2. SHELL AND TUBE HEAT EXCHANGER
20 C
1 Temperature detail:
water in Cold
fluid Hot
fluid 15
c
In let 15
c 97
c Waterout
Outlet 35
c 20
97
O
C
2. Heat load
Hot fluid: Aq. Ammonia Q
4
= m x Cp x ûW
Where m = 1453 kgnr = 3.98 kgsec. Cp = 2.57 KJ kg
∴ Q
H
= 3.98 x 2.57 [97- 20} Q
H
= 799.99 k.j B
a
+ Q
H
= Q
c
= Q = 799KJ
Where Q
H
= heat load of hot fluid Q
c
= heat load of cold fluid.
Cold fluid : Water Q
c
= M
C
x C
p
x. ût
Where M
C
= to be determined. C
p
= Sp heat of water. ∴ 799 = M
C
x 4.184 35.15
M
C
= 9.548 kg sec.
Mass of cold water required to remove the heat associated } = 9.548 kgsec.
2 LMTD Calculation , ût
T
1
= 97°c t
2
= 35°c T
2
= 20°c t
1
= 15°c ûT
lmtd
= 97°-35 – 20- 18 ln 97°-35
20- 15
Correction factor Fr
R = T
1
– T
2
S = t
2
- t
1
t
2
- t
1
T
1
– t
1
R = 97 – 20 S
= 35
– 15 35 – 15
97 – 15 R = 3.85
S = 0.2439
From perry 6
th.
cd. page 10.27 Considering 2-shell pass, 4 table pass i.e., 2.4 exchanger.
Corrected LMTD = 22.63 x 0.8 = 18.10° ûW
lmtd
= 18.10
3 Rounting of fluid:
Cleaner fluid is water -------- Shell side. Unclean fluid is Aq.Ammonia .---Tube side
4 Heat Transfer Area:
Reference perry, page 10-44U
d
for water in shell side, inorganic solvent in tube side is ranging between100-250 BTU F.Ft
2
.hr
Range is = 567.83 – 1419.57 J °C m
2
.s Total Heat Transfer Area HTA = 799 x 10
3
= 73.57 m
2
600 x 18.10 Choose l6 BWG tubes.
OD = 58 = 0.01587 m
ID = 0.495 P
Length of tube = 16ft = 4.8768 Heat transfer area = 0.1636 ft
2
ft
2
length = 0.04986 m
2
m.length. F
T
= 0.8
Heat transfer area of over tube = 0.04968 x 4.8768 =
0.2431 m
2
Tube ∴No of tube = 73.57 = 302
0.2431 Nearest tube count from perry, page 11-13 is 274. and corresponding shell diameter
inner = 438 mm. ∴Shell ID = 438mm.
Corrected heat transfer area = 274 x 0.243 = 66.60m
2
corrected U
d
= 799 x 10
3
66.60 x 18.1 = 662.0 Wm
2.
°k
5 Fluid velocity check
a Tube side: aq ammonia
Number of passes = 4 Available flow area = π x d
i 2
x N
T
4 N
P
= π x 0.01257
2
x 274 4 4
a
t
= 0.0085m
2
∴ Velocity of fluid in the tube V
t
= M
t
f x d
t
V
t
= 3.98 x 1 x 1 832 0.0085
V
t
= 0.563 ms
b Shell side: water
shell I D , D
s
= 438mm L
c
baffle cut = 0.25 x Ds L
s
, baffle spacing = 0.5 D
s
= 0.219m S
m
= [ p
1
– DoL
s
] x D
s
p
1
S
m
= Gross sectional area at centre of shell Nb = No of baffles , L = length of tube
p
1
= 13 inches square pitch = 0.0206m 16
S
m
= 0.0206 – 0.0158 x 0.219 x 0.438 0.0206
S
m
= 0.02235m
2
Shell side velocity, V
s
= M
s
S
s
x S
m
= 9.548 997.04 x 0.02235
= 0.4284 ms No. of baffles
N
b
+ 1 = Total length of tube
Baffle spacing
= 4.8768 0.2 = 22.26 ∼ 23
∴ N
b
= 22
6 Film transfer co-efficient
a Tube side
Richardson coulson Page no: 270 – 297
A
t
55 c
S = 832 kg m
3
Cp = 2.57 KJ kg k
M = 1.26 mN
s
m
2
= 1.26 x 10
–3
N
S
K = 0.219 wm k m
2
N
R
C
= fV
t
D M
= 832 x 0.563 x 0.01257 1.26 x 10
–3
= 4673 N
P
r
= MC
p
K = 2.57 x 10
–3
x 1.26 x 10
–3
14.78
= 14.78 From perry page 10-29 j
H
= 0.02 ∴ N
N
h
= j
H
x N
R
C
x N
P
r
N
N
h
= 0.02 x 4673 x 14.78
N
N
h
= 229.35 But, h
i
d
i
= N
N
h
k ∴h
i
= 229.35 x 0.219 0.01257
= 3995.83 wm
2
k h
i
= 3995.83 wm
2
k
b Shell side at 25
c C = 995.045 kg m
3
Cp = 4.184 kj kg k
M = 0.95 x 10
-3
poise k = 1.42 w m k
N
R
C
= f x V
s
x D D = tube outside dia
M = 995.04 x 0.4254 x 0.01587
0.095 x 10
–3
= 7620
N
Pr
= M x C
p
K
= 0.95 x 10
–3
x 4.18 x 10
3
1.42 = 2.8
From perry, page 10-29 , j
H
= 1 x 10
–3
∴ N
Nh
= 1 x 10
–3
x 7620 x 2.8
= 10.74 but, h
o
d
o
= N
Nh
k ∴ h
o
= 10.74 x 1.42 = 960 w m
2
k 0.01587
h
o
= 960 w m
2
k
Overall heat transfer coefficient
1 = 1 + D
o
x 1 + D
o
ln D
o
D
i
+ 1 U
o
ho D
i
hi 2k
w
h
dirt
For stainless steel k
w
= 45 h
dirt
= 0.003
1Uo= 1 960 + 0.01587 0.012257 x 1 3995.53 + 0.01587 ln 0.01587 0.01257 2x 45 + 1 0.003
U
o
= 243.096 w m
2
k
7 Pressure drop calculation :
a Tube side û P
L
= 4fLV
2
x f x g 2
g
D
i
but, f = 0.079 x R
c –0.25
= 0.079 x 4673
–0.25
= 0.0095 û P
L
= 4 x 0.0095 x 4.8768 x 0.563
2
x 832 2 x 0.01257
û P
L
= 1.943a Kp
a
∴ û P
t
= 2.5 f x Vt
2
2 û P
t
= 2.5 832 x 0.563
2
= 0.3796 KP
a
2 û P
total
1S [ û P
L
+ û P
t
= 4 x [ 1.9439 + 0.3796 ] û P
total
= 9.294 KPa û P
total
is less than 70 Kpa hence design is satisfactory.
b shell side Shell side pressure drop is calculated using bell ‘s method
Perry : page 10-26 to 10-31 N
R
C
= 7620
From figure 10.25 a page 10-31 friction factor f
k
f
k
= 0.19
i Pressure drop across cross flow section P
c
û P
c
= b x f
k
x w
2
x N
c
x Mw Mb
0.4
f . f
2 m
b= 2 x 10
-3
w = 9.54 kgs S
m
= 0.02235m
2
N
c
= D
s
x 1 – 2Lc D
s
P
p
Where Ds = shell 1D = 0.438m
Lc = 0.1095
P
p
= pitch parallel cross flow = 13 in = 0.0206m 16
N
c
= 0.438 1 – 2 0.1096 0.438 0.0206
N
c
= 16 ∴
ûP
c
= 2 x 10
-3
x 0.19 x 9.54
2
x 16 [1]
0.4
997.04 x 0.02235
= 0.0252 K
Pa
ii End zone pressure drop û3
c
û3
c
û P
L
1 + N
c
w
N
c
N
c
w
= 0.8l
c
P
p
= 0.8 x 0.1095 0.0206
= 4 ∴
ûP
c
= 0.0252 x 1+ 4 16
û3
c
= 0.0315 K
P
a
iii Pressure drop in window zone, û
P
w
û P
w
= b x w
2
[2+0.6 kl
cw
] fm x S
w
x f b = 5 x 10
-4
S
m
= 0.02235 m
2
N
c
w
= 4 w = 9.84 kg s
S = 997.045 kg m
2
Area for flow though window Sw = Sw
g
– Sw
t
Sw
g
, from fig 10-18 , page 10-29, perry hand book. Sw
g
= 0.029
Sw
t
= N
T
1-Fc π D
o
8 = 274 x 1 - 0.68 x π 0.0158
2
8 ∴∴∴∴
∴S
w
= 0.029 – 0.0029
S
w
= 0.0205 m
2
Pressure drop at window zone û
P
w
= 5 x 10
-4
x 9.54
2
x 2 + 0.6 x 4 0.02235x0.0205x997.045 û P
w
= 0.386 Kp
a
Total pressure drop at shell side, û P
T
would be given by û P
T
= 2 x û P
c
+ û N
b
– 1 x û P
c
+ N
b
û P
w
û P
T
= 2 x 0.0315 + 22 – 1 x 0.0252 + 22 x 0.386 û P
T
= 9.08 Kp
a
Total pressure drop at shell side is less than 70 Kp
a
hence, shell heat exchanger design is satisfactory.
Sw
t
= 0.0085
8. MECHANICAL DESIGN OF PROCESS EQUIPMENTS
