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CHAPTER 6
SAMPLE CALCULATIONS FOR SHELL-AND-TUBE HEAT
EXCHANGERS AND GASKETED-PLATE HEAT EXCHANGERS
6.1 ANALYSIS OF SHELL-AND-TUBE HEAT EXCHANGER
6.1.1 Objective-Process
Distilled water with a mass flow rate of 50 kg/s enters a baffled shell-and-
tube heat exchanger at 32ºC and leaves at 25ºC. The heat will be transferred to
150 kg/s of raw water coming from a supply at 20ºC. It is required to design the
heat exchanger for this purpose. A single-shell and single-tube pass is preferable.
The tube diameter is ¾ in. (19-mm O.D. with 16-mm I.D.) and tubes are laid out
on 1-in. square pitch. The maximum length of the heat exchanger is 8 m because
of space limitations. The tube material is 0.5 Cr alloy (k=42.3 W/m.K). Assume
total fouling resistance of 0.000176 m2.K/W. Note that the surface over design
should not exceed 30%. The maximum flow velocity through the tube is also
suggested to be 2 m/s to prevent erosion. Assume baffle spacing of 0.5 m.
Perform thermal and hydraulic analysis of the heat exchanger.
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6.1.2 Process Specifications
Tube-Side Data
Cold Fluid (raw water)
Inlet temperature (ºC) 20
Mass flow rate (kg/s) 150
Density (kg/m3) 998.20
Thermal conductivity (W/m.K) 0.6044
Dynamic viscosity (N.s/m2) 9.832 x 10-4
Specific heat (J/kg.K) 4181.6
Prandtl number 6.808
Shell-Side Data
Hot Fluid (distilled water)
Inlet temperature (ºC) 32
Outlet temperature (ºC) 25
Average temperature (ºC) 28.5
Mass flow rate (kg/s) 50
Density (kg/m3) 996.41
Thermal conductivity (W/m.K) 0.6144
Dynamic viscosity (N.s/m2) 8.292 x 10-4
Specific heat (J/kg.K) 4178.7
Prandtl number 5.641
6.1.3 Design of Shell-and-Tube Heat Exchanger
6.1.3.1 Sample Calculation by Kern Method
As given in section 6.1.1, a single-shell and single-tube pass, shell-and-
tube heat exchanger, where tubes are laid out on 1-inch square pitch, is to be
designed. The hot fluid (distilled water) is in the shell and the cold fluid (raw
water) is flowing in the tubes.
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Properties of hot water (shell-side):
Water properties are calculated at ( ) 5.2822532 =+ ºC and are given in section
6.1.2.
Properties of cold water (tube-side):Water properties are calculated at 21 ºC and are given in section 6.1.2.
Tube:
31016 −×=id m tube inside diameter
31019 −×=od m tube outside diameter
1= p N number of tube passes
The tube material is 0.5 Cr alloy:
3.42=t k W/m.K tube thermal conductivity
Tube layout:
Tubes are laid out on 1-inch square pitch:
21054.2 −×=T P m
Total fouling:
000176.0= ft R m2.K/W
TUBE-SIDE CALCULATIONS:Number of tubes:
From Eq. (4.2),
692.373016.022.998
11504422=
×××
××==
π π ρ icc
pc
t d u
N m N
Number of tubes is rounded off as 374.
374=t N tubes
Tube-side Reynolds number:
From Eq. (4.3),
4
410249.3
10832.9
016.022.998Re ×=
×
××==
−c
icct
d u
µ
ρ
Since , the flow is turbulent.10000Re >t
Tube-side Nusselt number:
The Nusselt number can be calculated from Eq. (4.4) as
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−
+
=
1Pr 2
7.1207.1
Pr Re2
3
22
1
c
ct
t
f
f
Nu
where can be calculated from Eq. (4.5) as f
( ) 3210797.528.3Reln58.1 −−
×=−= t f
Therefore,
552.225
1808.6
2
10797.57.1207.1
808.610249.32
10797.5
3
22
13
43
=
−
×+
×××
×
=−
−
t Nu
Tube-side heat transfer coefficient:
From Eq. (4.7),
2.8520016.0
6044.0552.225 =×==
i
ct i
d
k Nuh W/m2.K
SHELL-SIDE CALCULATIONS:
The total flow area through the tubes is:
0752.0374016.044
22 =××== π π t it N d A m
2
Shell inside diameter:
The tube pitch ratio is given by Eq. (4.11) as
337.1019.0
0254.0===
o
T
d
P PR
From section 4.2.2.1, the tube count calculation constant for one-tube pass is
93.0=CTP
and the tube layout constant for square pitch is
0.1=CL
Then the shell diameter is estimated from Eq. (4.12) as
( ) ( )2
122
2
122 374337.1019.0
93.0
0.1637.0637.0 ×××××== π π t o s N PRd
CTP
CL D
575.0= s D m
Shell inside diameter is rounded off as 580 mm.
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580.0= s D m
Shell equivalent diameter:
From Eq. (4.15), the shell equivalent diameter for square pitch is:
0242.0019.0
4019.00254.04
44
22
22
=×
×−
=
−
=π
π
π
π
o
oT
ed
d P
D m
The baffle spacing is assumed to be
5.0= B m
From Eq. (4.19), the tube clearance is:
3104.6019.00254.0 −×=−=−= oT d P C m
The bundle cross flow area is calculated from Eq. (4.18) as
0731.05.0104.60254.0
580.0 3 =×××== − BC P
D A
T
s s m2
Shell-side Reynolds number:
Shell-side Reynolds number is calculated from Eq. (4.21) as
4
410999.1
10292.8
0242.0
0731.0
50Re ×=
××=
=
−h
e
s
h s
D
A
m
µ
Therefore, the flow is turbulent.Shell-side heat transfer coefficient:
Shell-side heat transfer coefficient is calculated from Eq. (4.22) as
( ) 3
155.043
1
55.0 641.510999.10242.0
6144.036.0Pr Re
36.0×××
×== h s
e
ho
D
k h
9.3769=oh W/m2.K
Overall heat transfer coefficient:
Clean overall heat transfer coefficient based on the outer surface area is calculated
from Eq. (4.23) as
11
3.42
016.0
019.0ln
2
019.0
2.8520
1
016.0
019.0
9.3769
1ln
2
11
−−
+×+=
++=t
i
o
o
ii
o
o
ck
d
d
d
hd
d
hU
2.2256=c
U W/m2.K
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Fouled overall heat transfer coefficient based on the outer surface area is
calculated from Eq. (4.24) as
1
ln
211
−
+++=t
i
o
o ft
ii
o
o
f k
d
d
d Rhd
d h
U
1
4
3.42
016.0
019.0ln
2
019.01076.1
2.8520
1
016.0
019.0
9.3769
1
−
−
+×+×+= f U
9.1614= f U W/m2.K
Outlet temperature calculation:
The cold water outlet temperature is calculated from Eq. (4.25) as
( ) ( )33.295293
6.4181150
2983057.4178501
21
2 =+×
−××=+
−= c
pcc
hh phh
c T cm
T T cmT
K
LMTD calculations:
For counter flow:
67.933.295305211 =−=−=∆ ch T T T K
5293298122 =−=−=∆ ch T T T K
From Eq. (4.27), the log-mean temperature difference is:
08.7
5
67.9ln
567.9
ln2
1
21 =
−=
∆
∆
∆−∆=
T
T
T T LMTD K
From Eq. (3.25),
19.0293305
29333.295
11
12 =−
−=
−
−=
ch
cc
T T
T T P
From Eq. (3.26),
00.329333.295
298305
12
21 =−
−=
−
−=
cc
hh
T T
T T R
Therefore, the correction factor is assumed to be as LMTD
94.0= F
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The mean temperature difference is therefore
66.608.794.0)( =×==∆ LMTD F T m K
Heat balance for fluids:
The heat duty of the heat exchanger is:
( )( ) ( )( ) 6
21 10463.12983057.417850 ×=−×=−= hh phh T T cmQ W
From Eq. (4.26a), the surface area of the heat exchanger for the fouled condition
is:
09.13666.69.1614
10463.1 6
=×
×=
∆=
m f
f T U
Q A m2
Similarly, from Eq. (4.26b), the surface area of the heat exchanger for the clean
condition is:
41.9766.62.2256
10463.1 6
=×
×=
∆=
mc
cT U
Q A m2
The over surface design is:
40.1==c
f
A
AOS (40%)
The over surface design may not be more than 30% for economical reasons.
Assume 20% over surface design; then cleaning scheduling must be arranged
accordingly:
1.188020.1
2.2256
20.1=== c
f
U U W/m2.K
From Eq. (4.24), the corresponding total resistance will be:
510865.82.2256
1
1.1880
111 −×=−=−=c f
ft U U
R m2.K/W
For 20% over surface design, the surface area of the heat exchanger becomes:89.11641.9720.120.1 =×== c f A A m2
Length of the heat exchanger:
From Eq. (4.30), the length of the heat exchanger is:
24.5019.0374
89.116=
××==
π π ot
f
d N
A L m
Length of the heat exchanger is rounded off as 5.5 m.
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5.5= L m
Shell diameter:
Shell diameter can be calculated from Eq. (4.31) as
5.02
5.02
5.5019.034.189.116
93.00.1637.0637.0
×××=
=
Ld PR A
CTP CL D
o f
s
56.0= s D m
Shell diameter is rounded off as 0.60 m.
62.23m60.0 == s D in
Tube-side pressure drop:
The tube-side pressure drop can be calculated from Eq. (4.32):
24
4 2
cc p
i
pt
t
u N
d
LN f p ρ
+=∆
where ( )( ) ( )( ) 324210797.528.310249.3ln58.128.3Reln58.1 −−−
×=−××=−= t t f
and 1= p N
423
1039.22
22.99814
016.0
15.510797.54×=××
×+
××××=∆
−
t p Pa
9.23=∆ t p kPa
Shell-side pressure drop:
The shell-side pressure drop can be calculated from Eq. (4.33):
( )
seh
sb s s s
D
D N G f p
φ ρ 2
12 +=∆
where the friction factor is calculated from Eq. (4.34) as s f
( )( ) ( )( ) 271.010999.1ln19.0576.0expReln19.0576.0exp 4 =××−=−= s s f
The bundle cross flow area is calculated from Eq. (4.18) as
0756.05.0104.60254.0
6.0 3 =×××== − BC P
D A
T
s s m
2
The shell-side mass velocity is:
46.6610756.0
50===
s
h s
A
mG
kg/m2.s
The wall temperature is:
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83.2972
298305
2
33.295293
2
1
222
1 2121 =
++
+×=
++
+= hhcc
w
T T T T T K
The dynamic viscosity at the wall temperature is:
410001.9 −×=w
µ kg/m.s
989.010001.9
10292.814.0
4
414.0
=
×
×=
=
−
−
w
h s
µ
µ φ
The number of baffles is:
1015.0
5.51 =−=−=
B
L N b
Hence, the pressure drop in the shell-side is:
( ) 4
2
1064.1989.00242.041.9962
60.011046.661271.0 ×=×××
×+××=∆ s p Pa
4.16=∆ s p kPa
Tube-side pumping power:
Assume pump efficiency as 0.80.
80.0= pη
Tube-side pumping power is:
341049.4
80.02.998
1039.2150 ×=×
××=∆= pc
t ct
pm P η ρ
W
49.4=t P kW
Shell-side pumping power:
Assume pump efficiency as 0.80.
80.0= pη
Shell-side pumping power is:
34
1003.180.041.996
1064.150×=
×
××=
∆=
ph
sh s
pm P
η ρ
W
03.1= s P kW
The selected shell-and-tube heat exchanger for this purpose has the following
parameters which will be rated by the Bell-Delaware method:
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Table 6.1 Results of Kern Method
Shell diameter (m) 0.600
Number of tubes 374
Length of the heat exchanger (m) 5.5
Tube outside diameter (m) 0.019
Tube inside diameter (m) 0.016
Baffle spacing (baffle cut 25%) (m) 0.5
Tube pitch (m) 0.0254
Number of passes 1
Tube-side heat transfer coefficient (W/m2.K) 8520
Shell-side heat transfer coefficient (W/m2.K) 3770
Clean overall heat transfer coefficient (W/m2.K) 2256
Fouled overall heat transfer coefficient (W/m2.K) 1880
Tube-side pressure drop (kPa) 23.9
Shell-side pressure drop (kPa) 16.4
Tube-side pumping power (kW) 4.5
Shell-side pumping power (kW) 1.0
According to the given specifications, the size of the heat exchanger was
estimated. Such a heat exchanger may also be available. Then this heat exchanger
can be rated for design by the use of Bell-Delaware method which will be given
next.
6.1.3.2 Rating Problem by Bell-Delaware Method
Input data:
Heat exchanger to be rated by Bell-Delaware method is given in the following
table:
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Table 6.2 Input data in Bell-Delaware method
Tube inside diameter (m) 0.016 Specific heat of the shell-side
fluid (J/kg.K)
4178.7
Tube outside diameter (m) 0.019 Dynamic viscosity of the shell-
side fluid (kg/m.s)
410292.8
−×
Tube pitch (m) 0.0254 Dynamic viscosity of the shell-
side fluid evaluated at wall
surface temperature (kg/m.s)
410001.9
−×
Shell inside diameter (m) 0.600 Thermal conductivity of the
shell-side fluid (W/m.K)
0.6144
Length of the heat exchanger
(m)
5.5 Prandtl number of the shell-side
fluid
5.641
Baffle spacing (m) 0.5 Density of the shell-side fluid
(kg/m³
)
996.41
25% baffle cut as percent of
the shell inside diameter
25 Specific heat of the tube-side
fluid (J/kg.K)
4181.6
Clearance between two
adjacent tubes (m)
3104.6
−×
Dynamic viscosity of the tube-
side fluid (kg/m.s)
410832.9
−×
Inside shell diameter to tube
bundle bypass clearance, for
fixed tube sheet (m) [3]
31015
−×
Thermal conductivity of the
tube-side fluid (W/m.K)
0.6044
Central baffle spacing (m) 0.5 Thermal conductivity of the
tube material (W/m.K)
42.3
Inlet baffle spacing (m) 0.5 Prandtl number of the tube-side
fluid
6.808
Outlet baffle spacing (m) 0.5 Density of the tube-side fluid
(kg/m³)
998.20
Number of passes 1 Velocity of the tube-side fluid
(m/s)
2
Number of tubes 374 Inlet temperature of the hot fluid
(K)
305
Over surface design 0.20 Outlet temperature of the hot
fluid (K)
298
Mass flow rate of the shell-
side fluid (kg/s)
50 Inlet temperature of the cold
fluid (K)
293
Mass flow rate of the tube-
side fluid (kg/s)
150 Outlet temperature of the cold
fluid (K)
295.3
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Shell-side Reynolds number:
From Eq. (4.52), the diameter of the circle through the centers of the tube located
within the outermost tubes is:
( ) ( ) 566.0019.0015.06.0 =+−=+−=obb sctl
d L D D m
From section 4.3.4 (1), for 90º layout,
0254.0, == T eff T P P m
From Eq. (4.53), the cross flow area at the shell centerline with one baffle spacing
is:
( ) ( )
2
,
m079.0
019.00254.00254.0
566.0015.05.0
=
−+×=
−+=
m
oT
eff T
ctl bbm
S
d P P
D L BS
The maximum shell-side cross flow mass velocity is:
46.634079.0
50===
m
s s
S
m M
kg/m2.s
Then the Reynolds number on the shell-side is:
4
41045.1
10292.8
46.634019.0Re ×=
×
×==
− s
so s
M d
µ
Ideal shell-side heat transfer coefficient:
From Table (4.2),
370.01 =a 395.02 −=a 187.13 =a 370.04 =a
203.0Re14.01 4
3 =+
=a
s
aa
The Colburn j-factor for an ideal tube bank from Eq. (4.101) is:
( ) 3395.04
203.0
1 10386.81045.1
019.0
0254.0
33.1370.0Re33.1 2 −−×=××
×=
= a
s
a
o
T
i
d
P a j
The cross flow area at the centerline of the shell for one cross flow between two
baffles is, from Eq. (4.18):
0756.05.0104.60254.0
6.0 3 =×××== − BC P
D A
T
s s m2
From Eq. (4.100), the ideal tube bank-based coefficient is:
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14.03
2
=
w
s
s ps
s
s
s psiid
c
k
A
mc jh
µ
µ
µ
14.0
4
43
2
43
10001.910292.8
10292.87.41786144.0
0756.0507.417810386.8
×
××
×××
×××= −
−
−−id h
4.7232=id h W/m2.K
Segmental baffle window correction factor,c
J :
From Eq. (4.61), the angle intersecting the diameter of the circle through the
centers of the outermost tubes is:
1161002521
566.06.0cos2
10021cos2 11
=
×−××=
−= −− c
ctl
sctl B
D Dθ deg
From Eq. (4.64), the fraction of number of tubes in one baffle window is:
( ) ( )179.0
2
116sin
360
0.116
2
sin
360=
×−=−=
π π
θ θ ctl ctl w F
From Eq. (4.65), the fraction of number of tubes in pure cross flow between the
baffle cut tips is:
642.0179.02121 =×−=−= wc F F
From Eq. (4.79), the segmental baffle window correction factor is:
012.1642.072.055.072.055.0 =×+=+= cc F J
Baffle leakage effect correction factor,l
J :
From Eq. (4.60), the centri-angle of the baffle cut intersection with the inside shell
wall is:
120
100
2521cos2
100
21cos2 11 =
×−×=
−= −− c
ds
Bθ deg
From Eq. (4.50), the inside shell-to-baffle clearance (diametral) is:
333 105.56.0004.0101.3004.0101.3 −−− ×=×+×=+×= s sb D L m
From Eq. (4.76), the shell-to-baffle leakage area is:
( ) ( )120360105.56.000436.036000436.0 3 −××××=−= −ds sb s sb L DS θ
310453.3 −×= sbS m2
From Fig. (4.14), the diametral tube-to-baffle hole clearance for the maximum
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unsupported length greater than 36 in (the maximum unsupported length is greater
than 36 in, from Eq. (4.45) or from Fig. (4.8)) is:
64
1=tb L in = m4
10969.3 −×
From Eq. (4.77), the tube-to-baffle hole leakage area for one baffle is:
( )[ ] ( )
−−+= wt otbotb F N d Ld S 14
22π
( )[ ] ( ) 32 10675.3179.01374019. −×=
−××24 010969.3019.0
4
−
−×+×=π
tbS m2
From Eq. (4.80), the ratio of both leakage areas to the cross flow area is:
090.0
079.0
10675.310453.3 33
=×+×
=+
=−−
m
tb sblm
S
S S r
From Eq. (4.81), the ratio of the shell-to-baffle leakage area to the sum of both
leakage areas is:
485.010675.310453.3
10453.333
3
=×+×
×=
+=
−−
−
tb sb
sb s
S S
S r
From Eq. (4.82a), the baffle leakage heat transfer correction factor is:
( ) ( )[ ] ( )lm s sl r r r J 2.2exp144.01144.0 −−−+−=
( ) ( )[ ] ( ) 861.0090.02.2exp485.0144.01485.0144.0 =×−×−×−+−×=l J
Bundle bypass effect correction factor, :b
J
The effect of the tube lane partition bypass width is taken as the half of the tube
outside diameter:
3105.92
019.0
2
−×=== o pl
d L m
From Eq. (4.74), the bypass area within one baffle is:
( ) pl otl sb L D D BS +−=
where 585.0019.0566.0 =+=+= octl otl d D D m
( ) 0123.0105.9585.06.05.0 3 =×+−×= −bS m2
From Eq. (4.75), the fraction of the bypass area to the overall cross flow area is:
155.00788.0
0123.0===
m
b sbp
S
S F
From Table (4.1), the length for square pitch layout is: p P
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0254.0== T p P P m
From Eq. (4.70), the number of effective rows crossed in one cross flow section,
that is, between the baffle tips, is:
811.111002521
0254.06.0
10021 =
×−×=
−= c
p
stcc
B P D N
ss N is the number of sealing strips (pairs) in one baffle and it is assumed to be 1:
1= ss N
From Eq. (4.83), the ratio of the number of sealing strips (pairs) in one baffle to
the number of tube rows crossed between baffle tips in one baffle section, is:
0847.0811.11
1===
tcc
ss
ss N
N r
From Eq. (4.84), the correction factor for bundle bypass is:
)3 21exp ss sbpbhb r F C J −−=
where for turbulent flow. Therefore,25.1=bhC
) 917.00847.021155.025.1exp 3 =×−××−=b J
Correction factor for variable baffle spacing at the inlet and/or outlet
sections, :s J
The inlet baffle spacing and the outlet baffle spacing are equal to the central baffle
spacing:
oi B B B ==
From Eqs. (4.92) and (4.93), the dimensionless length ratios are:
B
B L i
i = and B
Boo = L
The number of baffles is:
1015.0
5.51 =−=−=
B
L N b
From Eq. (4.95), the correction factor for unequal baffle spacing at inlet and/or
outlet, is:
( ) ( ) ( )
( ) oib
n
o
n
ib s
L L N
L L N J
++−
++−=
−−
1
1 11
where 6.0=n for turbulent flow. Therefore,
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( ) ( ) ( )
( )0.1
11110
11110 6.016.01
=++−
++−=
−−
s J
Correction factor for adverse temperature gradient in laminar flow,r
J :
From Eq. (4.72), the effective number of tube rows crossed is:
189.42
566.06.0
100
256.0
0254.0
8.0
2100
8.0=
−−
××=
−−
= ctl sc
s
p
tcw
D D B D
P N
From Eq. (4.87), the total number of tube rows crossed in the entire heat
exchanger is:
( )( ) ( ) ( ) 176110189.4811.111 =+×+=++= btcwtccc N N N N
Since , the correction factor for adverse temperature gradient in laminar
flow, is:
100Re > s
0.1=r J
Shell-side heat transfer coefficient (Bell-Delaware method):
From Eq. (4.99), the actual shell-side heat transfer coefficient is:
( ) ( ) 57750.10.1917.0861.0012.14.7232 =×××××== r sbl cid o J J J J J hh W/m2.K
Overall heat transfer coefficient:
Clean overall heat transfer coefficient based on the outer surface area is:
11
3.42
016.0
019.0ln
2
019.0
2.8520
1
016.0
019.0
5775
1ln
2
11
−−
×+×+=
++=k
d
d
d
hd
d
hU
i
o
o
ii
o
o
c
Table 6.3 Correction factors in Bell-Delaware method
Correction factors Values
c J 1.012
l J 0.861
b J 0.917
s J 1.0
r J 1.0
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9.2847=cU W/m2.K
1
+=d d
hU
o
f
5775
1
= f U
1.1897= f
U
10463.1 ×=Q
∆=
m f
f T U
Q A
∆=
mc
cT U
Q A
==c
f
A
AOS
2.1== c
f
U U
510023.79.2847
1
3.2373
111 −×=−=−=c f
ft U U
R
Fouled overall heat transfer coefficient based on the outer surface area is:
1
ln
21
−
++k d
d
d Rh
i
o
o ft
ii
o
1
4
3.42
016.0
019.0ln
2
019.01076.1
2.8520
1
016.0
019.0
−
−
×+×+×+
W/m2.K
The heat duty of the heat exchanger and the mean temperature difference were
already calculated as:
6 W and 66.6=∆ mT K
Therefore, the surface area of the heat exchanger for the fouled condition is:
85.11566.61.1897
10463.1 6
=×
×= m2
The surface area of the heat exchanger for the clean condition is:
17.7766.69.2847
10463.1 6
=×
×= m2
The over surface design is:
50.117.77
85.115= (50%)
The over surface design should not be more than 30% for economical reasons.
Assume 20% over surface design; then cleaning scheduling must be arranged
accordingly:
3.23732.1
9.2847= W/m2.K
The corresponding total resistance will be:
m2.K/W
For 20% over surface design, the surface area of the heat exchanger becomes:
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60.9217.7720.120.1 =×== c f A A m2
Length of the heat exchanger:
15.4
019.0374
60.92=
××
==
π π ot
f
d N
A L m
Length of the heat exchanger is rounded off as 4.5 m.
5.4= L m
Tube-side pressure drop:
24
4 2
cc p
i
pt
t
u N
d
LN f p ρ
+=∆
where ( )( ) ( )( ) 324210797.528.310249.3ln58.128.3Reln58.1 −−−
×=−××=−= t t f
Therefore, the tube-side pressure drop is:
423
1010.22
22.99814
016.0
15.410797.54×=××
×+
××××=∆
−
t p Pa
0.21=∆ t p kPa
Shell-side pressure drop (Bell-Delaware method):
From Table (4.2),
391.01 =b 148.02 −=b 30.63 =b 378.04 =b
( )009.1
10454.114.01
30.6
Re14.01378.04
3
4
=××+
=+
=b
s
bb
From Eq. (4.102),
( ) ( ) 0942.010454.1
019.0
0254.0
33.1391.0Re
33.1 148.04
009.1
12 =××
×=
=
−b
s
b
o
T
i
d
P b f
From Eq. (4.104), the pressure drop in an equivalent ideal tube bank in one baffle
compartment of central baffle spacing, is:
811.1110292.8
10001.9
41.9962
46.6340942.04
24
14.0
4
4214.0
2
×
×
××
×××=
=∆
−
−
tcc
s
w
s
sibi N M
f pµ
µ
ρ
98.908=∆ bi p Pa kPa91.0=
From Eq. (4.82b), the correction factor for baffle leakage effects for pressurel R
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drop, is:
( )[ ] p
lm sl r r R +−= 133.1exp
where ( )[ ] ( )[ ] 577.08.0485.0115.08.0115.0 =++×−=++−= sr p
Therefore,
( ) 611.00904.0485.0133.1exp 577.0 =×+×−=l R
From Eq. (4.85), the correction factor for bundle bypass effects for pressure
drop, is:
b R
)3 21exp ss sbpbpb r F C R −−=
where for turbulent flow. Therefore,7.3=bpC
) 773.00847.021155.07.3exp 3=×−××−=b R
The number of baffles is:
815.0
5.41 =−=−=
B
L N b
From Eq. (4.103), the combined pressure drop of all the interior cross flow
sections (baffle tip to baffle tip), is:
( ) ( ) 31001.3773.0611.01898.9081 ×=××−×=−∆=∆ bl bbic R R N p p Pa
01.3=∆ c p kPa
From Eq. (4.63), the gross window flow area, that is, without tubes in the window,
is:
( ) ( )0553.0
2
120sin
360
1206.0
42
sin
3604
22 =
×−××=
−=
π
π
π
θ θ π dsds swg DS m2
From Eq. (4.66), the segmental baffle window area occupied by the tubes, is:
019.0019.04
179.03744
22 =
×××=
=
π π owt wt d F N S m2
From Eq. (4.68), the net cross flow area through one baffle window, is:
0363.0019.00553.0 =−=−= wt wg w S S S m2
From Eq. (4.105),
04.9350363.0079.0
50=
×==
wm
sw
S S
mm
kg/m2.s
From Eq. (4.106), the combined pressure drop in all the windows for turbulent
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flow, is:
( ) ( ) 611.041.9962
04.935189.46.028
26.02
22
×
×××+×=
+=∆ l
s
wtcwbw R
m N N p
ρ
31068.9 ×=∆ w p 68.9=Pa kPa
From Eq. (4.98), the pressure drop correction factor for unequal baffle spacing
at inlet and/or outlet, is:
s R
n
i
n
o
s B
B
B
B R
−−
+
=
22
where 2=n for turbulent flow. Therefore,
25.0
5.0
5.0
5.02222
=
+
=
−−
s R
Since all the baffle spacings are equal, 2= s R .
From Eq. (4.108), the pressure drop in the entrance and exit sections (in the two
end zones), is:
31090.12773.0811.11
189.4198.9081 ×=××
+×=
+∆=∆ sb
tcc
tcwbie R R
N
N p p Pa
90.1=∆ e p kPa
From Eq. (4.109), the total shell-side pressure drop is:
59.1490.168.901.3 =++=∆+∆+∆=∆ ewc s p p p p kPa
Tube-side pumping power:
Assume pump efficiency as 0.80.
80.0= pη
Tube-side pumping power is:
3
4
1094.380.02.998101.2150 ×=
×××=∆=
pc
t ct
pm P η ρ
W
94.3=t P kW
Shell-side pumping power:
Assume pump efficiency as 0.80.
80.0= pη
Shell-side pumping power is:
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8.91480.041.996
10459.150 4
=×
××=
∆=
ph
sh s
pm P
η ρ
W
91.0= s P kW
Results of the rating analysis of the shell-and-tube heat exchanger that was
predicted by the Kern method (or available) are given in the following table:
Table 6.4 Results of the Bell-Delaware Method
Shell diameter (m) 0.600
Number of tubes 374
Length of the heat exchanger (m) 4.5
Tube outside diameter (m) 0.019
Tube inside diameter (m) 0.016
Baffle spacing (baffle cut 25%) (m) 0.5
Tube pitch (m) 0.0254
Number of pass 1
Surface area of the heat exchanger (m2) 92.6
Tube-side heat transfer coefficient (W/m2.K) 8520
Shell-side heat transfer coefficient (W/m2.K) 5775
Clean overall heat transfer coefficient (W/m2.K) 2848
Fouled overall heat transfer coefficient (W/m2.K) 2373
Tube-side pressure drop (kPa) 21.0
Shell-side pressure drop (kPa) 14.6
Tube-side pumping power (kW) 3.9
Shell-side pumping power (kW) 0.9
There are two kinds of rating program. In the first case, the inlet and the
outlet temperatures (3 temperatures) and the mass flow rates are given. Therefore,
the heat transfer (heat load or heat duty) is fixed. Then the length of the heat
exchanger and the pressure drops for both streams are calculated, using log-mean
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temperature difference method, as done in this thesis. In the second case, the heat
duty is not known because only inlet temperatures are given while the outlet
temperatures are not. On the other hand, the heat exchanger length is fixed and
outlet temperatures and the pressure drops are to be calculated. In this case, outlet
temperatures can be calculated by using ε-NTU method. One of the constraints in
this problem may be a required outlet temperature and the pressure drop ∆P. For
instance, if it is a water heater, outlet temperature may be less than 40ºC. The
outlet temperatures must then satisfy this process requirement. Note that in both
cases, the pressure drop requirement must be satisfied. Do not forget that Heat
Exchanger is a part of the existing system, part of a piping circuit!
6.2 ANALYSIS OF GASKETED-PLATE HEAT EXCHANGER
6.2.1 Objective-Process
Distilled water with a mass flow rate of 50 kg/s enters a gasketed-plate
heat exchanger at 32ºC and leaves at 25ºC. The heat will be transferred to 150
kg/s of raw water coming from a supply at 20ºC. The maximum permissible
pressure drop for each stream is 70 kPa. It is required to design the heat exchanger
for this purpose.
6.2.2 Process Specifications
Cold Fluid Data
Cold Fluid (raw water)
Inlet temperature (ºC) 20
Flow rate (kg/s) 150
Density (kg/m3) 998.20
Thermal conductivity (W/m.K) 0.6044
Dynamic viscosity (N.s/m2) 9.832 x 10
-4
Specific heat (J/kg.K) 4181.6
Prandtl number 6.808
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Hot Fluid Data
Hot Fluid (distilled water)
Inlet temperature (ºC) 32
Outlet temperature (ºC) 25Average temperature (ºC) 28.5
Flow rate (kg/s) 50
Density (kg/m3) 996.41
Thermal conductivity (W/m.K) 0.6144
Dynamic viscosity (N.s/m2) 8.292 x 10-4
Specific heat (J/kg.K) 4178.7
Prandtl number 5.641
Constructional Data of the Proposed Plate Heat Exchanger
Plate material SS316
Plate thickness (mm) 0.6
Chevron angle (degrees) 45
Total number of plates 105
Enlargement factor 1.25
Number of passes single pass
Overall heat transfer coefficient (clean/fouled) (W/m2.K) 8000/4500
Total effective area (m2)
Fouling resistance (hot fluid/cold fluid) (m2.K/W)
110
0.00005/0
All port diameters (mm) 200
Compressed plate pack length, Lc, (m) 0.38
Vertical port distance, Lv, (m) 1.55
Horizontal port distance, Lh, (m) 0.43
Effective channel width, Lw, (m) 0.63
Thermal conductivity of the plate material (SS304) (W/m.K) 17.5
After giving the process requirements, this proposed heat exchanger may
be given by the manufacturer; or this heat exchanger may be available. So this
proposed one must be rated for the final design.
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6.2.3 Rating/Design of a Gasketed-Plate Heat Exchanger
The properties of hot water and cold water are given in section 6.2.2.
Since the inlet temperature and the outlet temperature of the hot fluid are known,
from Eq. (5.32), the required heat duty of the gasketed-plate heat exchanger is:
h phhrh T cmQ ∆=
where 729830521 =−=−=∆ hhh T T T K
Therefore,
610463.177.417850 ×=××=rhQ W
The outlet temperature of the cold fluid is therefore:
33.2956.4181150
10463.1293
6
12 =×
×+=+=
pcc
rhcc
cm
QT T
K
From Eq. (5.28), for counter flow arrangement the logarithmic mean temperature
difference is:
∆
∆
∆−∆=∆
2
1
21,
ln
T
T
T T T cf lm
where 5293298121 =−=−=∆ ch T T T K
and 67.933.295305212 =−=−=∆ ch T T T K
Therefore,
08.7
67.9
5ln
67.95, =
−=∆ cf lmT K
The effective number of plates is:
10321052 =−=−= t e N N
Effective flow length between the vertical ports is:
55.1== veff L L m
From Eq. (5.7), the plate pitch is:
31062.3
105
38.0 −×===t
c
N
L p m
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From Eq. (5.6), the mean channel flow gap is:
3331002.3106.01062.3 −−− ×=×−×=−= t pb m
From Eq. (5.8), one channel-flow area is:
33
1090.163.01002.3−−
×=××==wch bL A m
2
The single plate heat transfer area is:
068.1103
1101 ===
e
e
N
A A m
2
From Eq. (5.4),
35.12.055.1 =−=−= pv p D L L m
From Eq. (5.3), the projected plate area is:
851.063.035.11 =×== w p p L L A m2
Enlargement factor has been specified by the manufacturer, but it can be verified
from Eq. (5.2):
256.1851.0
068.1
1
1 === p A
Aφ
From Eq. (5.11), the channel equivalent diameter is:
33
1081.4256.1
1002.322 −−
×=××
==φ
b
De m
From Eq. (5.22), the number of channels per pass is:
5212
1105
2
1=
×
−=
−=
p
t cp
N
N N
Heat transfer analysis:
The mass flow rate per channel is:
96.052
50===
cp
h
ch N
mm
kg/s
From Eq. (5.21), the mass velocity is:
54.5051090.1
96.03=
×==
−ch
chch
A
mG
kg/m2.s
From Eq. (5.20), the hot fluid Reynolds number is:
7.293110292.8
1081.454.505Re
4
3
=×
××==
−
−
h
echh
DG
µ
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Similarly, the cold fluid Reynolds number is:
5.247210832.9
1081.454.505Re
4
3
=×
××==
−
−
c
echc
DG
µ
Therefore, both fluids are in turbulent flow.
From Eq. (5.18), the hot fluid Nusselt number is:
17.0
3
1
Pr Re
=
w
bh
y
hhh C Nuµ
µ
where from Table (5.2), 3.0=hC and 663.0= y . Assuming wb = ,
( ) 24.1061641.57.29313.017.03
1
663.0 =×××=h Nu
From Eq. (5.19), the hot fluid heat transfer coefficient is:
135741081.4
6144.024.1063
=×
×==
−e
hhh
D
k Nuh W/m
2.K
Similarly, the cold fluid Nusselt number is:
( ) 03.1011808.65.24723.0Pr Re17.03
1
663.0
17.0
3
1
=×××=
=
w
bc
y
chc C Nuµ
µ
The cold fluid heat transfer coefficient is:
126981081.4
6044.003.101 3 =×
×== −
e
ccc
Dk Nuh W/m2.K
Let’s calculate the heat transfer coefficients from Eq. (5.14) just to see and
compare the results with the above ones.
[ ] ( )[ ]14.0
3
17.390/2sin0543.0728.025 Pr Re10244.7006967.02668.0
×+−=
++−
w
b
hhh Nuµ
µ β β
πβ
Again assuming wb = ,
[ ]( )[ ] ( ) 14.03
1
7.390/785.02sin0543.0728.0
25
1641.52931.7
785.010244.7785.0006967.02668.0
××
×××+×−=
+×××+
−
π
h Nu
18.121=h Nu
The hot fluid heat transfer coefficient is:
154831081.4
6144.018.1213
=×
×==
−e
hhh
D
k Nuh W/m2.K
Similarly, the cold fluid Nusselt number is:
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[ ] ( )[ ]14.0
3
17.390/2sin0543.0728.025 Pr Re10244.7006967.02668.0
×+−=
++−
w
b
ccc Nuµ
µ β β
πβ
Again assuming wb = ,
( )[ ] ( ) 14.07.390/785.02sin0543.0728.0
25
1641.52472.5
785.010244.7785.0006967.02668.0
××
×××+×−=+×××+
−
π
c Nu
58.114=c Nu
The cold fluid heat transfer coefficient is:
144011081.4
6044.058.1143
=×
×==
−e
ccc
D
k Nuh W/m2.K
The hot fluid and the cold fluid heat transfer coefficients calculated from two
different correlations differ from each other by %14 and %13, respectively (the
second ones are %14 and %13 higher). The first results will be used here.
From Eq. (5.30), the clean overall heat transfer coefficient is:
5356
5.17
106
13574
1
12698
1
1
11
14
=
×++
=
++
=−
platehc
c
k
t
hh
U W/m2.K
From Eq. (5.31), the fouled (service) overall heat transfer coefficient is:
4225
01055356
11
1
1
5
=
+×+
=
++
=−
fc fh
c
f
R RU
U W/m2.K
From Eq. (5.31), the corresponding cleanliness factor is:
79.05356
4225===
c
f
U
U CF
From Eq. (5.33), the actual heat duty for fouled surface is:
6
, 1029.308.71104225 ×=××=∆= cf lme f f T AU Q W
Similarly, the actual heat duty for clean surface is:
6
, 1017.408.71105356 ×=××=∆= cf lmecc T AU Q W
From Eq. (5.34), the safety factor is:
25.21046.1
1029.36
6
=×
×==
r
f
sQ
QC
By replacing U in Eq. (4.28) by U from Eq. (4.29), the percentage over surface f f
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design is:
ft c RU OS 100=
where m55 1050105 −− ×=+×=+= fc fh ft R R R2.K/W
Therefore,
78.261055356100 5 =×××= −OS
The surface area of the gasketed-plate heat exchanger is:
9.4808.74225
1046.1 6
,
=×
×=
∆=
cf lm f
r
T U
Q A m2
Pressure drop analysis:
From Table (5.2), the fluid friction coefficients are:
441.1= p K and 206.0=m
From Eq. (5.24), the hot fluid friction coefficient is:
278.07.2931
441.1
Re 206.0===
m
h
p
h
K f
Similarly, the cold fluid friction coefficient is:
288.05.2472
441.1
Re 206.0===
m
c
p
c
K f
Let’s calculate the fluid friction coefficients from Eq. (5.15) just to see and
compare the results with the above ones.
[ ] [ ]{ }1.290/2sin0577.02.023 Re10016.21277.0917.2++−−×+−=
πβ β β hh f
[ ] [ ]{ }1.290/2sin0577.02.023 7.2931785.010016.2785.01277.0917.2 +×××+−− ×××+×−= β π
h f
389.0=h f
Similarly, the cold fluid friction coefficient is:[ ] [ ]{ }1.290/2sin0577.02.023 Re10016.21277.0917.2
++−−×+−=πβ
β β cc f
[ ] [ ]{ }1.290/2sin0577.02.023 5.2472785.010016.2785.01277.0917.2 +×××+−− ×××+×−= β π
c f
406.0=c f
The hot fluid and the cold fluid friction coefficients calculated from two different
correlations differ from each other by %40 and %41, respectively (the second
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ones are %40 and %41 higher). The first results will be used here.
From Eq. (5.23), the frictional pressure drop from hot and cold streams are
respectively:
( ) 17.03
217.02
141.99621081.4
54.505155.1278.042
4 −−
−
××××
××××=
=∆
w
b
he
ch peff hh
DG N L f p
µ µ
ρ
41060.4 ×=∆ h p Pa kPa0.46=
( ) 17.0
3
217.02
12.99821081.4
54.505155.1288.04
24
−
−
−
××××
××××=
=∆
w
b
ce
ch peff
cc D
G N L f p
µ
µ
ρ
41076.4 ×=∆ c p Pa kPa6.47=
From Eq. (5.26), the mass velocities of the hot fluid and cold fluid in the port are
respectively:
1592
4
2.0
50
4
22=
×
=
=
π π p
h ph
D
mG
kg/m2.s
4775
4
2.0
150
4
22=
×
=
=
π π p
c pc
D
mG
kg/m2.s
From Eq. (5.25), the port pressure drop for the hot fluid is:
322
108.141.9962
159214.1
24.1 ×=
×××==∆
h
ph
p ph
G N p
ρ Pa
8.1=∆ ph p kPa
Similarly, the port pressure drop for the cold fluid is:
422
1060.12.9982
477514.1
24.1 ×=
×××==∆
c
pc
p pc
G N p
ρ Pa
0.16=∆ pc p kPa
From Eq. (5.27), the total pressure drop for the hot fluid is:
8.478.10.46 =+=∆+∆=∆ phhth p p p kPa
Similarly, the total pressure drop for the cold fluid is:
6.630.166.47 =+=∆+∆=∆ pcctc p p p kPa
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Pumping powers:
Assume pump efficiency as 0.80.
80.0= pη
The pumping power for the hot fluid is:
33
100.380.041.996
108.4750×=
×
××=
∆=
ph
thhh
pm P
η ρ
W
0.3=t P kW
The pumping power for the cold fluid is:
43
1019.180.02.998
106.63150×=
×
××=
∆=
pc
tccc
pm P
η ρ
W
9.11=t P kW
Results of the design of the gasketed-plate heat exchanger are given in the
following table:
Table 6.5 Results of the rating analysis of the proposed gasketed-plate heat
exchanger
Heat transfer area (m2
) 110
Length of the heat exchanger (m) 1.5
Total number of plates 105
Plate thickness (mm) 0.6
Number of pass 1/1
Hot fluid heat transfer coefficient (W/m2.K) 13574
Cold fluid heat transfer coefficient (W/m2.K) 12698
Clean overall heat transfer coefficient (W/m2.K) 5356
Fouled overall heat transfer coefficient (W/m2.K) 4225
Over surface design (%) 26.78
Hot fluid pressure drop (kPa) 47.8
Cold fluid pressure drop (kPa) 63.6
Hot fluid pumping power (kW) 3.0
Cold fluid pumping power (kW) 11.9
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The over surface design value shown in Table 6.5 is acceptable. The
calculations show that the proposed unit satisfies the process required and
pressure drop constraint, but a smaller heat exchanger can also be used. There are
2 alternatives to do this: (1) one can fix the size of the Chevron plate, then the
number of the plates will be reduced. (2) the number of the plates can be fixed as
105, but the plate size will be different, then one can select a smaller Chevron
plate.
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