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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,
qAg cNa +
2.5 − x 2.5 − x
q Na cAg +
6.25 − 5x + x 2
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
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
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:
(40 / 60) R + 262.5
0.6667 R + 262.5
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
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.
- Spring '11
- The Land