thermal design of heat exchangers

1

Condensed version

Condensed version

, PhD,

Fellow ASME, Fellow ASHRAE

e-Mail [email protected]

Tel: 03-571212 ext. 55105

2

UA-LMTD-F/-NTU

3

2007

There are no foolish questions and no man becomes

a fool until he has stopped asking questions

Charles P. Steinmetz quotes (Prussian Engineer and Inventor,

1865-1923)

4

() - ()

- ()

-

()

1p

p

mW P

5

1 1 1 1

22 2 2

(A)(C)

(B)

(D)

(G)

(F)

(E)

P = P = PA + PB + PC + PD + PE + PF + PG

Copyright ITRI 2009

6

Pa Pf Pg

L

Q

P = Pa + Pf + Pg

2 21 1

2 2a o o i iP u u

singP g h gL

7

PfPf 85%Pf

Pf udie

Pf

ff Re

( , , , , )f

cn i

PF u d e

L

2, Re,

42

icn cn

i i

i

udP e ef F F

d dL u

d

Copyright ITRI 2009

8

-Moody Diagram

9

30 L/min10T = 20C

SI

( )

di = 2 cm = 0.02 m

L = 10 m

s/m 0005.0L/s 5.0s60

L 30min/L 30 3Q

222 m 000314.0)02.0(44

ic dA

uAQ c

0.0005 m3/s = (0.000314 m

2)u

u = 1.59 m/s

20C ( )

= 1002106 Ns/m2

3kg/m 2.998

31680101002

02.059.12.998Re

6

1-3 f Re = 31680 f

0.0058

2

2

14uf

d

LP

i

2)59.1(2.9982

10058.0

02.0

104

= 14636.6 Pa

L = 10 m

d = 2 cmiQ = 30 L/min

10

()(hydraulic diameter)

4 4( )ch

AD

P

do

b

a

Di

oi

oi

oi

h dDdD

dDD

)

44(4 22

,

ba

ab

ba

baDh

2

22

)(4,

11

b

A

A

Section AAHH-Horizontal planeVV-Vertical plane

R

V

H

V

xH

r

2r

(b)(a)

R2r

b

2r

Rmin

maxR

u

D=2r

u

R

R

D=2r

R

u

D=2r

12

1-1-1A1/A0= 41 m/s

1.2 kg/m3

1-1-1

i

C

u

u

A

A

0

1

A1/A0 = 4

uC = 4 m/s

Co1-10.41

Pa 9.32

42.141.0

2

iP

ui = 1 m/s

i = 1.2 kg/m3

uc Ao A1

13

1-2

( Kaka and Liu 1998

)

a

Blasius 25.0Re0791.0 f 3 103 Re 105

Drew, Koo, and McAdams f = 0.00140 + 0.125Re0.32 3 103 Re 5 106 Karman-Nikuradse

4.0)ln(Re737.11

ff

4.0)(Relog41

10 ff

f = 0.46 Re0.2

3 103 Re 3 106

Filonenko f = (3.64 log10Re 3.28)2 3 104 Re 106

a (bulk temperature)

14

)TT(L

kAQ 21wall,cond

k = thermal conductivity (W/m.C)

(heat conduction)

15 (Thermal Resistance)

kL

r

rln

TTQ

TTkLr

rlnQ

dTkLr

drQ

i

o

oicond

oi

i

ocond

oT

iT

or

ir

cond

2

2

2

.dr

dTrLkQcond 2

.

dx

dTkAQcond

kL

r

rln

Ri

o

2

Copyright ITRI 2009

16

.L

TTkA

dx

dTkAQ

L,T

,T

cond21

2

01

(Thermal Resistance)

R

VI

kA

L

TQ

kA

LRthermal

Fouriers Law

17

(Cont.)

(at 0)

k,(W/m.K)

Silver (pure) 410

Copper (pure) 385

Aluminum (pure) 202

Nickel (pure) 93

Iron (pure) 73

Carbon steel, 1%C 43

Lead (pure) 35

Chrome-nickel steel

(18%Cr, 8% Ni)

16.3

Copyright ITRI 2009

18

(heat convection)

)TT(hAQ fsconv

h = heat transfer coefficient (W/m2.C)

19

).( bwconvection TThAQ

(Thermal Resistance)

hA

TQ

1

hARthermal

1

q

T

T

w

b

Newtons Law of Cooling

20

(Cont.)

(heat transfer coefficient)

fluid state hW/m2.K

Gas 1 bar 80 - 125

Gas 10 bar 250 - 400

Water Single phase 5000 - 7500

Water Boiling < 5 bar 3000 - 10000

Steam Condensation

1 Bar 10000 - 15000

21

(Cont.)

(heat transfer coefficient)

fluid hW/m2.K

Air (Natural Convection) 5-25

Air/ superheated steam

(Forced Convection)

20-300

Oil (Forced Convection) 60-1800

Water (Forced Convection) 300-6000

Water (Boiling) 3000-60000

Steam (Condensing) 6000-120,000

22

() 1-7

1. Nu = 0.023 Re0.8Pr0.4 ()

Nu = 0.023 Re0.8Pr0.3 ()

Re > 104Dittus-Boelter

2.

2

32

28.3Reln58.1

1Pr2

7.1207.1

Pr1000Re2

f

f

f

Nu

2300 < Re < 105

Gnielinski

3. Nu = 0.022 Re0.8Pr0.5 Re > 5000

23

()

()

()

24

T inwT T >T w in

U in

( )

T T >T ( )ww in

(a)

(b)

25

a

c

b

a(b)(c)

14.0)(w

b

b(viscosity)w

Copyright ITRI 2009

26 Geometry (L/Dh > 100 ) NuT NuH1 NuH2 fRe

3.657 4.364 4.364 16.00

2b/2a = 0.5 3.742 4.558 3.802 16.83

2b/2a = 0.25 3.792 4.88 2.333 18.24 2b

2a 2b/2a = 0.125 3.725 5.085 0.9433 19.146

3.34 4.002 3.682 16.06

2120 , 0.289

2

b

a 2.0 2.68 0.62 12.744

2 190 ,

2 2

b

a 2.34 2.982 1.34 13.153

2 360 ,

2 2

b

a 2.47 3.111 1.892 13.333 2a

2b

2

30 , 1.8662

b

a 2.26 2.91 0.851 13.065

21

2

b

a 2.976 3.608 3.091 14.23

2 1

2 2

b

a 3.391 4.123 3.017 15.55

2 1

2 4

b

a 3.66 5.099 4.35 18.7 2a

2b

2 1

2 8

b

a 5.597 6.490 2.904 20.59

2b

20

2

b

a 7.541 8.235 8.235 24.0

0b

a 4.861 5.385 -- 24.0

27

Question 10.5 L/min10T = 20C

Question 230 L/min

L = 10 m

d = 2 cmi

Q = 0.5 & 30 L/min

28

-

R=RA+RW+RB

q q=(TA-TB)/R

Q=qA

RW

q

B A TA

RA RB TB

29

(Cont.)

Fig.

hT

T w1

h1

X

w2T

Tc

Q

h2

1

1Q

2

wQ

Q

Q2

Q=qA

R=RA+RW+RB

q=(Th-Tc)/R

q

RW

2 1 Th

RA

Tc RB

30

(Cont.)

1,11

1,11 whw

whw TTAh

QTTAhQ

2,1,

2,1,

ww

wwTT

X

kA

Q

X

TTkAQ

cww

cww TTAh

QTTAhQ 2,

22

2,22

ch

ww

TTQAh

X

kAAh

2211

111

(steady state)Q1 = Qw = Q2 = Q

)(

111

1

2211

ch

UA

ww

TT

Ah

X

kAAh

Q

hT

T w1

h1

X

w2T

Tc

Q

h2

1

1Q

2

wQ

Q

Q2

31

(Cont.)

Q

U( W/m2.K)A

Rt = R1+Rw+R2

2211

111

ww AhkA

X

AhUA

mTUAQ

oopp

p

ii

t

iioo

AhAk

X

Ah

RAUAU

11

11

)(

111

1

2211

ch

UA

ww

TT

Ah

X

kAAh

Q

)1

(t

R

mT

UA

mT

Q

Lk

rrR

Ak

XR

p

iow

pp

p

w

2

)/ln(

For plane wall

For tube wall

32

(Cont.)

soopp

p

ii AhAk

X

Ah

iioo AUAU

11

1

UoAoAi Ap

kps(fin

surface effectiveness, ) hohi

For fin-and-tube HX

Uoho

hi sAoAi

33

(Cont.)

Water to water 850-1700

Water to oil 110-350

Steam condenser (water in tubes) 1000-6000

Ammonia condenser (water in tubes) 800-1400

Alcohol condenser (water in tubes) 250-700

Finned tube heat exchanger (water 25-50

in tubes, air in cross flow)

FLUID COMBINATION U(W/m2.K)

(overall heat transfer coefficient)

(Cont.)

1,11

1,11 whw

whw TTAh

QTTAhQ

2,1,

2,1,

ww

wwTT

X

kA

Q

X

TTkAQ

cww

cww TTAh

QTTAhQ 2,

22

2,22

ch

ww

TTQAh

X

kAAh

2211

111


)(

111

1

2211

ch

UA

ww

TT

Ah

X

kAAh

Q

hT

T w1

h1

X

w2T

Tc

Q

h2

1

1Q

2

wQ

Q

Q2

36

Recuperator

Direct Contact Heat Exchanger

Regenerator

37

Wall/Interface

Heat flow

Stream A Stream B

38

Recuperator

(Fin-and-tube Heat Exchanger)

(Shell and Tube Heat Exchanger)

(Plate Heat Exchanger)

(Spiral Heat Exchanger)

(Tube in Tube Heat Exchanger)

Cross flow Heat Exchanger

39

Typical fin-and-tube heat exchanger (a) individual circular fin; (b) continuous wavy fin; (c) conventional air-cooled heat exchanger

(a) (b)

(c)

Slit fin, one-sided

(g)

Slit fin, double-sided

(h)

Convex-louver fin

(i)

Herringbone wavy fin

Louver fin, one-sidedSmooth wavy fin, type (II)

+(d)

Plain fin

(a)

Fp Fs

Pd

pF

Smooth wavy fin, type (I)

(e)

(b)

Louver fin, with re-direction louver

(f)

(c)

2X

f

Pd

pF

2X

Pd

pF

Lh

pF

L

Lh

pF

L PP

f

S h

Fp

h

wS

S

pF

wS

pF

Copyright ITRI 2009

40

Shell and Tube Heat Exchanger

41

(a) brazed type; (b) working principle; (c) cross

section

(a) (b) (c)

42 Spiral Heat Exchanger

Copyright ITRI 2009

43

(Tube-in-Tube Heat Exchanger)

(a)(b)

(c) (d)

44

Air-to-air Cross Flow Heat Exchanger

indoor outdoor

45

Stream B

Heat Flow

Time Period (b)

CooledGas Out

ColdGas In

CooledGas Out

HotGas In

HeatedGas Out

HeatedGas Out

Valve Open/closed

Valve closed/open

Cold period

Hot period

Stream A

Heat Flow

Time period (a)

Copyright ITRI 2009

46

Hot blast stove Fixed Bed type regenerator

47

(a) Working Principle & (b) Rotary wheel

Rotary heat exchanger

Motor

48

Cooling Tower

Tray-type direct HX & Spray condenser

Copyright ITRI 2009

49

(b)(a)

Air Flow

(h)

Pt

S

Pt

(e)

h t

bV

(c)

S

Pt

Po

Louveres Turned Thro'90

Pa

ha

S1

ah

c

Pt

(i)

bP

PackedTightlySheets

th

Alternate Layers of

H (f)

ah

(g) ah

cP

Pt

tP

(d)

S

tP Pt

cP...........

Pt

Pt

tP

DISTRIBUTION-IMPACTREGION

FILM-FILLREGION

Splash Type Film Type

50

UA-LMTD-F -NTU

1.

2. (one-dimensional)

3.(Cp)

4. U

5.

6.

7.()

A

B

51

(Cont.)

)( 12 iimQ

)()( 1221 ccchhh iimiimQ

)()()()( ,,,,,, ohihhhpicocccp TTCmTTCmQ

hot fluid

cold fluid

chcchhph cmCandcmC )()( ,,

Copyright ITRI 2009

52

A m2

Cc ( pcm )c W/K

Ch ( pcm )h W/K

Q W

Qmax W

(Q/Qmax)

m kg/s

cp J/kgK

q = Q/A W/m2K

Cmin Cc Ch W/K

U W/m2K

NTU UA/Cmin

C* Cmin/Cmax

Cmax Cc Ch W/K

Th,i CK

Th,o CK

Tc,i CK

Tc,o CK

F

P (Tc,o Tc,i)/(Th,i Tc,i)

R Cc/Ch(Th,i Th,,o)/(Tc,o Tc,i)

LMTD CK

o (Th,i Tc,i) CK

53

(1)(parallel flow)

(2)(counter flow)

(3)(cross flow)

(a) fin with both fluid unmixed (b) unfin with one fluid mixed and the other unmixed

54

(Cont.)

(1) (parallel flow)

(2) (counter flow)

Tc,i

Th,i

Th Tc

Th,o

Tc,o

0 L x

Tc,i

Th,o

Th,i

Tc,o

0 L x

55

(Cont.)

(3) (crossflow) (pass)

L2

L1 0 x

Tc,i

Tc,o

(4) (mixed flow)

56

(Cont.)

() ()1 Pass 2 Pass

()3 Pass ()4 Pass

57

(Cont.)

(1) U

(2) A

(3) Tm

mTUAQ

58

Th,i

c,oT

(g)

(d)

h,iT

Tc,i

h,iT

(a)

Tc,i

T

T

T

T

h,oT T

(e)

h,oT

c,iT

( )h

c,oT

Th,i

Tc,o

c,o

T

h,o

c,i

c,iT

h,o T

(b)

Th,o h,iT

h,iT

c,oT c,iT

h,o

Tc,o

T

h,o

c,i

(f)

c,o

T

T

h,o

c,o

(c)

h,i

c,i

T

T

h,i

evaporation

condensation condensation

heating

cooling

cooling

heating

cooling

evaporation

superheat

T

heating

heating

condensation

superheat

coolingsubcooling

condensation

multi-component

subcooling

evaporation

heating

(Cont.)

59

(Cont.)

1,11

1,11 whw

whw TTAh

QTTAhQ

2,1,

2,1,

ww

wwTT

X

kA

Q

X

TTkAQ

cww

cww TTAh

QTTAhQ 2,

22

2,22

ch

ww

TTQAh

X

kAAh

2211

111


)(

111

1

2211

ch

UA

ww

TT

Ah

X

kAAh

Q

hT

T w1

h1

X

w2T

Tc

Q

h2

1

1Q

2

wQ

Q

Q2

60

(rating problem)

()

(sizing problem) (

) ()

(Energy balance equation)

Tm

U & Tm?

Rating/Sizing?

TCpmQ

mTAUQ

)entialDrivingPotUAQ ,)((


61 UA-LMTD-F Method

Tm= ?

lmm TT

T

T

TTLMTD

2

1

21

ln

)(

mTUAQ

)(

)(

,,2

,,1

icoh

ocih

TTT

TTT

c

h

c,i

h,i

c,o

h,o

Area

T

Tem

per

atu

re

T

(x)

dT

dT

dx T

T

T

T

2

1

62

Counter Flow heat exchanger

Fig2.


Ch Cc

63 UA-LMTD-F Method(Cont.)

LMTD

The total heat transfer rate for all single-pass flow arrangements

( CF HX ) Ch=Cc

( PF HX )

1 2

1 2ln( )lm

T TT

T T

lmTUAQ

21)()( TTTTandTTUAQ chch

)()( ,,2,,1 ocohicih TTTTTT

T1

T2

21,,,, )()( TTandTTTT ohihicoc

64 UA-LMTD-F Method(Cont.)

(Multipass and Crossflow HX) Tm Tlm FF PR F = F(PR)

F1 for cross flow & multipass

arrangement0.9~1.0

F=1 For pure counterflow

2pass (1pass)

max,,

,,

T

T

TT

TTP c

icih

icoc

icoc

ohih

h

c

TT

TT

C

CR

,,

,,

)()(

ln

)()(,

,,

,,

,,,,

icoh

ocih

icohocih

cflm

TTTT

TTTTT

),

()(cflm

TFUAQ

Copyright ITRI 2009

65

F for Pure Cross Flow FPR

max,,

,,

T

T

TT

TTP c

icih

icoc

icoc

ohih

h

c

TT

TT

C

CR

,,

,,

))((

))((

,,

,,

ohihphh

icocpcc

TTCm

TTCmQ

UA-LMTD-F Method(Cont.)

66

LMTD correction factor F for a shell-and-tube heat exchanger one shell pass and two or multiple of two tube passes

FPR (one shell pass and two tube pass)

UA-LMTD-F Method(Cont.)

67

-NTU Method

(hot stream) (cold stream) (heat capacity)

Cmin

(effectiveness) NTU (number of transfer unit)

Q Qmax

hph CmC )( cpc CmC )(

maxQ

Q

Q

T h,i T h,o

T c,i T c,o

m h

m c

68 What is Qmax?

()

hph CmC )( cpccm ,

)(minmaxminmax cihi TTCTCQ Th,i

Th,o Tc,o

Tc,i )( ,,minmax icih TTCQQ

Cmin (heat capacity)Th,iTc,i(effectiveness)

Th,i=Tc,o Th,o

Tc,i Cmin

Cmax

Copyright ITRI 2009

69 What is Qmax(Cont.)?

If Cc < Ch (Tc,o-Tc,i) > (Th,i- Th,o)

If Cc > Ch (Tc,o-Tc,i) < (Th,i- Th,o)

hhphccpc

ohihhhpicocccp

CmCCmC

TTCmTTCmQ

)()(

)()()()(

,,

,,,,,,


Case1

hc

hhpccp

TT

CmCm

)()( ,, Cold fluid=minimum fluid

hc

icihccp

CCif

TTcmQ

)()( ,,,max

Cmin


Case2

hc

hhpccp

TT

CmCm

)()( ,, hot fluid=minimum fluid

hc

icihhhp

CCif

TTcmQ

)()( ,,,max

Cmin


)( ,,minmax icih TTCQQ

hcicihccp CCifTTcmQ )()( ,,,max

chicihhhp CCifTTcmQ )()( ,,,max

minminmax c,ih,i

c,ic,oC

c,ih,i

h,oh,ih

- TTC

- TTC=

- TTC

- TTC

Q

Q =

(effectiveness)

max min , ,( )h i c iQ C T T

73

1.(effectiveness)NTU (number of transfer units)C* (Cmin/Cmax)

= (NTU C*)

2.

NTU NTU UA/Cmin

(thermal size)A U Cmin

NTU 0.0 ~ 4.0 NTU NTU A NTUC*

B NTU NTU 2-30(a)

What is ?

mohihhhpicocccp TUATTCmTTCmQ )()()()( ,,,,,,

Copyright ITRI 2009

74

A NTU

=Q/Qmax

minC

UANTU =

75

(pass)

T1, in

(a) (b)in1,

T

Tout1,

out2,T

2,T

in

2,T

in

out2,T

outT

2,T

out1,

2,T

out

Tin2,

Tin2,

in1,T

in1,T

1,T

out

1,T

out

B 2-Pass

76

1. Capacity heat ratio C* Cmin/Cmax 1 C* 0

(evaporation)

(condensation) C*= 0

C* 0(

heat sink heat source? Cmax Cmin

)

NTU ()

C* 0 NTU

95%

Capacity Heat Ratio C* ?

CpmC P

T

iCp

PT

iCp

77

-NTU Method(Cont.)

Single pass heat exchanger Cc>Ch Ch=Cmin and Cc=Cmax

(1).For Cmin/Cmax=1 and Counter Flow

(2) .For Cmin/Cmax=1 and Parallel Flow

(3) For Cmin/Cmax=0, both CF and PF HXs

min max

min max min max

1 exp NTU(1-

1 ( )exp NTU(1-

C C

C C C C )

NTU

1 NTU

2NTU1 (1 )2

e

NTU1 e

Copyright ITRI 2009

78

NTU ()

79 NTU ()

Type (NTU,C*) NTU(,C*)

Counter

Flow

Parallel

Cross flow,

Cmin mixed

Cross flow,

Cmax mixed

1-2 shell-

and-tube

heat

exchanger

NTU1exp1

NTU1exp1

CC

C

1

1ln

1

1NTU

C

C

NTU1exp11

1

C

C

C

C NTUexp1exp1

NTUexp1exp11

C

C

2/121NTUexp1

2/121NTUexp1

2/1211

2

C

C

CC

C

C

11ln

1

1NTU

1ln1ln1

NTU C

C

C

C

1ln1

1-lnNTU

2/1

2112

2/12112

ln2/12

1

1NTU

CC

CC

C

80 Summary for -NTU & UA-LMTD-F

1C0)(

)( *

max

min

max

min , cm

cm

C

C= C

p

p*

minminmax c,ih,i

c,ic,oC

c,ih,i

h,oh,ihactual

- TTC

- TTC=

- TTC

- TTC

Q

Q =

minC

UANTU =

)()(lm

TFUAQ

2

1

21

ln

)(

T

T

TTTLMTD lm

icoh

ocih

TTT

TTT

,,2

,,1

max,,

,,

T

T

TT

TTP c

icih

icoc

icoc

ohih

h

c

TT

TT

C

CR

,,

,,

F = fn (PR) = fn (NTU C*)

)T(Tcm(Q c,ih,ip min)

UA-LMTD-F -NTU

81

Rating by -NTU Method

1.()

NTU C*

NTU UA/CminC*=Cmin/Cmax

2. NTUC* -NTU

= (NTU C*)

Note 0 <

82

Rating by UA-LMTD-F

1. Th,oTc,o

2. R = Cc/Ch R P = (Tc,o Tc,i)/(Th,i Tc,i) P

Tlm

3. PR F

4.

5.

6.

(2)-(5)

Rating for UA-LMTD-F Method need to iterate a outlet

temperature.

Q

hihoh CQTT /,, cicoc CQTT /,,

)()(lm

TFUAQ

Q

83

Rating

R-22 5 C

12 C 0.1 kg/s 4180

J/kgKU 2000 W/m2K

3 m2 cm

d

= 3 m = 0.02 m

L

R-22 R-22

84

Rating

R-22 5 C

12 C 0.1 kg/s 4180

J/kgKU 2000 W/m2K

3 m2 cm

d

= 3 m = 0.02 m

L

R-22 R-22

5

12

?

85

(1) Solution By UA-LMTD-F

TcmFLMTDUA p

Counter flow F = 1 A = 0.02 3 = 0.189 m2

1 = 12 5 = 7 C 2 = x 5 (assume outlet water is x C)

5

7

)5(7

2

1

21

xn

x

T

Tn

TTLMTD

)12(41801.0

5

7

12189.02000 x

xn

x

9043.05

7

xn x = 7.83 C

W1741.6

)83.712(41801.0

Q

(2) Solution by -NTU W/K41841801.0 minC

9043.0418

189.02000

minC

UANTU

0* max

min

C

CC

5952.0)exp(1 NTU

W2926)512(418max TCQ minmax

W6.174129265952.0 maxQQ

C 83.7418/6.174112, outcT

UA-LMTD-F -NTU

A Rating Example

86

Sizing Problem

Sizing by -NTU Method

1.Q

Th,o

Tc,o = Q/Qmax

C*=Cmin/Cmax

2. -NTU

NTU (

)

3A=NTUCmin/U

Sizing by UA-LMTD-F

1.Q

P R

2. PR

F

3.

LMTD Tlm

4.A=Q/(UFTlm)

87

Sizing

R-22 5 C

0.1 kg/s 12 7 C

4180 J/kgKU 2000 W/m2K

= 0.02 mL

d

R-22 R-22

= ? m

88

(1) Solution By UA-LMTD-F (2) Solution by -NTU

A Sizing Example

W209071241801.0 TcmQ p

TcmFLMTDUA p

1 = 12 5 = 7 C

2 = 7 5 = 2 C

C 99.3

2

7

27

2

1

21

nT

Tn

TTLMTD

counterflow F = 1

QLMTDUA

2m 2618.02000/99.3/2090

//

ULMTDQA

W209071241801.0 TcmQ p

W/K41841801.0 minC

W2926)512(418max TCQ minmax

7143.02926

2090

maxQ

Q

Also

7143.0)exp(1 NTU

253.1)7143.01ln()1ln( NTU

2618.02000

418253.1

U

CNTUA

C

UANTU min

min

A = 0.02 L L = 4.167 m

A = 0.02 L L = 4.167 m

RatingSizing UA-LMTD-F -NTU

89

UA-LMTD-F -NTU

1. -NTU

2. F

F

3. -NTU Cmin

4. -NTU F

5.ratingUA-LMTD-F-NTU

6.sizingUA-LMTD-F

A = Q/(UFTlm)

90

Sizing case: (1.5 mm) Rating case12 C5C1 kg/s & 2 kg/s7CU 2000 W/m2K

1 kg/s & 2 kg/s (U)

(-NTU chart)

Dd

D = 3 cm

d = 2 cmo

o

= 1.5 mm

92

700 m2/m3

m /m

20 10 560 40

10060 1000500200

0.2 0.152 1 0.5

1050002000

( ), m /m2 3

42x10 3x10

4 4

, D , mmh

()32

93

c,out

c,inT

h,inT

T

Th,out

94

A:

Ac: (free flow area, minimum flow area)

Afr:

L:

V:

: (contraction ratio Ac/Afr)

:

Dh:

cA4

Vc:

Vfr: (face velocity or frontal

velocity)

Gc Vc

ReDh GcDh/ (Gc Ac)

, Dh

Dh = (4Ac/P)(L/L) = 4LAc/A (A = PL)

PDh

Dh = 4(Ac/A)(L/L)

= 4(Ac/Afr)(Afr/A)(L/L)

= 4(Afr/A)(L/L)

= 4((AfrL)/(AL))

= 4V/AL

= 4/L

Dh = 4L(/L) Dh = 4/ (3-2)

(secondary area)

95

?

(thermal conductivity)

(compactness)

(b) circular(a) plain (c) continuous fin

96

(a) L (b)L (c) (d) (e) (f)

L-Foot(a) (c)(b) (d) (f)(e)

97

(fin efficiency)

(fin efficiency)f

Air

)( bf

T

)()(

TThAQTThA

Q

Q

Qbff

bfmax

f

(fin surface effectiveness)s

)( bs

T

hAos(Tb T) = hAb(Tb T) + hAff(Tb T)

& Ao = Ab + Af

)1(1 fo

f

sA

A

(a)f = s = 1

(b)s > f (c)1

hAo(Tb T)

(hAb(Tb T) + hAff(Tb T))

98

k = 204 W/mKf

y= 0.003 m

l = 0.07 m

2.h = 20 W/ m

k = 204 W/m K.

5.02

)tanh(

ky

hm

ml

ml

o

f

m = (220/204/0.003)0.5 = 8.085 m1

f = tanh(8.0850.07)/(8.0850.07) = 0.905

99

k = 20 W/mK f

m = (220/20/0.003)0.5 = 25.82 m1

f = tanh(25.820.07)/(25.820.07) = 0.524

Tb

42%

100

Stop & Think

Questions

()

101

1 1 2 2

1 1 1

w w

X

UA h A kA h A

(3-6)

1 1 1 2 2 2

1 1 1

w w

X

UA h A kA h A

(3-7)

102

hi 2000 W/m2Kho

50 W/m2Ks 0.7 ()

Ai ( = 1 m2) Ao/Ai =10 Ao =10 m

2 U Ao

oioioiooAA

.

AAA.AAU

0286.000050

35

1

2000

1

5070

1

2000

11

210.005( ) 0.0286( ) 0.0336 /

o

K m WU

Uo = 29.77 W/m2K

15% (0.005/0.0336)

85%

:

103

Case 1 hi 2000 W/m2Kho 100 W/m

2Ks 0.65 (()?)

Ai ( = 1 m2) Ao/Ai =10 Ao =10 m

2 U Ao

1 1 1 1 1 0 0005 0.0154

2000 0 65 100 2000 65t

i oo o i o i o

.

A AU A A . A A AR

2

2 2

10.0005 0.00154 0.00) ( 204 /( )

10.0204 / 49.02 /

t

o o

o

o

R K m WU A

K m W U W m KU

49.02/29.77=1.6565%

25% (0.0005/0.00204)

75%

: (Conti.)

104

Case 2 hi 4000 W/m2Kho 50 W/m

2Ks 0.70 (()?)

Ai ( = 1 m2) Ao/Ai =10 Ao =10 m

2 U Ao

1 1 1 1 1 0 00025 0.0286

4000 0 70 50 4000 35t

i oo o i o i o

.

A AU A A . A A AR

2

2 2

10.00025 0.00286 0.00) ( 311 /( )

10.0311 / 32.15 /

t

o o

o

o

R K m WU A

K m W U W m KU

32.15/29.77=1.088%

8.0% (0.00025/0.00311)

92%

: (Conti.)

105

f ( )

P

iP

1

1

entrance

entrance

eP

a a

()

exit2

exit

2

P

P + P P =

in

Pout

0

Flow

(a) Pi (b) Pf (c) Pa (d) Pe

106

Pi

(1 2)

Kc

2 2 2

1 1

1 1 1

2 2 2fr entrance entrance c c entrance cP V P V K V (3-22)

1 = entrance 3-22

2 2 21

1

( ) 1 1( )

2 2

i entrancec fr c c

i

P P PV V K V

(3-23)

Vfr2

= 2Vc2

22

1

(1 )2

i cc

P VK

(3-24)

107

Pf ()

2

42

c

hm

f V

D

Lf

P

fFanningDarcy

2

2c

hm

f V

D

Lf

P

Darcy fD 4f4

Fanning

108

Pa

entrance 1exit 2Pa =

11

)(

)(

2

12

1

2

2

2

1

21

2

222

12

2

c

ccentranceexitentranceexit

entranceentranceentranceexitexitexitentranceexita

G

GGVVVV

VVVVVA

mV

A

mP

2

2 1

1 1a cP G

(3-25)

Pe

(d)

Ke

22 2

2

1( )

2 2

e cc fr e

P VV V K

22(1 )

2

ce

VK

(3-26)

109

P = Pi + Pf + Pa + Pe

m = (1+2)/2

2 2 2

1 2 1 2

(1 ) (1 )1 12( )

2

c c e

m c

G K Kf A

A

110

KeKc(Kays and

London, 1984)

111

Kays and London

(3-18)KcKeKays

and London

Kc = Ke= 03-27

2

21 1

1 2

(1 ) 12

c

c m

G AP f

A

(3-29)

(a)

(b)

112

112

3-3-1strip3-19(Kays and

London, Fig. 10-56, strip fin)

Fin pitch = 782 fins/m

Plate spacing, b = 2.49103 m

Fin length = 3.175103 m

Hydraulic diameter, 4rh = 1.54103

m

Fin thickness = 0.10203 m

= 2254 m2/m3

Fin area/total area = 0.785

L = 0.4 m

Afr = 0.4 m2

Vfr = 10 m/s

1 = 1.145 kg/m3 2 = 1.1 kg/m

3

viscosity = 188.7107 Ns/m2

113

113 3-3-1

Dh

3-5

3-2Dh = 4/

= DH/4 = 0.001542254/4 = 0.868

Gc = 1Vfr/ = 1.14510/0.868 = 13.19 kg/m2s

ReDh = GcDh/ = 13.190.00154/(188.7107

) = 1076.4

f 0.044

2

2

121

22 )1()

11(2

)1(

2

e

cm

cc K

A

AfKGP

(a)

strip finKe Kc ReDh 3-15

Kc 0.1Ke 0.02m = (1.145+1.1)/2 1.123 kg/m3

Pa 3.261.145

1.0868.01

2

13.19)1(

2

22

1

22

cci

KGP

(b)

D h = 4AcL/AL = 0.4 m A/Ac = 4L/Dh = 40.4/0.00154 1039

Pa 354110391.123

0.044

2

13.19

2

22

cm

cf

A

AfGP

(c)

Pa 22.6145.1

1

1.1

12

2

13.19 )

11(2

2

2

12

2

ca

GP

(d)

Pa 92.171.1

02.0868.01

2

13.19

)1(

2

22

1

22

ece

KGP

P = 26.3+3541+6.2217.92 = 3555.6 Pa

Pf 99%

114

114

(Reynolds analogy)(M)(M)(Q)Qmax

M/M Q/Qmax (3-30)

w

p

dA h TdA

mV mc T

(3-31)

wV

2

w

p

h

Vc V

(3-32)

115

115

21

2w f V (3-33)

2

2

1/ 2

p

h f V

Vc V

(3-34)

2p

h f

Vc

(3-35)

Stanton numberpVc

hSt

St = f/2 (3-36)

3-361930Colburn

(Pr 0.6 ~ 60)3-36

StPr2/3

= f/2 (3-37)

Colburn j factor

j StPr2/3 (3-38)

j = f/2 (3-39)

116

116

j = f/2()

j

(3-19)2

3Prp

hj

Vc

117

117

(j & f)3-20

W = 595 mm

H =355 mm

N = 1

dc = 10.34mm

f = 0.12 mm

Pt = 25.4 mm

Pl = 22 mm

4 m/s(1) (Afr)(2)

()(3)(4) (5) (Vfr

= 4 m/s)(6) (35Ca = 1.145 kg/m3a

= 188.7107 Ns/m2cp,a = 1007 J/kgKPra = 0.71)

118

118

119

(a) (inline)(b) (staggered)(c)

V

(a)

Pl

1

Pt

1

(b)

1

1

(c)

1

ldP

o

dX

1

120

Eu/

Eu/

21

2c

PEu

u N

* tt

o

PP

d

* ll

o

PP

d

121

(1) Reb*

tP *

lP 1/1 ** lt PP (2) 5-45-5

(3) Eu/

(4)

1P

wb

b

Eu Eu

(5-12)

wb

0 1000

1 0.0018Re 0.28

0.0026Re 0.43

P

(5-13)

(5) NuEuP c

2

2

1

122

o

d r

Fre

ro

e d re o o

L

f

F = r - rL e

= 2= 2

0.110.2

0.3190.134Reo

s sd

L f

F Fj

F

(5-20)

0.927 0.11

0.3169.47Reo

t td

o d

P Pf

d X

(5-21)

2 2

2

t l

d

P PX

(5-22)

Rabas et al. (1981)

0.770.671.12 0.470.26

0.292Reo

fn s s e ed

o L s o f

F F d dj

d F F d

(5-23)

0.25 0.76 0.73 0.71 0.38

0.2343.805Reo

s o o tLd

o s e t l

F d d PFf

d F d P P

(5-24)

0.415 0.0346 e

s

dn

F

(5-25)

5-20 5-21 4 5-23

5-246do

o

d r

Fre

ro

e d re o o

L

f

F = r - rL e

= 2= 2

123

HXNr=?

(1)HX

(2) hi

(3) ho

(4)

(5) U

(6) LMTD

(7) Q=Uo AoF*LMTD Ao Nr

(8)

()

i

ii

D

kNuh

12

7.121

Pr1000Re2

322

1

r

fc

Pf

f

NumberNusseltNu

32

Pr

apacso CGjh

mr

mrf

)tanh(

soii

o

o

hhA

AU

11

1

2

1

21

ln

)(

T

T

TTLMTD

124

(a)50%(b)50%(c)

(b)(a) (c)

125

2

32

28.3Reln58.1,

1Pr2

7.1207.1

Pr1000Re2

ff

f

Nu

0.140,

,0.45,

,

n

b b

w w

nT

nT

Dh = 4 /Dh = (D i do) (

)

Dh,e = 4Ac/(do) = (D i2 do

2)/do

Taborek (1998) Hewitt et al. (1994)

Kaka Liu (1998) Kern (1950)Dh,e

Nu (ReDh)

Dh = (D i do)

Dittus-BoelterGnielinski

8000(HEDH, 2002)

126

0.8

0.8 0.5

0.467

0.19 RePr

3.66 1.2 1 0.14

1.07 0.117 RePr

h

i i

o o h

D

D D LNu

d d D

L

(8000 > Re > 2000)Taborek (1998)

,Re 2000 ,Re 8000

Re1.33

6000tr laminar turbulentNu Nu Nu

Taborek (1998)Re < 2000

127 Ti

oT

w,oT

w,iT

d i

Outer Wall

Inner Wall

o

od

Di

D

Th, i

Th,o

Tc , i

Tc ,o

D i

d i

do

L

ln1 1 1

2

o

i

i i o o w

d

d

UA h A h A k L

ln1 1 1

2

oo

o i

i i i o o w

dd

d d

U h d h k

128

Example:

, Water

, Engine oil370 K

od d

0.0

409

4 m

0.0

89 m

0.0

75 m

0.0

483

m

L

323 K

i

oD

Di

303 K

340 K

K 340K, 370 kg/s, 1 oil, outoil, inoil TTm

K 303kg/s, 0.767 water,inwater Tm

Carbon Steel (C 0.5% kw 53 W/mK)

T (K) o il (kg/m3) coil (kJ/kgK) o il (Pas) koil

(W/mK)

Proil

370 841.8 2.20 0.019 0.136 305

360 848.2 2.16 0.025 0.137 395

350 854.0 2.12 0.036 0.138 550

340 859.8 2.08 0.053 0.139 795

T

(K)

wate r

(kg/m3)

cp,wa ter

(kJ/kgK)

water

(Pas)

kwater

(W/mK)

Prwater

303 995.6 4.182 0.000798 0.603 5.4

313 992.2 4.179 0.000654 0.618 4.33

323 988.0 4.181 0.000548 0.631 3.56

333 983.3 4.185 0.000467 0.643 2.99

129

129

= (cp,oi l , in + cp,oi l ,out)/2 = (2200 + 2080)/2 = 2140 J/kgK

W6420034037021401,,, outhinhoilpoil TTcmQ

K 03.3234180767.0

64200303

,

,,

waterpwater

incoutccm

QTT

222, m 001316.0)04094.0(44

iic dA

skg/m 7.582001316.0/767.0/ 2, icwaterwater AmG

! 230036474000654.0

04094.07.582Re

water

iwateri

dG

Gnielinski (1-8)

00564.028.3Reln58.1 2 bif

198

133.42

00564.07.1207.1

33.41000364742

00564.0

1Pr2

7.1207.1

Pr1000Re2

3232

i

ii

if

f

Nu

K W/m9.298804094.0

198618.0 2

i

iii

d

Nukh

130

130

22222, m 002586.0)0483.0075.0(44

oiac dDA

Dh,o = D i do =0.075 0.0483 = 0.0267 m

skg/m 8.386002586.0/1/ 2, acoiloil AmG

! 20006.341030233.0

0267.08.386Re

,

oil

ohoilo

DG

8-3

467.0

8.0

5.08.0

PrRe117.007.1

PrRe19.0

14.012.166.3

L

D

L

D

d

D

d

DNu

h

h

o

i

o

ioil

131

131

L

350 m

59.5oilNu

K W/m9.280267.0

59.5138.0 2

,

oh

oiloiloil

D

Nukh

C 8.41

)()(

,,

,,

,,,,

ocih

icoh

ocihicoh

TT

TTn

TTTTLMTD

8-12w

i

oo

ooi

o

ii k

d

dd

hd

d

hU 2

ln111

i = o = 1U = 28.5

W/m2K

A = doL= Q/U/T lm L = 355 m 355.7 m

132

132

Example:

Carbon

Steel (C 0.5% kw 53 W/mK)H = 0.013 m

N = 60 f = 0.0009 m

W

f

A = A + A

D do

di

i

o

f

b

L

r

H

bA

f

A = NLH2

133

133

Q = 64200 W

h i = 2988.9 W/m2K

222

22,

m 001884.0013.00009.060)0483.0075.0(4

4

HNdDA foiac

Dh,o = 4Ac ,a/Pw

Pw = = 2NH + do + D i = 1.947 m

Dh,o = 4Ac ,a/Pw = 0.00387 m

skg/m 8.530001884.0/1/ 2, acoiloil AmG

! 20009.67030233.0

00387.08.530Re

,

oil

ohoilo

DG

L = 20 m8-3

18.5oilNu

K W/m7.18400387.0

26.5138.0 2

,

oh

oiloiloil

D

Nukh

C 8.41

)()(

,,

,,

,,,,

ocih

icoh

ocihicoh

TT

TTn

TTTTLMTD

134

134 8-13

8-11

i

ow

oooiii

d

dLk

AhAhUAln2

1111

A = Ao = P fLP f

(= 2NH + do) ( )

P f = 2NH + do = 1.7117 m

1 1

2 ln

f f

oi i o ow

i

P P

dU h d hk

d

(8-14)

o = 1 A f/Ao(1 f)

A f/Ao = (P f do)/P f

f = tanh(mH)/mH ( )

1-m 880009.053

7.18422

ff

oil

k

hm

f = tanh(mH)/mH = 0.713

o = 1 A f/Ao(1 f) = 0.738

8-14U = 41.68 W/m2K

A = P fL = Q/U/T lm = 36.85 m2

L = 21.53 m21.56 m

136

(Shell-and-tube HX)

(non-compact)

137

(d)(Kettle)

25 26 215 14 15 21

25

2

24 7 6 26 5

526 20

17

10231 3 13 27 11

12

29

22

619

8

(a)

20 21 321 20

9

7

17

2

18 234 1529 1 14 165 26 2710 13

12

17

1122

(Kettle)

138 27

11

(c)

520 26 1521 21 19

11

1018 23 14 1 10 29 13

5

(b)U

20 26 21 14 21 3 20

17

2

15 28 1 5 26 16 10 29 27

11

13

U

139

N

Special high pressure closure

sheet and removable coverChannel integral with tube-

BONNET (integral cover)

Channel integral with tube-sheet and removable cover

D

Removable

N

C

Only

TubeBundle

Kettle type reboiler

Cross flow

X

K

W

U

Divided flow

Double split flow

Split flow

J

H

G

T

S

P

and removable cover

Stationary head types

A

B

Channel

with longitudinal baffleTwo pass shell

One pass shell

F

E

M

L

floating tubesheetExternally sealed

U-tube bundle

Pull through floating head

with backing device

like "N" stationary headFixed tubesheet

Floating head

Outside packedfloating head

LIKE "B" STATIONARY HEAD

LIKE "A" STATIONARY HEADFIXED TUBESHEET

FIXED TUBESHEET

Head types

140

E shell

E shell

F shell ()F shell

J shell

F shellG H shell

X shell

141

(a) Bonnet (B )

(b) Channel head removable (A)

(c) Channel head integral with tube sheet

(C)

142

(L, M, N)100 F(56K)

U

tubes

gasket

tubesheetfloating

backing ringshell

cover

cover

tubesfloating-head

shell

gasketfloatingtubesheet

cover

flangeflange

floating-head

shell

shell stationaryhead

floating-head

gasket

backing ring

tubes

shell

skirt

gland followerstuffing-box flangelantern ring

packing

floating tubesheet

stiffomg-box flange

tubessnap ring

floating tubesheet

cover

(a) Packed lanternring exchanger (b) Outside-packed floating head exchanger

(d) Pull-through floating head exchanger(c) Internal floating head exchanger

143

/5~10 (baffle spacing Lb)

,

,(stationary)

flangeChannel

Window

Shell flange

Tube sheet

144

9-1 (Lb,max)

(mm)

do

Carbon & high alloy steel

Low alloy steel

Nickel-Copper

Nickel

Nickel-Chromium-Iron

Aluminum & aluminum alloys

Copper & copper alloys

Titanium & zirconium at code

max.

Allowable temperature

19 1520 1321

25 1880 1626

32 2240 2210

38 2540 2930

50 3175 2794

145

Orifice

Segmental baffle

Drilling

Disc and doughnut baffle

Orifice baffleBaffle

Doughnut

Doughnut

O.D. of tubes

Disc

2(50.8 mm)0.5Ds1Ds (baffle cut)

13 67

80 400

bc b

s s

L L

D D

146

baffle cut ()

Baffle

bc

L

Eddies

bL

Main flow

Main flow

Eddies

Idealized Shell-side Flow Model

147

Baffle edge

No. of Row

SectionWindowin one

SectionCrossflow

Bundle bypass area

No. of Rowsin one

L bc

Ds

2

bbW p

ct

ds

sb

Dot

LD s

bc

Ds

bb =

Bundle edge Baffle bypass area

D =D -dct

ot

baffle cut

o

(Bypass lane)

(Tube bundle)

Crossflow

(1-2 )

Window

ot

148

Nc (number of tube rows crossed in one

crossflow section, between baffle tips) Fc (fraction of total tubes in crossflow)

window (the number of effective crossflow rows in each window Ncw)Bell

149

Nb (crossflow area at or near the

centerline for one crossflow

section) Am

Atb Asb (shell to tube bundle diametrical clearance)

bb (=2bb)

Fsbpfraction of crossflow area for bypass flow

sb

Ds

Db

sD - bD sbsb

tbd

o

sb

Ds

Db

sD - bD sbsb

150

Roated Triangular PitchTriangular Pitch

Pitch

Rotated Square Pitch Square Pitch

Pitch

45

tP tPt

P

tP

P

P = PX t

PP

P

XP = PtXP = Pt

P = PX t

X

X

X

XPp

Pn

P = PP = P

p tP

P = P P = Pt p tp t

P = PP = P

nn t tn

P = PP = Pn t

Pp

Pitch nP n

P

nP

pP

Pitch

1.732

0.866

0.5

0.707

0.707

1.4140.5

0.866t

(30 ) (45 ) (60 ) (90 )

2

14

ctt

DN CTP

A

(9-1)

0.93 (1 Pass)

0.9 (2 Pass)

0.85 (3 Pass)

CTP

(9-2)

A1 = (CL)P t2

(9-3)

o o

o o

1.0 45 90

0.87 30 60CL

(9-4

151

Recrit = Gsdo/s = 100 Recrit < 100 Recrit > 100

152

( impinging plate impinging rod )

Impingement

(d)

(c)

Vapor belt

(b)

rod

plateExtra tube support

baffleFirst

Impingement

(a)

plate

baffleFirst

Tube plate

Tube plate

,

2250 m/s,

750 1 1 m/s, ,

n max

m m G L

Vx x

153

,max

4500 m/s,

m/s, ,0.8

( )

1500 1 1 m/s, ,

n

m m G L

cV c

x x

154

U( W/m2.K)AoAi

oi(fin surface effectiveness)

o=i =1

ho()hi

Rt=1/Uo

() Ro=1/(o *ho)

Rf,o=Rfo/ o

Ri=(Ao/Ai)*1/i Rfi= (Ao/Ai) *Rfi

Rwall=Ao*Rw

moom TAUQTUAQ

ln1

2

o

i

i i i o o o w

dA

dA A

U h A h A k L

155

(Kern method)

Kern (1950)

0.14

0.55 1 30.36Re Pro eshell shellw

h DNu

k

(9-36)

2

0.14

4 1

2

shell s s b

shell

e

w

f G D NP

D

(9-37)

0.576 0.19lnRe4 shellshellf e

(9-38)

Re s eshellG D

(9-39)

Ts

s

MG

A

(9-40)

As = CLbDs/P t (9-41)

C = P t do (9-42)

22

2

44

for square pitch

4

14 0.86

2 2 4 for triangular pitch

2

ot

o

et o

t

o

dP

d

DP d

P

d

(9-43)

Kern Method

t

t60

C'

P

P

C'

60 60

Kern Method PtC

156

9-4-1

Ds = 0.508 m

do = 0.01905 m

d i = 0.016 m

Lb = 0.5 m

= 5 m

P t = 0.0254 mP t* = 0.0254/0.01905 = 1.33

30 E shell363 K

(90C)25 kg/sKern

T (K) water

(kg/m3)

cp,water

(kJ/kgK)

water

(Pas)

kwater

(W/mK)

Prwater

283 999.6 4.194 0.00130

4

0.587 9.32

323 988.0 4.181 0.00054

8

0.631 3.56

363 965.3 4.207 0.00031

6

0.676 1.96

Carbon Steel (C 0.5% kw 53 W/mK)

K 283 kg/s, 50, water, iniwater Tm

157

157 C = P t do = 0.0254 0.01905 = 0.00635 m

As9-41

As = CLbDs/P t = 0.006350.50.508/0.0254 = 0.0635 m2

skg/m 7.3930635.0/25/ 2 sss AmG

2 21 0.0254 1 0.019054 0.86 4 0.86 0.0254

2 2 4 2 2 4

0.01905

2 2

0.018 m

t ot

eo

P dP

Dd

! 22468000316.0

018.07.393Re

water

esshell

DG

property index (/w)0.141

0.14

0.55 1 3 0.55 0.3330.36Re Pr 0.36 22468 1.96 111.4o eshell shellw

h DNu

k

ho = 0.676111.4/0.018 = 4185 W/m2K

0663.0

2651.04 22468ln19.0576.0Reln19.0576.0

shell

shell

f

eef shell

5 m0.5 m

Nb = 5/0.5 1 = 9

Pa 7.6006

018.03.9652

19508.07.3932651.0

2

14 2

14.0

2

w

e

bssshellshell

D

NDGfP

158

Bell-Delaware Method

Window

effects

Bypass

Tube-to-baffle

leakage

Shell-to-baffle

leakage

159

Idealized Shell-side Flow Model

160

B

B

BB

AF

BBB

EE

A

CC

C

E

BB

C

A stream B stream C stream E stream F stream

161

Bell-Delaware Method

Bell-Delaware

hs = hoJcJlJbJsJr (JcJlJbJsJr0.4~0.6)

ho:

JcJlJbJsJr

ho: Bell (1963) Coburn j

s

ps

o

cG

hj 32Pr

,15.273

15.273

, 1

25.0

14.0

s,av

w

s

s

T

s

Ts,avs 1

162

jf

163

JcFc

164

J (Atb + Asb)/Am

165

Jb Ab/Am

Sealing Strip

Sealing Strip

Cro

ssfl

ow

str

eam

Penetration area

Bypass area

166

Lbi Lb A Lb

Region of Central

Baffle Spacing

Lbo

L

167

Jr (adverse temperature gradient)

168

Bell-Delaware Method(Cont.)

(a)()

(b)

(c) Window

Ptot=Pc+Pw+Pe

169

Pb,i

2

,

4

2

i s cb i s

i

f G NP

mTs AMG / s

170

(E-shell )

Rs

Rb: (C & F stream) Illustration of a Typical E-Shell

Pressure drop at entrance and exit section

Pe = Pb, i(1 + Ncw/Nc)RsRb

171

Rb Fsbp

172

Pc Illustration of a Typical E-Shell

Pressure drop at interior crossflow section

Pc=Pb, i(Nb 1)RRb

173

R

174

Pwwindow

Pressure drop at window section

Illustration of a Typical E-ShellPw =Pw, iNbR

175

9-4-1Bell-Delaware split ring and

floating head baffle cut = 25%

sealing stripLbi = Lbo = 0.75

m9-4-1Lb = 0.5 m

176

(1)

split ring and floating head9-29

bb 0.035 m9-56

Dot = Ds bb = 0.508 0.035 = 0.473 m

Dct = D o t do = 0.473 0.01905 = 0.45395 m

30Pp = 0.022 mPn = 0.0127 m (9-15)

9-48

2m 0742.0

)01905.00254.0(0254.0

01905.0473.0)473.0508.0(5.0

)()(

ot

t

ootlotlsbm dP

P

dDDDLA

m/s 349.00742.03.965

25

m

Tmax

A

MV

20300000316.0

01905.0349.03.965Re max

s

odV

Bell-DelawaredoReKern

Ds

177

9-3

a1 = 0.321a2 = 0.388a3 = 1.45a4 = 0.519

0578.02030014.01

45.1

Re14.01519.0

3

4

a

aa

00684.02030033.1

33.1321.0Re

33.1 388.00578.0

*12

a

a

tPaj

Prs = 1.96property index 1 ( 1

14.0

w

ss

)9-59

K W/m618696.142077.33600684.0Pr 2667.032 spso cjGh

(2)Jc

baffle cut = 25%Lbc = 0.25Ds = 0.127 m

ot

bcs

ot

bcs

ot

bcsc

D

LD

D

LD

D

LDF

2cos2

2cossin

22

1 11

537.0473.0

128.020508.02

ot

bcs

D

LD

649.0537.0cos2537.0cossin537.021 11

cF

9-65

Jc = 0.55 + 0.72 Fc = 0.55 + 0.720.649 = 1.017

178

(2) J

9-5

260

01905.033.1

45395.0

87.0

9.0785.0785.0

22

2

22*

2

ot

ctt

dP

D

CL

CTPN

JA tbAsb tbsb

9-27Lb,max < 900 mm do = 0.01905 m < 0.032 m

tb = 0.8 mm = 0.0008 m t b = tb/2 = 0.0004 m

9-51

2m 005133.0

260649.012

10004.001905.01

2

1

tctbotb NFdA

9-55

sb = 3.1 + 0.004Ds ( mm) = 3.1 + 0.004508

= 5.132 mm = 0.005132 m

sb = sb/2 = 0.00257 m

21

1

m 00273.0508.0

127.021cos

1100257.0508.0

21cos

11

s

bcsbssb

D

LDA

(A tb+Asb)/Am = (0.005132 + 0.00273)/0.0742 = 0.1059

Asb/(A tb+Asb) = 0.00273/(0.005132 + 0.00273) = 0.3472

9-66

515.0)3472.01(44.0)3472.01(44.0

144.0144.0

1059.02.2

2.2

e

eAA

A

AA

AJ m

tbsb

A

AA

tbsb

sb

tbsb

sb

179

(4) Jb

sealing stripNss = 0

9-44

55.110212.0

508.0

127.021508.021

P

s

bcs

cP

D

LD

N

Ns+ = 0/11.55 = 0

E-shell Np = 0

236.0

0742.0

5.0473.0508.0)5.0(

m

bppots

sbpA

LWNDDF

1009-68Cbph = 1.259-67

3 3

3

12 1 2 1.25 0.236 1 2 0

1 212

1

0.745

bph sbp s

bph sbp s

s C F N

b C F N

s

N

J e e

e N

180

(5) Js

9-47

815.0

75.075.051

b

bobib

L

LLLN

L i+ = Lbi/Lb = 0.75/0.5 = 1.5

Lo+ = Lbo/Lb = 0.75/0.5 = 1.5

Re > 100n = 0.6 (9-71)

9-70

935.0

5.15.118

5.15.118

1

14.04.011

oib

n

o

n

ibs

LLN

LLNJ

(5)Jr

Res > 1009-72Jr = 1

(6)hs

hs = hoJcJJbJsJr = 61861.0170.5150.7450.9351.0

= 2363.1 W/m2K

Kern (4185 W/m2K)

JcJJbJsJr 0.382 0.6

181

(1) crossflow pbi

9-3

b1 = 0.372b2 = 0.123b3 = 7.0b4 = 0.5

334.02030014.01

0.7

Re14.015.0

3

4

b

bb

1098.02030033.1

33.1372.0Re

33.1 123.0334.0

*12

b

b

tPbf

9-74

Pa 7.2973.9652

55.117.3361098.04

2

4 22

,

i

csiib

NGfP

(2) Pe

Res > 1009-77 n = 1.89-76

964.05.15.1 8.18.1

n

b

bo

n

b

bis

L

L

L

LR

9-79Cbp = 3.79-78

418.0

13

3

021236.07.3

21

21

21

e

Ne

N

R

s

NFC

s

bssbpbp

9-46Ncw = 0.8(Lbc/PP) = 0.8(0.127/0.022) = 4.62

9-75

Pe = Pbi(1 + Ncw/Nc)RsRb = 297.7(1 + 4.62/11.55)0.9640.4182

= 168 Pa

182

(3) Pc

9-83 rs = Asb/(Asb + A tb) = 0.3472

9-84 r lm = (Asb + A tb)/Am = 0.1059

9-82 z = 0.15rs + 0.65 = 0.597

9-81

626.0605.01059.0)3472.01(33.1133.1 eeR

zlms rr

9-80

Pc =Pb, i(Nb 1)RRb = 297.7(8 1)0.6260.418 = 545.6 Pa

183

(4) Pw

1 2Lbc/Ds = 1 20.127/0.508 = 0.5

9-89

22

1 2

221 2

2

2 2 2cos 1 1 1 1 1

4 8

0.508 260 cos 0.5 0.5 1 0.5 1 0.649 0.01905

4 8

0.02663 m

s bc bc bc tw c o

s s s

D L L L NA F d

D D D

9-86

skg/m 3.562

02663.00742.0

25 22121

wm

sw

AA

mG

9-88

)120( rad. 094.2508.0

127.021cos2

21cos2 11

s

bcds

D

L

9-90

m 0327.0

2

094.2508.0

2

649.0126001905.0

0234.04

22

1

4

dsscto

ww DF

Nd

AD

9-85

Pa 3.7813.9652

3.56262.46.02

26.02

100 Re 2

002.026

100 Re 2

6.02

22

2

2

2

,

wcw

sw

w

bc

ot

cww

sw

cw

iw

GN

G

D

L

dP

NG

GN

P

window 9-91

Pa 2.3914626.03.7818, RPNP iwbw

(5) P to t=Pc+Pw+Pe = 545.6 + 3914.2 + 168 = 4627.7 Pa

184

Stop & Think

Leakage & Bypass

U

Sizing Rating (UA-LMTD-F -NTU)

ln1

2

o

i

i i i o o o w

dA

dA A

U h A h A k L

186

(foulant)

11-1

(food & kindred

products)

(milk processing) (gas side) (spray drying)

/

(textile mill products)

(cooling water) (cooling water)

Lumber & wood

products including

paper & allied

products

(liquid, cooling water) (process side, cooling water) (cooling water) (process side)

chemical & allied

(process side, cooling)

187

(precipitation fouling)

(particulate fouling)

(chemical reaction fouling)

(polymerization)(cracking)

(corrosion fouling)

(solidification fouling)

(biological fouling)

189

Periodic change of operation condition (d)

Linear (a)

, Time, t

delayt

fR

Falling rate (b)

Asymptotic (c)

190

hT

Tr

h Ts,hr

s,h

q

w,cTw,h wr Ts,cs,c

r

Tc

rc

w

cT

s,c

hT

s,h

rc

T - TTc h c

T

Trs,c

Ts,c

rw w,h

w,c

Th

T

rs,h

hr

s,h

Ai

oA

wA

1 1

ln ln1 1

2

fo o

o o

o o i o o ifi fo fi fo

i i o i w i i o i w

Ud d

A dA A d d d d

r r r rh A h A k L h d h d k

o fi

f fo

i

A RR R

A

1 1f

f C

RU U

191

QC = Q f = UCACTm = U f A f Tm11-12

1C C ff

UU R

U (11-13)

1 1C C C m C C CC ff f f m f f f

Q U A T U A AU R

Q U A T U A A

(11-14)

1f

C f

C

AU R

A (11-15)

192

(%) (HEDH, 2002 (%) (HEDH, 2002)

193

TEMA

194

24 1

2C C C

i

LP f u

d

24 1

2f f f

f

LP f u

d

f

i

d

d

C f

( fffiiifC ududmm

22

44 )

5

f f i

C C f

P f d

P f d

(11-25)

ln

2

s

f

f

f i f

f

kr

d d d

k

(11-26)

195

(11-10)

(3m/s)

- (Courtesy of Consolidated Restoration

Systems Inc.)

196

()

197

Ground water

Cooling water flow

Return cooling water Surface water

Normal

Cooling tower

ReversePump

Heat exchanger

198

(oxidization techniques)

(thermal treatments)

(irradiation techniques)

(ion deposition techniques)

199

Two-phase Flow & heat transfer

200

Wavy

flow

Plug

flow

Bubbly

flow

x = 0

Single

phase

liquid

Slug

flow

Annular flow

Intermittently dry

x = 1

Tube wall dry

Single Phase Liquid

Flow

Bubbly

x=0

Flow

Annular

Drop Flow

Slug Flow

x=1

Vapour

Single Phase

202

stratified flow (wavy flow)

203

intermittent flow

(slug flow)

204

annular flow

entrainment

205

A m2 AG m

2

AL m2

AG/A m =

G Lm m kg/s

Gm kg/s

Lm kg/s

i (enthalpy) kJ/kg iG (enthalpy) kJ/kg iL (enthalpy) kJ/kg ifg (latent heat) kJ/kg x = /Gm m

xth = (i iL)/ifg G = GG + GL = /m A kg/m

2s

GG = Gx kg/m2s GL = G(1 x) kg/m2s X Martinelli

(dPL/dPG)0.5

uG m/s uL m/s

206

qualityx() x

207

Question, ? (ifg)

mr

inx

L

out

id

x

= Q

208

(void fraction)

A=A +A ,

LA

AG

=G L

GA

A

AG

12 311

1

n nn

GLB

G L

xB

x

209

Martinelli parameter X

LL

L

L

L

LLL

xGxGd

d

L

xG

d

LGf

d

LP

2

110791.0

4

2

1Re0791.0

4

2

4

225.0

225.0

2

0.252 22

0.254 4 40.0791Re 0.0792 2 2

GG G G

G G G G

Gx GxL G L L GdxP f

d d d

0.520.25

0.5

0.25 2

11

2

2

L LL

G

G G

G xGd x

PX

P GxGdx

0.125 0.50.8751 GL

G L

x

x

(dPL/dPG)0.5

210

-

Cooper (1984)

0.67 0.5 0.55

10

0.67 0.5 0.55

10

55 ( log ) ( )

90 ( log ) ( )

m

r r

o m

r r

q M P Ph

q M P P

100.12 0.2log pm R

q(W/m2)Pr

reduced pressureRp

(m)M

M

kg

kmole

Pcr i t

bar

hre f (W/m2K)

Pr = 0.1

qre f = 20000 W/m2

Fk Kandlikar

R-11 137.37 44.7 2690 1.3

R-12 120.91 41.8 3290 1.5

R-13 104.46 38.7 3910 -

R-113 187.38 34.6 2180 1.3

R-114 170.92 32.5 2460 1.4

R-22 86.47 49.9 3930 2.2

R-134a 102.03 40.7 3500 1.63

R-123 152.93 36.6 2600 -

R-404A 97.6 37.8 - -

R-502 111.6 40.8 - -

R-410A 72.56 48.5 4400 1.4

R-407C 86.2 46.5 - -

R-125 120.02 36.3 - -

R-32 52.02 57.95 - -

211

(Boiling Curve)

212

R-22512 C6

R-22 Pcrit = 49.9 bar

=2.5 kg/s

= 4180 J/kg K

=12 C,

water,in

= 6 m

=

16m

m

water,out

= ? C

=5 C

o

=

14m

mi

c

T

d

d

T

L

R-22

.p

213

-NTU method

(1) 7C

(2) Q = 2.54180(127) = 52.25 kW = 52250 W

(3) q = Q/Ao

= Ao = doLN = 0.01665 = 1.508 m2

q = 52250/1.508 = 34649 W/m2

(4) 55.0105.067.0 )log(90 r

m

ro PPMqh

5CR-225.83 bar

49.9 bar

Pr = 5.83/49.9 = 0.117

Rp = 1.0 m

12.0log2.012.0 10 pRm

ho = 90346490.6786.470.5(0.117)0.12(0.932)0.55 = 8559 W/m2K

(5)

2

32

28.3Reln58.1

1Pr2

7.1207.1

Pr1000Re2

b

b

bbii

f

f

f

k

dhNu

d i = 0.014 mL = 6 m kg/s 5.2totalm

= m = 2.5/5 = 0.5 kg/s

214

222 m 000154.0)014.0(44

ic dA

G = cAm / = 0.5/0.000154 = 3247 kg/m2s

9.5Ck f 0.585 W/mK

f 1350106

Ns/m2Prb 10

33670101350

014.03247Re

6

f

iGd

0057474.028.3Reln58.1 2 bf

f /2 = 0.002874

2 3

2

0.667

Re 1000 Pr2

1.07 12.7 Pr 12

0.002874 33670 1000 100.585 11052 W/m K

0.014 1.07 12.7 0.002874 10 1

b bf f

i

i ib

fk k

h Nufd d

(6) 1/UA

1/UA 1/h iA i +1/hoAo

= A i = d i LN = 0.01465 = 1.32 m2

1/UA 1/h iA i +1/hoAo = 1/11052/1.32+1/8559/1.508 = 0.000146 K/W

UA = 6849.3 W/K

215

(7)

W/K1045041805.2 minC

65543.010450

3.6848

minC

UANTU

0* max

min

C

CC

4807.0)exp(1 NTU

W73150)512(10450 ominmax CQ

W8.35165731504807.0 maxQQ

C 63.810450/8.3516512, outcT

(8) 8.63C7C

-NTU

7C ho

-NTU

(9) 8.5C (1)(8)

Q = 2.54180(12 8.5) = 36575 W

q = Q/A = 36575/1.508 = 24253 W/m2

ho = 90242530.6786.470.50.117)0.12(0.932)0.55 = 6739.7 W/m2K

10Ch i 11000 W/m2K

1/UA 1/h iA i + 1/hoAo = 1/11000/1.32 + 1/6739.7/1.508 =

0.0001672 K/W

UA 5979 W/K

NTU = 5979/10450 = 0.572

= 1 eNTU = 0.4357

Q = Qmax = 0.435773150 = 31869 W

Tc ,out = 12 31869/10450 = 8.95C

8.5C

9.0C

216

-

(nucleate boiling)(forced convective evaporation)

217

(superposition model)

Chen (1966)

ChenhNBhCV

hCVhL

q = qNB + qCV (4-11)

q = h(Tw Ts) (4-12)

qNB = hNB(Tw Ts) (4-13)

qCV = hC V(Tw Ts) (4-14)

h = hNB + hCV

h = ShNB+EhL

S(suppression)E

(enhancement)Chens model

218

E = 2.35/(1/X t t +0.213)0.736

(4-17)

S = 1/(1+2.53106Re1.17) (4-18)

Re = ReLE1.25

(4-19)

ChenE S

Gungor amd Winterton (1986)3700

E =1+24000 Bo1.16

+1.23X t t0.86

(4-20)

S = (1+0.00000115E2ReL

1.17)1

(4-21)

0.125 0.50.8751 GL

tt

G L

xX

x

(4-22)

(boiling number)fg

qBo

Gi (4-23)

219

Chens correlation R 22q = 10

kW/m2K Ts = 5 C x = 0.5G = 200 kg/m

2sdi = 13 mmkL = 94 mW/(mK)L = 199 PasG

= 12 Pas L = 1265 kg/m3G = 25 kg/m

3PrL = 2.51ifg = 200 kJ/kgKP = 583 kPaPcrit = 4990

kPa ()Pr = 583/4990 = 0.1168 (reduced pressure) h = S hNB + E hL

(SI )

ReL = Gd i(1 x)/L = 2000.013(1 0.5)/0.000199 = 6533

hLDittus-Boelter 1-8

Nu = 0.023Re0.8

Pr0.4

hL = 0.094/0.0130.02365330.82.510.4 = 251.6 W/m2K

hNB Cooper (4-6) (Cooper

9055) 55.0

105.067.0 )log(55 r

m

ro PPMqh

ho = 55100000.6786.470.50.1170.120.9320.55 = 2275 W/m2

4-2-1

SE

0.125 0.50.875 0.875 0.125 0.51 1 0.5 0.000199 1265

10.10.5 0.000012 25

L Gtt

G L

xX

x

E = 2.35/(1/X t t + 0.213)0.736

= 0.696

Re = ReLE1.25

= 65330.6961.25 = 4153.5

S = 1/(1+2.5310 -6Re1.17) = 0.958

h = ShNB + EhL = 0.9582275 + 0.696251.6 = 2355.7 W/m2K

220

(enhanced model)

( )cnL

hf

h

(asymptotic model)

qNB qCV

q = qNB + qCV

n n nNB CVq q q (4-40)

n n nNB CVh h h (4-41)

221

q = qnb + qcv

h = hnb + hcv

222

-

Nusselt (1916)

R

Tw

o

g

d = R2

13 4( )

0.728( )

L L G fg oc oc

L L s w L

gi dh dNu

k T T k

41

3

4/3

4/1

41

3

41

341

3

)(728.0

)(728.0

)(

)(728.0

)(

)(728.0

oL

LfgGLL

c

c

oL

LfgGLL

owscL

cLfgGLL

owsL

LfgGLL

c

qd

kgih

hqd

kgi

dTTh

hkgi

dTT

kgih

13 3( )

0.655L L G fg L

c

L o

gi kh

qd

(4-49)

223

4-4-146C38C19 mmR-134a

(kL = 72.2 mW/mKL = 1120

kg/m3G = 59.21 kg/m

3 i fg =156.67 kJ/kgL = 165.7 Pas)

4-4-1

K W/m1664019.0/8.4370722.0

8.4370722.0)3846(0001657.0

019.0156670806.921.5911201120728.0

2

41

3

c

o

Lc

c

Nuk

h

Nu

d

224

waterm

Twater , in ( )

(1) Twater ,outh i

LMTD

(2)

inwateroutwaterwaterpwater

TTcmQ,,,

(3) q = Q/A

(4) 3

13

655.0

oL

LfgGLL

cqd

kgih

hc

(5) hcU

(6) UALMTD = Q

(7) (6) Q (2) (1)-(6)

225

Splashing,

TurbulenceIdealized Model

Model

Side-DrainageNusselt

(a)

Nusselt

(b)

Ripples,

(c)

Vapor

Shear

High

(d)

226

NhN,m

, 1 4

1

N mhN

h

(4-54)

Kern (1958)Nusselt

Kern

5 65 6

1

1Nh

N Nh

(4-55)

N hN,m

, 1 6

1

N mhN

h

(4-56)

227

Nusselt

Butterworth (1977)

(4-13)

1 2

1 2

1 20.416 1 1 9.47

Re

c

G

NuF

(4-50)

2o L fg

G L s w

gd iF

u k T T

(4-51)

Re L G oGL

u d

(4-52)

uGL

oGLG

du

eR

~

4-50

0.0010.1

0.01 0.1

1.0~N

u

10

10001.0

F

10 100

Data

Nusselt Theory (4-49)

Re

-1/2

4-13

228

- Shah (1979)

3.8

1c Lh hZ

(4-60)

0.8

0.41r

xZ P

x

(4-61)

0.8

0.4(1 )0.023 PriLLi L

G x dkh

d

(4-62)

hc local

(x = 1) (x = 0)

, 0.38

2.090.55c m L

r

h hP

(4-63)

ShahPrPrandtl numberPr reduced

pressure

229

P = Pa + Pf + PgPaPfPg

22 2 sin1

(1 / )

fgTP Lfg

i L L fg L

vdP dx f G v gG v x

dz dz d v v xv v

Blasius

f = 0.0791Re0.25

230

Homogeneous Model

Separated Flow Model

231

ReRe = Gd i/T P

TPTP

(x = 0TP = Lx = 1TP = G)TP

11

TP G L

xx

(4-67)

TP = (1 x)L + x G (4-68)

1 LG

TP

G L

xx

(4-69)

1

1

G L

xx

(4-70)

232

7 mm 0.5mhomogeneous modelL = 998.3 kg/m

3G = 1.098 kg/m3L = 0.0661

N/mL = 0.00046 PasG = 0.0000203 Pas

m = 0.012 kg/s

m = 0.003 kg/s

L = 0.5 m

d = 0.007 miG

L

kg/s 015.0012.0003.0 LG mmm

x = 0.003/(0.003 + 0.012) = 0.2

Ac = d i2

/4 = 3.848105

m2

G = m /Ac = 390 kg/m2s

31 1 5.47 kg/m

1 1 0.20.2

1.098 998.3G L

xx

1 1

0.0000863 Pa s1 1 0.20.2

0.0000203 0.00046

TP

G L

xx

84.316330000863.0

007.0390Re

TP

iTP

GD

turbulent

fm = 0.079ReTP0.25

= 0.00593

kPa 56.2347.52

390

007.0

00593.05.04

2

4 22

hom

G

D

LfdP m

233

GLLO

22

21

sin 11

G L

fG L

xdP d x dPG g

dz dz dz

2

,

f

Gf G

dP

dzdP

dz

(4-72)

2

,

f

Lf L

dP

dzdP

dz

(4-73)

2

,

f

LOf LO

dP

dzdP

dz

(4-74)

234

dP f

dPLO2LO

(dP f =2LO dPLO)dPLO

2LO

235

Friedel (1979) 2 2 3

1 0.045 0.035

3.24LO

A AA

Fr We (4-75)

2 2

1 1L GO

G LO

fA x x

f

(4-76)

fLO fGO

Blasius

0.2240.78

2 1A x x (4-77)

0.91 0.19 0.7

3 1G GL

G L L

A

(4-78)

2

2

GFr

gd (4-79)

2G dWe

(4-80)

1

1

G L

xx

(4-81)

236

7 mm 0.5mFriedelL = 998.3 kg/m

3G = 1.098 kg/m3

L = 0.0661 N/mL = 0.00046 PasG = 0.0000203 Pas kg/s 015.0012.0003.0 LG mmm

x = 0.003/(0.003 + 0.012) = 0.2

Ac = d i2/4 = 3.84810

5 m

2

G = m /Ac = 390 kg/m2s

ReG = Gd i/G = 1.34105

ReL = Gd i/L = 5.93103

fGO = 0.0791ReG0.25

= 0.00413

fLO = 0.0791ReL0.25

= 0.00901

A1 = 17.31

A2 = 0.271

A3 = 263.8 3kg/m 47.5

Fr = 7.41104

We = 2.94103

1.12324.3

035.0045.032

12

WeFr

AAALO

PLO = 4L/d ifLOG2/2/L =195.9 Pa

kPa 1.249.1951.1232 LOLOf PP

m = 0.012 kg/s

m = 0.003 kg/s

L = 0.5 m

d = 0.007 miG

L

25%50%Webb and Kim (Principles of enhanced heat transfer, 2005)

237

239

(Conti.) (ohoAo ~ ihiAi)

240

(absolute)

Air-cooled 500 bar ()

600 C

()

5 ~350 m2 (

)

16~25 bar (40 bar)

-25~ 200C.

1 ~to 1200 m2

1 bar ~ 600C.

300 bar () 1400 bar

().

100 to 600C

()

0.25 ~ 200 m2 per

unit multiple units are often used.

High thermal efficiency, standard

modular construction.

Heat-pipe ~ 1 bar 200C

Low pressure

gases.

100 ~ 1000 m2.

Plate-fin

100 bar

() 200 bar

()

273~150C () ~ 600C

()

Low fouling. 9 m3

Very small possible. Incorporation

of multiple streams. Very large

surface area per unit volume. DT

Printed-circuit 1000 bar 800C ()

Low fouling 1 to 1000 m2 Very large surface area per unit

volume. Stainless steel or higher

alloys normal construction

material.

Rotary regenerators ~ 1 bar 980C. Low pressured gases.

Inter-stream leakage must be

tolerated

Shell-and-tube 300 bar (). 1400 bar

().

25 ~ 600C ()

10 to 1000 m2 (per

shell multiple shells can be used).

Very adaptable and can be used for

nearly all applications.

Spiral

18 bar ~ 400C Subject only to

materials of

construction.

Often used for

fouling duties.

~ 200 m2. High heat transfer efficiency.

Cylindrical geometry useful as

integral part of distillation tower.

241

() 15.88, 19.05 25.4 mm 19.05 mm 6.35 mm250~1200 0.9~2.4 m/s(3 ~ 8 ft/s, )0.6~1.5 m/s (2~5 ft/s) (Pt/dopitch ratio)1.25~2.01.25(Pt do)1/8 (3.175 mm) 1/4 (6.35 mm)

242

304560903060304590

45o90o

Roated Triangular PitchTriangular Pitch

Pitch

Rotated Square Pitch Square Pitch

Pitch

45

tP tPt

P

tP

P

P = PX t

PP

P

XP = PtXP = Pt

P = PX t

X

X

X

XPp

Pn

P = PP = P

p tP

P = P P = Pt p tp t

P = PP = P

nn t tn

P = PP = Pn t

Pp

Pitch nP n

P

nP

pP

Pitch

1.732

0.866

0.5

0.707

0.707

1.4140.5

0.866t

(30 ) (45 ) (60 ) (90 )

243

/5~10

,

,(stationary)

flangeChannel

Window

Shell flange

Tube sheet

Orifice

Segmental baffle

Drilling

Disc and doughnut baffle

Orifice baffleBaffle

Doughnut

Doughnut

O.D. of tubes

Disc

244

1/5~1 shell dia.

2/5~1/2 shell dia.

2 in. or 1/5 shell dia. ()

245

20~49%

20~25%

45~50%

25%

window()

Baffle

bc

L

Eddies

bL

Main flow

Main flow

Eddies

246

Rod baffle

Q/P

Plate Baffle Example Rod Baffle Example

248

Impingement baffle

249

E shell,

()(effectiveness, why?)

F shell (longitudinal baffle)(why?)

G & H shell : Pphase change

J shell E shell 1/8

X shell ()

251

Shell < 100 mm

(Leakage & bypass loss)

Guideline (ESDU 92013): 25%

252

Highly flexible (), sealing may be a problem.

Maximum temperature/pressure constraints.

()

253

Cross floweffectiveness Cooling tower Induced draft fan A-frame Steam condensation

1 in. with 2 in. wound fins. Fin pitch ~ 11 fins/in.

2~ 5 m/s. Row number: 3~8.

254

() Maximum Pressure

Temperature range

Fluids limitation

Size range available

Fouling & cleanability

Plot area available

Design life

Location (maintenance)

Is there a temperature cross?. If so, HX approaches counter flow is more appropriate

255

Extended surfaces

Which is better?

(wavy)

(louver)

(louver)

(slit)

(slit)

convex

louver

256

Recap & Quick overview

Q = UATm

Q :

A :

Tm:

257

1.

2. (LMTD)

3. 4.

(Pumping power)

259

()

260

Shell side boiling Shell side condensation

261

Microfin tube

Benefits: increase 100% more heat transfer coefficients with

only 10~50% increase of pressure drops

262

1970 1980 1990 2000

12.7mm(1/2) 9.52mm

(3/8) 7.94mm

(5/16)

6.35mm (1/4 )

5mm,4mm

7mm

Fig.

264

-NTU

NTU

()

12

95%

265

NTU

1 2

air,T

air,T

air,T

Tair,

air,T

T

air,T

air,T

air,

air,T

7

From ESDU

98004

266

267

P

h

(V )

P ~V1.75

0.8h ~ V

268

(Performance evaluation criteria)

(LMTD) (pumping power)

269

(1) Handbook of heat exchanger design (HEDH)

(2) VDI Heat Atlas

(3) Principles of enhanced heat transfer

(4) Fundamentals of heat exchanger design

(5)

Thank you

270

thermal design of heat exchangers

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