
out
h
out
c
out
h
out
c
lm
S
T
T
T
T
T
UA
Q
    35000
15
4180 ,
.
,
,
.
.












 out
c
water
fresh
in
c
out
c
p T
m
T
T
C
m
Q
    35000
75
4295 ,
.
,
,
.
.












 out
h
water
dyeing
out
h
in
h
p T
m
T
T
C
m
Q
kg/s
0.317

m


Example 6.4
A cross-flow air-to-water heat exchanger with an
effectiveness of 0.65 is used to heat water
(Cp=4180J/kgK) with hot air (Cp=1010J/kgK).
Water enters the heat exchanger at 200
C at a rate of
4kg/s, while air enters at 1000
C at a rate of 9kg/s. If
the overall heat transfer coefficient based on the
water side is 260W/m2
K, determine the heat transfer
surface area of the heat exchanger on the water side.
Assume both fluids are unmixed.
46

47
Hot Air
100C
9 kg/s
Water
20C, 4 kg/s

Solution
 The heat capacity rates of the hot and cold fluids
are
Therefore,
and
48
C m C
C m C
h h ph
c c pc
   
   


(4 kg / s)(4.18 kJ / kg. C) = 16.72 kW/ C
(9 kg / s)(1.01 kJ / kg. C) = 9.09 kW/ C
C Cc
min .
  
9 09 kW/ C
C
C
C
  
min
max
.
.
.
9 09
16 72
0544

 Then the NTU of this heat exchanger corresponding
to C = 0.544 and = 0.65 is determined to be
NTU = 1.5
 Then the surface area of this heat exchanger
becomes
49
2
m
52.4








C
.
kW/m
260
.
0
)
C
kW/
09
.
9
)(
5
.
1
(
NTU
NTU 2
min
min U
C
A
C
UA
s
s

6.6 FOULING FACTOR
 After a period of operation the heat transfer
surfaces for a heat exchanger may
 become coated with various deposits present in the flow
systems, or
 the surfaces may become corroded as a result of the
interaction between the fluids and the material used for
construction of the heat exchanger.
 In either event, this coating represents an additional
resistance to the heat flow, and thus results in
decreased performance.
50

6.6 FOULING FACTOR…
 The overall effect is usually represented by a fouling
factor, or fouling resistance, Rf, which must be included
along with the other thermal resistances making up the
overall heat transfer coefficient.
 Fouling factors must be obtained experimentally by
determining the values of U for both clean and dirty
conditions in the heat exchanger. The fouling factor is
thus defined as
(6.19)
 Fouling factors for some fluids are given in table 6.2.
51
clean
dirty
f
U
U
R
1
1



52
Table 6.2 Fouling factors of some fluids

53
Table 6.2 Fouling factors of some fluids…

Example 6.5
Water at an average temperature of 107°C and an average
velocity of 3.5m/s flows through a 5m long stainless steel tube
(k=14.2W/mK) in a boiler. The inner and outer diameters of the
tube are Di=1.0cm and Do=1.4cm, respectively.
a) If the convection heat transfer coefficient at the outer surface of
the tube where boiling is taking place is ho=8400W/m2
K,
determine the overall heat transfer coefficient Ui of this boiler
based on the inner surface area of the tube.
b) Repeat the solution assuming a fouling factor Rf,i =
0.0005m2
K/W on the inner surface of the tube
54

Solution
a) The properties water at 107C  110C are
 The Reynolds number is
 which is greater than 10,000. Therefore, the flow is
turbulent.
55
58
.
1
Pr
K
.
W/m
682
.
0
/s
m
10
268
.
0
/
2
2
6




 
k



600
,
130
s
/
m
10
268
.
0
m)
m/s)(0.01
5
.
3
(
Re 2
6



 

h
m D
V

 Assuming fully developed flow,
and
 The total resistance of this heat exchanger is then
determined from
56
342
)
58
.
1
(
)
600
,
130
(
023
.
0
Pr
Re
023
.
0 4
.
0
8
.
0
4
.
0
8
.
0




k
hD
Nu h
C
.
W/m
23,324
=
(342)
m
01
.
0
C
W/m.
682
.
0 2



 Nu
D
k
h
h

57
C/W
00157
.
0
=
]
m)
m)(5
(0.014
C)[
.
W/m
8400
(
1
m)]
C)(5
W/m.
2
.
14
(
2
[
)
1
/
4
.
1
ln(
]
m)
m)(5
(0.01
C)[
.
W/m
324
,
23
(
1
1
2
)
/
ln(
1
2
2

















 o
o
i
o
i
i
o
wall
i
total
A
h
kL
D
D
A
h
R
R
R
R
R
and
C
.
W/m
4055 2








]
m)
m)(5
(0.01
C/W)[
00157
.
0
(
1
1
1

i
i
i
i RA
U
A
U
R

b) The thermal resistance of heat exchanger with a
fouling factor of Rf,i=0.0005m2
.0
C/W is
determined from
58
C/W
00476
.
0
m)]
m)(5
014
.
0
(
[
C)
.
W/m
8400
(
1
m)
C)(5
W/m.
2
.
14
(
2
)
1
/
4
.
1
ln(
m)]
m)(5
01
.
0
(
[
C/W
.
m
0005
.
0
m)]
m)(5
01
.
0
(
[
C)
.
W/m
324
,
23
(
1
1
2
)
/
ln(
1
2
2
2
,



















R
A
h
kL
D
D
A
R
A
h
R
o
o
i
o
i
i
f
i
i

Then,
59
C
.
W/m
1337 2








]
m)
m)(5
(0.01
C/W)[
00476
.
0
(
1
1
1

i
i
i
i RA
U
A
U
R

Heat transfer / Heat exchanger Chapter6.ppt

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