Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Thus methanol purity 30 64630 x 100 785 mol however

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Unformatted text preview: s next. We note that the distillate flow rate for the abnormal operation is almost exactly the same as that for the normal operation. A flow rate equal to that of he leakage passes out the bottom of the column. In normal operation, the water passing out in the distillate = 0.1(53) = 5.3 kmol/h, while for the abnormal operation, the water passing out in the distillate = 0.2(53) = 10.6 kmol/h. Thus, an additional 5.3 kmol/h of water leaves in the distillate. For the abnormal operation, the overhead vapor rate = 53 + 94 = 147 kmol/h and, therefore, 53/147 x 100% = 36% of the overhead vapor (total condensate) is distillate. Thus, if 15 kmol/h of water leaked into the overhead vapor, then, we would expect 0.36(15) = 5.4 kmol/h would be expected to leave with the distillate. This compares very well with the 5.3 kmol/h additional water calculated above by material balance. If the degree of fractionation within the column is about the same as for the normal operation, it could be concluded that a condenser cooling water leak is to blame. To check the cooling water leak, could meter the cooling water in and out of the condenser and see if there is a difference. If the vapor rate is kept constant and the reflux 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 eliminate the leak. Exercise 7.20 Effect on reflux and boilup compositions of reducing the feed rate to a distillation Subject: column when the reflux and boilup rate are held constant Given: Column with 3 theoretical plates, a total condenser, and a partial reboiler. Feed is a saturated liquid of 50 mol% A and 50 mol% B, fed to the bottom tray. At a feed rate of 100 kmol/h, desired products of distillate with 90 mol% A and bottoms of 20 mol% A can be achieved, when a reflux corresponding to L/V = 0.75 is used. Relative volatility of A to B is constant at 3.0. Assumptions: Constant molar overflow. Saturated liquid reflux. Find: Compositions of reflux and boilup when feed rate is inadvertently reduced to 25 kmol/h, while keeping the reflux and boilup flow rates constant. A...
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