# This still does not account for the effect of

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Unformatted text preview: ontinued) Exercise 7.41 (continued) (b) From Eq. (7-42), Eo = 13.3 - 66.7 log Take the viscosity as that of the feed = 0.34 cP Eo = 13.3 - 66.7 log (0.34) = 44.6% This is poor agreement with the performance data. (c) From Eq. (7-43), Eo = 50.3()-0.226 At the feed composition, x = 0.36 and y = 0.71. Therefore, from Eqs (2-19) and (2-21 combined, the relative volatility is, = ( y/x)/[(1 - y)/(1 - x)] = (0.71/0.36)/(0.29/0.64) = 4.4 Exercise 7.41 (continued) Analysis: Eo = 50.3[(4.4)(0.34)]-0.226 = 45.9% Now correct for length of liquid path from Fig. 7.5. Column diameter = 6 ft. Assume length of liquid path = 70% of column diameter = 0.7(6) = 4.2 ft. From Fig. 7.5, correction to be added = 10%. Therefore corrected Eo = 45.9 + 10 = 55.9% This also appears to be low. (d) From Eq. (6-56), NOG = - ln (1 - EOV). Therefore, EOV = 1 - exp(-NOG) Use Eqs. (6-62, (6-64), (6-66), and (6-67) as in Example 6.7. Carry out the calculations at the bottom tray based on methanol diffusion. Conditions are: Molar flow...
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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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