Page 47 - Applied Process Design For Chemical And Petrochemical Plants Volume II

36 Applied Process Design for Chemical and Petrochemical Plants

Graphically, read 13 steps or theoretical plates from the
top plate through bottom reboiler (assuming a total con- so: -=-- ‘ 2216 - 1-29
V,
171.6
denser).
-=-- 50 -0.291
rectifjmg section = 5 V, 171.6
feed plate -1
stripping section = 2 (includes reboiler) Stripping section operating line:
13 Plates including reboiler

To calculate this stepwise: ym 1.29 X, + 1 - 0.291 XB
XB = 0.05
Operating line of rectifying section:
ym = 1.29 X, + 1 - 0.01455
Use this equation as described above following down
from the feed plate cross-over from the rectifylng equation
to the stripping equation.
L/V = 0.59
L/D = 1.432, D = 50 mols product
L = (50) 1.432 = 71.6 mols liquid reflux B. Fmq= 0
V, = L, + D = 71.6 + 50 = 121.6 mols
This represents feed as all vapor (not superheated).
then: yn + 1 = 0.59 xn + 50 (.95)/121.6 Slope of “q” line:
yn + 1 = 0.59 xn + 0.39, operating line equation
At top:yn + 1 = XD = 0.95 -q -0
=-=-- -0
1-q 1-0
So: From equilibrium curve at yn +1 = 0.95, read the liquid
in equilibrium, which is x, (or top plate in this case) x, = This represents no change in overflow from the feed
xtop = 0.88. plate, and the increase in vapor flow is equal to the mols
Now substitute this value x = 0.88 into the equation and of feed.
calculate the vapor coming up from the first plate below
the top (t - 1). Thus, if x, = top plate, y, + 1 = vapor from Minimum reflux : (k)
plate below top. Now, read equilibrium curve at y(t - 1) and =- XD -yc,
get x(, + 1) or xt - 1 which is liquid on plate below top. Then min YC -XC
using xt - 1, calculate yt - 2 (second plate below top, etc.). where: XD = 0.95
Then, read equilibrium curve to get corresponding liquid yc = 0.50
xt - 2. Continue until feed plate composition is reached, xC = 0.29 ]read from graph
then switch to equation of stripping section and continue
-
as before until desired bottoms composition is reached. - 0.95 - 0.50
Operating line of stripping section: 0.50 - 0.29
= 2.14 min. reflux ratio, reflux/product
Slope of operating line at minimum reflux:

Because the feed is a super cooled liquid, L,/V, is not
equal to LJV,. From definition of “q”:

L, = L, + qF Slope from graph = 0.688
L, = 71.6 + (1.5) (100)
L, = 221.6 Operating reflux ratio = (1.5) (2.14) = 3.21, reflux/product,
v* -vs (L/D),,
Also: -= 1- q
F Slope of operating line = (L/V),,p = 3.21/(3.21 + 1) = 0.763
121.6 - V, = 1 - 1.5 No. of theoretical plates from graph = 11
100 No. plates rectifjmg section =5
feed plate =1
121.6 - V, = -50 stripping section = 2 (includes reboiler)
LrS = 171.6 total = 11 (includes reboiler)