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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.
 Spring '11
 Levicky
 The Land

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