chapter 21 performance of heat exchangers

Ch t 21Chapter 21 Performance Curves for

Individual Unit Operations(H E h E i )(Heat Exchange Equipment)

Department of Chemical Engineering

West Virginia University

Copyright J.A. Shaeiwitz and R. Turton - 2012 1

OutlineOutline

• Heat Exchanger Design ReviewHeat Exchanger Design Review

• T‐Q Diagrams

f C ffi i• Heat Transfer Coefficients

• Heat Exchanger Regulation

• Heat Exchanger Performance


OutlineOutline


• T‐Q Diagrams





Heat Exchanger DesignHeat Exchanger Design

TT

( ) QTTCQ && λ

Tc,out

TT

Tc,in

( )( )or

or

,,,

,,,

hhouthinhhph

ccincoutccpc

FTUAQ

mQTTCmQ

mQTTCmQ&&

&&

Δ

=−=

=−=

λ

λ Th,inTh,out

( ) ( )( )

,,,,

ith

outcinhincouthlm

lm

TTTTTT

T

FTUAQ

−−−−

=Δ

Δ=

( )( )

),,,,,(

ln,,

,,

hphcpcincoutcinhouth

outcinh

incouth

CmCmTTTTfF

TTTT

&&=

−

),,,,,( ,,,,,, hphcpcincoutcinhouthf


Log‐mean Correction FactorLog mean Correction Factor

• F is log‐mean correction factorF is log mean correction factor

• Correction for not pure countercurrent flow in multiple pass exchangersmultiple‐pass exchangers

• “Default” exchanger is 1 shell, 2 tube pass


Log‐mean Correction FactorLog mean Correction Factor

T T T

Q Q Q

F > 0.9 F < 0.9 F < 0.7

As temperatures approach, F decreases. When F gets too small, must addadditional shell passes to increase F.


Additional Shell PassesAdditional Shell Passes

one shell passone shell passtwo tube passes two shell passes

four tube passes


OutlineOutline


• T‐Q Diagrams





T‐Q DiagramsT Q Diagrams

T T T

Q Q Q

no phase change hot fluid condensing hot fluid desuperheating,phase change condensing, subcooling

Always sketch T-Q diagram to make sure there is no temperature cross.


T‐Q DiagramsT Q Diagrams

T T

Q Q

incorrect – caused by just correct

Always sketch T Q diagram to make sure there is no temperature cross

y jlooking at end points (red line)and not sketching T-Q diagram.

Copyright J.A. Shaeiwitz and R. Turton - 2012

Always sketch T-Q diagram to make sure there is no temperature cross.

10

OutlineOutline


• T‐Q Diagrams





Heat Transfer CoefficientsHeat Transfer Coefficients

ooi

oRRR

R11

ln111

++⎟⎟⎠

⎞⎜⎜⎝

⎛

++=iiifiofoo hRhRkhhU ,,

++++=

For class problems assume no fouling thin walls and negligible wall resistance

111

For class problems, assume no fouling, thin walls, and negligible wall resistance.If given detailed data (like in major), then use all terms.

ioo hhU+≈


Heat Transfer CoefficientsHeat Transfer Coefficients

change phase no annulus,in or in tubes fluid - )or toequivalent(

606060

8.08.08.0 mvvhi ∝ &&

function)very weak (actually changes phasefor )(

changephasenoflow,crossin shellin fluid-)or toequivalent( 6.06.06.0

vfh

mvvho

≠

∝ &&

F50F10if4to3 F10 if 25.0 where)( boiling,for Tn

TnTfh n

°≤Δ≤°≈°≤Δ≈Δ=

surfacecoldandfluidcondensingbetweendifferenceis 25.0 where)( ,condensingfor

fluid boiling and surfacehot between difference is

TnTfh

Tn

Δ−≈Δ=

Δ

surface coldandfluidcondensingbetween difference is TΔ


OutlineOutline


• T‐Q Diagrams





Heat Exchanger RegulationNo phase change

UNREGULATED REGULATEDUNREGULATEDif process fluid flowrate increases

h increases, get increased heat transfersomewhat self-regulating

REGULATEDcontrol utility rate in response to

measured outlet temperature


however, h does not increase linearly with flowrate

15

Heat Exchanger RegulationPhase change

Stream b is two phase mixtureCV-1 lowers steam pressurechanges driving forceflowrate not regulatedsteam trap removes condensate as it accumulates

Stream b is two-phase mixturebfw rate adjusted by measuring temperatureof outlet and changing rate of steam removalif level of liquid in accumulator drum changes,condensate recycle rate is adjusted


steam trap removes condensate as it accumulates condensate recycle rate is adjusted

16

OutlineOutline


• T‐Q Diagrams





Heat Exchanger PerformanceHeat Exchanger Performance

150°C – shell – condensation150°C

C93.114 °=Δ lmT40°C30°C – tubes – cooling water

problem must scale down by 50%problem – must scale down by 50%

case 1: all resistance on tube side

MUST USE ENERGY BALANCES AND DESIGN EQUATION



150°C

C93.114 °=Δ lmT150°C – shell – condensation

150°C

40°C30°C – tubes – cooling water

problem – must scale down by 50%


11

22

1

2mm

QQ

=&&λλ

40 C

2 = new1 = old

( )( )1,,1,,1,1,

2,,2,,2,2,

1

2

111

TTCmTTCm

QQ

mQ

incwoutcwpcw

incwoutcwpcw

−

−=

&

&

&λ

( )

11,11

22,22

1

2

1,,1,,1,1,1

FTAUFTAU

QQ

Q

lm

lm

incwoutcwpcw

Δ

Δ=



150°C


150°C




( )305.0

TMQ ==

define

40 C2 = new1 = old

( )( )3040

30

22

2,,

TU

TMQ

lm

outcwcw

Δ−

−=

2,

1

2

cwmM

QQQ

&=

=

( )93.1141

2,2

UTU

Q lmΔ=

1,

2,

1,

stm

stm

cwcw

mm

M

mM

&

&

&

=

=

,



150°C

C93.114 °=Δ lmT

150°C – shell – condensation150°C




111211

11

1

1oi hhU

UU

+≈=define

40 C2 = new1 = old

all resistance on tube side

222

1 111oi hhU

U +

2

1

2

define

mQQQ

&

=

11

2,

1,

2,

stm

cw

cwcw

mM

mm

M

&

&

&

&

=

= 8.0

1

2

2

1

2

1

1

211 cw

i

i

i

i Mhh

h

h

U

UUU

==≈=

1,stmm&



150°C


150°C




( )−=

outcwTQ

2 305.0

define

40 C2 = new1 = old

( )( )

( )( ) ( )

( ) ( )( )

( ) ⎤⎡

−=

⎤⎡

−−−=

−=

outcwcwoutcwcw

outcwcw

TMTM

TM

,2,8.0

,2,8.0

2,,

30150

309311430150

1503015093114

5.0

304030

5.0

2

1

2

define

mQQQ

&

=

( ) ( )( )

( ) ( )( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

outcwoutcw TT ,2,,2, 15030150ln

93.114150

30150ln93.114

2,

1,

2,

stm

cw

cwcw

mM

mm

M

&

&

&

&

=

=

1,stmm&



150°C


150°C




2 new

( )( )

( ) ( )−

−=

=

outcwcw

TM

Q

2,,

304030

5.0

5.0

1

2

define

QQQ =

40 C2 = new1 = old

( )( ) ( )

( )( )

( )( )

( )⎥⎥⎦⎤

⎢⎢⎣

⎡

−

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

−−−=

outcw

outcwcw

outcw

outcwcw

T

TM

T

TM

,2,

,2,8.0

,2,

,2,8.0

150120ln

3093.114

15030150ln

1503015093.114

5.0

2,

1,

2,

1

stm

cw

cwcw

mM

mm

M

Q

&

&

&

=

=

1,stmm&

CTM

outcw

cw

°==

77.41425.0

,2,solve second and third equations:



150°C


150°C




CTM

outcw

cw

°==

77.41425.0

,2,

40 C

solve second and third equations:

T 150°CT 150°C

41.77°C40°C

30°C

40 C


Q

24

Class ExamplesClass Examples

1. Re‐do previous problem if all resistance on shell1. Re do previous problem if all resistance on shell side.

2. Re‐do previous problem if equal resistances in base p p qcase on shell and tube sides.

3. Re‐do previous problem if 150°C stream is utility p p yand 30°C stream is process stream that must be scaled down by 50%. All resistance is on the tube dside


Reboiler PerformanceReboiler Performance

222 mQ stmstm λ&254°C

condensing steam

want to scale up 20%

11

22

1

2

11

22

1

2

mm

QQ

mQQ

orgorg

stmstm

stmstm

=

=

λλλ

&

&&T

234°Cboiling organic

11

22

111

222

1

2

111

TUTU

TAUTAU

QQ

mQ

lm

lm

orgorg

ΔΔ

=ΔΔ

=

λ&Q



22 mQ stm=& 22.1 mstm=

&&254°C

condensing steam

want to scale up reboiling organic by 20%

1

2

1

2

11

2.1mm

QQ

mQ

org

org

stm

==&

&&

1

2

1

2.1mmm

org

stm

=&

&&T 234°C

boiling organic

11

22

1

2

11

TUTU

QQ

Q org

ΔΔ

=

11

22

1

2.1TUTU

morg

ΔΔ

=

&Q

h not strong function of flow for phase change



TΔ22.1 mstm=

&254°Ccondensing steam

want to scale up reboiling organic by 20%

T

T

=Δ

Δ=

2420

2.1

2

2

2

1

2.1

2.1

mmm

org

stm

=&

&&T 234°C

boiling organic

CTCT

org

stm

°=°=⇒

230or258

2

1

2

1

2.1TT

morg

ΔΔ

=

&Q

1TΔ


DesuperheaterDesuperheater

reduces steam pressure, which lowers temperaturespray in bfw to resaturate at lower pressurep y phence, get saturated steam at lower pressure

can get any saturated steam between available steam levels


More Complicated ExampleMore Complicated Example

• Must scale up or downMust scale up or down



222

reactor

rxnHQ Δξ( )( )

2,,2,,2,,2,2

exchangerheat

incwoutcwcwpcw TTCmQ −=

&

( )( )

2,42,32,22

1,1

2,2

1

2

DpD

rxn

rxn

TTCTTCm

QQ

HQQ

−=

Δ=

&

&

ξξ

sameequation

( )( )( )141311

2,42,32,2

1

2

1,,1,,1,11

DD

DpD

incwoutcwcwpcw

TTCmTTCm

QQ

TTCmQ

−

−=

−

&

&

&

( )

1,,1,1,

2,,2,2,

1

2

1,41,31,11

rxrlmrxrrxr

rxrlmrxrrxr

DpD

TAUTAU

QQ

TTCmQ

Δ

Δ=

−& ( )

1,,1,1,

2,,2,2,

1

2

1,41,31,11

hxlmhxhx

hxlmhxhx

DpD

TAUTAU

QQ

TTCmQ

Δ

Δ=

&



• Result is 5 equationsResult is 5 equations

• Must be 5 unknowns



• 6 Unknowns: T3 T4 T Q M MD6 Unknowns: T3 T4 Tcw,out Q Mcw MD

ratios



• Have control of one variable– required to maintain

control– otherwise, only one

ibl f ipossible set of operating conditions

• Usually coolant flowrate• Determined by pump

curves


OutlineOutline


• T‐Q Diagrams



