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

# 8 days 94646 the lead bed must be replaced every 138

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Unformatted text preview: 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 ) 6.46 × 10 −5 c ( j ) − 0.223 − c ( j ) 5.72 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 C, mg/L 0 0.223 500 0.216 1000 0.209 1500 0.202 2000 0.196 2500 0.189 3000 0.183 3500 0.177 4000 0.172 4500 0.166 5000 0.161 The time required is found to be 51,500 s = 858 minutes = 14.3 h. A plot of c vs. t is given on the next page, where semi-log coordinates gives almost a straight line. Exercise 15.23 (continued) Analysis: (b) (continued) C o nc entratio n o f C hloro fo rm in so lution , m g/L 1 Slurry Batch Adsorption 0.1 0.01 0 10000 20000 30000 40000 Time, seconds 50000 60000 Exercise 15.23 (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(0.286) = 0.572 g carbon/h. A rearrangement of Eq. (15-86) gives the residence time as: t res = 0.223 − 0.01 cF − cout = * 6.46 × 10 −5 0.01 − c* k L a cout − c (4) The quantity c* is the ficticious concentration in equilibrium with qout, which is obtained by material balance, Eq. (15-87), which after rearrangement is: qout = Q cF − cout 1(0.223 − 0.01) = = 0.372 mg / g S 0.572 Then, from a rearrangement of the given Freundlich adsorption isotherm, q c* = out 10 1/ 0.564 0.372 = 10 1.773 = 0.00292 mg / L Substitution into Eq. (4) gives, tres = 0.223 − 0.01 = 466, 000 s = 7,770 minutes = 130 h = 5.4 days. 6.46 × 10−5 ( 0.01 − 0.00293) This is a large residence time compared to the batch time in Part (b). (d) Semicontinuous mode. Average residence time of the solution in the tank = 1.5(7,700) = 11,600 min. Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min. The volume of solution in the tank = 189(11,600) = 2,190,000 L or 2,200 m3. For a cylindrical tank with height = diameter, this gives a tank diameter of 14.1 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 = 6.46 × 10−5 (60) cout − c * (11,600)(189) = 8,500 cout − c * (5) dt where S is in g, q is in mg C/g, and t is in min. From Eq. (15-86), cF + k L at resc * 0.223 + 6.46 × 10 −5 (60)(11,600)c * cout = = = 0.00485 + 0.97826c * (6) 1 + k L at res 1 + 6.46 × 10 −5 (60)(11,600) Exercise 15.23 (continued) Analysis: (d) (continued) From the given Freundlich equation, c* = (q/10)1.773 (7) = 0.01687 q1.773 Combining Eqs. (5), (6), and (7), gives, dq S = 8,500 0.00485 + 0.97826c * − c * = 412 − 184.79c* = 412 − 31...
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