7 5 becomes l d y m 1 x m 2 x d 2 v v solving eq

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Unformatted text preview: M +1 = x M + 2 + x D (2) V V Solving Eq. (2) for xD , L y M +1 - x M +2 0.740 - (0.763)(0.680) V xD = = = 0.933 D /V 0.237 Therefore the composition of the distillate is 93.3 mol% A and 6.7 mol% B. Test 2: Because y M +1 = x M + 2 and y M = x M +1 , operation is at total reflux, i.e. reflux ratio = infinity. In this case, Eq. (2) can not be solved for xD because D/V = 0. We can not determine the composition of the distillate. We do know that it must at least 75 mol% A. Exercise 7.11 Subject: Five procedures for continuous distillation of a mixture of benzene (A) and toluene (B). Given: Operation at 1 atm to produce a distillate of 80 mol% benzene (i.e. yD = 0.8) from a saturated liquid feed of 70 mol% benzene (xF = 0.7) . Procedures are: 1. No column. Just a partial condenser on top of a partial reboiler. Feed is to the reboiler. Reflux ratio, L/D = 0.5. Vapor distillate is totally condensed. 2. Same as 1 except that one equilibrium stage sits between the condenser and reboiler. 3. Same as 1 except...
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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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