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

# This equation represents an initial value problem in

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Unformatted text preview: t, which is obtained by material balance, Eq. (15-87), which after rearrangement is: qout = Q cF − cout 1(0.324 − 0.002) = = 1118 mg / g . S 0.288 Then, from a rearrangement of the given Freundlich adsorption isotherm, c* = qout 32 1/ 0.428 = 1118 . 32 2 .336 = 0.000395 mg / L Substitution into Eq. (4) gives, 0.324 − 0.002 = 6, 640, 000 s = 110,700 minutes = 1,845 h = 76.9 days. 3.02 × 10 −5 ( 0.002 − 0.000395 ) This very large residence time is due to the very low exit concentration of benzene in the solution, making the continuous mode impractical. tres = (d) Semicontinuous mode. Average residence time of the solution in the tank = 1.5(110,700) = 166,100 min. Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min. The volume of solution in the tank = 189(166,100) = 31,400,000 L or 31,400 m3. For a cylindrical tank with height = diameter, this gives a tank diameter of 34.2 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, dq S = k L a cout − c * t resQ = 3.02 × 10 −5 (60) cout − c * (166,100)(189) = 56,900 cout − c * (5) dt where S is in g, q is in mg B/g, and t is in min. Exercise 15.22 (continued) Analysis: (d) (continued) From Eq. (15-86), c + k L at resc * 0.324 + 3.02 × 10−5 (60)(166,100)c * cout = F = = 0.00107 + 0.99669c * 1 + k L at res 1 + 3.02 × 10−5 (60)(166,100) From the given Freundlich equation, c* = (q/32)2.336 = 0.000305 q2.336 (7) Combining Eqs. (5), (6), and (7), gives, dq S = 56,900 0.00107 + 0.99669c * − c * = 60.88 − 188.34c* = 60.88 − 0.0574q 2.336 dt (6) (8) Let the volume of solids in the tank = 2.5 vol%. The solids-free volume of liquid in the tank = 31,400 m3. Therefore, the solids volume = (2.5/97.5)(31,400) =805 m3. Assume a particle density = 850 kg/m3. Therefore, S = 805 (850) = 684,000 kg = 6.84 x 108 g and Eq. (8) becomes, dq = 8.90 × 10 −8 − 8.39 × 10 −11 q 2.336 (9) dt Now, compute the run time to achieve a cumulative cout = 0.002 mg/L, where, t ccum = 0.002 = cout dt (10) t Eq. (9) 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, with j as the iteration index, is, 0 q ( j +1) = q ( j ) + ( ∆t ) [ 8.90 × 10−8 − 8.39 × 10 −11 q 2.336 ( j) (11) The total run time is specified to be not less than 10 times the average liquid residence time; thus, 10(166,100) = 1,661,000 minutes. Try a ∆t of 500,000 minutes. From the spreadsheet, the first 6 and last 6 values are: t, min. q, mg/g c*, mg/L c out c cum 0 0 0 0.00154 0 500000 0.04450 2.12E-07 0.00154 0.00154 1000000 0.08900 1.07E-06 0.00154 0.00154 1500000 0.13500 2.76E-06 0.00155 0.00154 2000000 0.17800 5.41E-06 0.00155 0.00154 2500000 0.22250 9.11E-06 0.00155 0.00154 20000000 20500000 21000000 21500000 22000000 22500000 1.77815 1.82249 1.86682 1.91114 1.95545 1.99974...
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