Unformatted text preview: ge have the same composition.
Find: The percent recovery of NaOH in the final overflow (extract) and the wt% NaOH in the
dried CaCO3. Also, is it worthwhile to add a third washing stage?
Analysis: Because the slurry composition and flow rate are given, it is not necessary to consider
the precipitation reaction, which is:
CaO (s) + Na2CO3 (aq) + H2O (l) = CaCO3 (s) + 2 NaOH (aq)
First compute the component flow rates in the slurry leaving the precipitation vessel and entering
washing Stage 1.
Component
Wt%
Flow rate, lb/h
CaCO3
5.0
5,000
NaOH
0.1
100
Water
94.9
94,900
Now consider the underflows. Each underflow will contain all of the solid CaCO3 or
5,000 lb/h of solids. Therefore, the liquid in each underflow of 20% solids will be 20,000 lb/h of
aqueous NaOH. Because the underflow basis is on total solution (not just the water), use mass
fractions to express the NaOH composition in the overflow and the liquid in the underflow.
Let: yi = mass fraction of NaOH in the overflow from Stage i
xi = mass fraction of NaOH in the underflow from Stage i
Vi = lb/h of liquid in the overflow from Stage i.
Li = lb/h of liquid in the underflow from Stage i.
S = lb/h of fresh wash water entering Stage 2.
Then: xi = yi
Li = 20,000 lb/h
Overall mass balance for the two washing stages gives:
100,000 + 20,000 = 5,000 + 20,000 + V1
Solving, V1 = 95,000 lb/h
Total mass balance around Stage 2 gives:
20,000 + 5,000 + 20,000 = 5,000 + 20,000 + V2
Solving V2 = 20,000 lb/h Exercise 16.3 (continued)
Mass balances for NaOH for each stage:
Stage 1 (to which the slurry from precipitation enters):
100 + y2V2 = y1V1 + x1L1
or,
100 + y2(20,000) = y1(95,000) + y1(20,000) (1) Stage 2 (to which the fresh wash water is sent):
0 + x1L1 = y2V2 + x2L2
or,
y1(20,000) = y2(20,000) + y2(20,000) (2) We have two linear equations in two variables: y1 and y2.
Solving (1) and (2), gives: y1 = x1 = 0.000952 and y2 = x2 = 0.000476
Flow rate of NaOH in final overflow = y1V1 = 0.000952(95,000) = 90.44 lb/h
Therefore, the recovery of NaOH in the extract = 90.44/100 = 0.9044 or 90.44%
Flow rate of NaOH in the dried solids = 100 – 90.44 = 9.56 lb/h
Therefore, the wt% NaOH in the dried solids = 9.56/(9.56 + 5,000) = 0.00191 or 0.191 wt%.
If a third washing stage is added, the three mass balance equations for NaOH are:
Stage 1:
20,000 y2 + 100 = 20,000 y1 + 95,000 y1
Stage 2:
20,000 y3 + 20,000 y1 = 20,000 y2 + 20,000 y2
Stage 3:
0 + 20,000 y2 = 20,000 y3 + 20,000 y3
Solving these three linear equations, y1 = x1 = 0.000984
y2 = x2 = 0.000656
y3 = x3 = 0.000328
The % recovery of NaOH in the extract = (95,000)(0.000984)/100 = 0.9348 or 93.48%
The lb/h NaOH in the dry solids = 100 – 93.48 = 6.52 lb/h
The wt% NaOH in the dry solids = 6.52/(6.52 + 5,000) = 0.00130 or 0.130 wt%
The use of three stages does give a significantly increased recovery of NaOH or a significantly
increased purity of CaCO3. Exercise 16.4
Subject: Recovery of zinc from a ZnS ore by oxidizing the ZnS to ZnO, followed by leaching
with aq. H2SO4 to produce a decanted sludge of water soluble ZnSO...
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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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