Solutions Manual for Heat Exchangers: Selection, Rating, and Thermal Design (4th Edition) - Kakaç, Sadık

ALL 13 CHAPTERS COVERED
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SOLUTIONS MANUAL
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,TABLE OFCONTENTS
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1. Classification of Heat Exchangers j j j




2. Basic Design Methods of Heat Exchangers
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3. Forced Convection Correlations for the Single-
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Phase Side of Heat Exchangers
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4. Heat Exchanger Pressure Drop and Pumping Power
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5. Micro/Nano Heat Transfer j j




6. Fouling of Heat Exchangers
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7. Double-Pipe Heat Exchangers j j




8. Design Correlations for Condensers and Evaporators
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9. Shell-and-TubeHeatExchangers j j




10. Compact Heat Exchangers j j




11. Gasketed-Plate Heat Exchangers j j




12. Condensers and Evaporators j j




13. Polymer Heat Exchangers j j

,Problemj2.1

StartingjfromjEq.j(2.22),jshowjthatjforjajparallelflowjheatjexchanger,jEq.j(2.26a)jbecomes

T 2 jT  j 1 1j  
2
jexpj     UA 
Tj j jT  jC Cj  




SOLUTION:



ThejheatjtransferredjacrossjthejareajdAjis:
Qj=jUThjjTcjdA (1)
Thej heatj transferj ratej canj alsoj bej writtenj asj thej changej inj enthalpyj ofj eachj fluidj (withj thejcorre
ctjsign)jbetweenjthejareajAjandjA+dA:
* forjthejhotjfluidj(dTh<0)
Qj=j-mjhcp,hdTh (2)
* forjthejcoldjfluidj(dTc>0)
Qj=jmjccp,cdTc (3)
Thejnotionjofjheatjcapacityjcanjbejintroducedjas:
Cj=jmjcp
(4)
Thisjparameterjrepresentsjthejratejofjheatjtransferredjbyjajfluidjwhenjitsjtemperaturejvariesj withjo
nejdegree.
Thejequationj(2)jandj(3)jgive:
Qj=j-ChdThj jCcdTc (5)
Equationsj(1)jandj(5)jgive:
j dThj Uj
j
jj dAjT (6)
hj jTc Ch
dTcj Uj
jj dAj (7)
Thj jTc Cc
Subtractingjequationj(7)jfromj(6):
j
d(ThjjTcj)j j 1 -j 1j UdA (8)
j 
j

ThjjTc jCcj j Chj
ConsideringjthejoveralljheatjtransferjcoefficientjU=constant,jequationj(8)jcanjbejintegrated:
j 1 j 1 j j
lnTj jTj jj -j UAj+jlnB (9)
h  C C 
c
j j c hj

j 1 1j  
Thj jTcj jBexpj -j jUA
Cj j Cj 
j c hj  (10)

, Thejconstantjofjintegration,jKjisjobtainedjfromjthejboundaryjconditionjatjthejinlet:
atjA=0, Thj jTcj =jTh1j jTc2 (11)
K=jTh1j jTc2 (12)
Introducingjequationj(12)jinj(10)jwejhave:
Thj jTcj j j 1 
Tj j jT jexp j - 1j  UA 
j
(13)
C Cj 
h1 c2 j c hj 
AtjthejoutletjthejheatjtransferjareajisjAt=AjandjTh-Tc=Th2-Tc2j and:
j 1j j 1j 
T jTcj2jj (14)
 j j+ jUA
j h2j

j Chjj j jCcj 
ej
Th1j jTc1