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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 ionexchange 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/(1000262.5) = 0.1186
The wt% NaCl in the solution outside the resin is: 0.06(1000)/(1000262.5) 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 gastoparticle masstransfer and heattransfer coefficients
in a fixedbed adsorption column.
Given: Fixed bed of 2foot diame...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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