Unformatted text preview: 0.001170
0.001239
0.001311
0.001385
0.001461
0.001539 0.00270
0.00277
0.00284
0.00292
0.00299
0.00307 0.00189
0.00191
0.00193
0.00195
0.00198
0.00200 Thus, the run time is 22,500,000 minutes (15,600 days), which is much greater than the above
1,661,000 minutes. Unfortunately, the vessel size is enormous. Exercise 15.23
Subject:
Comparison of batch, continuous, and semicontinuous modes of slurry adsorption
for removal of chloroform (C) from water with activated carbon.
Given: Feed of water at 25oC containing 0.223 mg/L of C. Activated carbon with an average
particle diameter of 1.5 mm = 0.15 cm and the following Freundlich adsorption isotherm for C at
25oC: q = 10 c0.564 where q = mg C adsorbed/g carbon and c = mg C/L solution.
Sherwood number = 30. Particle external surface area = 5 m2/kg.
Assumptions: Masstransfer resistance in the pores is negligible compared to the external
resistance.
Find: For an effluent of 0.01 mg C/L,
(a) Minimum amount of adsorbent needed per liter of feed solution.
(b) For batch mode with twice the minimum amount of adsorbent, the contact time.
(c) For continuous mode with twice the minimum amount of adsorbent, the required
residence time.
(d) For semicontinuous mode at a feed rate of 50 gpm and a liquid residence time of 1.5
times that in Part (c), the amount of activated carbon to give a reasonable vol% solids in the tank
and run time of not less than 10 times the liquid residence time.
Analysis: (a) The minimum amount of adsorbent corresponds to equilibrium with a solution of
the effluent concentration of 0.01 mg TCE/L. From the Freundlich equation, the equilibrium
loading is given by: q = 10(0.01)0.564 = 0.745 mg C/g carbon.
By material balance on the C,
cF Q = cfinal Q + q Smin
Solving for Smin, for a basis of 1 L of solution,
Q ( cF − cfinal ) 1.0(0.223 − 0.01)
S min =
=
= 0.286 g carbon/L solution
q
0.745
(b) For batch mode, use 2(0.286) = 0.572 g carbon/L solution = 0.572 kg/m3 = S/Q
Combining Eqs. (1577), (1578), and (1579), to eliminate q and c*,
Q cF − c
dc
− = k La c −
dt
Sk n 0.223 − c
= k La c −
(0.572)(10) 1/ 0.564 = k La c − 0.223 − c
5.72 From Eq. (1565), kL in cm/s = NSh Di /Dp = 30 Di /0.15 = 200 Di (in cm2/s)
From the WilkeChang Eq. (339),
DA,B 7.4 × 10 −8 φ B M B
=
0
µ Bυ A.6 1/ 2 T (2) 1.773 (1) Exercise 15.23 (continued)
Analysis: (b) (continued)
From Table 3.2, υA = υTCE = 14.8 + 3.7+ 3(21.6) = 83.3 x 103 m3/kmol = 83.3 cm3/mol
Solvent water viscosity at 25oC = 0.94 cP, MB = 18, and φB = 2.6. Substitution into Eq. (2) gives,
1/ 2 7.4 × 10−8 2.6 × 18 298
DA,B =
= 113 × 10−5 cm2 / s = Di
.
0 .6
0.94(83.3)
Therefore, kL = 200(1.13 x 105) = 0.00226 cm/s
The units of a in Eq. (1) are cm2 of external surface area of adsorbent/ cm3 of solution.
Therefore, a = 5 m2/kg (S/Q) = 5(0.572) = 2.86 m2/m3 of solution = 0.0286 cm2/cm3.
Therefore, kLa = 0.00226(0.0286) = 6.46 x 105 s1
Eq. (1) becomes: dc
− = 6.46 × 10−5 c −
dt 0.223 − c
5.72 1.773 where c is in mg C/L solution,...
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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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