# The equilibrium curve is computed from eq 7 3 3x x y

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Unformatted text preview: 3x x y= = 1 + x ( - 1) 1 + 2 x The given mole fractions are: xF = 0.50 xD = 0.90 xB = 0.20 The rectification section operating line has a slope of 0.75 and passes through point {0.9, 0.9}. The q-line is vertical at x = 0.5. The plot shows almost perfect agreement with the desired separation for a feed rate of 100 kmol/h. Analysis: (continued) Exercise 7.20 (continued) For the base case of F = 100 kmol/h, the material balance equations are F = D + B and xFF = 50 = xDD + xBB = 0.90D + 0.20B. Solving, these equations, along with V = L + D, V/L = 0.75, V = V , and L = L + F , gives the following results: Stream Feed Distillate Bottoms Reflux, L Overhead vapor, V Liquid to reboiler, L Vapor from reboiler, V Flow rate, kmol/h 100.00 42.86 57.14 128.58 171.44 228.58 171.44 Mol% A 50 90 20 Mol% B 50 10 80 When the feed rate is reduced to 25 kmol/h, the reflux rate, L, is maintained at 128.58 kmol/h and the boilup, V , is maintained at 171.44 kmol/h. Therefore by material balances, L = L + F = 128...
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