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

# B composition of the two products analysis from the

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Unformatted text preview: 3 98.90 98.50 98.35 Raoult’s law xA yA -0.018 -0.026 0.276 0.350 0.459 0.547 0.589 0.672 0.729 0.793 0.876 0.910 0.954 0.967 0.992 0.995 1.007 1.005 Experimental xA yA 0.025 0.048 0.129 0.205 0.250 0.349 0.354 0.454 0.497 0.577 0.692 0.742 0.843 0.864 0.940 0.948 0.994 0.993 The Raoult’s law values are in very poor agreement with the experimental values. Analysis: (e) (continued) Exercise 4.9 (continued) Analysis: (e) (continued) Exercise 4.9 (continued) Comparison with Experimental Data Exercise 4.10 Subject: Continuous, single stage distillation of A and B to produce a distillate and bottoms. Given: Saturated liquid feed of 50 mol% A and 50 mol% B fed to a still at 40 mol/h. Relative volatility = αA,B = 2. Bottoms rate = 30 mol/h (a) Total condenser with a reflux ratio = 1. (b) No reflux. Assumptions: Still is an equilibrium stage. Find: (a) Composition of the two products. (b) Composition of the two products. Analysis: From the definition of the relative volatility, α A,B = yA x B yA 1 − xA = =2 xA y B xA 1 − yA (1) (a) Distillate = D = F - W = 40 - 30 = 10 mol/h Material balance for A: 0.5(40) = 20 = yA(10) + xA(30) (2) Solving Eqs. (1) and (2) simultaneously by eliminating yA , we obtain: 2 3 xA + 3 xA − 2 = 0 (3) Solving Eq. (3), a quadratic equation, get only one postive root: xA = 0.4575, xB = 0.5425 for the bottoms Substitution into Eq. (2), gives, yA = 0.6275, yB = 0.3725 for the distillate (b) Note that the solution to Part (a) was independent of the reflux ratio. Accordingly, the solution to Part (b) is the as for Part (a) Exercise 4.11 Subject: Distillation of an acetone (A) - water (B) mixture that is partially vaporized. Given: Feed is 57 mol% A and 43 mol% B as a liquid at 125oC and 687 kPa. It is flashed across a valve to the column pressure of 101.3 kPa, with a resulting temperature of 60oC. Vapor-liquid equilibrium data at column pressure. Enthalpy data at column conditions. Compositions of the distillate and bottoms. Assumptions: Feed is at equilibrium downstream of the feed valve. Will have to check if feed valve operates adiabatically. Given heat capacities are for the liquid and are constant. Heats of vaporization are constant. No effect of pressure on enthalpy. Find: Mole ratio of liquid to vapor in the feed downstream of the valve. Construct an H-x-y diagram. Analysis: From the equilibrium data, at 60oC, xA = 0.50 and yA = 0.85 . Take a basis of F = feed rate = 1 kmol/s. Total material balance around feed valve: F = 1 = V + L (1) Acetone material balance around feed valve: 0.57(1) = 0.85V + 0.50L Solving Eqs. (1) and (2) simultaneously, V = 0.2 kmol/s and L = 0.8 kmol/s (2) Therefore, after the valve, moles L/moles V = 0.8/0.2 = 4 Now check whether valve is operating adiabatically. Enthalpy of liquid entering valve = 0 (as given) Enthalpy of feed after the valve, using given enthalpies = 27,200(0.2) + (-5,270)(0.8) = 1224 kJ/s Therefore, the enthalpy increases across the valve by 1224 kJ/s To construct an enthalpy diagram for 1 atm pressure, take as an enthalpy datum, A and B as liquids a...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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