# Assume the minimum reflux is controlled by the upper

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: agram below, where the equilibrium curve is plotted from the data, the q-line for Feed 1 is vertical through the point, x = 0.75, the q-line for Feed 2 has a slope of -1 starting from x = 0.50, and the operating line for the upper section between Feed 1 and the condenser is drawn through the two points, {x=0.98, y=0.98} and the intersection of the equilibrium curve and the q-line for Feed 1. From the plot, for the upper section, L/V = (0.98-0.828)/(0.98-0.75) = 0.661. From Eq. (7-27), R = L/D = (L/V)/[1 - (L/V)] = 0.661/(1 - 0.661) = 1.95. Therefore, L = 1.95D = 1.95(123.66) = 241.1 kmol/h and V = L + D = 241.1 + 123.66 = 364.8 kmol/h. In the middle section, between the two feeds, L' = L + F1 = 241.1 + 100 = 341.1 kmol/h and V' = V = 364.8 kmol/h. Therefore, the slope of the middle section operating line = L'/V' = 341.1/364.8 = 0.935. As seen in the diagram below, this line does not cause a pinched region at Feed 2. Therefore, the assumption is correct and Rmin = 1.95. For an operating reflux ratio of 1.2 times mi...
View Full Document

Ask a homework question - tutors are online