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

Using data in the table that accompanies example 49

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Unformatted text preview: flow rates, of overflow and underflow, and the percentage of oil in the feed that is recovered in the overflow. Analysis: The flakes contain (0.19)(100,000) = 19,000 kg/h of oil and (100,000 – 19,000) = 81,000 kg/h of oil-free solids. However, all of the oil is not leached. For convenience in the calculations, lump the unleached oil with the oil-free solids to give an effective solids. The flow rate of unleached oil = (81,000)(0.5/99.5) = 407 kg/h. Therefore, take the solids flow rate as (81,000 + 407) = 81,407 kg/h. Take the oil in the feed as just the amount that is leached or (19,000 – 407) = 18,593 kg/h. Therefore, in the feed, F, YA = (81,407/18,593) = 4.38 XB = 1.0 The sum of the liquid solutions in the underflow and overflow include 200,000 kg/h of hexane and 18,593 kg/h of leached oil. Therefore, for the underflow and overflow, XB = YB = [18,593/(200,000 + 18,593)] = 0.0851 This is a case of variable solution underflow. Using data in the table that accompanies Example 4.9, and the conversion shown there of that table to XA as a function of XB, we obtain XA = 2.09. Using the algebraic method, since the flow rate of solids in the underflow = 81,407 kg/h, the flow rate of liquid in the underflow = 81,407/2.09 = 38,950 kg/h. The total flow rate of underflow is U = 81,407 + 38,950 = 120,357 kg/h. By mass balance, the flow rate of overflow = 300,000 – 120357 = 179,643 kg/h. Exercise 4.52 (continued) Now compute the compositions of the underflow and overflow. In the overflow, with XB = 0.0851, the mass fractions of solute B and solvent C are, respectively, 0.0851 and (1 – 0.0851) = 0.9149. In the underflow, using XA = 2.09 and XB = 0.0851, the mass fractions of solids, B, and C, are, respectively, [2.09/(1 + 2.09)] = 0.676, 0.0851(1 – 0.676) = 0.0276, and (1 – 0.676 – 0.0276) = 0.2964. From above, the oil flow rate in the feed is 19,000 kg/h. The oil flow rate in the overflow = YBV = 0.0851(179,643) = 15,288 kg/h. Thus, the percentage of the oil in the feed that is recovered in the overflow = 15,288/19,000 = 0.805 or 80.5%. Exercise 4.53 Subject: Leaching of Na2CO3 from a solid by water. Given: Lo = 3,750 kg/h of a solid containing 1,350 kg/h Na2CO3, contacted with So = 4,000 kg/h of water. Assumptions: Ideal leaching stage so that compositions of overflow and underflow liquid are equal. Underflow contains 40 wt% water on a solute-free basis. Find: Compositions and flow rates, L1 and S1, respectively, of overflow and underflow. Analysis: Let x = mass fraction of Na2CO3 (solute). Lo + So = 3,750 + 4,000 = 7,750 = L1 + S1 (a) Total balance: Na2CO3 balance: x Lo Lo = 1,350 = x L1 L1 + xS1 S1 Insoluble solids balance: (3,750 − 1,350) = 2,400 = (1 − 0.40) 1 − xS1 S1 = 0.60 1 − xS1 S1 Equilibrium: x L1 = xS1 S1 S1 − (0.60) 1 − xS1 S1 = (1) (2) (3) xS1 (4) 0.4 + 0.6 xS1 Eqs. (1) to (4) can be reduced to one quadratic equation in xS1 . Solving, xS1 = 01189 ; x L1 = 0.2523 ; S1 = 4,540 kg / h ; and L1 = 3,210 kg / h . The material balance may be summari...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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