College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Solution to Quiz Ten – Heat Exchangers
A thin-walled double-pipe parallel-flow heat exchanger is used to heat a chemical whose
specific heat is 1800 J/kgoC with hot water (cp = 4180 J/kgoC). The chemical enters at
20oC at a rate of 3 kg/s, while the water enters at 110oC at a rate of 2 kg/s. The heat
transfer surface area of the heat exchanger is 7 m2 and the overall heat transfer coefficient
is 1200 W/m2oC. Determine the outlet temperatures of the chemical and the water.
We compute the maximum heat transfer by first computing the products of mass flow rate times
heat capacity and finding which is the smaller.
Cc  m c c p ,c 
3 kg 1800 J 1 W  s 5400 W
 o
s kg o C J
C
C h  m h c p ,h 
2 kg 4180 J 1 W  s 8360 W
 o
s kg o C kJ
C
So Cmin = Cc = 5400 W/oC. We use this to compute the maximum heat transfer.


5400 W
Q max  C min Th ,in  Tc ,in   o
110 o C  20 o C  486,000 W
C
In order to find the heat transfer we have to find the heat exchanger effectiveness. We do this by
computing the NTU and using the charts that give the effectiveness as a function of NTU and the
ratio of Cmin/Cmax.
1200 W

7 m2 
2 o
UAs
NTU 
 m  C
 1.556
5400 W
C min
o
C
5400 W
o
C min
C  0.646
c

C max 8360 W
o
C
For this parallel-flow heat exchanger we can find the effectiveness from the first equation in Table
11.4 on page 636 of the text..

1  e  NTU 1c  1  e 1.55610.646

 0.56
1 c
1  0.646
The actual heat transfer is the product of the maximum heat transfer and the heat exchanger
effectiveness.


Q  Q max  0.56 4.86 x10 5 W  = 2.72x104 W
We find the outlet temperatures from the first law energy balance for each stream.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Quiz ten solutions
ME 375, L. S. Caretto, Spring 2007
Tc ,out  Tc ,in 
Th ,out  Th ,in 
Q
Cc
 20 o C 
2.72 x10 4 W
 70.4 o C
5400 kW
o
C
Q
2.72 x10 4 W
 110 o C 
 77.4 o C
8360 W
Cc
o
C
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