Unformatted text preview: vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2),
yR
0.710
xR =
=
= 0.495
yR + α (1 − yR ) 0.710 + 2.5(1 − 0.710)
Overall total material balance,
F = 100 = D + B
(6)
Overall benzene material balance, xFF = yCD + xRB or 70 = 0.8D + 0.495B (7)
Solving Eqs. (6) and (7), D = 67.2 mol/s or 67.2 mol / mol feed, and B = 32.8 mol/s.
Therefore, vapor generated = V = 1.5D = 1.5(67.2) = 100.8 mol/s
The operating line for the yx diagram passes through the (y, x) point (0.8, 0.8) with a slope,
L/V = 1/3, as shown in the diagram below. Solving for yR, yR = y1 + ( x1 − xC ) Exercise 7.11 (continued) Analysis: (continued)
Procedure 3:
The slope and top point of the operating line are the same as for Procedures 1 and 2. We
just have to extend Procedure 2 by stepping off a second equilibrium stage. From above, the
results for the condenser and stage 1 are:
yC = 0.80
xC = 0.615
y1 = 0.738
x1 = 0.530
y2 = 0.710
x2 = 0.495
Benzene material balance around Stage 2,
y RV + x1 L = y2V + x2 L
(8)
L
1
Solving for yR,
y R = y2 + ( x2 − x1 )
= 0.710 + (0.495 − 0.530)
= 0.698
V
3
The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2),
yR
0.698
xR =
=
= 0.480
yR + α (1 − yR ) 0.698 + 2.5(1 − 0.698)
Overall total material balance,
F = 100 = D + B
(10)
Overall benzene material balance, xFF = yCD + xRB or 70 = 0.8D + 0.480B (11)
Solving Eqs. (10) and (11), D = 68.8 mol/s or 68.8 mol / 100 mol feed, and B = 31.2 mol/s.
Therefore, vapor generated = V = 1.5D = 1.5(68.8) = 103.2 mol/s
The operating line for the yx diagram passes through the (y, x) point (0.8, 0.8) with a slope, L/V = 1/3, as shown in the diagram below. Exercise 7.11 (continued) Analysis: (continued)
Procedure 4:
If the reflux bypasses the top stage, the vapor and liquid pass through that stage without
change. Therefore, this procedure is the same as Procedure 2, i.e. just one stage in the column.
Procedure 5:
The slope and top point of the operating line are the same as for Procedure 1. We just
have to add the feed to the stage in the column. Therefore from the results above, we have:
yC = 0.80
xC = 0.615
y1 = 0.738
x1 = 0.530
Benzene material balance around Stage 1, which now includes the feed,
xFF + y RV + xC L = y1V + x1 L
(12)
Solving for yR,
V
L
L
F
V
L
L
100
y R = y1
+ x1
− xC
− xF
= 0.738
+ 0.530
− 0.615
− 0.70
(13)
V
V
V
V
V
V
V
V
Because the feed is a saturated liquid, , V =V and L = L + 100 From above, V = 1.5D and L/V = 1/3. Also, L / V = L / V + 100 / V = 1 / 3 + 100 / V
Therefore, Eq. (13) becomes,
1 100
1
100
17
11.33
(14)
y R = 0.738 + 0.530 +
− 0.615
− 0.70
= 0.710 −
= 0.710 −
3V
3
V
V
D The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2),
yR
xR =
(15)
y R + 2.5(1 − y R )
Overall total material balance,
F = 100 = D + B
(16)
(17)
Overall benzene material balance, xFF = yCD + xRB or 70 =...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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