Page 48 - Applied Process Design For Chemical And Petrochemical Plants Volume II

Distillation 37

Rectzfjing Section Equation for Operating Line:
Minimum reflux : (L/D)~, = -
-Yc
XD
Yc - xc
where XD = 0.95
yc
xC = 0*277]read from graph
0.138
=
L/V = 0.763
L/D = 3.21
L = (3.21.) (50 mol product, D) = 160.6 mols (reflux liquid) .95 - .277
V, = L, + D = 160.6 + 50 = .277 - .138
V, = 210.6 mols vapor up column
(L/D)min = 4.84 reflux/product
then : yn+l = 0.763 x, + - 50 (0.95)
210.6 Slope of operating line at minimum reflux:

yn + 1 = 0.763 X, + 0.225 (L/V) . =-=- L/D 4.84
mm L/D + 1 4.84+ 1
Liquid Down Stripping Section: = 0.830 (graph reads 0.844)

L, = L, + qF Actual Operating Line:
L, = 160.6 + (0) (100 mols feed)
L, = 160.6 = L, Operating reflux ratio = (1.5) (4.84) = 7.26 reflux/product

Vapor Up Stripping Section: 7 26
Slope of actual operating line = (L/V) = -
0.879
=
7.23 i- 1
Graphically we read 8.5 total plates thru bottom reboiler
210.6 - V, =1-0
100 recwng section = 5
feed plate =1
210.6 - V, = 100 stripping section = 2.5 (includes reboiler)
V, = 110.6 mols total = 8.5 (includes reboiler)

Stripping Section Equation for Operating Line Equations for Stqise Tray to Traj Calculations RectiJjing
Section Operating Line

160.6
ym =- 50 (0.05)
110.6 Xm+’-- 110.6 L,/V, = 0.879

ym = 1.452 X, + 1 - 0.0226 L/D = 7.26
L, = (7.26) (50) = 363 mols liquid reflux
Use these equations as described for the (a) part of prob V, = Lr + D = 363 + 50 = 413
lem in solving for number of theoretical plates stepwise. 50
+
yn+l = 0.879~~ - (0.95)
413
C. Fmq= -1.5
yn + 1 = 0.879 X, + 0.115
This represents feed as a superheated vapor, and there
is a decrease in liquid overflow from feed plate.
Liquid Down Stripping Section:

-9 -
-
Slope of “q” line = - = 0.60 L, = L, + qF
1-q 1-(-1.5) L, = 363 + (-1.5) (100) = 213 mols liquid