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

033 and ww0 1 02 08 20 mol vaporized thus eq 13 3

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Unformatted text preview: istillate from Eq. (2), until the value of 0.45 is reached. xW yD - xW 1/(yD - xW) Increment of integral 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.10 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.4681 0.4691 0.4654 0.4573 0.4446 0.4274 0.4057 0.3794 0.3486 0.3132 0.2733 0.2517 0.2289 0.2050 0.1800 0.1538 0.1265 0.0980 0.0685 2.1362 2.1320 2.1485 2.1867 2.2491 2.3396 2.4651 2.6358 2.8688 3.1926 3.6584 3.9730 4.3683 4.8777 5.5565 6.5022 7.9062 10.2000 14.6067 0.0427 0.0428 0.0434 0.0444 0.0459 0.0480 0.0510 0.0550 0.0606 0.0685 0.0382 0.0417 0.0462 0.0522 0.0603 0.0720 0.0905 0.1204 ln(Wo/W) 0.0427 0.0855 0.1288 0.1732 0.2191 0.2671 0.3181 0.3732 0.4338 0.5023 0.5405 0.5822 0.6284 0.6806 0.7409 0.8129 0.9034 1.0275 W/Wo Cum yD 0.9582 0.9181 0.8791 0.8410 0.8033 0.7656 0.7275 0.6885 0.6480 0.6051 0.5825 0.5587 0.5334 0.5063 0.4767 0.4436 0.4052 0.3579 0.7587 0.7482 0.7363 0.7231 0.7083 0.6919 0.6738 0.6537 0.6314 0.6065 0.5930 0.5785 0.5630 0.5462 0.5277 0.5073 0.4839 0.4561 From this table, the residue is 35 lbmol at a benzene mole fraction in the residue of about 0.02. Exercise 13.11 Subject: Batch distillation of a mixture of benzene (B) and toluene (T) in a column with 2 equilibrium stages and a reboiler (3 total stages), under conditions of a constant reflux ratio. Given: Feed of 1 kmole containing 50 mol% B and 50 mol% T. Distillation with a reflux of L/D = R = 4. Constant relative volatility of 2.5 for B with respect to T. Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser. Find: kmoles and composition of cumulative distillate when the instantaneous distillate is 55 mol% B. Analysis: Make calculations in terms of B, the more volatile component. Eq. (13-2) applies, where yD is the mole fraction of B in the instantaneous vapor leaving the top stage, and xW is the mole fraction of B in the liquid in the reboiler. 0.50 W 1 dxW ln 0 = ln = (1) xW y − x W W D W From Eq. (4-8), the vapor-liquid equilibrium curve is given by, αx 2.5x y= = (4) 1 + x (α − 1) 1 + 15x . This is the equilibrium curve for applying the McCabe-Thiele method. The relationship between yD and xW in Eq. (1) is obtained from a McCabe-Thiele diagram by drawing a series of operating lines of slope = L/V = R/(1+R) = 4/5 = 0.8 For each operating line, starting from the intersection with the 45o line, which is yD, 3 stages are stepped off to determine the corresponding xW. A typical construction that starts from yD = 0.90 is shown on the next page, where the xW = 0.465. Other sets of values are given in the following table, which also includes values of the integrand, f = 1/(yD - xW), from which the integral is evaluated by the trapezoidal rule with variable increments of ∆xW. For example, the increment of the integral from x1 to x2 is (x1 - x2)(f1/2 + f2/2). The increments are summed over the region from xW = 0.40 to the value corresponding to 70 mol% distilled. xW yD from McCabe-Thiele f = 1/(yD - xW) 0.500 0.465 0.365 0.233 0.172 0.125 0.91 0.90 0.85 0.75 0.65 0.55 2.439 2.300 2.062 1.934 2.092 2.353 Increment of Integral Cumulative ln(W0/W) 0.083 0.218 0.264 0.123 0.105...
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