9 ldxd 675 ldxd 6 assume equilibrium in the partial

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Unformatted text preview: reboiler. Using the given with its definition in Eq. (7-2). For the partial condenser, = 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, = 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 Analysis: (continued) Graphical Method: On the McCabe-Thiele diagram below, the equilibrium curve is obtained from Eq. (7-3), x 2.5x y= = 1 + x ( - 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 =...
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