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Separator Design 343

Substituting X and Y into Equation 6.27.5 we obtain

RO (1.6-1.171)
——— = ——————— (4.399 - 7.5) + 1.6
1.589 6.5

Solving, for RQ, we find that RQ = 2.217.
To obtain the number of actual stages for the separation, first calculate the
minimum number of equilibrium stages from Equation 6.27.2.

log ( 0.3465 / 0.0015) (0.1485 / 0.0035)
N M = ————————————————————— = 13.14
log 2.013

Next, solve Equations 6.27.8 to 6.27. 1 1 to obtain the number of equilibrium
stages. From Equation 6.27.10,

2.217-1.589
X e = ———————— = 0.1952
2.217+1

and from Equation 6.27. 1 1 ,

B e = 0.105 log 0.1952 + 0.44 = 0.3655

Next, calculate the value of Y from Equation 6.27.8.

Y e= 1-0. 1952° 3655 = 0.4496

Finally, from Equation 6.27.9,

N e- 13.14
—————— = 0.4496
N e +l

The number of equilibrium stages, N, equals 24.68. Rounding off N to the next
e e
highest stage, N e = 25.
The feed point location is calculated from the Kirkbride equation, Equation
6.27.12.

Nn |~ f 0.15 "1 ( 5.819xlO" 3 V ( 0.6015 ° 206

N L L I 0.35 ) U.764xlO~ 3 J ( 0.3984 J J

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