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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. (15181), 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. (15139)
321.9
From Eq. (15181), 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 wavelike front.
From Eq. (15181):
*
*
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
fixedbed 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 50WX8 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. (15183), 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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 Spring '11
 Levicky
 The Land

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