Page 310 - Chemical process engineering design and economics

Separator Design 289

noncorrosive service

Follow the procedure outlined in Table 6.10. Assume that a vertical separator
with a mist eliminator will be used. From Equation 6.9.3, k = 0.35 ft/s (0.107
v
m/s).
From Equation 6.9.2, the maximum vapor velocity,

(31.15-1.372 Y /2
v v = 0.35 ———————— I = 1.631 ft/s (0.497 m/s)
I 1.372 )

From Equation 6.9.1, the cross-sectional area of the separator,

2
2
A = 200.7 / 60 (1.631) = 2.051 ft (0.191 m )
From Equation 6.9.4, the separator diameter,
1/2
D = [ (4 / 3.142) (2.051) ] = 1.616 ft (0.493 m)
Because the separator diameter is below 30 in (0.762 m), select standard pipe.
From the chemical engineering handbook (6.66), the closest pipe size is 20 in
(0.508 m), Schedule 10 pipe, which has an inside diameter of 19.50 in (1.625 ft,
2
2
0.495 m), an inside cross-sectional area of 2.074 ft (0.193 m ), and a wall thick-
ness of 0.25 in (6.35 mm). From piping tables, the allowable pressure for carbon
steel at 200 °F (93.3 °C) is 186 psig (12.8 barg), which is above the design pres-
sure of 50 psig (3.45 barg).
Now, calculate the length of the separator. First, calculate the height of the
liquid from Equation 6.9.5. Use an average of the residence times given by Equa-
tion 6.9.6.

5.0 gal/min 4 min 1
L L = ——————— ——— ———— = 1.289 ft (0.393 m)
7.481 gal/ft 3 1 2.074ft 2

The minimum liquid level is 2.0 ft (0.610 m).

From Equation 6.9.7,

L = 2.0+1.5 (1.625) +1.5 = 5.938 ft (1.81m)
Round off the length in three-inch intervals. Therefore, L = 6.0 ft (1.83 m), but
according to Figure 6.4, the minimum length is 8.5 ft (1.83 m).

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