Unformatted text preview: nimum, L = 1.2(241.1) = 289.3 kmol/h. The vapor rate in the upper section = V = L + D = 289.3 + 123.66 = 413 kmol/h. Therefore the upper section operating line has a slope, L/V = 289.3/413 = 0.700 and passes through the point y = x = 0.98. It intersects the vertical qline at x = 0.75 and, for the slope of 0.700 = (0.98  y)/(0.98  0.75), at y = 0.819. For the middle section, L' = L + F1 = 289.3 +100 = 389.3 kmol/h and V' = V = 413 kmol/h. Therefore, the middle section operating line has a slope of L'/V' = 389.3/413 = 0.943 and intersects the qline for x = 0.75 at y = 0.819. It intersects the qline for Feed 2 at y = 0.543. For the lower section, L" = L' + 0.5F2 = 389.3 + 50 = 439.3 kmol/h and V"=V'  0.5F2= 413  50 = 363 kmol/h. Therefore, the lower section operating line has a slope of L"/V" = 439.3/363 = 1.210 and intersects the qline for Feed 2 at y = 0.543 and the 45o line at xB = 0.05. In the McCabeThiele diagrams below and on the next page for the high, middl...
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 Spring '06
 selebi
 Distillation, pH, Mass Transfer, Vaporliquid equilibrium, section operating line, reflux, feed, constant molar overflow

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