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

442 gas is air containing water vapor which flows

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Unformatted text preview: uilibrium. Since we begin with 2.5 eq of each ion, then, also, x = eq of Na+ in the solution at equilibrium. Therefore, xx qAg cNa + x2 1 50 K= = = (1) 2.5 − x 2.5 − x q Na cAg + 6.25 − 5x + x 2 1 50 Analysis: To solve Eq. (1), the need the molar selectivity coefficient, which depends on xNa+ in the table above. Assume a value of 12. Eq. (1), which is a quadratic equation in x becomes: 11 x2 - 60 x + 75 = 0. Solving, x = 1.94 eq of Na+ in solution. Therefore, the equivalent fraction of Na+ in solution = 1.94/2.5 = 0.776. So we need to assume a higher value of K. Using the above table of data, assume a value of 15.2 for the molar selectivity coefficient. Then Eq. (1) becomes: 14.2 x2 - 76 x + 95 = 0. Solving, x = 2.0 and the equivalent fraction of Na+ in solution = 2/2.5 = 0.80. From the above table, K = 15.6. Then Eq. (1) becomes: 14.6 x2 - 78 x + 97.5 = 0. Solving x = 1.995. The value of K = 15.6 is close enough. Therefore, the equivalent fractions at equilibrium are: In the resin, yNa = 0.202 and yAg = 0.798 In the methanol solution, xNa+ = 0.798 and xAg+ = 0.202 Exercise 15.15 Subject: Recovery of glycerol from an aqueous solution of NaCl by ion exclusion. Given: Feed solution of 1,000 kg containing 6 wt% NaCl, 35 wt% glycerol, and 47 wt% water. Dowex 50 ion-exchange resin in the sodium form, prewetted to contain 40 wt% water. The equilibrium distribution coefficient for glycerol is given by: mass fraction glycerol in solution inside the resin Kd = mass fraction glycerol in solution outside the resin Values of Kd depend on the mass fraction of glycerol in the solution outside the resin and on the wt% NaCl in the solution outside the resin, as given in a table in the problem statement. Assumptions: The remainder of the feed solution (12 wt%) is inert material. During ion exclusion, no additional water or NaCl will transfer to the solution inside the resin. Find: The kg of resin on a dry basis needed to transfer 75% of the glycerol to the solution inside the resin. Analysis: Let R = kg of dry resin needed. Then the water in the prewetted resin = (40/60) R. The original solution contains 0.35(1000) = 350 kg of glycerol. For a transfer of 75%, the equilibrium solution inside the resin will contain 0.75(350) = 262.5 kg of glycerol, leaving 87.5 kg of glycerol in the equilibrium solution outside the resin. Therefore, we can write Kd as: 262.5 3000 (40 / 60) R + 262.5 Kd = = 87.5 0.6667 R + 262.5 100 Solving, R = 3000 − 262.5Kd = 4500 − 393.75Kd 0.6667 (1) At equilbrium, the mass fraction of glycerine in the solution outside the resin is: 87.5/(1000-262.5) = 0.1186 The wt% NaCl in the solution outside the resin is: 0.06(1000)/(1000-262.5) x 100% = 8.13 wt% Interpolate the given table for Kd to obtain: (0.759)(0.467) + (0.914)(0.533) = 0.842 = Kd Therefore, from Eq. (1), R = 4500 - 393.75(0.842) = 4,170 kg dry resin. Exercise 15.16 Subject: Estimation of external gas-to-particle mass-transfer and heat-transfer coefficients in a fixed-bed adsorption column. Given: Fixed bed of 2-foot diame...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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