Unformatted text preview: ontinued)
Analysis: (continued)
Try increasing the reflux ratio for Step 1 from 5 to 15, noting that although it will
increase the time for rectification, the increase will be small because the time for Step 1 is small.
The result for this change in reflux ratio is as follows:
(a) The total time for rectification = 2.50+1.76+12.25+1.32+5.70 = 23.53 hours, a small
increase.
(b) The total amount of distillates = 8.89+4.10+24.45+3.81+9.37 = 50.62 lbmol.
Of more importance is the amount of C3 cut (Step 1) is increased from 5.70 to 8.89 lbmol.
The chart below shows the mole fractions in the instantaneous distillate with time. Exercise 13.28
Subject: Stability and stiffness of batch rectification calculations
Given: Compositions, holdups, and total exiting flow rates for the top three equilibrium stages
of a column separating a mixture of benzene (B), monochlorobenzene (MCB), and odichlorobenzene (DCB) at a time of 0.60 hour after total reflux conditions are established.
Holdups in the reboiler and condenserreflux drum.
Find: Using Eqs. (1336) and (1339) to estimate the liquidphase mole fraction of benzene
leaving Stage 2 from the top at 0.61 hour, using the explicit Euler method with a ∆t = 0.01 h.
If result is unreasonable, explain why using stability and stiffness considerations.
Analysis: Assume that over this small period of time (0.01 h) that the molar holdups on the
stages do not change. Then, Eq. (1336) becomes:
dxB,2
L + KB,2V2
KV
L1
=
(1)
xB,1 − 2
xB,2 + B,3 3 xB,3
dt
M2
M2
M2
From the given data, the Kvalues are: KB,2 = 0.0449/0.0121 = 3.71 and KB,3 = 0.0331/0.00884 =
3.74. Using other data in the given tables, with the explicit Euler discretization of Eq. (1349),
dxB,2 ∆xB,2 ∆xB,2
158.0 + 3.71 209.0
3.74 209.5
157.5
=
=
=
0.0276 −
0.0121 +
0.00884
dt
∆t
0.01
0.01092
0.01092
0.01092
= 398.077 − 1034.251 + 634.286 = −1.8880
Therefore, xB,2 at t = 0.61h = 0.0121  1.8880(0.01) = 0.00678
This negative value is unacceptable, so first determine stability. The criterion for stability is ∆t <
or = 2 divided by the maximum eigenvalue. A rough estimate of the maximum eigenvalue is
given by Eq. (1352), where in this case, taking benzene, which has the highest Kvalue and
using Stage 3 data where the holdup is the smallest,
L + KB,3V3
1581 + 3.74(209.5)
.
λ max = 2 3
=2
= 173,300
0.01087
M3
Therefore, max. ∆t = 2/173,300 = 0.0000115 h, which is much smaller than the value of 0.01 h.
Therefore, we have instability.
Now check the stiffness ratio. Use Eq. (1354), which requires estimations of λ min and λ max .
From Eq. (1353), where N+1 refers to the boiler and other quantities are estimated,
L + KDCB, NVN +1
158 + 1(210)
λ min = N
=
= 5.54
66.4
M N +1
From Eq. (1354), SR = λ max / λ min = 173,300/5.54 = 31,300, which is quite stiff.
Thus, in this case the explicit Euler method gives an unreasonable answer because we have
instability and stiffness. It is probably best to switch to an implicit Euler method or a Gear
method. Exercise 13.29
Subject...
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 Spring '11
 Levicky
 The Land

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