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

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Unformatted text preview: 1 1 1 + 0.0388 0.0118 = 0.226 1 = 0.226(0.00905) = 0.00205 cm 2 /s 25.8 + 84.7 Note that Knudsen diffusion is somewhat more important than molecular diffusion. Exercise 15.21 Subject: Comparison of batch, continuous, and semicontinuous modes of slurry adsorption for removal of trichloroethylene (TCE) from water with activated carbon. Given: Feed of water at 25oC containing 3.3 mg/L of TCE. Activated carbon with an average particle diameter of 1.5 mm = 0.15 cm and the following Freundlich adsorption isotherm for TCE at 25oC: q = 67 c0.564 where q = mg TCE adsorbed/g carbon and c = mg TCE/L solution. Sherwood number = 30. Particle external surface area = 5 m2/kg. Assumptions: Mass-transfer resistance in the pores is negligible compared to the external resistance. Find: For an effluent of 0.01 mg TCE/L, (a) Minimum amount of adsorbent needed per liter of feed solution. (b) For batch mode with twice the minimum amount of adsorbent, the contact time. (c) For continuous mode with twice the minimum amount of adsorbent, the required residence time. (d) For semicontinuous mode at a feed rate of 50 gpm and a liquid residence time of 1.5 times that in Part (c), the amount of activated carbon to give a reasonable vol% solids in the tank and run time of not less than 10 times the liquid residence time. Analysis: (a) The minimum amount of adsorbent corresponds to equilibrium with a solution of the effluent concentration of 0.01 mg TCE/L. From the Freundlich equation, the equilibrium loading is given by: q = 67(0.01)0.564 = 4.99 mg TCE/g carbon. By material balance on the TCE, cF Q = cfinal Q + q Smin Solving for Smin, for a basis of 1 L of solution, S min = Q ( cF − cfinal ) 1.0(3.3 − 0.01) = = 0.66 g carbon/L solution q 4.99 (b) For batch mode, use 2(0.66) = 1.32 g carbon/L solution = 1.32 kg/m3 = S/Q Combining Eqs. (15-77), (15-78), and (15-79), to eliminate q and c*, Q cF − c dc − = k La c − dt Sk n 3.3 − c = k La c − (132)(67) . 1/ 0.564 = k La c − 3.3 − c 88.4 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 = µ Bυ 0.6 A 1/ 2 T (2) 1.773 (1) Exercise 15.21 (continued) Analysis: (b) (continued) From Table 3.2, υA = υTCE = (2)14.8 + 3.7+ 3(21.6) = 98.1 x 10-3 m3/kmol = 98.1 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 = = 102 × 10−5 cm2 / s = Di . 0 .6 0.94(981) . Therefore, kL = 200(1.02 x 10-5) = 0.00204 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(1.32) = 6.60 m2/m3 of solution = 0.066 cm2/cm3. Therefore, kLa = 0.00204(0.066) = 1.35 x 10-4 s-1 Eq. (1) becomes: dc − = 1.35 × 10 −4 c − dt 3.3 − c 88.4 1.773 where c is in mg TCE/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...
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