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

The results are as follows on the next page the mole

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Unformatted text preview: flux drum holdups were zero. Thus, the separation into a nearly pure light component is more difficult. However, column (plate or packed) holdup causes a flywheel-like inertia effect, which decreases the rate of change of stage compositions. This can be observed in Eq. (13-36), where the rate of change in stage composition is inversely proportional to the stage holdup. Thus, the degree of separation is improved because the composition does not as rapidly as it would with zero holdup. Thus, these two effects oppose each other and it is difficult to predict what the overall effect of holdup will be. Exercise 13.23 Subject: Calculation of batch rectification of a binary mixture by the shortcut method of Sandaram and Evans. Given: Conditions in Example 13.7. Charge of 100 kmol of an equimolar mixture of n-hexane (A) and n-heptane (B). Reflux rate of 10 kmol/h and a distillate rate of 10 kmol/h. Column has one equilibrium plate, a total condenser, and a partial reboiler. Rectification at 15 psia. Assumptions: Zero condenser and column holdups. Constant molar overflow. Constant Kvalues for A of 1.212 at the plate and 1.420 in the reboiler, while for B, 0.4838 at the plate and 0.5810 in the reboiler. No pressure drop. Find: Mole fractions of A and B for t = 0.05 h at the condenser and the reboiler. Analysis: With a constant reflux rate of 10 kmol/h and a distillate rate of 10 kmol/h, the vapor rate is 20 kmol/h and the reflux ratio is 1. Assuming startup conditions of total reflux, use Eq. (13-27) to compute the distillate composition. For the relative volatility, use the arithmetic average value based on the above K-values. 1.420 1212 . + α A,B = 0.5810 0.4838 = 2.47 2 0.50 From Eq. (13-27), with N = 2, x DB = = 0141 and x DA = 1 − 0141 = 0.859 . . 2 2 0.5 2.47 + 0.5 1 Take a time increment = ∆t = 0.05 h. Note that V = 20 kmol/h and R = 1. 20 From Eq. (13-22), W (1) = 100 − 0.05 = 99.5 kmol 2 99.5 − 100 (1) (1) From Eq. (13-23), xW = 0.5 + (0.859 − 0.5) = 0.498 and xWB = 1 − 0.498 = 0.502 A 100 From Eq. (13-28), 2 − N min 1 − Rmin = 0.75 1 − 3 2 0.5668 (1) 2.47 N min − 2.47 (2) (2.47 − 1)[0.498(2.47) N min + 0.502(1) N min ] Unfortunately, when Eqs. (1) and (2) are solved simultaneously, the results are Rmin = -0.0714, which is impossible, and Nmin = 0.92. Ignoring the impossible Rmin value and using the value of 0.92 for Nmin in Eq. (13-27), 0.502 xDB = = 0.305 and xDA = 1 − 0.305 = 0.695 0.92 0.92 0.498 ( 2.47 ) + 0.502 (1) From Eq. (13-30), Rmin = This result does not compare well with that of Example 13.7, where x DA = 0.8116. It appears that the shortcut method may not work well for a very small number of stages. Exercise 13.24 Subject: Calculation of batch rectification of a ternary mixture by the shortcut method of Sandaram and Evans. Given: Charge of 100 kmol of an equimolar mixture of an equimolar mixture of A, B, and C. Reflux ratio, R = 5 and vapor rate, V = 100 kmol/h. Column has three equilibrium plates, a total condenser, and a partial reboiler. Thus, N = 4. αA,C...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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