If the vapor rate is kept constant and the reflux

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Unformatted text preview: ux rate is increased, then the distillate rate must be decreased. Assume a vapor rate of 147 kmol/h, with 30 kmol/h to distillate and 117 kmol/h to reflux. Then, 30/117 x 100% = 25.6% of the overhead vapor is distillate. Therefore, the water leak to the distillate would be 0.256(15) = 3.84 kmol/h. If the fractionation were otherwise the same as for normal operation so that the overhead vapor was 90 mol% methanol, the dilution with leakage would result in 0.1(30 - 3.84) + 3.84 = 6.46 kmol/h of water in 30 kmol/h. Thus, methanol purity = (30 - 6.46)/30 x 100% = 78.5 mol%. However, the higher reflux ratio would increase the fractionation, so as to increase the purity above this value. A further increase in fractionation could be achieved, if the feed were condensed to a saturated liquid and additional heat was transferred in the reboiler. But, even if a pure methanol overhead vapor were achieved, the methanol purity after dilution with the water leakage would be : (30 - 3.84)/30 x 100% = 87.2 mol% methanol. Must elimina...
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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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