Unformatted text preview: mixture of 5 mol% A and 95 mol% B. Distillate to contain 35
mol% A and a bottoms to contain 0.2 mol% A. Relative volatility, αA,B = 6 = a constant.
Column equipped with partial condenser and partial reboiler.
Assumptions: Constant molar overflow.
Find: Using algebraic methods,
(a) Minimum number of equilibrium stages.
(b) Minimum boilup ratio, VB = V / B .
(c) Number of equilibrium stages for a boilup ratio = 1.2 times minimum. Analysis: From a rearrangement of the equilibrium equation, Eq. (73),
y
y
x=
=
(1)
y + α (1 − y ) 6 − 5 y
(a) For minimum stages, have total reflux, so that y = x for passing streams. Begin
calculations from the top. yD = y1 = 0.35. From Eq. (1), x1 = 0.35/[6  5(0.35)] = 0.0824.
Therefore, y2 = x1 = 0.0824. From Eq. (1), x2 = 0.0824/[65(0.0824)] = 0.0147. Therefore,
y3 = x2 = 0.0147. From Eq. (1), x3 = 0.0147/[65(0.0147)] = 0.0025. This is close to but not
quite equal to the desired value of 0.002. Thus, we need just slightly more than 3 minimum
equilibrium stages.
(b) For minimum boilup ratio, the stripping section operating line connects the two
points for {y, x} of {0.002, 0.002} and {y in equilibrium with x = 0.05}. From a rearrangement
of Eq. (1), the y in equilibrium with x = 0.05 is: y = αx/[1 +
x(α − 1)] = 6(0.05)/[1 + 0.05(6 − 1)] = 0.24. The slope of the operating line = ( L / V ) = (0.24 0.002)/(0.05  0.002) = 4.96. From a rearrangement of Eq. (712), (VB)min = 1/ [( L / V )  1] =
1/(4.96  1) = 0.253.
(c) The boilup ratio = VB = 1.2(0.253) = 0.3036. From Eq. (712), the slope of the
stripping section operating line = L / V = (VB + 1)/VB = (0.3036 + 1)/0.3036 = 4.294. This line
intersects the vertical qline (xF = 0.05) at 0.002 + 4.294(0.05  0.002) = 0.2081. Therefore, the
slope of the rectifying line = L/V = (0.35  0.2081)/(0.35  0.05) = 0.4730. From a rearrangement
of Eq. (78), R = (L/V)/[1  (L/V)] = 0.473/(1  0.473) = 0.8975. The equation for the rectifying
section operating line, using Eq. ((79), with a modification for a partial condenser as determined
from Fig. 7.18, is,
y n +1 = R
1
xn +
y D = 0.473xn + 0.184 5
R +1
R +1 (2) Analysis: (c) (continued) Exercise 7.22 (continued) The equation for the stripping section operating line, using Eq. (712) is,
1
V +1
y m +1 = B
xm −
x B = 4.294 xm − 0.00659
VB
VB (3) We can now calculate stage by stage down from the top, starting from yD = 0.35, alternating
between the equilibrium curve, Eq. (1) and the appropriate operating line, Eq. (2) or (3). We
begin using Eq. (2), but switch to Eq. (3), when x < xF = 0.05. The calculations are terminated
when x < xB = 0.002. The calculations can be done with a spreadsheet, with the following
results, given as mole fractions of A leaving an equilibrium stage. The optimal feed stage is the
top plate. Equilibrium
stage
Partial condenser
1
2
3
4
5
6
7
8
Partial reboiler yA xA 0.350
0.223
0.190
0.155
0.121
0.0894
0.0626
0.0407
0.0235
0.0106 0.0824
0.0458
0.0376
0.0296
0.0224
0.0161
0.0110
0.00701
0.00400
0.00178 The calculations show that besides the partial condenser and partial reboiler, 8 equilibrium stages
are needed in the col...
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 Spring '11
 Levicky
 The Land

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