# It shows a q line with a slope of 034 slope q q 1

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Unformatted text preview: lope = q /(q - 1) Therefore, q = slope/(slope-1)=-0.34/(-0.34-1.0)=0.25 From Eq. (7-19), molar fraction vaporized = 1 - q = 1 - 0.25 = 0.75 Analysis: Normal operation (continued) Exercise 7.19 (continued) The material balance for the normal operation is as follows, using the overall balances, F = 100 = D + B and 0.5F = 0.5(100) = 50 = xDD + xBB = 0.90D + 0.05B. Stream Feed Bottoms Distillate Reflux kmol/h 100 47.06 52.94 52.94 mol% methanol 50 5 90 90 Analysis: Abnormal operation For the abnormal operation, first check the overall total material balance using the given data. F = 100 kmol/h. D + B = 53 + 62 = 115 kmol/h. Therefore, it appears that we have 115 - 100 = 15 kmol/h more flow out of the distillation system. Now check the methanol overall material balance using the given data. Methanol flow rate in = 0.51(100) = 51 kmol/h. Methanol flow rate out = 0.80(53) + 0.12(62) = 49.84 kmol/h. Therefore, the methanol balance is close, with only about a 2% discrepancy. Now check the water overall material balance us...
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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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