Chapter 3.2: Heat Exchanger Analysis
Using -NTU method
The Effectiveness-NTU Method:
The rate of heat transfer in parallel-flow as well as counter-flow heat
exchangers is given by same heat balance.
In. case. of a counter-flow heat
exchanger,
.
Q  m h c p ,h Th,i  Th,o   mc c p ,c Tc ,o  Tc ,i 
Q  C h Th ,i  Th ,o   C c Tc ,o  Tc ,i 
If C h  C c , i .e . C h  C min , C c  C max
Th,i  Th,o   Tc ,o  Tc ,i 
.
The maximum temperature difference in a heat exchanger is the
difference between the inlet temperatures of the hot and cold fluids.
That is, ΔT = Th,i - Tc,i . The heat transfer in a heat exchanger will reach
its maximum value when
(1) the cold fluid is heated to the inlet temperature of the hot fluid or
(2) the hot fluid is cooled to the inlet temperature of the cold fluid.
These two limiting conditions will not be reached simultaneously
unless the heat capacity rates of the hot and cold fluids are identical.
If the heat capacity rates are not the same, then the fluid with
the smaller heat capacity rate will experience a larger
temperature change, and thus it will be the first to reach the
maximum temperature, at which point the heat transfer will
come to a halt.
Therefore, the maximum possible heat transfer rate in a heat
exchanger is
 Q max  C min Th ,i  Tc ,i 
.
where
.
Q max  maximum possible heat rate,
 . 
Cmin   m c p   minimum heat capacity rate,

 min
Th,i  TC ,i   maximum possible temperatur e change.
Example 1
• Cold water enters a counter-flow heat
exchanger at 10°C at a rate of 8 kg/s, where it
is heated by a hot water stream that enters
the heat exchanger at 70°C at a rate of 2 kg/s.
Assuming the specific heat of water to
remain constant at Cp = 4.18 kJ/kg · °C,
determine the maximum heat transfer rate
and the outlet temperatures of the cold and
the hot water streams for this limiting case.
Solution
The first step in the ε–NTU method is to determine the heat capacity
rates of the hot and cold fluids and identify the smaller one:
Ch = m·h Cph = (2 kg/s)(4.18 kJ/kg · °C) = 8.36 kW/°C
Cc = m·c Cpc = (8 kg/s)(4.18 kJ/kg · °C) = 33.4 kW/°C
Therefore,
Cmin = Ch = 8.36 kW/°C
( and cr = Cmin / Cmax = 0.25 )
Then the maximum heat transfer rate is
Q·max = Cmin(Th, in -Tc, in ) = (8.36 kW/°C)(70-10)°C = 502 kW
That is, the maximum possible heat transfer rate in this heat
exchanger is 502 kW. This value would be approached in a counterflow heat exchanger with a very large heat transfer surface area.
The maximum temperature difference in this heat exchanger is
ΔTmax = Th, in -Tc, in = (70 - 10)°C = 60°C.
continued
Therefore, the hot water cannot be cooled by more than 60°C (to 10°C) in
this heat exchanger, and the cold water cannot be heated by more than 60°C
(to 70°C), no matter what we do. The outlet temperatures of the cold and the
hot streams in this limiting case are determined to be
Q· = Cc (Tc, out - Tc, in )
then, Tc, out = Tc, in Q· /Cc = 10°C + 502/33.4 = 25°C , and
Q· = Ch (Th,in – Th,out)
then, Th, out = Th, in - Q· / Ch = 70°C - 502/8.38 = 10°C
Note that
The hot water is cooled to the limit of 10°C (the inlet temperature of the cold
water stream), but the cold water is heated to 25°C only when maximum
heat transfer occurs in the heat exchanger. This is not surprising, since the
mass flow rate of the hot water is only one-fourth that of the cold water,
and, as a result, the temperature of the cold water increases by 0.25°C for
each 1°C drop in the temperature of the hot water.
Also note that if Cmax is used instead of Cmin to evaluate Q·max then,
Q·max = (33.4)(60) = 2008 kW
Tc, out = 10°C + 2008/33.4 = 10 + 60= 70°C
And Th, out = 70°C – 2008/8.38 = 70 - 240 = -170°C impossible
Heat Exchanger Effectiveness ():
actual heat transfer rate
Heat exchanger effectiven ess 
.
maximum possible heat transfer rate
Q
  .
, 0    1
Q max
 Q   Q max   C min Th,i  Tc ,i 
.
.
If  , Th,i , Tc,i are known, the actual heat transfer rate can be
determined from the previous equation.
.
Since, Q  C h Th ,i  Th ,o   C c Tc ,o  Tc ,i 
.
and, Q max  C min Th ,i  Tc ,i 
C h Th,i  Th,o 
C c Tc ,o  Tc ,i 
 
, or  
C min Th,i  Tc ,i 
C min Th,i  Tc ,i 
C min  Tmin. capacity rate fluid
In general,  
C min  Ti

Tmin. capacity rate
Ti
fluid

Tmin. capacity rate
fluid
Th,i  Tc ,i
where, Tmin. Capacity rate fluid = temperature difference of the fluid
which has minimum capacity rate.
Heat Capacity Rate Ratio (Cr) :
 . 
 m cp 
C min 
 min
Cr 
 .
C max 

 m cp 

 max
Cr is a dimensionless number
0  Cr  1
C r  0 when c p   case of ev aporation or condens ation
C r  1 when C min  C max case of equal capacity rates 
Number of Transfer Units (NTU):
UA
UA
NTU 
 .
C min 

m
c

p

 min
,
dimensionl ess number 
 . 
Q  UA Tlm   m c p   Tmin. capacity rate

 min
.
Tmin
UA
 NTU  .

Tlm


m
c

p

 min
UA Tmin
 NTU 

C min Tlm
fluid
Effectiveness – NTU Relations:
a) Parallel-flow Heat Exchanger:
Assuming Cmin = Ch , the exchanger effectiveness may be given by
Tmin Th,i  Th,o


... 1
Ti
Th,i  Tc ,i
The heat capacity rate ratio is defined as
C min
Cr 
C max
 Q  C min Th,i  Th,o   C max Tc ,o  Tc ,i 
.
C min Tc ,o  Tc ,i
Cr 

... 2
C max Th,i  Th,o
The heat rate equation may be expressed as
.
T2  T1 
Q  UA Tlm PF  UA 
lnT2 T1 
 lnT2 T1   
UA
.
T1  T2 
Q
 Th,o  Tc ,o 
UA


 ln
  . Th,i  Tc ,i  Th,o  Tc ,o 

 Th,i  Tc ,i 
Q
T T

T

T
 1
1
h ,i
h ,o
c ,o
c ,i 

 UA

 UA

.
.


C min C max

Q
Q


C min
UA 


1

C min  C max
Th,o  Tc ,o

 e  NTU 1Cr 
Th,i  Tc ,i
Rearranging the left hand side

   NTU 1  C r 

... 3



Th,o  Tc ,o Th,o  Th,i  Th,i   Tc ,o  Tc ,i  Tc ,i 

Th,i  Tc ,i
Th,i  Tc ,i
 Th,i  Th,o   Th,i  Tc ,i   Tc ,o  Tc ,i 

Th,i  Tc ,i 
T

T
T

T
Tc ,o  Tc ,i 
 Th,o 
1
Th,i  Tc ,i 
h ,i  Tc ,i 
h,i
Tc ,o  Tc ,i  Th,i  Th,o 
 Th,o 
1

Th,i  Th,o  Th,i  Tc ,i 
h ,i  Tc ,i 
h,i
Making use of eqns.(1) and (2)


Tmin Th,i  Th,o

... 1
 
Ti
Th,i  Tc ,i



C min Tc ,o  Tc ,i
, C r 

C max Th,i  Th,o

Th,o  Tc ,o

   1  C r   1   1  C r 
Th,i  Tc ,i

2

Th,o  Tc ,o

 1   1  C r 
Th,i  Tc ,i
... 4
Th,o  Tc ,o
Equating eqn.(4) and eqn.(3) 
 e  NTU 1Cr 
 Th,i  Tc ,i
1   1  C r   e  NTU 1Cr 

... 3 

Solving for  , we obtain for the parallel-flow heat exchanger
1  e  NTU 1Cr 

1  Cr
5
Since the same result may be obtained if we assume that Cmin =
Cc , equation (5) applies for any parallel-flow heat exchanger
irrespective of whether the minimum heat capacity rate is
associated with the hot or cold fluid
b) Counter-flow Heat Exchanger:
A similar expression has been developed for a counter-flow heat
exchanger. It is in the form
1  e  NTU 1Cr 

, for C r  1
 NTU 1C r 
1  Cr e
... 6
In case of Cr = 1,  is indeterminate and the applying of L’Hospital
rule gives
NTU

,
1  NTU
for C r  1
... 7 
In case of Cr = 0, it is either parallel-flow or counter-flow regime.
Therefore, eqns.(5) and (6) yield the same expression:
  1  e  NTU ,
for C r  0
... 8
All the foregoing
results show that
for any heat exchanger
  f  NTU , Cr 
The following table
presents heat
exchanger – NTU
relations.
In heat exchanger design
calculations, it is more
convenient to work with
 - NTU relations in the
form:
NTU  f  , Cr 
Explicit relations for NTU
as a function of  and
NTU are provided in the
following table.
The previous expressions are represented graphically in the
following figures.
Effectiveness of a parallel-flow heat exchanger
Effectiveness of a counter-flow heat exchanger
Effectiveness of a shell-and-tube heat exchanger
with one Shell and any multiple of two tube passes
Effectiveness of a shell-and-tube heat exchanger with
two Shell passes and any multiple of four tube passes
Effectiveness of a single-pass, cross-flow heat exchanger
with both fluids unmixed
Effectiveness of a single-pass, cross-flow heat exchanger
with one fluids mixed and the other unmixed
Notes
1. The heat exchanger thermal effectiveness, ε,
increases with increasing values of NTU for a
specified cr .
2. The heat exchanger thermal effectiveness, ε,
increases with decreasing values of cr for a
specified NTU.
3. For ε ˂40%, the capacity rate ratio, cr , does not
have a significant influence on the exchanger
effectiveness.
4. For a small increase in the ε at high values of ε , a
significant increase in the NTU (size) is required.
5. The counter-flow exchanger has the highest ε for
a specified NTU and cr .
Example 2
• Hot oil is to be cooled by water in a 1-shell-pass and
8-tube-passes heat exchanger. The tubes are thinwalled and are made of copper with an internal
diameter of 1.4 cm. The length of each tube pass in
the heat exchanger is 5 m, and the overall heat
transfer coefficient is 310 W/m2·°C. Water flows
through the tubes at a rate of 0.2 kg/s, and the oil
through the shell at a rate of 0.3 kg/s. The water and
the oil enter at temperatures of 20°C and 150°C,
respectively.
• Determine the rate of heat transfer in the heat
exchanger and the outlet temperatures of the water
and the oil.
Solution
The first step in the ε–NTU method is to determine the heat capacity
rates of the hot and cold fluids and identify the smaller one:
Ch = m·h Cph = (0.3 kg/s)(2.13 kJ/kg · °C) = 0.639 kW/°C
Cc = m·c Cpc = (0.2 kg/s)(4.18 kJ/kg · °C) = 0.836 kW/°C
Therefore,
Cmin = Ch = 0.639 kW/°C
and
cr = Cmin / Cmax = 0.764
Then the maximum heat transfer rate is
Q·max = Cmin(Th, in -Tc, in ) = (0.639 kW/°C)(150-20)°C = 83.1 kW
The heat transfer surface area is
As = n(π /DL) = 8/(0.014 m)(5 m) = 1.76 m2
Then the NTU of this heat exchanger becomes
NTU = UAs / Cmin = 0.853
The effectiveness of this heat exchanger corresponding to cr = 0.764
and NTU = 0.853 is determined from the NTU figures and found to be
0.47
We could also determine the effectiveness from the relation in ε -table
more accurately but with more labor.
Then, the actual rate of heat transfer becomes
Q· = ε Q·max = (0.47)(83.1 kW) = 39.1 kW
Finally, the outlet temperatures of the cold and the hot fluid streams
are determined to be
Q· = Cc (Tc, out - Tc, in )
then, Tc, out = Tc, in Q· /Cc = 20°C + 39.1/0.836 = 66.8°C
Q· = Ch (Th,in – Th,out) =
then, Th, out = Th, in - Q· / Ch = 150°C - 39.1/0.639 = 88.8°C
Therefore, the temperature of the cooling water will rise from 20°C to
66.8°C as it cools the hot oil from 150°C to 88.8°C in this heat
exchanger.
Example 3:
Hot exhaust gases, which enter a finned-tube cross-flow heat
exchanger at 300oC and leave at 100oC, are used to heat water at
a flow rate of 1 kg/s from 35 to 125oC. The exhaust gas specific
heat is approximately 1000 J/kg.K, and the overall heat transfer
coefficient based on the gas-side surface area is Uh= 100 W/m2.K.
Calculate the rate of heat transfer and the mass flow rate of the
exhaust gases. Determine also the required gas-side surface area
Ah using: 1) the  - NTU method, and
2) the LMTD method.
Data: Cross-flow heat exchanger with both fluids unmixed.
hot fluid: exhaust gases Th,i = 300oC, Th,o = 100oC, Cp = 1000 J/kg.K.
.
cold fluid: water m c = 1 kg/s, Tc,i = 35oC, Tc,o = 125oC.
Uh = 100 W/m2.K.
.
.
Find: Q , m h , Ah
Solution: Properties: from the table of water properties at
T c  Tc ,i  Tc ,o  2  35  125 2  160 2  80o C
 c p  4197 J / kg. K
 Q  mc c p ,c Tc ,o  Tc ,i 
.
.
.
 Q  1  4197  125  35
 3.777  105 W
 Q  m h c p ,h Th,i  Th,o 
.
.
.
3.777  105
 mh 

 1.889 kg/s
c p ,h Th,i  Th,o  1000  300  100
.
Q
1)  - NTU
method:
.
C c  mc c p ,c  1  4197  4197 W/K
.
C h  mh c p ,h  1.889  1000  1889 W/K  C min
Q max  C min Th,i  Tc ,i   1889  300  35  5.006  105 W
.
.

Q
.
3.777  105

 0.75
5
5.006  10
Q max
C min 1889
Cr 

 0.45
C max 4197
It follows from  - NTU figure for the cross-flow heat exchanger
with both fluids unmixed that with   0.75 and Cr  0.45,
NTU  2.1
U h Ah
 NTU 
C min
NTU  C min 2.1  1889
 Ah 

Uh
100
 39.67 m 2
1) LMTD method:
.
 Q  U h Ah F.Tlm CF
Q
 Ah 
U h F Tlm CF
t o  t i Tc ,o  Tc ,i 125  35
P


 0.34
Ti  t i Th,i  Tc ,i 300  35
and
Ti  To Th,i  Th,o 300  100
R


 2.22
t o  t i Tc ,o  Tc ,i 125  35
It follows from the correction factor
figure for the cross-flow exchanger
with both fluids unmixed that
F  0.87

Th,i  Tc ,o   Th,o  Tc ,i 
Tlm CF 
lnTh,i  Tc ,o  Th,o  Tc ,i 
300  125  100  35
Tlm CF 
ln300  125 100  35
 111o C
in which case,
3.777  105
Ah 
 39.11 m 2
100  0.87  111
Example 4:
Consider the heat exchanger design of the previous example, that
is, a finned-tube, cross-flow heat exchanger with a gas-side
overall heat transfer coefficient and area of 100 W/m2.K and 40
m2, respectively. The water flow rate and inlet temperature
remain at 1 kg/s and 35oC. However, a change in operating
conditions for the hot gas generator causes the gases to enter the
heat exchanger with a flow rate of 1.5 kg/s and a temperature of
250oC. What are the effectiveness of the heat exchanger and the
rate of heat transfer, and what are the gas and water outlet
temperatures. Assuming the specific heat of gas and water under
constant pressure are 1000 J/kg.K and 4197 j/kg.K respectively.
Data: Cross-flow heat exchanger with both fluids unmixed.
Uh = 100 W/m2.K. , Ah = 40 m2,
.
hot fluid: gases m h = 1.5 kg/s, Th,i = 250oC, cp,h = 1000 J/kg.K.
.
cold fluid: water m c = 1 kg/s, Tc,i = 35oC, cp,c = 4197 J/kg.K.
.
Find:  , Q , Th ,o , Tc ,o
Solution:
.
C c  mc c p ,c  1  4197  4197 W/K
.
C h  m h c p ,h  1.5  1000  1500 W/K  C min
C min 1500
Cr 

 0.36
C max 4197
U h Ah 100  40
NTU 

 2.67
C min
1500
It follows from  - NTU figure for
the cross-flow heat exchanger
with both fluids unmixed that
with NTU = 2.67 and Cr = 0.36 ,
or from the equation:
 
 
 1 

0.22
0.78
  1  exp  NTU  exp  C r  NTU   1 
 C r 

   0.845
Q max  C min Th,i  Tc ,i   1500  250  35  3.225  105 W
.
.
 
.
Q
.
Q max
.
 Q   Q max  0.845  3.225  105  2.725  105 W
 Q  m h c p ,h Th,i  Th,o 
.
.
 Th,o  Th,i  Q C h  250  2.725  105 1500  68.3o C
.
 Q  mc c p ,c Tc ,o  Tc ,i 
.
.
 Tc ,o  Tc ,i  Q C c  35  2.725  105 4197  99.9o C
.
Example 5 :
The condenser of a large steam power plant is a heat exchanger in
which steam is condensed to liquid water. Assume the condenser
to be a shell-and-tube heat exchanger consisting of a single shell
and 30,000 tubes, each executing two passes. The tubes are of
thin wall construction with d = 25 mm, and steam condenses on
their outer surface with an associated convection coefficient of ho
= 11,000 W/m2.K. The heat transfer rate that must be rejected is
2109 W, and this accomplished by passing cooling water through
the tubes at a rate of 3104 kg/s (the flow rate per tube is
therefore 1 kg/s). The water enters at 20oC, while the steam
condenses at 50oC. What are the effectiveness of heat exchanger
and the temperature of the cooling water leaving the condenser?
and what is the mass flow rate of steam in the condenser?.
Calculate also the required tube length L per pass using the LMTD
method?. Take the properties of cooling water at T c  27o C .
Data: Steam condenser, shell
-and-tube heat exchanger,
single shell and tube passes,
N = 30,000 tubes
d = do ≈ di = 0.025 m (thin wall) , ho = 11,000 W/m2.K
.
.
9
Q  2 10 W , m c  3  104 kg / s , Tc,i  20o C , Th  Ts  50o C
.
Find:  , Tc,o , m s , L (using LMTU method)
Solution: from the table of water properties at T c  27o C = 300K
cp,c = 4179 J/kg.K,  = 85510-6 N.s/m2, k = 0.613 W/m.K , Pr = 5.83
.
C c  mc c p ,c  3  104  4179  1.254  108 W/K  C min
.
condensation 
 C min Th,i  Tc ,i   1.254  108  50  20  3.76  109
C h  m h c p ,h    C max
.
Q max
.
.
   Q Q max  2  109 3.76  109  0.532
W
 Q  mc c p ,c Tc ,o  Tc ,i   C c Tc ,o  Tc ,i 
.
.
 Tc ,o  Tc ,i  Q C c  20  2  109 1.254  108   36o C
The energy balance equation for the steam is
.
.
.
Q  m h h fg  m s h fg
But from the steam tables for Th =Ts = 50oC , h fg  2.3829 106 J/kg
.
.
Q
2  109
ms 

 839.3 kg/s
6
h fg 2.3829  10
Using the LMTD method , the heat rate equation is
.
.
Q  UA F Tlm CF
A   d  2L  N
,
U
1
1 1

hi ho
where, hi may be estimated from internal pipe flow correlation.
With,
um d 4 m c ,tube
4  3  104 30,000
Re 


 59567
6

 d
  0.025  855  10
.
thus, the flow is turbulent. Applying Dittus - Boelter correlation:
Nu  0.023Re 0.8 Pr 0.4  0.02359567 5.83  307.6
k
0.613
 hi  Nu  307.6 
 7542.4 W/m 2 . K
d
0.025
1
1
U 

 4474.4 W / m 2 . K
1 1  1   1 




hi ho  7542.4   11,000 
0.8
0.4
The correction factor may be obtained from the following figure:
with ,
t o  t i Tc ,o  Tc ,i 36  20
P


 0.53
Ti  t i Th,i  Tc ,i 50  20
and,
Ti  To Th,i  Th,o 50  50
R


0
t o  t i Tc ,o  Tc ,i 36  20
Accordingly, F = 1.
Tlm

T  T   T
 
lnT  T  T
h ,i
c ,o
CF
h ,i
c ,o
 Tc ,i 
h ,o  Tc ,i 
h ,o
50  36  50  20
Tlm CF 
ln50  36 50  20
 21o C
.
Q
2  109
A

 21284 .6 m 2
UF Tlm CF 4474 .4  1  21
 A   d  2L  N
A
21284.6
L 

 4.52 m
2 dN 2  0.025  30,000
Summary of LMTD
• In the LMTD method, the task is to select a heat exchanger that
will meet the prescribed heat transfer requirements. The
procedure to be followed by the selection process is:
1. Select the type of heat exchanger suitable for the application.
2. Determine any unknown inlet or outlet temperature and the
heat transfer rate using an energy balance.
3. Calculate the log mean temperature difference (ΔTlm) and the
correction factor F, if necessary.
4. Obtain (select or calculate) the value of the overall heat transfer
coefficient U.
5. Calculate the heat transfer surface area As.
• The task is completed by selecting a heat exchanger that has a
heat transfer surface area equal to or larger than As.
Summary of ε-NTU
• In the ε-NTU the determination of Q·max and the outlet temperature
of the hot and cold fluids is the task. Given the inlet temperature of
the hot and cold fluids and their mass flow rates and the surface
area, the procedure to be followed by the selection process is:
1. Find the effectiveness of the heat exchanger. The effectiveness of a
heat exchanger depends on the geometry of the heat exchanger as
well as the flow arrangement. Therefore, different types of heat
exchangers have different effectiveness relations.
2. The actual heat transfer rate Q· can be determined from
Q· = ε Q·max = ε Cmin (Th, in – Tc, in)
3. Then determine As if not explicitly given.
4. Find the NTU from the appropriate NTU relations.
5. Finally, find the outlet temperature of the hot and cold fluids.
Note
The ε-NTU method can be used to find the length or diameter of the
heat exchanger when one of the outlet temperatures is given.
Conclusion
Assumptions
1. The heat exchanger operates at steady-flow steady-state
conditions.
2. Heat transfer to the surrounding is negligible.
3. There is no heat generation in the heat exchanger.
4. The specific heat at constant pressure is constant for each fluid.
5. Longitudinal heat conduction in the fluid and the wall are
negligible.
6. The overall heat transfer coefficient between the fluids are
constant throughout the heat exchanger including the case of
phase change.
7. Phase change, if happens, occurs at constant temperature and
pressure.
8. In counter-flow and parallel-flow heat exchangers, the
temperature of each fluid is uniform over every cross-section.
9. In cross-flow heat exchangers, each fluid is considered mixed or
unmixed for every cross-section depending on the
specifications.