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

5 13 x 1 1 1 010 f x 1 e 1 00688s 2 solving

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Unformatted text preview: Liquid-liquid extraction of uranyl nitrate (UN) from water (W) with tributyl phosphate (TBP) Given: 100 grams of 20 wt% UN in water. Thus, FW = 80 g . Assumptions: Equilibrium batch contacts. W and TBP are mutually insoluble. Y ' . (1) Distribution coefficient is given in the form of Eq. (5-15): K DUN = UN = 55 X UN Find: Grams of TBP solvent needed to extract 90% of the UN 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 DUN X (i ) = X UN in aqueous liquid leaving Stage i Y (i ) = YUN in TBP liquid leaving Stage i From Eq. (5-14), the extraction factor = E = KS/FW = 5.5S/80 = 0.0688S For 90% extraction, X(N)/X(F) = 1 - 0.90 = 0.10. (a) For a single stage, using Eq. (5-13), X (1) 1 1 = 010 = . = (F) X 1 + E 1 + 0.0688S (2) Solving Eq. (2), S = 130.8 g of TBP. (b) For two crosscurrent stages with equal solvent additions, where S = total solvent, Eq. X (2) 1 1 (5-21) applies: = 010 = . = (3) 2 2 (F) X 1+ E / 2 1 + 0.0688S / 2 Solving Eq. (3), S = 62.9 g of TBP. Analysis (continued): Exercise 5.9 (continued) (c) For two countercurrent stages, Eq. (5-28) applies: X (2) 1 1 = 010 = . = (F) 2 X 1+ E + E 1 + 0.0688S + 0.0688S 2 (4) Solving Eq. (4), S = 36.9 g of TBP. (d) For an infinite number of crosscurrent stages, Eq. (5-23) applies: X (∞) 1 1 = 010 = . = (F) X exp E exp(0.0688S ) (5) Solving Eq. (5), S = 33.5 g of TBP. (e) For an infinite number of countercurrent stages, if E < 1, Eq. (5-32) applies: X ( ∞ ) / X ( F ) = 0.10 = 1 − E = 1 − 0.0688S Solving, S = 13.1 g of TBP Exercise 5.10 Subject: Liquid-liquid extraction of uranyl nitrate (UN) from water (W) with tributyl phosphate (TBP) Given: 2 kg of 20 wt% UN in water. Thus, FW =1.6 kg. S = 0.5 kg of TBP. Assumptions: Equilibrium batch contacts. W and TBP are mutually insoluble. Y ' . (1) Distribution coefficient is given in the form of Eq. (5-15): K DUN = UN = 55 X UN Find: Percent recovery of UN for: (a) Two-stage batch extraction. (b) Three batch extractions. (c) Two-stage cocurrent extraction. (d) Three-stage countercurrent extraction. (e) Infinite-stage countercurrent extraction. (f) Infinite-stage crosscurrent extraction. Analysis: Let: ' K = K DUN X (i ) = X UN in aqueous liquid leaving Stage i Y (i ) = YUN in TBP liquid leaving Stage i From Eq. (5-14), the extraction factor = E = KS/FW = 5.5(0.5)/1.6 = 1.72 (a) For a single stage, using Eq. (5-13), X (1) 1 1 = = = 0.368 (F) X 1 + E 1 + 1.72 (2) % extraction = (1 - 0.368) x 100% = 63.2% (b) Three batch extractions is equivalent to a crosscurrent arrangement with equal solvent additions, where S = total solvent. Eq. (5-21) applies: X ( 3) 1 1 = = = 0.257 3 3 (F) X 1+ E / 3 1 + 172 / 3 . (3) % extraction = (1 - 0.257) x 100% = 74.3% (c) Two-stage cocurrent is same as one stage. Therefore, % extraction = 63.2%. Analysis (continued): Exercise 5.10 (continued) (d) For three-stage countercurrent extraction, Eq. (5-28) applies: X ( 3) 1 1 = = = 0.093 (F) 2 3 X 1+ E + E + E...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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