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

# 2327 y2 x2 001927 y3 x3 0001482 sulfate in the final

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Unformatted text preview: 2 and 3 give, respectively, V3 = V2 =240,000 kg/day. The water balance around Stage 1 is: L0 + V2 = L1 + V1 or 68,320 + 240,000 = 20,000 + V1 Solving, V1 = 288,320 kg/day Because the underflow liquid rates are constant in water, rather than water plus dissolved sulfate, it is preferable to work with mass ratios of sulfate to water, instead of sulfate mass fractions. Sulfate material balances around Stages 1, 2, and 3, respectively, in terms of mass ratios, Yi = kg dissolved Al2(SO4)3 / kg H2O in overflow from Stage i Xi = kg dissolved Al2(SO4)3 / kg H2O in underflow from Stage i (2) are: 67,116 + 240,000Y2 = 20,000X1 + 288,320Y1 20,000X1 + 240,000Y3 = 20,000X2 + 240,000Y2 (3) 20,000X2 = 20,000X3 + 240,000Y3 (4) From equilibrium at each stage for the leaving liquids, Y 2 = X2 Y 3 = X3 (5, 6, 7) Y 1 = X1 Solving the six linear equations, (2) through (7), we obtain: Y1= X1 = 0.2327 Y2 = X2 = 0.01927 Y3 = X3 = 0.001482 Sulfate in the final extract = V1Y1 = 288,320(0.2327) = 67,092 kg/day The overall material balance for flow rates in kg/day is: Component Bauxite Acid Reactor Solvent Final ore effluent Extract Al2O3 solid 20,000 Inert solid 20,000 20,000 H2SO4 57,718 Dissolved Al2(SO4)3 67,116 67,092 H2 O 57,718 68,320 240,000 288,320 Total: 40,000 115,436 155,436 240,000 355,412 Percent extraction of sulfate = (67,092/67,116) x 100% = 99.96% Waste 20,000 24 20,000 40,024 Analysis: (continued) Exercise 5.6 (continued) If an additional stage is added, a sulfate material balance for Stage 4 is needed and eight equations in eight unknowns are solved to give Y1= X1 = 0.23277. The percent extraction of sulfate now = (288,320)(0.23277)/67,116 x 100% = 99.996%. For the small additional recovery of sulfate (only 20 kg/day out of 67,116 kg/day), it is doubtful that an additional stage could be justified. Exercise 5.7 Subject: Given: Efficient rinsing of clothes in a washing machine Fixed amount of rinse water. Assumptions: Perfect mixing such that the concentration of soluble dirt in water adhering to the clothes (underflow) is equal to that in the discharged water (overflow). Find: (a) Whether the rinse water should be divided into two rinses. (b) Devise an efficient machine cycle. Analysis: Use the nomenclature of Section 5.2, letting: FA = mass of dry insoluble clothing FB = mass of soluble dirt. S = mass of rinse water Y = mass of soluble dirt/mass of overflow water X = mass of soluble dirt/mass of underflow water R = mass of water/mass of insoluble clothing in the underflow (a) For one rinsing stage, a material balance on soluble dirt gives, FB = XRFA + Y(S - RFA) (1) For equilibrium between overflow and underflow liquids, Y =X (2) Combining Eqs. (1) and (2), X = Y = FB/S (3) Mass of solids still left in liquid adhering to the clothes = XRFA = FBRFB/S Consider two crossflow rinsing stages, as shown in the diagram below, where the rinse water to each stage is 50% of the total rinse water of S/2. A soluble dirt material balance around Stage 1, with X1 = Y1 gives, FB = X1RFA + Y1(S/2 - RFA) = Y1S/2 (4) Solving, Y 1 = X1 = 2...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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