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

At that time will the g and v waves be separated the

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Unformatted text preview: . The equation for this operating line is obtained from a C3= material balance around a portion of the upper section of stages, from the top stage (but below the desorber) to an intermediate stage above the feed, in a manner similar to that for distillation on p. 363, where S is the adsorbent flow rate in kg/h and 2S is the equilibrium loading of adsorbate in mol/h, 2 SyC3= + LR x DC3= = 2 Sy DC3= + LR xC3= Solving, yC3= = LR L xC3= − R x DC3= + y DC3= 2S 2S Thus, the slope of the operating line = LR/2S = 0.5706. Therefore, LR = 1.0152 S. Also, V = 2S and by adsorbate material balance around the desorber and divider, V = LR + D = LR + 17,900. Solving these three equations for the minimum adsorbent condition, S = Smin = 18,200 kg/h, V = 2(18,200) = 36,400 mol/h, and LR = 1.0152(18,200) = 18,500 mol/h. (a) For an adsorbent rate = 1.2 Smin, S = 1.2(18,200) = 21,850 kg/h. Then, V = 2S = 2(21,850) = 43,700 mol/h and LR = V - 17,900 = 43,700 - 17,900 = 25,800 mol/h. The slope of the resulting operating line = LR/2S = 25,800/[2(21,850)] = 0.590. (b) This operating line, together with resulting operating line for the lower section is shown in the second McCabe-Thiele diagram below, where 10 equilibrium stages are stepped for the optimal feed-stage location. Exercise 15.34 (continued) Analysis: (continued) McCabe-Thiele diagram for minimum adsorbent rate 1.00 y, mole fraction of C3= in the adsorbate 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 x, mole fraction of C3= in the gas Exercise 15.34 (continued) Analysis: (continued) McCabe-Thiele diagram for adsorbent rate Equal to 1.2 times minimum value 1.00 y, mole fraction of C3= in the adsorbate 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 x, mole fraction of C3= in the gas Exercise 15.35 Subject: Softening of hard water in a fixed-bed ion exchange column Given: Feed at 25oC containing 400 ppm (by wt) CaCl2 and 50 ppm (by wt) NaCl. Bed of 8.5 ft diameter by 10 ft high of gel resin with a cation capacity of 2.3 eq/L of bed volume. Wetted resin void fraction = εb = 0.38. 15 gal/min-ft2 for loading; 1.5 gal/min-ft2 for displacement, washing , and regeneration. Displacement and regeneration solutions are aq. saturated NaCl (26 wt%). Assumption: Because of dilute conditions, feed solution contains 1,000 g H2O/L Find: (a) (b) (c) (d) (e) (f) Feed solution flow rate, L/min. Loading time to breakthrough, h Loading wave front velocity, cm/min Regeneration solution flow rate, L/min Displacement time, h Additional time for regeneration, h Analysis: M of CaCl2 = 110.99 M of NaCl = 58.45 Concentration of CaCl2 = 400(1000)/[110.99(1,000,000)] = 0.00360 mol/L = 0.00720 eq/L Concentration of NaCl = 50(1000)/[58.45(1,000,000)] = 0.000855 mol/L = 0.000855 eq/L (a) Bed cross-sectional area = 3.14(8.5)2/4 = 56.7 ft2 Feed solution flow rate = 15(56.7) = 851 gpm = 3,219 L/min (b) Behind the loading wave front in the feed, eq...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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