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

# C two countercurrent stages d infinite number of

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Unformatted text preview: 1 + 172 + 1.72 2 + 172 3 . . % extraction = (1 - 0.093) x 100% = 90.7% (e) For infinite-stage countercurrent extraction, because E > 1, Eq. (5-31) applies. Therefore, X ( ∞ ) / X ( F ) = 0.0 and % extraction = 100% (f) For infinite-stage crosscurrent extraction, Eq. (5-23) applies: X (∞) 1 1 = = = 0179 . (F) X exp E exp(172) . % extraction = (1 - 0.179) x 100% = 82.1% (5) (4) Exercise 5.11 Subject: Liquid-liquid extraction of dioxane (B) from water (A) with benzene (C). Given: 1,000 kg of 30 wt% B in A. Thus, FA = 700 kg . Assumptions: Equilibrium batch contacts. A and C are mutually insoluble. Distribution coefficient from Example 5.2, in the form of Eq. (5-15) is: Y ' K DB = B = 12 . (1) XB Find: S = kg of C needed to extract 95% of the dioxane by the following schemes: (a) One stage (b) Two crosscurrent stages with 50% of solvent to each stage. (c) Two countercurrent stages. (d) Infinite number of crosscurrent stages. (e) Infinite number of countercurrent stages. Analysis: Let: ' K = K DB X (i ) = X B in aqueous liquid leaving Stage i Y (i ) = YB in benzene liquid leaving Stage i From Eq. (5-14), the extraction factor = E = KS/FA = 1.2S/700 = 0.001714S For 95% extraction, X(N)/X(F) = 1 - 0.95 = 0.05. (a) For a single stage, using Eq. (5-13), X (1) 1 1 = 0.05 = = (F) X 1 + E 1 + 0.001714 S (2) Solving Eq. (2), S = 11,090 kg of benzene. (b) For two crosscurrent stages with equal solvent additions, where S = total solvent, Eq. (5-21) applies: X (2) 1 = 0.05 = (F) X 1+ E / 2 2 Solving Eq. (3), S = 4,050 kg of benzene. = 1 1 + 0.001714 S / 2 2 (3) Analysis (continued): Exercise 5.11 (continued) (c) For two countercurrent stages, Eq. (5-28) applies: X (2) 1 1 = 0.05 = = (F) 2 X 1+ E + E 1 + 0.001714 S + 0.001714 S 2 Solving Eq. (4), S = 2,270 kg of benzene. (d) For an infinite number of crosscurrent stages, Eq. (5-23) applies: X (∞) 1 1 = 0.05 = = (F) X exp E exp(0.001714 S ) (5) Solving Eq. (5), S = 1,748 kg of benzene (e) For an infinite number of countercurrent stages, if E < 1, Eq. (5-32) applies: X ( ∞ ) / X ( F ) = 0.05 = 1 − E = 1 − 0.001714 S Solving, S = 554 kg of benzene (4) Exercise 5.12 Subject: Liquid-liquid extraction of benzoic acid (B) from water (A) using chloroform (C). Given: Q = 1,000 L/h of aqueous waste containing 0.05 mol/L of B. S = 500 L/h of chloroform solvent. Extraction with a distribution coefficient, cB C '' K DB = = 4.2 (1) cB A where concentrations are in mol/L. 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 crosscurrent stages with one-third of solvent to each stage. (c) Three countercurrent stages. Analysis: To simplify the nomenclature, let: '' K = K DB y = cB C x = cB A Thus, from Eq. (1) y = Kx = 4.2x (2) (a) For one stage, the acid material balance is, xFQ = y1S + x1Q or (0.05)(1,000) = 50 = y1500 + x11,000 Combining Eqs. (2) and (3), y1 = 0.06...
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