Unformatted text preview: mol/h and V' = V = 180.5 kmol/h Therefore, the middle section operating line has a slope of L'/V' = 163.8/180.5 = 0.908 and intersects the qline for x = 0.6 at y = 0.794. It intersects the vertical qline at x = 0.4 and for the slope of 0.908 = (0.794  y)/(0.6  0.4), y = 0.613. For the lower section, L" = L' + F1 = 163.8 + 125 = 288.8 kmol/h and V"=V' = 180.5 kmol/h Therefore, the lower section operating line has a slope of L"/V" = 288.8/180.5 = 1.60 and intersects the qline for x = 0.4 and y = 0.613. As seen in Fig. 7.27(c), the mole fraction of A in the bottoms, xB , is determined from the intersection of the operating line for the lower section with the yaxis. Thus, 1.60 = (0.613  0)/(0.4  xB). Solving, xB = 0.0169 for component A. Since the bottoms contains 5 kmol/h of A, the bottoms rate = B = 5/0.0169 = 295.9 kmol/h. Thus, the bottoms contains 290.9 kmol/h of water. But, the flow rate of water entering in the two feeds = 125 + 83.3  100 = 108.3 kmol/h. Therefore, the o...
View
Full
Document
 Spring '06
 selebi
 Distillation, pH, Mass Transfer, Vaporliquid equilibrium, section operating line, reflux, feed, constant molar overflow

Click to edit the document details