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8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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Heat ExchangersDr. Senthilmurugan S. Department of Chemical Engineering IIT Guwahati - CL204 - Part !
LMTD method
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide .
Temperature ro!ile and Enthalp" #alances in Heatexchangers
! onl" sensi+le heat is trans!erred and
constant speci!ic heats are assumed
&here cpc % speci!ic heat o! cold !luid
cph % speci!ic heat o! &arm !luid
arallel !lo&
+a ( ) ( )h ph ha hb c pc cb cam c T T m c T T − = −& &
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide 0
Temperature ro!ile and Enthalp" #alances in Heatexchangers
,ondenser The (apor enters the condenser as
saturated (apor no superheat and
lea(es the condensate lea(es at
condensing temperature &ithout +eing
!urther cooled3
$here % 4ate o! condensation o!
(apour' hlh % latent heat o! (aporiation
o! (apour
! the condensate lea(es at a
temperature that is less than the
condensing temperature o! the (apor
( )h lh c pc cb cam h m c T T q= − =& &
( )
( )
h lh ph hb ha
c pc cb ca
m h c T T
m c T T
+ − = −
&
&
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide 6
ntegration o! heat !lux across H-
To calculate heat !lux to the entire area o! a
heat exchanger 7o' the euation must +e
integrated3 This can +e done !ormall" &here
certain simpli!"ing assumptions are accepted3
The assumptions are
13 The o(erall coe!!icient U is constant'
23 The speci!ic heats o! the hot and cold!luids are constant'
.3 Heat exchange &ith the am+ient is
negligi+le'
03 The !lo& is stead" and either parallel or
counter current'
.
Moderate temperature ' the assumption o!constant U and Cp is not seriousl" in error3
. 7ssumptions 2 and 0 impl" that i! Tc and Th'
are plotted against gi(es straight line linear
relationship3 Since Th and Tc ' (ar" linearl"
&ith ' 8T
The heat trans!erred through an element
o! area d7 ma" +e &ritten
Th% a(g3 temperature o! hot stream at
area d7o T,% a(g3 temperature o! cold stream at
area d7i corresponding to d7o
9o% *(er all heat trans!er coe!!icient+ased on d7o
,on(ection
Energ" #alance
( )o o h cdq U dA T T = −
h ph hdq m c dT = − &
c pc cdq m c dT = &
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide :
ntegration o! heat !lux across H-
hh ph
dqdT
m c=
− &
c
c pc
dqdT
m c=&
1 1h c
c pc h ph
dT dT dqm c m c
− = − + ÷ ÷ & &
( )o o h cdq U dA T T = −
( ) 1 1h co o
h c c pc h ph
d T T U dA
T T m c m c
−= − + ÷ ÷− & &
( )
0
1 1hb cb o
ha ca
T T A
h c
o o
h c c pc h phT T
d T T
U dAT T m c m c
−
−
−= − + ÷ ÷− ∫ ∫ & &
1 1ln
hb cbo o
ha ca c pc h ph
T T U A
T T m c m c
−= − + ÷ ÷ ÷− & &
( ) ( )) )h ph ha hb c pc cb caq m c T T m c T T = − = −& &
( )ln hb cb o o ha hb cb caha ca
T T U AT T T T
T T q − −= − + − ÷−
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide ;
ntegration o! heat !lux across H-,o
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide =
Logarithmic Mean Temperature Di!!erence LMTD
Distance
Tha
Tc+
Th+
Tca
d
d7
,old !luid Tc
Hot !luid Th
8Ta8T+
Distance
Tha
Tc+
Th+
Tca
d
d7,old !luid Tc
Hot !luid Th
8T+
n
b a M
b
a
T T T T
l T
∆ − ∆∆ = ∆ ÷∆
n
b a M
b
a
T T T T
l T
∆ − ∆∆ = ∆ ÷∆
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide 1>
,orrection !actor !or LMTD
LMTD correction is reuired &hen
!ollo&ing assumptions are not hold?
13 The o(erall coe!!icient U is
constant'
23 The speci!ic heats o! the hot and
cold !luids are constant'.3 Heat exchange &ith the am+ient is
negligi+le' and
03 The !lo& is stead" and either
parallel or counter current'
53
@ (s T linear +eha(iours13 Hot stream is saturated
(apor'
23 Ao reaction occur during heat
exchange
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide 11
Tca
Tc+
Baria+le *(erall Heat Trans!er ,oe!!icient
$hen 9 (aries linearl" &ith the
temperature drop o(er the entire heating
sur!ace then total heat !lux is de!ined as
Distance
Tha
Tc+
Th+
Tca
d
d7
,old !luid Tc
Hot !luid Th
8Ta8T+
n
ob b oa ao
ob b
oa a
U T U T q A
U T l U T
∆ − ∆=
∆ ÷∆
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide 12
,orrection o! LMTD !or ,ross!lo& H-
$hen !lo& t"pes other than countercurrent
or parallel appear' it is customar" to de!ine
a correction !actor FG, &hich is so
determined that &hen it is multiplied +" the
LMTD !or countercurrent !lo&' the product
is the true a(erage temperature drop3
The !actor C is the ratio o! the !all in
temperature o! the hot !luid to the rise in
temperature o! the cold !luid3
The !actor ηH is the heating e!!ecti(eness'or the ratio o! the actual temperature rise o!the cold !luid to the maximum possi+le
temperature rise o+taina+le i! the &arm<
end approach &ere ero +ased on
countercurrent !lo&3
,orrection o! LMTD !or cross!lo&3
G o o M q F U A T = ∆
in out
out in
h h
c c
T T Z
T T −= −out in
in in
c h H
h c
T T
T T η −= −
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide 1.
,orrection o! LMTD
F is al&a"s less than unit"3 The mean
temperature drop' and there!ore the capacit" o! the
exchanger' is less than that o! a countercurrent
exchanger ha(ing the same LMTD3
$hen F G is less than a+out >3;' the exchanger
should +e redesigned &ith more passes or larger
temperature di!!erences? other&ise the heat<
trans!er sur!ace is ine!!icientl" used and there isdanger that small changes in operating conditions
ma" cause the exchanger to +ecome unsta+le
operation3
$hen F G is less than >3:5' it !alls rapidl" as ηH increases' so that operation is sensiti(e to small
changes3 n this region' also'
7n" de(iations !rom the +asic assumptions on&hich the charts are +ased +ecome important'
especiall" that o! a uni!orm thermal histor" !or all
elements o! !luid3 Leaage through and around the
+a!!les ma" partiall" in(alidate this assumption3
1
8/17/2019 Lect - 19 Heat Exchanger Lecture 3 of 4.pptx
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5/12/16 | Slide 10
Siing a Heat ExchangerF
,ase 1F *utlet temperature H- is no&n
*ne euation and one unno&n
Eas" to sol(e
,ase 2F *utlet temperature hot stream H- is
unno&n
*ne euation and three unno&ns
Hard to sol(e and use o! iterati(e techniues Example
,alculate @ and the unno&n outlet hot
stream temperature3
,alculate 8TM and o+tain the correction
!actor ) i! necessar"
,alculate the o(erall heat trans!ercoe!!icient3
Determine 73
The LMTD method is not as eas" to use !or
per!ormance anal"sis &hen +oth steam out let
temperature not no&n
LMTD method
,ase 2
,ase 1
( ) ( )
n
hb cb ha ca
o o
hb cb
ha ca
T T T T q U A
T T l
T T
− − − = − ÷−
( ) ( )
n
hb cb ha ca
o o
hb cb
ha ca
T T T T q U A
T T l T T
− − − =
− ÷−
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