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Unformatted text preview: derflow depends on concentration of solutes/kg solution as follows
(note the linear relationship):
Solute Concentration,
Retention of Solution,
kg solutes/kg solution
kg solution/kg TiO2
0.0
0.30
0.2
0.34
0.4
0.38
0.6
0.42
Analysis: Because the underflow basis is on total liquid (not just the water), use mass fractions
to express the solutes composition in the overflow and in the liquid in the underflow.
Let:
yi = mass fraction of solutes in the overflow from Stage i
xi = mass fraction of solutes in the underflow from Stage i
Vi = kg/h of liquid in the overflow from Stage i
Li = kg/h of liquid in the underflow from Stage i
(a) First compute the overall mass balance:
Final underflow from Stage N is 0.999(200,000) = 199,800 kg/h of TiO2 and 200 kg/h of
solutes. Solution in the final underflow = 0.4(199,800) = 79,920 kg/h. Water in the final
underflow = 79,920 – 200 = 79,720 kg/h. Therefore, xN = 200/79,920 = 0.00250.
Feed to the system is 199,800 kg/h TiO2, (20/50)199,800 = 79,920 kg/h of solutes, and
(30/50)199,800 = 119,880 kg/h of water.
Fresh water flow rate = total feed flow rate = 199,800 + 79,920 + 119,880 = 399,600 kg/h
Therefore, by overall mass balances,
Water in final extract = 119,880 + 399,600 – 79,720 = 439,660 kg/h
Solutes in final extract = 79,920 200 = 79,720 kg/h
Summary of overall mass balance in kg/h:
Component
Feed
Wash Water
Final Underflow
Final Extract
TiO2
199,800
199,800
Solutes
79,920
200
79,720
Water
119,880
399,600
79,720
439,760
Total
399,600
399,600
279,720
519,480 Exercise 16.7 (continued)
Referring to Figure 16.7, all underflow liquid flow rates = L = 79,920 kg/h. All overflow liquid
flow rates, except for the leaching stage = V = 399,600 kg/h. xL = yL = 79,720/519,480 = 0.153.
Calculate the solute mass balance for just the leaching stage:
79,920 + y1V1 = 79,720 + xL (79,920)
or,
79,920 + y1(399,600) = 79,720 + 0.153(79,920)
Solving, y1 = 0.0301
x − y N +1
0.00250 − 0
log N
log
yL − y1
0.153 − 0.0301
=
= 2.4 or 3 washing stages
From (168),
N=
L
79,920
log
log
V
399, 600
Therefore, need 1 ideal leaching stage and 3 ideal washing stages.
(b) Because the flow rate of liquid in the underflow depends on the solute concentration and,
therefore, varies from stage to stage, we can not apply the McCabeThiele algebraic method used
in Part (a). Instead, we must set up solute mass balances for each washing stage. The
relationship between the retention of solution on the solids = Ri in kg solution/kg TiO2 and the
mass fraction, xi, of solutes/kg solution is linear and we can develop the following equation from
the given data in the above table:
Ri = 0.2 xi + 0.30
(1)
First compute the overall mass balance:
Final underflow from Stage N is 0.999(200,000) = 199,800 kg/h of TiO2 and 200 kg/h of
solutes. From Part (a), the final underflow will likely have a xN < 0.01. Therefore, let RN = 0.3.
Therefore, solution in the final underflow = 0.3(199,800) = 59,940 kg/h. Water in the final
underflow = 59,940 – 200 = 59,7...
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 Spring '11
 Levicky
 The Land

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