Page 87 - Applied Process Design For Chemical And Petrochemical Plants Volume II

76 Applied Process Design for Chemical and Petrochemical Plants

Stripping: apply only to components heavier than For component No. 2,
heavy key, i = H.

(a1 - 1) (all
Read from Figure 8-48 value of Cmi for each com-
ponent. Because this is at minimum reflux, and adjacent keys system,
Calculate for each component:
FX, = DXDL
C,i CI' x Ips
1
Sum these values: FX, = BXBH

L ai xips Therefore, for component No. 4, lighter than light key,
i-H
FXF4 = (1) (0.225) = 0.225
1 2. Calculate:
Then, DXD~ = 0.225

For component No. 1, heavier than heavy key, this com-
ponent will not appear in the overhead.
E the two values of p are not very nearly equal, Bottoms:
this requires a retrial with a new (L/D)min, and a
follow through of the steps above.
When rps/rpr > p', the assumed (L/D)min is too high.
Note that rp,Jrpr changes rapidly with small changes in
(L/D),in, p changes slightly. When p = pf, the proper
(L/D)min has been found. Colburn reports the method
accurate to 1%. It is convenient to graph the assumed -- FxFi - (S,)i t I
(L/D)min versus p and pf in order to facilitate the selec- BxBi
tion of the correct (L/D)min.
(S, )i = -, by definition
Example 8-24: Using the Colburn Equation Calculate the BxBi
Minimum Reflux Ratio
Then, BXB~ -
FxFi
=
The mixture of four components is as listed below, (Sr)i +I
using n-butane as the base component.
Component No. 1:
Component Relative Vol. xf (Sr)i = DXD/BXB
1 0.25 0.10 -
=
Hvy Key 0.50 0.225 0.03 FXF~ = BXB~ (1) (0.1) = 0.10
n-butane 1.0 0.430 20.00
4 2.0 0.225 - Component No. 2:

(S ri ).=-- DxD2 -0.05
BxB2

S, = separation ratio = - Substituting in equation previously established,
DXDi
BxBi

Hall at top, S, = 1
If all at bottom, S, = 0
or, because DXD~ has been calculated,
For component No. 3; Basis; 1 mol feed,
DXD~ 0.01072 o.214
BxB2 = - -
=
=
(S,)i 0.05