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354 Chapter 6

(48.3 m/h). To calculate the column cross-sectional area from Equation
gal/h-ft2
6.32.7, requires the volumetric flow rates of both the light and heavy phases.

2185 33
m F
3
_ = ——— + —— = 27.19 ft /h (203.4 gal/h, 0.770 nrVh)
82.41 48.7
p F
3
ms / p s = (7.481) (2106 / 62.43) = 252.4 gal/h (0.995 m /h)
From Equations 6.32.7 and 6.32.8,
203.4 + 252.4
2
2
A = ——————— = 0.3821 ft (0.0355 m )
1193
1/2
1 2
D = (4 A / 7i) = [4 (0.3821) / 3.142] ' = 0.6975 ft (8.370 in, 0.213 m)
Next correct HETS for column diameter from Equation 6.32.9. Because D is
less than 30 in (0.762 m), select a standard pipe size. From piping tables [66],
select a Schedule 10S pipe, which has an inside diameter of 10.42 in (0.265 m).
(10.42 V' 38
HETS = 5.6 I ——— I = 5.307 in (0.135 m)
I 12 )
Because 5.307 in. is a minimum value, increase it by 20% to avoid flooding.
Therefore, the design HETS is 6.368 in (0.162 m). From Equation 6.32.6, the ex-
traction height,

Z E = 4 (6.368) = 25.47 in (0.6469 m)

Hounding the height to the nearest 3 in (0.0762 m), Z E = 27 in. (0.6858 m).
Because this is a short extractor and because of the assumptions made, increase the
extraction height to 6 ft (1.97 m). The extra cost would not be substantial.
Now, add top and bottom sections to separate the phases. The diameter of
both settlers is 50% greater than the extractor diameter, and the height of each
settler is equal to the settler diameter. Therefore, the height of both settlers,

Z s = 2 (1.5) (10.42) = 31.26 in (0.794 m)

To join the settlers to the extractor requires reducers, which are about a foot
long. The total height of the column,

+ reducers= 27.0 + 31.26 + 24.0 = 82.26 in (6.86 ft, 2.09 m)
Z = Z E + Z s

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