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Unformatted text preview: 4 and insoluble residue called
gangue.
Given: 20,000 kg/h of decanted sludge, consisting of 5 wt% water, 10 wt% ZnSO4, and 85 wt%
gangue. A leaching stage and a continuous, countercurrent washing system, using wash water to
produce an extract (strong solution) of 10 wt% ZnSO4 in water with a ZnSO4 recovery of 98%.
Assumptions: Each underflow consists of two parts of water (ZnSO4free basis) per part
gangue. No solids in the overflows. Liquid in the underflow and liquid in the overflow leaving a
stage have the same composition.
Find: The number of stages
Analysis: Because the underflow liquid is given as a mass ratio of ZnSO4free water to the solid
gangue in the underflow, use ZnSO4 compositions in the liquid as mass ratios, X, to water, and
liquid flow rates on a ZnSO4free basis.
Let: Yi = mass ratio of ZnSO4 to water in the overflow from Stage i
Xi = mass ratio of ZnSO4 to water in the underflow from Stage i
Vi = kg/h of water in the overflow from Stage i
Li = kg/h of water in the underflow from Stage i
S = kg/h of wash water entering the last stage
Decanted sludge consists of:
0.05(20,000) = 1,000 kg/h of water
0.10(20,000) = 2,000 kg/h of ZnSO4
0.85(20,000) = 17,000 kg/h of solid gangue
ZnSO4 in the extract = 0.98(2,000) = 1,960 kg/h
Water in the extract of 90% water = (9/1)(1,960 = 17,640 kg/h
Final underflow is: 17,000 kg/h of gangue
2(17,000) = 34,000 kg/h of water
2,000 – 1,960 = 40 kg/h of soluble ZnSO4
By overall water mass balance, flow rate of wash water = 34,000 + 17,640 – 1,000 = 50,640 kg/h
Use (168) of the McCabeThiele algebraic method to determine the number of stages. That
equation is in terms of mass fractions, but it also applies to the use of mass ratios. Thus,
log
N= X N − YN +1
YL − Y1
L
log
V (1) where the subscripts on the symbols refer to Figure 16.7. Thus, the countercurrent system
consists of one leaching stage, L, and N washing stages for a total of N+1 stages. Exercise 16.4 (continued)
From the calculations above, XN = 40/34,000 = 0.001176
Also, YN+1 = 0
YL = 10/90 = 0.1111
S = 17,000 kg/h leaving each stage
Therefore, L = 2(17,000) = 34,000 kg/h, constant from each stage
And, by mass balance for each washing stage, V = 50,640 kg/h, a constant
Obtain Y1 by a ZnSO4 mass balance around just the leaching stage, L:
Y1V1 + 2,000 = 1,960 + XLLL
But, V1 = V, LL = L, and XL = YL , and therefore,
Y1(50,640) + 2,000 = 1,960 + 0.1111(34,000)
Solving, Y1 = 0.0738
Now apply (1) to calculate the number of washing stages:
log
N= 0.001176 − 0
0.1111 − 0.0738
= 8.7
34, 000
log
50, 640 Therefore, need 9 ideal washing stages plus a leaching stage or 10 stages total. Exercise 16.5
Subject: Leaching of oil from flaked soybeans with nhexane in a continuous, countercurrent
system consisting of a leaching stage and three washing stages.
Given: 50,000 kg/h of flaked soybeans containing 20 wt% oil and 80 wt% insoluble solids.
50,000 kg/h of nhexane solvent for washing. From experiments, all underflows contain 0.8 kg
of liquid/kg of oilfree soybeans.
Assumptions: No solids in the overflows. Liquid in the underflow and liquid in the overflow
leaving a stage have the same composition.
Find: (a) The % recovery of o...
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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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