05 182 nog 10672 1488 thus the gas phase resistance

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Unformatted text preview: -56), EOV = 1 - exp(- N OG ) = 1 - exp(-1.488) = 0.774 or 77.4% This is in very good agreement with the performance data. Exercise 7.42 Subject: Estimation of efficiencies, EMV and Eo from EOV for methanol-water mixture, as measured with a small Oldershaw column. Given: Column conditions from Exercise 7.41. Find: EMV and Eo Analysis: Case 1: Assume complete mixing on the trays. EMV = EOV = 0.65 or 65% Case 2: Assume plug flow of liquid with no longitudinal diffusion. Take conditions at the top of the column. From Eq. (7-7), L/V = R/(R + 1) = 0.947/(1 + 0.947) = 0.486 From Eq. (6-33), = m/(L/V) From the vapor-liquid equilibrium data given in Exercise 7.41, m = dy/dx = (1 - 0.915)/(1 - 0.793) = 0.41 = 0.41/0.486 = 0.844 From Eq. (6-32), EMV = 1 1 [ exp(EOV ) - 1] = {exp [0.844(0.65)] - 1} = 0.754 or 75.4% 0.844 The actual value of EMV probably lies inbetween 65% and 75.4%, or say 70%. From Eq. (6-37), assuming that the equilibrium and operating lines are straight, Eo = log [1 + EMV ( - 1) ] lo...
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