{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

# Exercise 725 subject rectification of a mixture of

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online