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

2081 therefore the slope of the rectifying line lv 035

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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. (7-3), 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/[6-5(0.0824)] = 0.0147. Therefore, y3 = x2 = 0.0147. From Eq. (1), x3 = 0.0147/[6-5(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. (7-12), (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. (7-12), 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 q-line (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. (7-8), R = (L/V)/[1 - (L/V)] = 0.473/(1 - 0.473) = 0.8975. The equation for the rectifying section operating line, using Eq. ((7-9), 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. (7-12) 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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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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