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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 McCabeThiele diagram below, where 10 equilibrium stages are stepped for the optimal
feedstage location. Exercise 15.34 (continued)
Analysis: (continued) McCabeThiele 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) McCabeThiele 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 fixedbed 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/minft2 for loading; 1.5 gal/minft2 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 crosssectional 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.
 Spring '11
 Levicky
 The Land

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