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

fill tub with water agitate spin to small value of r

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Unformatted text preview: FB / S (5) Analysis: (a) (continued) Exercise 5.7 (continued) The underflow sent to Stage 2 is X1RFA = 2FBRFA/S A similar material balance around Stage 2, with X2 = Y2 , gives, 2FBRFA/S = Y2(S/2 - RFA) + X2RFA = X2S/2 Solving, (6) Y2 = X2 = 4FBRFA/S2 The mass of solubles still left in the underflow from Stage 2 = X2RFA = 4FB(RFA)2/S2 The ratio of the mass of soluble dirt in the final underflow for two stages to that for one stage is, 2 4 FB RFA 4 RFA S2 = FB RFA S S It is highly likely that the rinsing will be such that the amount of rinse water used, S, will be much greater than the amount of water, RFA, that adheres to the clothing leaving in the underflow. Therefore, S >> 4 RFA and two stages of rinsing is much better than one stage. (b) A possible clothes washing cycle for efficient use of rinse water might be as follows: Put clothes and detergent in tub.. Fill tub with water. Agitate. Spin to small value of R. Fill tub with half of the rinse water. Agitate. Spin to small value of R. Fill tub with other half of the rinse water. Agitate. Spin to small value of R. Remove damp clothes and put into dryer. Exercise 5.8 Subject: Liquid-liquid extraction of acetic acid (B) from water (A) using chloroform (C). Given: Q = 10 liters of aqueous solution with 6 moles of acetic acid per liter. S = 10 liters of chloroform solvent. Extraction at 25oC with a distribution coefficient, cB C '' K DB = = 2.8 (1) cB A where concentrations are in moles/liter. Assumptions: Water is insoluble in chloroform. Chloroform is insoluble in water. Equilibrium is achieved in each stage. No change in volumes of feed and solvent as acid transfers from feed to the solvent. Find: Percent extraction of acid for: (a) One stage. (b) Three stages (crosscurrent) with one-third of solvent to each stage. (c) Three stages (crosscurrent) with 5 liters of solvent to Stage 1, 3 liters to Stage 2, and 2 liters to Stage 3. Analysis: To simplify the nomenclature, let: '' K = K DB y = cB C x = cB A Thus, from Eq. (1) y = Kx = 2.8x (2) (a) For one stage, the acid material balance is, xFQ = y1S + x1Q or (6)(10) = 60 = y110 + x110 Combining Eqs. (2) and (3), y1 = 4.42 mol/L, x1 = 1.58 mol/L % extracted = y1S/ xFQ x 100% = (4.42)(10)/(6)(10) x 100% = 73.7% (b) For three crosscurrent stages, can use Eq. (5-21) in the following form, x3 1 = xF 1+ E / N 3 (4) where from Eq. (5-14), E = KS/Q = (2.8)(10)/10 = 2.8 and N = 3. Therefore, from Eq. (4), Percent extraction = 1 − x3 1 × 100% = 1 − × 100% = 86.2% 3 xF (1 + 2.8 / 3) (3) Exercise 5.8 (continued) (c) Use Eq. (5-13) one stage at a time, where from Eq. (5-14), Ei = KSi /Q = (2.8)Si /10 = 0.28 Si For Stage 1, S1 = 5 L, E1 = 1.4, and x1 = xF[1/(1+E1)] = 6[1/(1+1.4)] = 2.5 mol/L For Stage 2, S2 = 3 L, E2 = 0.84, and x2 = x1[1/(1+E2)] = 2.5[1/(1+0.84)] = 1.36 mol/L For Stage 3, S3 = 2 L, E3 = 0.56, and x3 = x2[1/(1+E3)] = 1.36[1/(1+0.56)] = 0.872 mol/L Percent extraction = (6 - 0.872)/6 x 100% = 85.5%, just slightly lower than in Part (b). Exercise 5.9 Subject:...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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