Unformatted text preview: ial balance on B,
cF Q = cfinal Q + q Smin
Solving for Smin, for a basis of 1 L of solution,
Q ( cF − cfinal ) 1.0(0.324 − 0.002)
S min =
=
= 0.144 g carbon/L solution
q
2.24
From the Freundlich equation for X, the equilibrium loading is given by: q = 125(0.002)0.333 =
15.78 mg X/g carbon.By material balance on X,
cF Q = cfinal Q + q Smin
Solving for Smin, for a basis of 1 L of solution,
Q cF − cfinal
1.0(0.630 − 0.002)
S min =
=
= 0.0398 g carbon / L solution
q
15.78
Therefore B controls and the minimum amount of adsorbent = 0.144 g carbon/L solution.
(b) For batch mode, use 2(0.144) = 0.288 g carbon/L solution = 0.288 kg/m3 = S/Q
Combining Eqs. (1577), (1578), and (1579) and applying them to B, to eliminate q and c*, Exercise 15.22 (continued)
Analysis: (a) (continued)
Q cF − c
dc
− = k La c −
dt
Sk n 0.324 − c
= k La c −
(0.288)(32) 1/ 0.428 = k La c − 0.324 − c
9.22 2 .336 (1) 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) From Table 3.2, υA = υB = (6)14.8 + 6(3.7)  15 = 96 x 103 m3/kmol = 96 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 =
= 104 × 10−5 cm2 / s = Di
.
0 .6
0.94(96)
Therefore, kL = 200(1.04 x 105) = 0.0021 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(0.288) = 1.44 m2/m3 of solution = 0.0144 cm2/cm3.
Therefore, kLa = 0.0021(0.0144) = 3.02 x 105 s1
Eq. (1) becomes: dc
− = 3.02 × 10 −5 c −
dt 0.324 − c
9.22 2 .336 where c is in mg B/L solution, and t is in seconds. This equation represents an initial value problem in ODEs, which can be solved by the Euler
method using a spreadsheet if a small time step is taken. The Euler form, where j is the iteration
number, is,
c ( j +1) = c ( j ) − ( ∆t ) 3.02 × 10−5 c ( j ) − 0.324 − c ( j )
9.22 2 .336 (3) If a time step, ∆t, of 1,000 s is used to find the time when c becomes 0.002 mg/L, the first 7
values of c for B are as follows:
Time, s c of B, mg/L
0
0.324
1,000
0.314
2,000
0.305
3,000
0.296
4,000
0.287
5,000
0.278
6,000
0.270 Exercise 15.22 (continued)
Analysis: (b) (continued)
The time required is found to be 172,000 s = 2,870 minutes = 47.8 h. A plot of c vs. t follows,
where semilog coordinates gives almost a straight line. C oncentration of benzene in solution, m g/L 1.000 0.100 0.010 0.001
0 50000 100000 150000 Time, seconds 200000 250000 Exercise 15.22 (continued)
Analysis: (continued)
(c) For continuous slurry adsorption, assume a solution feed rate, Q, of 1 L/h. Therefore,
the adsorbent feed rate, from Part (a) = S = 2(0.144) = 0.288 g carbon/h. A rearrangement of Eq.
(1586) gives the residence time as:
t res = 0.324 − 0.002
cF − cout
=
*
3.02 × 10−5 0.002 − c*
k L a cout − c (4) The quantity c* is the ficticious concentration in equilibrium with qou...
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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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