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

# It is expected that with mass transfer taken into

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Unformatted text preview: uivalent fraction of Ca 2+ = xCa 2+ = 0.0072 / (0.0072 + 0.000855) = 0.9883 Since no NaCl in the feed is exchanged, CL = 0.0072 eq/L From Table 15.5, KCa 2+ ,Na + = 5.2 / 2 = 2.6 From Eq. (15-181), 2.6 Q = 2.3 eq/L y 2+ 1 − 0.8938 yCa 2+ 2.3 = 830.6 = Ca = 0.1188 0.0072 0.8938 1 − yCa 2+ 1 − yCa 2+ Solving, yCa 2+ = 0.99986 equivalent fraction 3 Bed volume = (56.7)(10) = 567 ft or 16,060 L Total Bed Capacity = 2.3 (16,060) = 36,940 eq Ca2+ absorbed by resin = 0.99986(36,940) = 36,935 eq Ca2+ entering bed in feed solution = 0.0072(3,219) = 23.18 eq/min 36,935 Ideal loading time to breakthrough = t L = = 1, 593 min = 26.6 h 23.18 Exercise 15.35 (continued) Analysis: (continued) (c) Loading wave front velocity = uL = L/tL= 10/1,593 = 0.00628 ft/min or 0.191 cm/min 1.5 (d) Flow rate of regeneration solution = ( 3,219 ) = 321.9 L/min 15 (e) Displacement time = time for 321.9 L/min to displace liquid in the voids Void volume = 0.38 (16,060) = 6,103 L 6103 Displacement time tD = = 19 min 321.9 (f) For a 26 wt% NaCl solution at 25°C, the density from Perry's Chemical Engineers' 3 Handbook = 1.19443 g/cm 3219(1,000)(119443)(0.26) . . Flow rate of Na+ in regeneration solution = = 1,710 eq/min 58.45 1,710 NaCl concentration in regenerating solution = = 5.31 eq/L = cR = C in Eq. (15-139) 321.9 From Eq. (15-181), noting conditions in Fig. 15.50b: KCa 2+ , Na + Q 2.3 = 2.6 = 1126 . cR 5.31 Unfortunately, this is > 1. Therefore, equilibrium is not as favorable as desired. Also the regeneration wave may not sharpen quickly. Nevertheless, assume a shock wave-like front. From Eq. (15-181): * * 0.99986 1 − xCa 2+ 1 − xCa 2+ 1.126 = * = 7,142 * xCa 2+ 1 − 0.99986 xCa 2+ Solving: * xCa 2+ = 0.99984 So downstream of the regeneration wave front, but upstream of the displacement wave front, the liquid contains very few sodium ions. However, this will not be quite so because the wave will not be sharp. Exercise 15.36 Subject: Separation of glutamic acid, glycine, and valine, in an aqueous solution, by a fixed-bed chromatographic column, using equilibrium theory. Given: Aqueous solution, buffered to a 3.4 pH by sodium citrate and containing 20 mol/m3 each of glutamic acid (GA), glycine (G), and valine (V). Chromatographic column, packed with Dowex 50W-X8 in the sodium form to a depth of 470 mm = 47 cm. Resin is 0.07 mm in diameter and is packed to a void fraction, εb , of 0.374. All three solutes follow Henry's law, q = Kc, where the values of K are 1.18 for GA, 1.74 for G, and 2.64 for V. Superficial solution velocity = us = 0.025 cm/s. Assumptions: Equilibrium theory. Find: Pulse duration to achieve complete separation. Time duration of the elution step before the second pulse begins. Analysis: Interstitial velocity = u = us/εb = 0.025/0.374 = 0.0668 cm/s. u 0.0668 0.0668 From Eq. (15-183), solute wave velocity = ui = = = 1 − 0.374 1 + 1674 Ki . 1 − εb Ki 1+ Ki 1 + 0.374 εb Solute Ki ui , cm/s GA 1.18 0.0225 G 1.74 0.0171 V 2.64 0.0123 From the K values,...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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