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

15vbmin 1150407 0468 from eq 7 12 the slope of

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: ing from the upper line, which passes through the point x = 0.96 on the 45o line. Equilibrium stages are stepped off starting from the distillate composition and switching operating lines at the appropriate times to determine the optimal feed stage and sidestream stage locations. A separate McCabe-Thiele diagram is used for the upper section to achieve better accuracy. As seen 19 equilibrium stages plus a partial reboiler are required. The optimal feed stage is 9 from the top and the optimal sidestream stage is 17 from the top. Analysis: (continued) Exercise 7.36 (continued) Exercise 7.37 Subject: Distillation at 1 atm of a mixture of toluene and phenol for a given boilup ratio, with an alternative using an interreboiler. Given: 1,000 kmol/h of a saturated liquid feed of 25 mol% toluene. Distillate is 98 mol% toluene and bottoms is 2 mol% toluene. The base unit consists of a total condenser, a plate column, and a partial reboiler, with a boilup ratio, VB = V / B = 1.15 times minimum. The alternative unit adds an interreboiler midway in the stripping section to provide 50% of the boilup. A table of T-y-x phase equilibrium data Assumptions: Constant molar overflow. Find: For each unit, determine the number of theoretical stages. For the alternative unit, determine the temperature of the interreboiler stage. Analysis: Toluene is the more volatile unit. Base unit: From the McCabe-Thiele diagram below, the minimum boilup ratio is determined from the slope of the stripping section operating line that intersects the equilibrium curve at the feed composition of xF = 0.25. ( L /V ) min = 0.815 − 0.02 = 3.46 0.25 − 0.02 From Eq. (7-28), (VB )min = 1 ( L /V ) min −1 = 1 = 0.407 3.46 − 1 Exercise 7.37 (continued) Analysis: Base Case (continued) For column operation, VB = 1.15(VB)min = 1.15(0.407) = 0.468 From Eq. (7-12), the slope of the stripping section operating line is, V + 1 0.468 + 1 L /V = B = = 314 . VB 0.468 As shown in the McCabe-Thiele diagram, below, a line of this slope is drawn through the point x = y = 0.02 until it intersects the q-line. Equilibrium stages are stepped off, starting from the distillate point at y = x = 0.98. The optimal feed stage location is located as shown at stage 5 from the top. The total number of stages required is just less than 8, with one of those stages being the partial reboiler. Exercise 7.37 (continued) Analysis: Base Case (continued) Exercise 7.37 (continued) Analysis: (continued) Alternative unit with Interreboiler: For column operation with an interreboiler that provides 50% of the reboiler duty, the operating boilup ratio between the reboiler and the interreboiler is 50% of 0.468 or 0.234. From Eq. (7-12), the slope of the operating line in this region is: L /V = VB + 1 0.234 + 1 = = 5.27 VB 0.234 Between the interreboiler and the feed stage, the slope of the operating line remains at 3.14, based on a material balance around the column section from the bottoms to the region between the interreboiler and the f...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online