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

5 67 nd nv ne 3c 10 2c 6 c 4 total condenser have

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: 55 kmol/h (reflux ratio = 155/45), gives: kmol/h: Component Feed Distillate Bottoms C3 5 4.996 0.004 iC 4 15 14.987 0.013 nC 4 25 24.587 0.413 iC 5 20 0.354 19.646 nC 5 35 0.077 34.923 Total 100 45.001 54.999 Exercise 5.25 Subject: Degrees of freedom analyses for a partial reboiler and a total condenser. Given: A partial reboiler with exiting vapor and liquid in equilibrium. A total condenser. Find: Number of variables, number of independent equations, and degrees of freedom. Analysis: Partial reboiler: Have 3 streams with C + 3 variables each, plus the heat transfer rate. Therefore, NV = 3(C + 3) + 1 = 3C + 10 The independent relationships are: C Component material balances 1 Enthalpy balance 1 Equality of temperatures of exiting vapor and liquid 1 Equality of pressures of exiting vapor and liquid C Component phase equilibrium between exiting vapor and liquid 3 Mole fraction summations for 3 streams Therefore, NE = 2C + 6 From Eq. (5-67), ND = NV - NE = (3C + 10) - (2C + 6) = C + 4 Total condenser: Have 2 streams with C + 3 variables each, plus the heat transfer rate. Therefore, NV = 2(C + 3) + 1 = 2C + 7 The independent relationships are: C Component material balances 1 Enthalpy balance 2 Mole fraction summations for 2 streams Therefore, NE = C + 3 From Eq. (5-67), ND = NV - NE = (2C + 7) - (C + 3) = C + 4 Exercise 5.26 Subject: Given: Degrees of freedom analyses for a stream mixer and a stream divider. A steam mixer and a stream divider, each with heat transfer. Find: Number of variables, number of independent equations, and degrees of freedom. Analysis: Stream mixer with two inlet streams: Have 3 streams with C + 3 variables each, plus the heat transfer rate. Therefore, NV = 3(C + 3) + 1 = 3C + 10 The independent relationships are: C Component material balances 1 Enthalpy balance 3 Mole fraction summations for 3 streams Therefore, NE =C + 4 From Eq. (5-67), ND = NV - NE = (3C + 10) - (C + 4) = 2C + 6 Stream divider with two exiting streams: Have 3 streams with C + 3 variables each, plus the heat transfer rate. Therefore, NV = 3(C + 3) + 1 = 3C + 10 The independent relationships are: 1 Total material balance Component mole fraction equalities for exiting streams relative to the feed 2(C - 1) 1 Enthalpy balance 1 Equality of temperatures of exiting streams 1 Equality of pressures of exiting streams 3 Mole fraction summations for 3 streams Therefore, NE = 2C + 5 From Eq. (5-67), ND = NV - NE = (3C + 10) - (2C + 5) = C + 5 Exercise 5.27 Subject: Degrees of freedom analysis for the specifications of a distillation column. Given: Feed of 10 mol% benzoic acid and 90 mol% maleic anhydride. Column has a total condenser and a partial reboiler. Operation is at 100 torr with a reflux ratio of 1.2 times minimum. Desired products are a distillate of 99.5 mol% maleic anhydride and a bottoms of 0.5 mol% anhydride. Find: Whether problem is completely specified. Analysis: From operation (b) in Table 5.4, N D = 2N + C + 9 = 2N + 11 The given specifications are equivalent to: Pressure at each stage, including partial reboil...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online