This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 4 and insoluble residue called
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 (ZnSO4-free 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 ZnSO4-free 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 ZnSO4-free 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 (16-8) of the McCabe-Thiele 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,
N= X N − YN +1
YL − Y1
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:
N= 0.001176 − 0
0.1111 − 0.0738
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 n-hexane 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 n-hexane solvent for washing. From experiments, all underflows contain 0.8 kg
of liquid/kg of oil-free 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...
View Full Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land