thermal design of heat exchangers
DESCRIPTION
Thermal Design of Heat ExchangersTRANSCRIPT
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Condensed version
Condensed version
, PhD,
Fellow ASME, Fellow ASHRAE
e-Mail [email protected]
Tel: 03-571212 ext. 55105
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UA-LMTD-F/-NTU
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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)
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() - ()
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()
1p
p
mW P
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1 1 1 1
22 2 2
(A)(C)
(B)
(D)
(G)
(F)
(E)
P = P = PA + PB + PC + PD + PE + PF + PG
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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
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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
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Copyright ITRI 2009
8
-Moody Diagram
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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
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()(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,
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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
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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
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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)
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)TT(L
kAQ 21wall,cond
k = thermal conductivity (W/m.C)
(heat conduction)
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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
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.L
TTkA
dx
dTkAQ
L,T
,T
cond21
2
01
(Thermal Resistance)
R
VI
kA
L
TQ
kA
LRthermal
Fouriers Law
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(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
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(heat convection)
)TT(hAQ fsconv
h = heat transfer coefficient (W/m2.C)
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).( bwconvection TThAQ
(Thermal Resistance)
hA
TQ
1
hARthermal
1
q
T
T
w
b
Newtons Law of Cooling
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(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
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(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
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() 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
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()
()
()
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T inwT T >T w in
U in
( )
T T >T ( )ww in
(a)
(b)
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a
c
b
a(b)(c)
14.0)(w
b
b(viscosity)w
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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
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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
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Question 10.5 L/min10T = 20C
Question 230 L/min
L = 10 m
d = 2 cmi
Q = 0.5 & 30 L/min
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-
R=RA+RW+RB
q q=(TA-TB)/R
Q=qA
RW
q
B A TA
RA RB TB
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(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
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(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
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(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
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(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
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(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)
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(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
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Recuperator
Direct Contact Heat Exchanger
Regenerator
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Wall/Interface
Heat flow
Stream A Stream B
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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
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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
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Shell and Tube Heat Exchanger
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(a) brazed type; (b) working principle; (c) cross
section
(a) (b) (c)
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42 Spiral Heat Exchanger
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(Tube-in-Tube Heat Exchanger)
(a)(b)
(c) (d)
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Air-to-air Cross Flow Heat Exchanger
indoor outdoor
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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)
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Hot blast stove Fixed Bed type regenerator
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(a) Working Principle & (b) Rotary wheel
Rotary heat exchanger
Motor
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Cooling Tower
Tray-type direct HX & Spray condenser
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(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
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UA-LMTD-F -NTU
1.
2. (one-dimensional)
3.(Cp)
4. U
5.
6.
7.()
A
B
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(Cont.)
)( 12 iimQ
)()( 1221 ccchhh iimiimQ
)()()()( ,,,,,, ohihhhpicocccp TTCmTTCmQ
hot fluid
cold fluid
chcchhph cmCandcmC )()( ,,
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Copyright ITRI 2009
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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
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(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
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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
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(Cont.)
(3) (crossflow) (pass)
L2
L1 0 x
Tc,i
Tc,o
(4) (mixed flow)
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(Cont.)
() ()1 Pass 2 Pass
()3 Pass ()4 Pass
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57
(Cont.)
(1) U
(2) A
(3) Tm
mTUAQ
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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.)
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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
(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
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(rating problem)
()
(sizing problem) (
) ()
(Energy balance equation)
Tm
U & Tm?
Rating/Sizing?
TCpmQ
mTAUQ
)entialDrivingPotUAQ ,)((
)()()()( ,,,,,, ohihhhpicocccp TTCmTTCmQ
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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
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62
Counter Flow heat exchanger
Fig2.
)()()()( ,,,,,, ohihhhpicocccp TTCmTTCmQ
Ch Cc
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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
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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
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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.)
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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.)
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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
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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
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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
)()(
)()()()(
,,
,,,,,,
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70 What is Qmax(Cont.)?
Case1
hc
hhpccp
TT
CmCm
)()( ,, Cold fluid=minimum fluid
hc
icihccp
CCif
TTcmQ
)()( ,,,max
Cmin
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71 What is Qmax(Cont.)?
Case2
hc
hhpccp
TT
CmCm
)()( ,, hot fluid=minimum fluid
hc
icihhhp
CCif
TTcmQ
)()( ,,,max
Cmin
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72 What is Qmax(Cont.)?
)( ,,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
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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 )()()()( ,,,,,,
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Copyright ITRI 2009
74
A NTU
=Q/Qmax
minC
UANTU =
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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
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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
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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
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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
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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
-
91
&
-
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
-
135
-
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
-
185
-
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)
-
188
-
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
-
201
-
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
-
238
-
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
-
247
-
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 ()
-
250
-
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)
-
258
-
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.
-
263
-
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