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

must determine which solute controls from the

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Unformatted text preview: if a small time step is taken. The Euler form, where j is the iteration number, is, c ( j +1) = c ( j ) − ( ∆t ) 1.35 × 10 −4 c ( j ) − 3.3 − c ( j ) 88.4 1.773 (3) If a time step, ∆t, of 500 s is used to find the time when c becomes 0.01 mg/L, the first 10 values of c are as follows: Time, s c of TCE, mg/L 0 3.300 500 3.077 1000 2.870 1500 2.676 2000 2.495 2500 2.327 3000 2.170 3500 2.023 4000 1.887 4500 1.760 5000 1.641 The time required is found to be 44,000 s = 733 minutes = 12.2 h. A plot of c vs. t is given on the next page, where semi-log coordinates gives almost a straight line. Exercise 15.21 (continued) Analysis: (b) (continued) Slurry Batch Adsorption Concentration of TCE, mg/L 10.00 1.00 0.10 0.01 0 5000 10000 15000 20000 25000 Time, seconds 30000 35000 40000 45000 Exercise 15.21 (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(0.66) = 1.32 g carbon/h. A rearrangement of Eq. (15-86) gives the residence time as: cF − cout 3.3 − 0.01 = t res = (4) * k L a cout − c 135 × 10−4 0.01 − c* . The quantity c* is the ficticious concentration in equilibrium with qout, which is obtained by material balance, Eq. (15-87), which after rearrangement is: Q cF − cout 1(3.3 − 0.01) qout = = = 2.42 mg / g S 1.32 Then, from a rearrangement of the given Freundlich adsorption isotherm, 1/ 0.564 1.773 qout 2.42 c* = = = 0.00277 mg / L 67 67 Substitution into Eq. (4) gives, 3.3 − 0.01 tres = = 3,370, 000 s = 56,200 minutes = 936 h = 39.0 days. −4 1.35 × 10 ( 0.01 − 0.00277 ) This very large residence time is due to the very low exit concentration of TCE in the solution, making the continuous mode impractical. (d) Semicontinuous mode. Average residence time of the solution in the tank = 1.5(57,370) = 86,100 min. Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min. The volume of solution in the tank = 189(86,100) = 16,300,000 L or 16,300 m3. For a cylindrical tank with height = diameter, this gives a tank diameter of 27.5 m, which is probably impractical, although many smaller tanks in parallel could be used. In the tank, where the solid adsorbent charge resides, the variation of the loading, q, with run time is given by Eq. (15-91), which for time in minutes is, S dq = k L a cout − c * t resQ = 1.35 × 10 −4 (60) cout − c * (86,100)(189) = 132,000 cout − c * dt (5) where S is in g, q is in mg TCE/g, and t is in min. From Eq. (15-86), cout cF + k L at resc * 3.3 + 135 × 10−4 (60)(86,100)c * . = = = 0.00473 + 0.99857c * 1 + k L at res 1 + 1.35 × 10 −4 (60)(86,100) (6) Exercise 15.21 (continued) Analysis: (d) (continued) From the given Freundlich equation, c* = (q/67)1.773 (7) = 0.000579 q1.773 Combining Eqs. (5), (6), and (7), gives, dq S = 132,000 0.00473 + 0.99857c * − c * = 624.36 − 188.76c* = 624.36 − 01093q 1.773 . (8) dt Let the volume of solids in the tank = 2.5 vol%. The solids-free...
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