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Prof Andrew Tay ME3122 Heat Transfer
13. Heat Exchangers
A heat exchanger (HX) is a device which enables the
continuous transfer of heat between two fluid streams
separated by a solid wall.
Heat exchangers are used in numerous engineering systems
such as power generation, air conditioning, chemical
industries, food industries, aeroplanes, cars, etc. They are
usually classified according to flow arrangement and type of
construction.
1
Prof Andrew Tay ME3122 Heat Transfer
13.1 Basic Types of Heat Exchangers
(i) Concentric tube or double-pipe heat exchanger.
One fluid flows through the inner tube, the other fluid flows through the
annulus. Both fluid streams traverse the exchanger once only and this
arrangement is called a single-pass arrangement.
If both fluids flow in the same direction, HX is called a parallel-flow
type. If both fluids flow in opposite directions, HX is called a counter-
flow type.
Parallel-flow Counter-flow
2
Prof Andrew Tay ME3122 Heat Transfer
(ii) Cross-flow heat exchangers
When the two fluid flow paths along the heat transfer surface are at right
angles, the heat exchanger is of the cross-flow type.
(a) Both fluids unmixed
Each fluid is unmixed as it flows through the HX. The temperature of the
cross-flowing fluid is not uniform laterally and varies in x and y
directions.
3
Prof Andrew Tay ME3122 Heat Transfer
(b) One unmixed, the other fully mixed
Temperature of the cross-flowing fluid is laterally uniform and varies in
the x direction only.
4
Prof Andrew Tay ME3122 Heat Transfer
(iii) Shell-and-tube heat exchangers
In order to increase the effective heat transfer area per unit volume,
commercial HXs provide for multiple passes through the tubes; fluid
outside tubes in the shell may be routed back and forth by means of
baffles. The diagram above shows a shell and tube HX with one tube
passes and one shell pass in a cross-counter-flow arrangement.
One fluid flows in the tubes while the other flows in a shell enclosing
the tubes.
5
Prof Andrew Tay ME3122 Heat Transfer
(a) One shell pass and two tube passes.
(b) Two shell passes and four tube passes.
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Prof Andrew Tay ME3122 Heat Transfer
Baffle Types
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Prof Andrew Tay ME3122 Heat Transfer
(iv) Compact heat exchangers
Compact heat exchangers have very large area per unit volume.
8
Prof Andrew Tay ME3122 Heat Transfer
13.2 Overall Heat Transfer Coefficient
oo
w
iiooii AhR
AhAUAU
1111
Heat transfer between two fluids A and B across a tube wall is
Overall heat transfer coefficient, U is given by
(13.4)
(13.3)
(13.2)
(13.1)
q
ii Ah
1
wR
Thermal Resistance Network
oo
w
ii
BA
AhR
Ah
TTq
11
TA
hi
ho
TB
q
k
ro ri
oo Ah
1
resistance thermal walltube
2
ln
kL
rrR iow
where Ui , Uo are based on Ai , Ao , respectively.
BAoioi TTAUq //
From eqns (13.1) & (13.3):
where L = length of tube, and i/o refers to
inside/outside surface of the smaller tube.
9
Prof Andrew Tay ME3122 Heat Transfer
oo
w
ii
BA
AhR
Ah
TTq
11
For a double-pipe heat exchanger,
B
1 2
A (13.5)
(13.1)
Hence from eqn (13.4):
Rw is usually negligible for thin metallic tubes since ro ri
ln
= 0, and k is high.
oi hhU
111
UUUAAA oioi ; since
Furthermore, if one of the fluids is boiling or condensing, where h is
very high,
kL
rrR iow
2
ln (13.2)
oi horhU
10
Prof Andrew Tay ME3122 Heat Transfer
Approximate Values of Overall Heat Transfer Coefficients Physical Situation U (W/m2K)
Brick exterior wall, plaster interior, uninsulated 2.55
Frame exterior wall, plaster interior, uninsulated 1.42
Frame exterior wall, plaster interior, rock-wool insulation 0.4
Plate-glass window 6.2
Double plate-glass window 2.3
Steam condenser 1100-5600
Feedwater heater 1100-8500
Freon 12 condenser with water coolant 280-850
Water-to-water heat exchanger 850-1700
Finned-tube heat exchanger, water in tubes, air across tubes 25-55
Water-to-oil heat exchanger, steam in tubes, air over tubes 110-350
Steam to light fuel oil 170-340
Steam to heavy fuel oil 56-170
Steam to kerosene or gasoline 280-1140
Finned-tube heat exchanger, steam in tubes, air over tubes 28-280
Ammonia condenser, water in tubes 850-1400
Alcohol condenser, water in tubes 255-680
11
Prof Andrew Tay ME3122 Heat Transfer
13.3 Fouling Factors
Over time the surfaces of the tube may be pitted
due to corrosion and/or covered with deposits,
leading to increased thermal resistance to heat
transfer between the two fluids, A and B, which
can be accounted for by Fouling Factors,
B A
ooo
fo
w
i
fi
ii
BA
AhA
RR
A
R
Ah
TTq
11
(13.6)
.and fofi RR
q
ii Ah
1
wR
Thermal Resistance Network
oo Ah
1
o
fo
A
R
i
fi
A
R
12
Prof Andrew Tay ME3122 Heat Transfer 13
Prof Andrew Tay ME3122 Heat Transfer
There are generally two types of analysis performed on heat exchangers.
13.3 Heat Exchanger Analysis
(a) Heat Exchanger Design
In this analysis, usually the inlet temperatures of the heating and
cooling fluids are known, and the heat exchanger type and area is
required to be determined in order to bring the outlet temperature of
one of the fluids to a specified value. All the inlet and outlet fluid
temperatures are known or can be easily determined. For this
analysis, the Log Mean Temperature Difference (LMTD) Method
is best used.
(b) Heat Exchanger Performance/Rating
In this analysis, a heat exchanger is available and one is required to
determine its performance under various flow conditions. For this
analysis, the Effectiveness-NTU Method is best used.
14
Prof Andrew Tay ME3122 Heat Transfer
13.4 Log Mean Temperature Difference Method
As can be seen, T between the hot and cold fluid streams is not
constant over the HX area. Hence, the rate of heat transfer between
the two fluid streams is
where T is some suitable mean temperature difference.
(13.7) TUAq
2 2
T
A
Thi
Tho
Tco
Tci
T1 T2
T
A
Thi
Tho
Tci
Tco T1
T2
Parallel Flow Counter Flow 1 1
The figure below illustrates the variations in fluid temperature for
both parallel and counter flow heat exchangers.
T T
15
Prof Andrew Tay ME3122 Heat Transfer
Heat transfer across
elemental area dA is
Consider a parallel flow HX:
ccchhh dTcmdTcmdq
hh
hcm
dqdT
cc
ccm
dqdT
cchh
chcmcm
dqTdTTd
11)(
(13.8)
Parallel Flow Heat Exchanger
T
A
Thi
Tho
dq
Tco
Tci
dA
T T1 T2
dAcmcm
UT
Td
cchh
11Eliminating dq (13.11)
TUdATTUdAdq ch Now, dq is also given by (13.10)
(13.9)
16
Prof Andrew Tay ME3122 Heat Transfer
cchh cmcmUA
T
T
11ln
1
2
Integrating
(13.16)
(13.15)
(13.14)
(13.13)
(13.12)
1212 cccchhhh TTcmTTcmq
21 hh
hhTT
qcm
12 cc
ccTT
qcm
q
TT
q
TTUA
T
T cchh 1221
1
2ln
lm
chch TUAT/T
TTUA
T/T
TTTTUAq
12
12
12
1122
lnln
The total heat transfer as given by the energy balance is
or and
Substituting and gives
Log Mean Temp. Difference (LMTD)
17
Prof Andrew Tay ME3122 Heat Transfer
Special Operating Conditions
pcmC Rate,Capacity (13.17)
(a) Hot fluid condensing: Th is constant and Ch is infinite.
(b) Cold fluid evaporating: Tc is constant and Cc is infinite.
(c) Ch = Cc : T is constant.
18
Prof Andrew Tay ME3122 Heat Transfer
For other heat exchanger systems, the heat transfer is calculated using a
correction factor F applied to the LMTD for a counter flow double-
pipe system with the same hot and cold fluid temperatures. Thus
lmTUAFq
The above recognizes the fact that the counter flow double-pipe heat
exchanger is the most efficient flow arrangement.
(13.18)
13.5 Correction Factor Charts
Correction Factor, F, values are available for various heat exchanger
systems, usually in the form of charts.
19
Prof Andrew Tay ME3122 Heat Transfer
11
12
tT
ttP
12
21
tt
TT
2112 ttcmTTcm tpSp
Sp
tp
cm
cm
In the Correction Factor charts provided, the parameters P and R are
defined as
and R =
where t : tube side; T : shell side;
1 : inlet and 2 : outlet
R is also =
Correction Factor Charts
and because
(13.20)
(13.19)
20
Prof Andrew Tay ME3122 Heat Transfer
F can be seen as a correction factor for a heat exchanger, and is a
measure of the deviation of Tlm from the counter flow case. Also
Note:
When a phase change occurs, as in condensation or boiling
(evaporation), the fluid remains at essentially constant temperature in
the heat exchanger. Therefore,
P = 0 (t2 = t1)
or R = 1 (T2 = T1)
and F = 1 for boiling or condensation
values of P : 0 < P < 1
values of R : 0 R
For all cases, F 1.0
21
Prof Andrew Tay ME3122 Heat Transfer
Fig. 1: Correction Factor for Heat Exchanger with
One Shell Pass and Two (or Multiples of Two) Tube
Passes.
1 : in
2 : out
22
Prof Andrew Tay ME3122 Heat Transfer
Fig. 2: Correction Factor for Heat Exchanger
with Two Shell Passes and Four (or Multiples of
Four) Tube Passes.
23
Prof Andrew Tay ME3122 Heat Transfer
Fig. 3: Correction Factor for Single Pass Cross-
Flow Heat Exchangers with the Shell Side Fluid
Mixed, and the Other Fluid Unmixed.
24
Prof Andrew Tay ME3122 Heat Transfer
Fig. 4: Correction Factor for a Single Pass Cross-
Flow Heat Exchanger with Both Fluids Unmixed.
25
Prof Andrew Tay ME3122 Heat Transfer
Example
Determine the heat transfer surface area required for a heat exchanger
constructed from 25.4 mm OD tube to cool 25000 kg/h of ethyl
alcohol solution (cp = 3.822 kJ/kgK ) from 65.6 C to 39.4 C using
22727 kg/hr of water available at 10 C. Assume that the overall heat
transfer coefficient based on the outer tube area is 568 W/m2K and
consider each of the following arrangements
a) parallel-flow double-pipe HX
b) counter-flow double-pipe HX
c) Shell and tube HX with 2 shell passes and 18 multiples of 4 tube
passes, alcohol flowing through the shell and water through the
tubes, and
d) Cross-flow HX, one tube pass and one shell pass, shell side mixed.
26
Prof Andrew Tay ME3122 Heat Transfer
C36.1
102422727439665822325000
2
2
1221
c
c
cccchhhh
T
T....
TTcmTTcmq
W692778
4.396.65822.32500021
hhhh TTcmq
212
12
m65.9318.5568
692778
C18.51065.6/36.139.4ln
1065.636.139.4
ln
LMTDq/UA
T/T
TTLMTD
Solution:
Energy balance for determining the outlet temperature of water for each of the
4 arrangements:
Heat transfer rate,
a) Parallel flow double-pipe HX:
m826025409365 ./.D/AL (Too long, needs multiple tube passes.)
27
Prof Andrew Tay ME3122 Heat Transfer
TUAq
2m3541529568
692778.
.A
b) Counter flow double-pipe HX:
We note that = = 26542 W/K (special case) and in this case temperature difference T between hot and cold fluids is
constant, and T = 65.6 36.1 = 29.5 C
m518025403541 ./.D/AL
Too long still, needs multiple tube passes.
28
Prof Andrew Tay ME3122 Heat Transfer
47.0106.65
101.36
11
12
tT
ttP
112
21
tt
TT
cm
cmR
s
t
lmTUAFq
c) Shell and tube HX with multiple shell and tube passes:
From F vs P chart (Fig. 2, multiple of 4 tube passes), F = 0.97;
So A = 41.35 / 0.97 = 42.63 m2
There are n = 18 4 = 72 tube segments: A = DL n L = A / (72 D ) = 7.45 m
For 72 tube segments of 25.4 mm OD tubes, L of 7.45 m is meaningful
as compared with parallel-flow and counter-flow double-pipe cases.
29
Prof Andrew Tay ME3122 Heat Transfer
d) Crossflow HX, shell side mixed arrangement:
Again P = 0.47 R = 1.0
From chart F = 0.88
Hence A = 41.35 / 0.88 = 46.99 m2
Corresponding L is physically reasonable if number of tube
passes is sufficiently large.
30
Prof Andrew Tay ME3122 Heat Transfer
13.6 Effectiveness-NTU Method
The LMTD method is convenient when Tci, Tco, Thi, Tho are known or
easily determined. When either Tis or Tos need to be determined from
q, U or A, a trial & error procedure is needed. The -NTU method is
more convenient to use for this situation. Also, it provides a basis for
comparing various heat exchangers.
The effectiveness is defined as
* actual rate of heat lost or gained by either fluid
** rate of heat gained by one fluid if it were to undergo a change in
temperature equal to maximum temperature difference present in
the heat exchanger.
**
*
ratetransferheatpossibleMaximum
ratetransferheatActual (13.21)
31
Prof Andrew Tay ME3122 Heat Transfer
cihi TTcmq minmax
The fluid which attains this maximum change in temperature is the
one with the smaller/minimun value of capacity rate, ie .
(13.22)
mincm
This can be reasoned out as follows. If the temperature change for the
fluid with the minimum capacity rate is and that for the other fluid is , then for energy balance,
From the above equation,
BA TcmTcmq maxmin (13.23)
BA TT
Hence the fluid which can undergo a temperature change equal to the
maximum temperature difference in the HX, must be the one with the
minimum capacity rate . mincm
We define the Capacity Rate Ratio, max
min
max
min
C
C
cm
cmCr
(13.24)
32
Prof Andrew Tay ME3122 Heat Transfer
cihi
cico
cihicc
cicoccc
cihi
hohi
cihihh
hohihhh
TT
TT
TTcm
TTcm
TT
TT
TTcm
TTcm
For parallel flow heat exchanger:
T
A
Thi
Tho
Tci
The subscript on in the above equations denotes which fluid has the minimum value of .
(13.25)
(13.26)
Tco
mincm
2 1
33
Prof Andrew Tay ME3122 Heat Transfer
Thi
Tho
2 1
Tco
Tci
T
A
For counter flow heat exchanger:
cihi
cico
cihicc
cicoccc
cihi
hohi
cihihh
hohihhh
TT
TT
TTcm
TTcm
TT
TT
TTcm
TTcm
(13.28)
(13.27)
HXindifferenceetemperaturMaximum
fluidimumminT )(
In general,
(13.29)
The minimum fluid is always the one experiencing the larger T in the
HX and the maximum T is always (Thi Tci).
34
Prof Andrew Tay ME3122 Heat Transfer
For the parallel flow HX, we had
(13.30)
Derivation of Effectiveness of Heat Exchangers
cchh cmcmUA
T
T
11ln
1
2
if the cold fluid is the minimum fluid.
(13.12)
chcihi
coho
CCUA
TT
TTei
11exp..
r
ch
c
c
CC
UA
C
C
C
UA1exp1exp
cocirhihocicochohih TTCTTTTCTTC .Now
Substituting for Tho into LHS of eqn (13.30), we get
cihi
cocicocircihi
cihi
cococirhi
cihi
coho
TT
TTTTCTT
TT
TTTCT
TT
TT
rcihi
cocir C
TT
TTC
1111
Equating the above with the RHS of eqn (13.30), we get
rrc CCCUA 1/1/exp1 (13.31)
35
Prof Andrew Tay ME3122 Heat Transfer
If the hot fluid is the minimum fluid, it can be shown that the expression
for the effectiveness is the same as eqn (13.30), except that Cc and Ch are interchanged. Hence, the following expression will cater for both cases:
r
r
C
CCUA
1
1/exp1 min (13.32)
Similarly, for the counter flow HX, it can be shown that
rr
r
CNTUC
CNTU
1exp1
1exp1 (13.35)
nits)Transfer U of(Number / min NTUCUA
since it indicates the size of the HX. Hence,
(13.33)
r
r
C
CNTU
1
1exp1 (13.34)
36
Prof Andrew Tay ME3122 Heat Transfer 2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 37
Prof Andrew Tay ME3122 Heat Transfer 38
Prof Andrew Tay ME3122 Heat Transfer
13.7 -NTU Charts for Heat Exchangers
Expressions for (NTU, Cr), have been obtained for other heat exchanger systems.
Often, these are presented in graphical form (-NTU charts) which are more convenient to use.
39
Prof Andrew Tay ME3122 Heat Transfer 40
Prof Andrew Tay ME3122 Heat Transfer 41
Prof Andrew Tay ME3122 Heat Transfer 42
Prof Andrew Tay ME3122 Heat Transfer
-NTU Method for Boilers and Condensers
0 and
0 if finite, is As
max
minmaxm
m
C
CCCC
.C,TTCq
r
m
If one fluid is boiling or condensing, the fluid temperature remains
constant and the other fluid is the fluid.
-NTUe- 1Hence from eqn (13.34): (13.37)
(13.36)
Why?
43
-NTUcihiminmax e-TTCqq 1 and (13.38)
Eqns (13.36) to (13.38) apply for all types of heat exchangers if one of
the fluids is boiling or condensing.
Prof Andrew Tay ME3122 Heat Transfer
Example
Water at the rate of 68 kg/min is heated from 35C to 75C by an oil
having a specific heat of 1.90 kJ/kgK. The oil enters the counter
flow double-pipe heat exchanger at 110C and leaves at 75C. The
overall heat transfer coefficient U is 320 W/m2K. Calculate the heat
exchanger area.
Calculate the water exit temperature if the mass flow rate of water is
reduced to 40 kg/min with other parameters unchanged. Calculate
also the total heat transfer rate under the new conditions.
Prof Andrew Tay ME3122 Heat Transfer
ccchhh TcmTcm
min.mh kg/97170751101900
3575418068
J/min.K3249001711900 hhcm
J/min.K284240418068 cccm
533.075
40
35110
3575
cihi
cico
TT
TT
Solution
a) Energy balance to heat exchanger
Capacity rates:
So water is the minimum capacity rate fluid.
87.0324900
284240
max
min C
C
UA/Cmin = 1.05
A = 1.05284240/(32060) = 15.5 m2
From chart. NTU = 1.05 .
Prof Andrew Tay ME3122 Heat Transfer
b) = = 40 kg/min = 40 4180 = 167200 J/min.K = 2787 W/K
Now NTU = UA/Cmin = (320 15.5)/2787 = 1.78
5150324900
167200
max
min .C
CCr
and q = 2787(91.3 - 35) = 156.8 kW
C391
35110
35
.T
T
TT
TT
co
co
cihi
cico
Hence from chart, 750.
Prof Andrew Tay ME3122 Heat Transfer
Example
A shell and tube heat exchanger with one shell pass and four tube
passes has 4.83 m2 of heat-transfer area. The overall heat transfer
coefficient is 320 W/m2K. It is proposed to cool a stream of oil (cp
= 2.22 kJ/kgK) at 120C, flowing at 1.50 kg/s, with cooling water
available at 13C and a flow of 0.63 kg/s.
Determine the exiting temperature of the 2 streams.
Prof Andrew Tay ME3122 Heat Transfer
W/K3330222050.1 oocm
795.03330
2646
max
min C
CCr 584.0
2646
83.4320
min
C
UANTU
350
131202646max.
q
q
q act
Solution
So water is the minimum fluid.
-NTU chart gives : 0.35
q = 99093 W
99093 = 3330 (120 To2)
To2 = 120 29.8 = 90.2 C
99093 = 2646(Tw2 13)
Tw2 = 50.45 C
W/K2646420063.0 wwcm