Page 45 - Applied Process Design For Chemical And Petrochemical Plants Volume II

34 Applied Process Design for Chemical and Petrochemical Plants

A. For a Feed at its Boiling Point: From the equilibrium curve at yt = 0.95
then: xt= 0.88
-
Y(~ 1) = 0.631 (Xt) + 0.331
-
Y(~ 1) = 0.651 (0.88) + 0.331 = 0.903
= or fi. = m, 0.70 (from curve)
=
yF
Vn L/D+1 V XD-XF yt - 1 = 0.903, then xt - 1 from equilibrium curve = 0.788
- - 0.95 - 0.70 Now calculate yt - 2
0.95 - 0.30
yt - 2 = 0.651 (.788) + 0.331 = 0.844
At yt - 2 = 0.844, curve reads: xt - 2 = 0.69
Minimum L/V = 0.35 mol reflux/mol vapor up
Then: yt - 2 = 0.651 (0.69) + .331 = 0.780
At yt - 3, curve reads: xt - 3 = 0.60
L/ D
substituting : 0.55 = - Then: yt - 4 = 0.651 (0.60) + .331 = 0.722
L/D + 1 At yt - 4, curve reads: xt - 4 = 0.52 (Feed Tray)
Then: yt - 5 = 0.651 (0.52) +.331 = 0.669 (too far below feed).
0.55 L/D + 0.35 = L/D
0.45 L/D = 0.59 Now go to stripping section curve:
Reflux/Product = L/D = 0.55/0.45 = 1.22

The value of L/D minimum should be equal to:
The feed was at its boiling point:
L12 XD - yc - 0.95 - 0.70 .25
-=-- = - 1.25 V, = V, = 143.9
=
D yc -xC 0.70- 0.50 .20 B = Bottoms = 50
L,=B+V=50+ 143.9~193.9
The slight difference is probably due to inaccuracy in
143.9
reading yc = 0.70 from equilibrium curve. Ym = (E) (0.50)
50
Xm+l-
B. Theoretical Plates at L/D = 1.5 Times Minimum: = 1.35 X, + 1 - .01736
Operating Reflux Ratio = (1.5) (1.25) = 1.878 = L/D Starting at t - 4 = feed tray:

Slope of operating line at this reflux ratio: xt-4-0.52
y (feed - 1) = 1.33 xf- 0.0176 (f - 1)
L L/D y(f- 1) = 1.35 (0.52) - 0.01736 + 0.685
_=-
V L/D+1 At yt - 1 = 0.685, ~f- 0.475,
1
=
Note: This is not too accurate due to switched operating
line equations before the feed compositions were reached,
From Graph, L/V = 0.653 was plotted based on feed at yet, one more calculation on the stripping line would have
its boiling point, No. of theoretical plates (step-wise placed us below the feed plate composition. Hence a
graph) = 11.3 change in reflux ratio is necessary in order to split right at
Now, to calculate theoretical plates: the feed composition.
Rectifyng section:
continuing:
L, + 1 D
XD
Yn =- x,+~ + - operating lime yf-2 = 1.35 (0.475) - .01736 = 0.624
Vn Vn
From curve at yf- 2 = 0.624
At: L/D = 1.878, D = 50 mols overhead xf - 2 = 0.405
L = (1.878) (50) = 93.9 mols reflux to column yf- 3 = 1.35 (.405) - .01736 = 0.531
V = L + D = 93.9 + 50 = 143.9 mols to vapor overhead
From curve, xf - 3 = 0.32
93.9 50 yf- 4 = 1.35 (.32) - .01736 = 0.416
(0.95) = 0.652 xn+1 + 0.331
Yn =143.9Xn+l+145.9 xf-4'0.23
yf- 5 = 1.35 (.23) - .01736 0.294
For a total condenser: ytop = XD = XR = 0.95 xf- 5= 0.15