Lect - 19 Heat Exchanger Lecture 3 of 4.pptx

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


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3/15

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 − = −& &


4/15

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

+ − = −

&

&


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6/15

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 = &


7/15

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 − −= − + − ÷−


8/15

5/12/16 | Slide ;

ntegration o! heat !lux across H-,o


9/15

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

∆ − ∆∆ = ∆ ÷∆


10/15

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


11/15

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

∆ − ∆=

∆ ÷∆


12/15

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 −= −


13/15

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


14/15

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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Lect - 19 Heat Exchanger Lecture 3 of 4.pptx