Problems & Solutions to Heat Exchangers: Selection, Rating, and Thermal Design, 4th Edition by (Sadık Kakaç, 2020) | Chapters 2 - 13 Covered

ALL 13 CHAPTERS COVERED
m0 m0 m0




SOLUTIONS MANUAL

,TABLE OF CONTENTS
m0 m0




1. Classification of Heat Exchangers m0 m0 m0




2. Basic Design Methods of Heat Exchangers
m0 m0 m0 m0 m0




3. Forced Convection Correlations for the Single-Phase Side of
m0 m0 m0 m0 m0 m0 m0




m0 Heat Exchangers
m0




4. Heat Exchanger Pressure Drop and Pumping Power
m0 m0 m0 m0 m0 m0




5. Micro/Nano Heat Transfer m0 m0




6. Fouling of Heat Exchangers m0 m0 m0




7. Double-Pipe Heat Exchangers m0 m0




8. Design Correlations for Condensers and Evaporators
m0 m0 m0 m0 m0




9. Shell-and-Tube Heat Exchangers m0 m0




10. Compact Heat Exchangers m0 m0




11. Gasketed-Plate Heat Exchangers m0 m0




12. Condensers and Evaporators m0 m0




13. Polymer Heat Exchangers m0 m0

,Problem 2.1 m0




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
m0 m0 m0 m0 m0 m0



is:
m0
(1)
Q = U(Th − Tc )dA
m0 m0 m0 m0 m0




The heat transfer rate can also be written as the change in enthalpy of
m0 m0 m0 m0 m0 m0 m 0 m0 m0 m0 m 0 m 0 m0



m 0 each fluid (with the correct sign) between the area A and A+dA:
m 0 m 0 m0 m0 m0 m0 m0 m0 m0 m0 m0




* for the hot fluid (dTh<0)
m0 m0 m0 m0



Q = -m m0 m0 m0
hcp,hdTh (2)
* for the cold fluid (dTc>0)
m0 m0 m0 m0



Q = m m0 m0 m0
ccp,cdTc (3)
The notion of heat capacity can be introduced
m0 m0 m0 m0 m0 m0 m0



as: m0

(4)
C = m cp
m0 m0 m0




This parameter represents the rate of heat transferred by a fluid when its temperature
m0 m0 m0 m0 m0 m0 m0 m0 m0 m0 m0 m0 m0



m0 varies with one degree.
m 0 m0 m0




The equation (2) and (3) give:
m0 m0 m0 m0 m0




Q = -ChdTh = m0 (5)
m0
m 0


m0 CcdTc
Equations (1) and (5) give: m0 m0 m0 m0




m 0 dTh m 0


U
(6)
= −
m 0
m0
dA
m0 m 0

m0 Th − Tc Ch m 0
m0




dTc U
= − m0 m 0m 0 (7)
m 0
Th dA
Cc
− Tc
m0
m0
m0




Subtracting equation (7) from (6): m0 m0 m0 m0




d(Th − Tc ) 1  m 0


= - (8)
m
0 m0 m0


m0 m0 m 0

 1 m 0 m0

UdA

m0



Th − Tc  Cc m0 Ch  m0 m0 m 0 m 0 m0




Considering the overall heat transfer coefficient U=constant, equation (8) can be integrated:

m0 m0 m0 m0 m0 m0 m0 m0 m0 m0 m0

1
ln(1T  − T ) =
m 0 m 0

- m 0m (9) 0m0 m0 m 0 m0 m0

UA + lnB m 0 m0 m0


h c C C 
 ch  m 0 m 0 m0



 11 m 0 m 0  
Th m 0
− Tc m0
m 0
= Bexp m0 m0 - m0  UA
m0

C m 0 m 0 C m 0

,  m 0
c h m0
  (10)