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

Find check on mass balances around the leaching unit

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Unformatted text preview: eriods that result in the separation of the three solutes. Analysis: Interstitial velocity = u = us/εb = 0.025/0.374 = 0.0668 cm/s. From Eq. (15-140), the solute wave velocities, ui, can be computed for εb = 0.374 and the following values of K, giving: Solute GA G V K 1.18 1.74 2.64 De , cm2 /s 1.94 x 10-7 4.07 x 10-7 3.58 x 10-7 ui, cm/s 0.0225 0.0171 0.0123 Thus, the GA wave will leave first, followed by G and then V. From the results of Exercise 15.36 for the equilibrium theory, the pulse time is 660 s and the elution time is 1732 s. It is expected that with mass transfer taken into account, a shorter pulse time and a longer elution time will be required to avoid significant overlap of the solute waves. Calculate the fixed solute parameters in Carta's equation. Particle radius = Rp = 0.07/2 = 0.035 mm = 0.00 35 cm From Eq. (15-154), the overall mass-transfer coefficient = 1 1 1 k= = = (1) 2 2 Rp Rp 0.0035 0.0035 8167 × 10 −7 . + 0.778 + + −3 . 15De De 3kc 15De 3(15 × 10 ) From Eq. (15-194), with εb = 0.374, β = From Eq. (15-192), with z = 47 cm, n f = εb 0.374 0.597 = = 1 − ε b K (1 − 0.374) K K 1 − ε b kz (1 − 0.374) k (47) = = 1178k , εb u 0.374(0.0668) (2) (3) Exercise 15.37 (continued) Analysis: (continued) Calculations with the given values of K and De, and Eqs. (1), (2), and (3) give: Solute GA G V k, s-1 0.2005 0.359 0.327 k/K, s-1 0.170 0.206 0.124 β 0.506 0.343 0.226 nf 236 423 385 Now, we must choose a pulse time, tF, and elution time, tE. Assume a pulse time that is 50% of that computed by the equilibrium theory of Exercise 15.36, namely 0.5(660) = 330 s. For now, leave the elution time at 1732 s because it does not affect the first set of solute waves. Compute values of c/cF as a function of time using Carta's infinite-series equation (15-46) using a spreadsheet, being careful to include a sufficient number of terms in the series to obtain a converged result. In this exercise, that number in Eq. (15-46) is not more than m = 20. The result is that with tF = 330 s, the GA and G waves overlap significantly at z = 47 cm. Try tF = 0.5(330) = 165 cm. Now, as seen in the table on the next page and the plot on the following page, only a small overlap of the GA and G waves occurs. An even smaller overlap of the G and V waves is evident. However, although not shown, a significant overlap of the first V wave and the second GA wave occurs with tE = 1,732 s. Therefore, we must increase the value of tE. If it is increased to 2,732 s, the results are acceptable, as shown in the table and plot below, where the first set of solute waves and the second wave for GA are shown. Thus, a possible cycle, taking into account mass transfer is: Pulse: Elute: Pulse: Elute: etc. 165 s 2,732 s 165 s 2,732 s Because of the short pulse time and the long elution time, this is probably not a desirable cycle. Analysis: (continued) Exercise 15.37 (continued) Z = 47 cm, tF = 165 s, __ tE = 2,732 s Exercise 15.37 (continued) Analysis: (continued) Exercise 15.38 Subject: Separati...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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