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

572 switching time 10 minutes find using the steady

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Unformatted text preview: V is the most strongly adsorbed. Therefore, the solutes will leave the column in the order of GA first, then G, and V last. First, assume that the separation between GA and G controls, as analogous to Fig. 15.49, where 182 cm is replaced by 47 cm. Thus, at the end of the column, the trailing edge of GA will coincide with the leading edge of G. From Example 15.20, 47 47 1 1 tP + = Solving, t P = 47 − = 660 s 0.0225 0.0171 0.0171 0.0225 Thus, the time for the trailing edge of the GA wave to reach the end of the column at 47 cm = tP + 47/0.0225 = 660 + 2,089 = 2,749 s. At that time, will the G and V waves be separated? The trailing edge of the G wave will be at (2,089)(0.0171) = 35.7 cm, while the leading edge of the V wave will be at (2,749)(0.0123) = 33.8 cm. Therefore, the G and V waves will be separated. Now assume that the separation between G and V controls. Thus, at the end of the column, the trailing edge of G will coincide with the leading edge of V. From Example 15.20, 47 47 1 1 tP + = Solving, t P = 47 − = 1,073 s 0.0171 0.0123 0.0123 0.0171 Exercise 15.36 (continued) Analysis: (continued) Thus, the time for the trailing edge of the G wave to reach the end of the column at 47 cm = = tP + 47/0.0171 = 1,073 + 2,749 = 3,822 s. At that time, will the G and GA waves be separated? The leading edge of the G wave will be at a hypothetical distance outside of the column of (3,822)(0.0171) = 65.4 cm. The trailing edge of the GA wave will be at a hypothetical distance outside of the column at 2,749(0.0225) = 61.9 cm. Thus, the G and GA waves are not separated. Therefore, the pulse duration = 660 seconds. To determine the elution time, compute the time for the trailing edge of the slow V wave to reach 47 cm. This time = 660 + 47/0.0123 = 4,481 s. The time for the leading edge of the second pulse of GA to reach 47 cm so that V and GA are just separated = 47/0.0225 = 2,089 s. The difference is 4,481 - 2,089 = 2,392 s before the second pulse starts. But 660 s of this is the first pulse. Therefore, the elution time = 2,392 - 660 = 1,732 s. Thus, the ideal cycle is: Pulse: Elute: Pulse: Elute: etc. 660 s 1,732 s 660 s 1,732 s Exercise 15.37 Subject: Separation of glutamic acid, glycine, and valine, in an aqueous solution, by a fixed-bed chromatographic column, accounting for mass transfer resistances. 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 given below. Superficial solution velocity = us = 0.025 cm/s. Effective diffusivities of the three solutes are given below. External mass-transfer coefficient = 1.5 x 10-3 cm/s for each solute. Assumptions: Application of Carta's equation, which accounts for mass transfer. Find: A cycle of feed pulses and elution p...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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