Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Given charge of 100 kmol of 30 mol methanol 30 mol

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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 condenser-reflux drum. Find: Using Eqs. (13-36) and (13-39) to estimate the liquid-phase 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. (13-36) 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 K-values 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. (13-49), 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. (13-52), where in this case, taking benzene, which has the highest K-value 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. (13-54), which requires estimations of λ min and λ max . From Eq. (13-53), 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. (13-54), 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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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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