Solutions for Heat Exchangers Selection, Rating, and Thermal Design, Fourth Edition By Kakaç, Sadık

ALL 13 CHAPTERS COVERED
d d d




SOLUTIONS MANUAL
d

,TABLEOFCONTENTS
d d




1. Classificationof Heat Exchangers d d d




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




3. Forced Convection Correlations for theSingle-PhaseSideof
d d d d d d d




d Heat Exchangers
d




4. HeatExchanger PressureDrop and Pumping Power
d d d d d d




5. Micro/Nano Heat Transfer d d




6. Fouling of Heat Exchangers
d d d




7. Double-Pipe Heat Exchangers d d




8. DesignCorrelations for Condensers andEvaporators
d d d d d




9. Shell-and-TubeHeatExchangers d d




10. Compact Heat Exchangers d d




11. Gasketed-Plate Heat Exchangers d d




12. Condensers and Evaporators d d




13. Polymer Heat Exchangers d d

,Problem 2.1 d




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




T 2 −T   1 1 
= exp  −  + UA 
d d d
2

T −T  C C
d d



d d d

 d







SOLUTION:



The heat transferred across the area dA is:
d d d d d d d




Q = U(Th −Tc)dA d d d d d (1)
The heat transfer rate can also be written as the change in enthalpy of each fluid (with the
d d d d d d d d d d d d d d d d d




d correct sign) between the area A and A+dA:
d d d d d d d




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




Q = -ṁhcp,hdTh d d d
(2)
* for the cold fluid (dTc>0)
d d d d




Q = ṁccp,cdTc d d d
(3)
The notion of heat capacity can be introduced as:
d d d d d d d d




C = ṁc p d d d
(4)
This parameter represents the rate of heat transferred by a fluid when its temperature varies with
d d d d d d d d d d d d d d d




d one degree.
d




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




Q = -ChdTh = CcdTc (5) d d
d
d




Equations (1) and (5) give: d d d d




dTh
U
=− (6)
d
d d

d d



dA Th − Tc Ch
d


d d
d




dTc U
=− dA (7)
d




Th −
d d



d d
d Cc
d Tc
Subtracting equation (7) from (6): d d d d




1 
d(Th − Tc)  1 - UdA
d



(8)
d




= 
d




d d d
d d d

d



Th − Tc  Cc Ch  d d d d d d




Considering the overall heat transfer coefficient U=constant, equation (8) can be integrated:
 1 1 
d d d d d d d d d d d




ln(T − T )=
d d


- UA +lnB
d d



(9) d d d d d d d d




C C  h c
 c h d d d




 1 1   d d



Th −Tc = Bexp - UA d d d d d


C C
d d




 c h   (10)
d d d

d
d

, The constant of integration, K is obtained from the boundary condition at the inlet:
d d d d d d d d d d d d d




at A=0, Th − Tc =Th1 −Tc2
d
d
d
d
d
d
(11)
d




K=Th1 −Tc2 d
d
d (12)
Introducing equation (12) in (10) we have:
d d d d d d




Th − Tc  1 
T − T = exp  - 1 UA 
d d d d d d d

d
(13)
C C 
d
d d d




 c 
d


h1 c2 h d
d




At the outlet the heat transfer area is At=A and Th-Tc=Th2-Tc2 and:
d d d d d d d d d d d



 1 1 
Th2 −Tc2 d d

UA
d d


(14)
= − +
d d
d d d d d d d



C h C c 
e
d d
d d d d



Th1 − Tc1
d


d
d