9 1254 836 molh o2 and 791 0025 79075 molh of

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: h O2 and 79.1 - 0.025 = 79.075 mol/h of N2. The mol% N2 in the overhead vapor = 79.075/(8.36 + 79.075) x 100% = 90.4 %. (b) Assume constant molar overflow. Then because the feed is assumed to be a saturated liquid, the moles of vapor generated in the reboiler per 100 moles of feed = mol/h of overhead vapor = 8.36 + 79.075 = 87.435 moles per 100 moles of feed. (c) Use a y-x diagram for N2 because it is the more volatile. The slope of the operating line is L/V = 100/87.435 = 1.14. At the top of the column, the operating line terminates at a (y-x) of (0.904, 0.791). At the bottom of the column, with a total reboiler, the operating line terminates at a (y-x) of (0.002, 0.002). To determine the number of equilibrium stages, it is convenient to use two diagrams, the usual one and a second one for just the very low mole fraction region so as to gain accuracy in the region of the lower end of the operating line. From the two diagrams, it is seen that just less than 8 equilibrium stages are required. Analysis (d) (contin...
View Full Document

This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

Ask a homework question - tutors are online