Page 116 - Process Equipment and Plant Design Principles and Practices by Subhabrata Ray Gargi Das

4.7 Design illustration 113

X 10:849
¼ 0:8866ð> 0:8Þ.
Y 12:236
This gives F T ¼ ¼
From heat balance equation (Eq. 2.1)
3 6
Q ¼ m h C p;h T h;in T h;out ¼ 10:8472 2:345 10 ð130 50Þ¼ 2:0349 10 W
Q
¼ 40:57 kg s.

and mass flow rate of cooling water required m CW ¼ m c ¼
C p;w T w;out T w;in
As mentioned in text, we select cooling water as the tube-side fluid and kerosene is the shell-side
fluid. Henceforth, all properties pertaining to cooling water will be denoted by subscript ‘t’ and those
referring to kerosene will be subscripted by ‘s’.
From Table 2.5, we assume U D ¼ 700 W (closer to the higher value to arrive at a compact
m K
2
exchanger with a smaller heat exchange area).
Using Eq. (2.7), we obtain heat transfer area, i.e., effective tube outside area A o as
Q 2:0349 10 6 2
¼ 77:31 m
A o ¼ ¼
F T U D DT LMTD;counterflow 0:89 700 42:25
From point 6 in Section 4.4, we adopt D 0 ¼ 25:4 mm and tube length ¼ 6 m arranged in a
0
triangular pitch of layout 60 with P T ¼ 1:25D 0 ¼ 31:75 mm ¼ 0.03175 m.
From Table 4.5, we select BWG number ¼ 10 which gives wall thickness ¼ 3.4 mm, tube inner
2
diameter D i ¼ 18:59 mm ¼ 0:01859 m and A i ¼ 2:714 10 4 m .
Assuming two tube passes with fixed tube sheet (TEMA AJW), we adopt TS ¼ 50 mm as a first
guess and estimate the effective tube length L e from Eq. 4.31(a) as
L e ¼ L 2 TS ¼ 6 ð2 0:05Þ¼ 5:9m
This gives shell inner diameter D s from Eq. (4.6) for CL ¼ 0:87 and CTP ¼ 0:9 (from Table 4.7).

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
CL A o P t 0:87 77:31 ð0:03175Þ
2 2
D s ¼ 0:637 ¼ 0:637 ¼ 0:4516 m
CTP D o L e 0:9 0:0254 5:9
Referring to Table 4.8b, the next higher shell diameter is D s ¼ 489 mm ¼ 0.489 m and the cor-
responding number of tubes N t ¼ 152.
r Number of tubes per pass ¼ 152/2 ¼ 76, and
4 2
Flow area per pass ¼ 2:714 10 76 ¼ 0:0206 m
This gives cooling water flow velocity (inside tubes)
m CW 40:57
¼ 1:985 m s (lies between 0.9 and 2.4 m/s, so Ok).
u t ¼ ¼
r CW A i;per pass 992:22 0:0206
2
Considering the operating pressure to be 6 kg/cm (higher of the two operating pressures), the
2
design pressure p design ¼ 1:3 6 ¼ 7:8kg cm and from Eq. (4.32),
p ffiffiffiffiffiffiffi
7:8

¼ 23:425 mm
p ffiffiffiffiffiffiffiffiffiffiffiffiffi 489
TS ¼ D s p design =58:3 ¼
58:3
Since D s < 500 mm, TS is higher of D s =10 ¼ 49mm or the value calculated from Eq. (4.32).
Accordingly, our initial guess of TS ¼ 50 mm is Ok.