This preview shows page 1. Sign up to view the full content.
Unformatted text preview: = 4, αB,C = 2, and αC,C = 1.
Assumptions: No condenser and column holdups. Constant molar overflow. No pressure drop.
Startup under total reflux.
Find: Variation of instantaneous residue and distillate compositions with time following
startup.
Analysis: For startup conditions of total reflux, use Eq. (1327) to compute the distillate
0.333
(
composition using N = 4, x D0C) =
= 0.00366
4
4
0.333 4 + 0.333(2) 4 + 0.333 1
0.00366 4
(
4 = 0.9370 and x D0B) = 1 − 0.0037 − 0.9370 = 0.0593
0.333
Take a time increment = ∆t = 0.25 h.
100
From Eq. (1322), W (1) = 100 −
0.25 = 9583 kmol
.
1+ 5
9583 − 100
.
(
From Eq. (1323), x W1) = 0.3333 + (0.9370 − 0.3333)
= 0.3081
A
100
95.83 − 100
(
x W1) = 0.3333 + (0.0593 − 0.3333)
= 0.3447
B
100
(
From Eq. (1325), x D0A) = 0.333 (
x W1) = 1 − 0.3081 − 0.3448 = 0.3471
C 4 − N min
5 − Rmin
From Eq. (1328),
= 0.75 1 −
4 +1
5 +1 0.5668 (1) 4 N min − 4
From Eq. (1330), Rmin =
(2)
(4 − 1)[0.3081(4) N min + 0.3448(2) N min + 0.3471(1) N min ]
When Eqs. (1) and (2) are solved simultaneously, the results are Rmin = 0.917 and Nmin = 3.265.
Using the latter value in Eq. (1327),
0.3471
x DC =
= 0.0108
3.265
3.265
0.3081 4
+ 0.3447 2
+ 0.3471(1) 3.265
From Eq. (1325),
0.0108 3.265
0.0108 3.265
x DA = 0.3081
4
= 0.8861 and x DB = 0.3447
2
= 0.1031
0.3471
0.3471 Exercise 13.24 (continued)
Analysis:
The calculations above for the first time increment are repeated for subsequent time
increments, using a spread sheet, up to 5 hours. The results are as follows. On the next page,
the mole fraction compositions are plotted as a function of time. Analysis: (continued) Exercise 13.24 (continued) Exercise 13.25
Subject: Calculation of batch rectification of a ternary mixture by the shortcut method of
Sandaram and Evans.
Given: Charge of 200 kmol of 40 mol% A, 50 mol% B, and 10 mol% C. Reflux ratio, R = 10
and vapor rate, V = 100 kmol/h. Column has two equilibrium plates, a total condenser, and a
partial reboiler. Thus, N = 3. αA,C = 2, αB,C = 1.5, and αC,C = 1.
Assumptions: No condenser and column holdups. Constant molar overflow. No pressure drop.
Startup under total reflux.
Find: Variation of instantaneous residue and distillate compositions for 1hour following
startup.
Analysis: For startup conditions of total reflux, use Eq. (1327) to compute the distillate
0100
.
(
composition using N = 3, x D0C) =
= 0.0199
3
3
0.4 2 + 0.5(15) 3 + 0.1333 1
.
0.0199 3
(
2 = 0.6368 and x D0B) = 1 − 0.0199 − 0.6368 = 0.3433
01
.
Take a time increment = ∆t = 0.10 h.
100
From Eq. (1322), W (1) = 200 −
0.10 = 199.09 kmol
1 + 10
199.09 − 200
(
From Eq. (1323), x W1) = 0.4 + (0.6368 − 0.4)
= 0.3989
A
200
199.09 − 200
(
x W1) = 0.5 + (0.3433 − 0.5)
= 0.5007
B
200
(
From Eq. (1325), x D0A) = 0.4 (
x W1) = 1 − 0.3989 − 0.5007 = 01004
.
C 3 − N min
10 − Rmin
From Eq. (1328),
= 0.75 1 −
3 +1
10 +1 0.5668 (1) 2 N min − 2
From Eq. (1330), Rmin =
(2)
(2 − 1)[0.3989(2) N min + 0.5007(15) N min + 0.1004(1) N min ]
.
Whe...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details