CHEE 3369
Transport Processes
LP15. Heat Exchangers
Department of Chemical and Biomolecular Engineering
Heat Exchanger Design & Analysis
Heat Exchanger Configurations
Textbook reading 7th ed. Ch. 22
Sections 22.1, 2, 3, 4
2
Shell-and-tube heat exchanger
Fig_22-3
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Compact heat exchangers
Fig_22-4
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Two kinds of crossflow heat exchangers:
Fluid A is unmixed. Fluid B can be mixed or unmixed
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Left: A very large shell-and-tube heat exchanger. Right: Shelland-tube heat exchanger baffles.
A worker
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Double – Pipe Heat Exchanger
Fig_22-1
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Double – Pipe Heat Exchanger
Fig_22-1
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Temperature Profiles – Single Pass, Double –Pipe
Heat Exchangers
9
Temperature profile in a condenser with subcooling
Fig_22-6
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Diagram of temperature vs. contact area for singlepass counterflow heat exchanger
Fig_22-7
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Heat Exchanger Design & Analysis: Single Pass Counterflow
“1”
Nomenclature:
TH 1
T=
TH 2 TH in
H out ,
TC1
T=
TC 2 TC out
C in ,
TH2
TC2
TH1 z
Thermal Energy Balances
Across dA:
=
dQ U (TH − TC )dA
“2”
TC1
(1)
dQ = m H c pH dTH
(2)
dQ = m C c pC dTC
(3)
“H”
z
dA=wdz
“C”
H
m
C
m
Flat plate heat exchanger
 1
1 
−
d (TH − T=
 dQ

From (2) and (3):
C)
 m H c pH m C c pC 
 1
1 
Using (1):
d (TH − T=
−
U (TH − TC ) dA

C)
 m H c pH m C c pC 
(4)
12
Heat Exchanger Design & Analysis: Single Pass Counterflow
Integrating Eq. (4) assuming constant U
z=L


d (TH − TC )
1
1
−
U ∫ wdz

∫∆T (TH=


−
T
m
c
m
)
C
H
pH
C c pC 
0
z=

1
∆T2
(5,6)
 ∆T2   1
1 
−
ln 
UA,

=
 ∆T1   m H c pH m C c pC 
A=
Lw
Overall thermal energy balances:
=
Q m H c pH (TH 2 − T
=
m C c pC (TC 2 − TC1 )
H1)
(7,8)
13
Using Eqs. (7) and (8) in Eq. (6):
 ∆T2   (TH 2 − TH 1 ) (TC 2 − TC1 ) 
−
ln 
UA =
= 
Q
Q
 ∆T1  

or
=
Q
∆T2 − ∆T1 )
(
=
UA
ln ( ∆T2 ∆T1 )
UA
( ∆T2 − ∆T1 )
Q
UA ∆Tlm
(9)
∆Tlm is called “logarithmic-mean temperature difference (LMTD).”
Even though eq. (9) was derived for single-pass counterflow parallelplate exchangers, Eq. (9) is also applicable to single-pass tubular heat
exchangers for both counterflow and parallel flow. In fact, it applies to
all types of heat exchangers shown in fig. 22.5 of the textbook.
14
Heat Exchanger Design & Analysis (Cont.)
Problem arises in using eq. (9) when ∆T2 ≈ ∆T1
Solution:
If ∆T2 ∆T1 < 1.5 then the simple arithmetic mean is within 1%
of the logarithmic mean temperature difference and
=
Q UA ∆TAM
can be used, where
1
∆TAM ≡ ( ∆T1 + ∆T2 )
2
If U = a(1 + bT), it can be shown that: Q = A
(U1∆T2 − U 2 ∆T1 )
 U1∆T2 
ln 

U
∆
T
 2 1
15
Summary – Heat Exchanger Design & Analysis
For an exchanger of any configuration with hot and cold streams
and with the heat transfer coefficient and the heat capacities
specified, the 8 variables are:
TH in , TH out , TC in , TC out , Q , A, m H , m C ,
(1)
or equivalently
TH in , TH out , TC in , TC out , Q , A, CH , CC ,
(2)
where
CH
ˆ
m=
CC m C cˆ pC
H c pH ,
(3), (4)
16
Basic Heat Exchanger Equations
Stream equations for any exchanger:
Q=
CH (TH in − TH out ) , Q =
CC (TC out − TC in )
For any single pass, double-pipe heat exchanger, (counterflow or
parallel flow), condenser, or evaporator:
Q = UA∆Tlm = UA
∆T1 − ∆T2
∆T − ∆T1
= UA 2
 ∆T 
 ∆T 
ln  1 
ln  2 
 ∆T2 
 ∆T1 
(5), (6)
(7)
For parallel flow heat exchanger
∆T=
1
(T
H in
− TC in ) , ∆T=
2
(T
H out
− TC out )
For counterflow heat exchanger
∆T=
1
(T
H in
− TC out ) , ∆T=
2
(T
H out
(7a,b)
− TC in )
17
Heat Exchanger Design & Analysis (Cont.)
Structure of Simple Heat Exchangers (one pass each side)
• Variables
• Four end temperatures
4
• Two flow rates
2
• Area of heat exchanger
1
• Heat duty
1
Exchanger is completely described by these variables.
Key to problem is determining what is given and what is unknown.
• Equations
• Energy balance – stream 1
• Energy balance – stream 2
• Performance equation
=
Q UA
( ∆T2 − ∆T1=
)
ln ( ∆T2 ∆T1 )
1
1
1
UA ∆Tlm
No. degrees of freedom = No. variables – No. equations = 8 – 3 = 5
Exchanger is uniquely defined only if 5 of the variables (above) are specified.
18
Overall Heat Transfer Coefficient: Double Pipe Heat Exchanger
From the lecture on conduction through
multiple resistances, the total heat transfer
rate between the two fluids is given by:
Q=
or
=
Q
2π L(T fi − T fo )
1  Ro 
1 
 1
+ ln   +


 hi Ri k p  Ri  ho Ro 
2π Ri L(T fi − T fo )
= U i Ai ∆T
 1 Ri  Ro  Ri 
 + ln   +

h
k
R
h
R
 i
p
 i  o o 
1
Ui ≡
 1 Ri  Ro  Ri 
 + ln   +

h
k
R
h
R
p
 i  o o 
 i
Inside
Fluid
(1)
Ri
Ro
Outside
Fluid
Ri = inside radius
Ro = outside radius
(2,3)
(4)
Based on the inside
area of the inner
tube
Ai = 2π Ri L
19
Similarly,
Inside
Fluid
Q
=
2π R0 L(T fi − T fo )
= U o Ao ∆T ,
Outside
 R0 Ro  Ro  1 
+ ln   + 
Fluid

 hi Ri k p  Ri  ho 
U0 ≡
1
Ri
Ro
Ri = inside radius
Ro = outside radius
 R0 Ro  Ro  1 
+ ln   + 

 hi Ri k p  Ri  ho 
Based on the outside area of the inner tube, A0 = 2π R0 L
Ui and Uo are different from one another, but the product UA
20
is the same for the inside and outside radius.
Fouling Resistance
In practical applications, the performance of a
heat exchanger diminishes with time because of
the buildup of scale on the tube walls.
The thermal resistance of the scale is given by
1
1
R
=
−
S
U f U0
where U0 is the overall heat transfer coefficient of
the clean exchanger and Uf is that of the fouled
exchanger. The overall coefficient can be
monitored with time and the exchanger cleaned
when the fouling resistance becomes sufficiently
high. See Table 22.1 for fouling resistances and
Table 22.2 for estimates of typical exchanger
overall heat transfer coefficients.
Time in
service
Scale Buildup
21
Design of More General Types of Heat Exchangers
Use modified equation for Q:
=
Q
UA ( F ∆Tlm )
where F is a correction factor which must be specified for each
exchanger. F values for different shell-and-tube, as well as crossflow
heat exchangers are given in the textbook. ∆Tlm has to be calculated on
the basis of counterflow.
The correction factor is represented in terms of two variables: Y and Z
F = F (Y , Z )
Subscripts “t” or “s” mean
“tube” or “shell” side,
respectively
Y≡
Tt , out − Tt , in
Ts , in − Tt , in
m t c p , t Ct Ts , in − Ts , out
Z≡
==
m s c p , s Cs Tt , out − Tt , in
This approach is used to design a heat exchanger to perform a specific task
of heating a stream from a known temperature to another known
temperature, and cooling a different stream from a known temperature to
22
still another known temperature.
Finding the correction factor F
Z≡
Ts ,in − Ts ,out
Tt ,out − Tt ,in
TH 1 − TH 2
Tt ,out − Tt ,in TC 2 − TC1
=
Y =
=
Ts ,in − Tt ,in TH 1 − TC1
TC 2 − TC1
23
The factor F is defined such that the LMTD should always be
calculated for the equivalent counterflow heat exchanger with
the same hot and cold temperatures.
24
For various types of exchangers for which correction
factors are available:
T
(
Q = UAF
H in
− TC out ) − (TH out − TC in )
 TH in − TC out 
ln 
 TH out − TC in 


(8)
(9)
where
F = F (Y , Z )
 Tt ,out − Tt , in 
and Y ≡ 
,
 Ts , in − Tt , in 


 Ts , in − Ts , out
Ct
Z ≡
=

Cs
 Tt, out − Tt , in



(10),(11)
25
Fig_22-9c
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
The NTU method for heat exchanger design
The LMTD design equation Q
= UA∆Tlm for heat exchangers is
useful, provided that the inlet and exit temperatures of both
hot and cold streams are known. If only the inlet temperatures
are known a trial-and-error procedure is necessary to find the
exit temperatures.
There is another, namely, the number of transfer units (NTU),
method to find the exit temperatures without trial-and-error.
However, this requires that we know (or we can determine) the
heat exchanger effectiveness (ε).
The effectiveness is defined as the actual heat transfer divided
by the maximum possible heat transfer that would take place if
infinite surface area were available for heat exchange.
27
Observe that the fluid with a smaller capacity coefficient
(designated as Cmin), suffers a greater total temperature change.
If Cc=Cmin (left figure), and if there is an infinite area available for
energy transfer, the exit temperature of the cold fluid will equal
the inlet temperature of the hot fluid. In such case,
Fig_22-11
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
CH (TH in − TH out ) Cmax (TH in − TH out )
=
CC (TC out − TC in ) max
Cmin (TH in − TC in )
ε
(12)
If the hot fluid has the minimum heat capacity coefficient (right
fig. of previous page) the effectiveness equation becomes:
ε
CC (TC out − TC in )
Cmax (TC out − TC in )
=
CH (TH in − TH out ) max Cmin (TH in − TC in )
(13)
Note that the above two equations have the same denominator
and that the numerator is the actual heat transfer load, Q.
29
Use either eq. (12) or eq. (13) to get
=
Q ε Cmin (TH in − TC in )
(14)
Thus, if ε and the hot and cold stream inlet temperatures are
known, one can calculate the heat transfer rate. In practice, this
situation arises when someone wants to use an existing heat
exchanger (that was designed for a different process).
The heat exchanger effectiveness ε has been calculated for a
variety of heat exchanger geometries and designs and can be
found in tables (see for example W. M. Kays and A. L. London,
Compact Heat Exchangers 3rd edition McGraw-Hill, New York
1984) or in the form of graphs, see for example next page.
30
For example, for a single-pass counterflow heat exchanger,
the effectiveness is given by (see textbook)
Cmin
)]
Cmax
C
exp[− NTU (1 − min )]
Cmax
1 − exp[− NTU (1 −
ε=
1−
Cmin
Cmax
UA
where the number of transfer units NTU =
Cmin
31
Heat exchanger effectiveness
for a shell-and-tube
configuration with one shell
pass and two (or multiples of
two) tube passes.
Fig_22-12c
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Example 1
A shell-and-tube heat exchanger must be designed to heat 2.5 kg/s of water
from 15 °C to 85 °C. The heating is to be accomplished by passing hot engine oil
which is available at 160 °C, through the shell side of the heat exchanger. The
oil is known to provide an average heat transport coefficient of h0=400 W/m2-K
on the outside of the tubes. Ten tubes pass the water through the shell. Each
tube is thin walled, of diameter D=25 mm and makes eight passes through the
shell. If the oil leaves the exchanger at 100 °C, what is its flow rate? How long
must the tubes be to accomplish the desired heating?
At the average temperature of 130 °C, the oil heat capacity is 2350 J/kg-K. At
the average temperature of 50 °C the properties of water are: Cp=4181 J/kg-K,
µ=548 x 10-6 N-s/m2, k=0.643 W/m-K, Pr=3.56
Solution
An overall energy balance on the water stream gives the heat load
required of the exchanger
Q
mC C pC (TC out −=
TC in ) 2.5 x 4181 x (85-15)=7.317 x 105 W
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Schematic for Example 1.
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
This amount of heat comes by cooling of the oil.
=
Q mH C pH (TH in − TH out ) ⇒ 7.317 x 105 W=
5.19 kg/s.
mH 2350 x (160-100) ⇒ mH =
The required tube length is found using
Now
1
U=
1/hi +1/h0
=
Q UAF ∆Tlm
Assumes no conduction resistance through the
tube wall (thin wall, eq. 1 with Ri≈R0).
hi is calculated using a heat transfer correlation for flow inside tubes.
m1 = mC / N =2.5/10=0.25 kg/s per tube (note N=10 tubes).
4m1
4 x 0.25
=
Re D =
=
23, 234 ⇒ turbulent flow
-6
π Dµ 3.14 x 0.025 x 548 x 10
=
Use NuD = 0.023 ReD4/5 Pr 0.4 0.023
=
x (23,234) 4/5 x (3.56)0.4 119
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Finding the correction factor F
Z≡
Ts ,in − Ts ,out
Tt ,out − Tt ,in
TH 1 − TH 2
Tt ,out − Tt ,in TC 2 − TC1
=
Y =
=
Ts ,in − Tt ,in TH 1 − TC1
TC 2 − TC1
36
0.643
k
119 3060 W/m 2 -K
=
hi =
NuD
=
0.025
D
1
U=
= 354 W/m 2 -K
1/400+1/3060
The correction factor F maybe obtained from ther fig. on slide # 16.
Z
TC 2 − TC1 85 − 15
TH 1 − TH 2 160 − 100
Y
= = 0.48
= = 0.86=
TH 1 − TC1 160 − 15
TC 2 − TC1
85 − 15
From the figure one reads F =0.87
(TH in − TC out ) − (TH out − TC in )
75 − 85
=
ΔTlm
= = 80 C
Now
ln(75 / 85)
 (TH in − TC out ) 
ln 

T
T
−
(
)
 H out C in 
37
The total area of the 10 pipes is A = N π DL
Finally
=
L
Q
=
UN π DF ∆Tlm
7.317 x 105
= 37.8 m
354 x 10 x 3.14 x 0.025 x 0.87 x 80
Observations:
1. With L/D=37.8/0.025=1516 the assumption of fully developed
conditions throughout the tube is justified.
2. With 8 passes the length of the shell is approximately 37.8/8=4.7 m
Assumptions made:
1. Negligible heat losses to the environment and negligible kinetic and
potential energy changes.
2. Constant physical and transport properties.
3. Negligible thermal resistance of tube wall.
4. No fouling resistance.
5. Fully developed velocity and temperature profiles in the tubes.
38
Example 2
Hot exhaust gases enter a finned-tube , crossflow heat exchanger at 250 °C with a
flow of 1.5 kg/s. The hot gaseous stream is used to heat pressurized water at a
flow of 1 kg/s, which enters the heat exchanger at 35 °C. The gas side overall
heat transfer coefficient is 100 W/m2-K and the corresponding surface area is 40
m2. What is the heat transfer load of the exchanger and what is the exit
temperature of each of the two streams?
Data: CpC=4197 J/kg-K, CpH=1000 J/kg-K
Solution
Use the NTU method since the exit temperatures are not known and you can’t
calculate the LMTD. You can guess the exit temperature of one of the streams
and use the LMTD method (and iterate, as necessary, i.e., trial-and-error), but
the NTU approach is more convenient in this case.
The heat capacity rates are:
39
=
CH m=
1.5 x 1000=1500 W/K
H C pH
CC =mC C pC = 1.0 x 4197=4197 W/K
Cmin 1500
Cmin =CH and = = 0.357
Cmax 4197
UA 100 x 40
= =
= 2.67
The number of transfer units is NTU
Cmin
1500
Use Fig.22.13(a) of textbook to get ε = 0.82
Then use eq. (14) to calculate the heat load
5
=
Q ε Cmin (TH in − T=
)
0.82
x
1500
x
(250-35)=2.65
x
10
W
C in
The exit temperatures are then easily calculated using
5
Q
= CH (TH in − TH out ) ⇒ 2.65 x 10=
1500 (250-TH out ) ⇒ TH out= 73.3 °C
=
Q CC (TC out − TC in ) ⇒ 2.65 x =
105 4197 (TC out − 35) ⇒ T=
98.1 °C
C out
40
Heat exchanger effectiveness for cross flow with
both fluids unmixed
Fig_22-13a
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.