# 13xn 1 xn 1 1 313xn 1 at total reflux numbering

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Unformatted text preview: 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 from a spreadsheet are, Equilbrium stage 1 (still) 2 (bottom plate 3 4 5 (top plate) reflux drum x 0.00100 0.00412 0.01679 0.06587 0.22554 0.54603 y 0.00412 0.01679 0.06587 0.22554 0.54603 By overall total material balance, F = 100 = D + B (3) By overall benzene material balance, FxF = (100)(0.20) = 20 = DxD + BxB = D(0.54603) + B(0.001) (4) Solving Eqs. (3) and (4), D = distillate in reflux drum = 36.51 kmoles B = bottoms in still = 63.49 kmoles Exercise 7.25 (continued) Graphical Method: On the McCabe-Thiele plot on the next page, the equilibrium curve is computed from Eq. (1). The rectification section operating line is the 45o line. Five stages are stepped off from...
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## This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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