Page 287 - Separation process engineering

from the Gilliland correlation, and then find N. (L/D) min was found in Example 7-2, and N min in

Example 7-1. The feed plate location is estimated from Eqs. (7-41) and (7-40).
D. Do It.

The corresponding ordinate (N − N min )/(N + 1) = 0.27 using Liddle’s curve. Since N min = 3.77, N =

5.53. From Eq. (7-40a), N F,min is calculated as

Where x was found from Example 7-2 as
LK,dist

and

x HK,dist = x cum,dist = 0.021
Then, from Eq. (7-40b),

E. Check. A check of the Gilliland correlation can be obtained from Eq. (7-42b). With x = 0.455 this
is

or (1 − 0.283) N = N + 0.283, which gives N = 5.65. The 2% difference between these two results
min
gives an idea of the accuracy of Eq. (7-42) in fitting the curve.

A check on the value of N can be obtained with Kirkbride’s Eq. (7-41). To use this equation we need
f
to know the terms on the RHS. From Example 7-2, F = 100 and D = 69.92. Thus, B = 100 − 69.92 =
30.08. The HK = cumene and the LK = toluene. The feed mole fractions of both are 0.30. From
example 7-2:
Dx cum,dist = 1.5. Then x HK,dist = (Dx cum,dist )/D = 1.5/69.92 = 0.02145.

Dx tol,dist = 28.5. Then Bx tol,bot = Fz − Dx tol,dist = 30.0 − 28.5 = 1.5, and

x LK,bot = Bx tol,bot /B = 1.5/30.08 = 0.04987.

Then, Eq. (7-41) becomes,

and (N − 1)/(N − N ) = 1.1898
f
f
which gives N = 3.46 if we use N = 5.53 or N = 3.7 if we use N = 6. Thus, the best estimate is to use
f f
either the 3rd or 4th stage for the feed. This agrees rather well with the previous estimate.