Solutions Manual for Heat Exchangers: Selection, Rating, and Thermal Design (4th Edition) by Sadık Kakaç – Step-by-Step Solutions, Thermal Design Calculations & Exam Prep Guide

ALL 13 CHAPTERS COVERED




SOLUTIONS MANUAL

,TABLE OF CONTENTS

1. Classification of Heat Exchangers

2. Basic Design Methods of Heat Exchangers

3. Forced Convection Correlations for the Single-Phase Side of
Heat Exchangers

4. Heat Exchanger Pressure Drop and Pumping Power

5. Micro/Nano Heat Transfer

6. Fouling of Heat Exchangers

7. Double-Pipe Heat Exchangers

8. Design Correlations for Condensers and Evaporators

9. Shell-and-Tube Heat Exchangers

10. Compact Heat Exchangers

11. Gasketed-Plate Heat Exchangers

12. Condensers and Evaporators

13. Polymer Heat Exchangers

,Problem 2.1

Starting from Eq. (2.22), show that for a parallelflow heat exchanger, Eq. (2.26a) becomes

T 2 −T 2   1 1  
= exp −  +  
UA
T −T  C C 



SOLUTION:



The heat transferred across the area dA is:
Q = U(Th − Tc )dA (1)
The heat transfer rate can also be written as the change in enthalpy of each fluid (with the
correct sign) between the area A and A+dA:
* for the hot fluid (dTh<0)
Q = -mhcp,hdTh (2)
* for the cold fluid (dTc>0)
Q = mccp,cdTc (3)
The notion of heat capacity can be introduced as:
C = mcp (4)
This parameter represents the rate of heat transferred by a fluid when its temperature varies
with one degree.
The equation (2) and (3) give:
Q = -ChdTh = CcdTc (5)
Equations (1) and (5) give:
dTh U
=− dA (6)
Th − Tc Ch
dTc U
=− dA (7)
Th − Tc Cc
Subtracting equation (7) from (6):
1 
d(Th − Tc )  1 - UdA (8)
= 
Th − Tc  c
C Ch

Considering the overall heat transfer coefficient U=constant, equation (8) can be integrated:
 1 1 
ln(T − T ) = - UA + lnB (9)
h c C C 
 c h

 1 1  
Th − Tc = Bexp -  UA
C C 
 c h  (10)

, The constant of integration, K is obtained from the boundary condition at the inlet:
at A=0, Th − Tc = Th1 − Tc2 (11)
K= Th1 − Tc2 (12)
Introducing equation (12) in (10) we have:
Th − Tc  1 
T − T = exp  - 1  UA  (13)
C C 
h1 c2  c h 
At the outlet the heat transfer area is At=A and Th-Tc=Th2-Tc2 and:
 1 1 
Th2 − Tc 2 + UA (14)
= −
 C h Cc 
e
Th1 − Tc1