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

# 77815 182249 186682 191114 195545 199974 0001170

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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. (15-77), (15-78), and (15-79) 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. (15-65), kL in cm/s = NSh Di /Dp = 30 Di /0.15 = 200 Di (in cm2/s) From the Wilke-Chang Eq. (3-39), 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 10-3 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 10-5) = 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 10-5 s-1 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 semi-log 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. (15-86) 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.

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