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

Exercise 725 subject rectification of a mixture of

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Unformatted text preview: tial reboiler and partial condenser, using subscripts B for streams leaving the reboiler and D for streams leaving the partial condenser. Partial reboiler: Total material balance: F + LD = B + VB or 100 + LD = 75 + VB (3) Benzene material balance: FxF + LDxD = BxB + VByB (4) or 69.4 + LDxD = 75(0.625) + VByB = 46.9 + VByB Partial condenser: Total material balance: VB = D + L D = 7 5 + L D (5) Benzene material balance: VByB = DyD + LDxD = 75(0.9) + LDxD = 67.5 + LDxD (6) Assume equilibrium in the partial condenser and partial reboiler. Using the given α with its definition in Eq. (7-2). For the partial condenser, α = 2.5 = yD (1 - xD)/xD(1 - yD) = 0.9(1 - xD)/xD(1 -0.9) = 9(1 - xD)/xD (7) Solving Eq. (7), xD = 0.783 For the partial reboiler, α = 2.5 = yB (1 - xB)/xB(1 - yB) = yB (1 - 0.625)/0.625(1 - yB) = 0.6 yB/(1 - yB) (8) Solving Eq. (8), yB = 0.806 Eqs. (3) through (6), are 4 equations in 2 unknowns, VB and LD. We only need 2 of the 4 equations. Using Eqs. (3) and (4), 100 + LD = 75 + VB (3) 69.4 + LD(0.783) = 46.9 + VB(0.806) (9) Solving Eqs. (3) and (9), LD = 98 moles /100 moles of feed and VB = 123 moles/100 moles of feed Exercise 7.24 (continued) Analysis: (continued) Graphical Method: On the McCabe-Thiele diagram below, the equilibrium curve is obtained from Eq. (7-3), αx 2.5x y= = 1 + x (α − 1) 1 + 15x . The rectification section operating line is located, as shown, so that two equilibrium steps, one for the partial condenser and one for the partial reboiler, are stepped off between xC = 0.9 (from the total condenser) and xB = 0.625. The measured slope of the operating line = LD/VB = 0.8. Combining this with Eq. (3), VB = 125 moles/100 moles of feed, which is close to the analytical value. Exercise 7.25 Subject: Rectification of a mixture of benzene and chlorobenzene at total reflux. Given: Feed of 100 kmol of 20 mol% benzene and 80 mol% chlorobenzene. Column has 4 theoretical plates, a total condenser, a reflux drum, and a still to vaporize the feed. At an operating pressure of 1 atm, relative volatility of benzene with respect to chlorobenzene = α = 4.13. Operate at total reflux with holdups only in the reflux drum and the still. Want liquid in the still with 0.1 mol% benzene. Assumptions: Perfect mixing to give uniform compositions in the reflux drum and the still. Find: Moles of liquid in the still at steady state. Analysis: This exercise can be solved analytically or graphically. Since benzene is the more volatile component, Eq. (7-3) gives the equilibrium relation at the still or any of the 4 plates, n, as, yn = αxn 4.13xn = 1 + xn (α − 1) 1 + 313xn . (1) At total reflux, numbering stages up from the bottom, xn+1 = yn (2) Analytical Method: Start at the bottom, stage 1, with x1 = 0.001. Solve for y1 (vapor leaving the still) with Eq. (1). Then, from Eq. (2), x2 = y1. Continue in this manner, solving alternately Eq. (1) and then Eq. (2), until y5 (vapor leaving the top plate) is reached. Then, x in the reflux drum = y5. The results...
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