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

Thus methanol purity 30 64630 x 100 785 mol however

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online