Unformatted text preview: r unknowns of yL, y1, y2, and y3,
yL = 0.2736, y1 = 0.2616, y2 = 0.1873, y1 = 0.0731
Therefore % recovery of oil to the final extract = 0.2736(28,000)/10,000 = 0.766 or 76.6% Exercise 16.6
Subject: Leaching and washing of Na2CO3 from a solid mixture with insoluble solids in a
continuous, countercurrent system.
Given: 100 tons/h of feed of 20 wt% watersoluble Na2CO3 and 80 wt% insoluble solids.
Final extract contains 15 wt% solute. Recovery of solute is 98%. Underflow contains 0.5 lb
solution/lb insoluble solids. Washing is with pure water.
Assumptions: One ideal leaching stage. No solids in the overflows. Liquid in the underflow
and liquid in the overflow leaving a stage have the same composition.
Find: Number of ideal washing stages required.
Analysis: First compute the overall mass balance from the given data:
Feed is 20 tons/h of Na2CO3 and 80 tons/h insoluble solids.
Final extract contains 0.98(20) = 19.6 tons/h of Na2CO3.
Water rate in final extract = (85/15)19.6 = 111.1 tons/h
Final underflow contains 20 19.6 = 0.4 tons/h, 80 tons/h of insoluble solids, and 0.5(80)
= 40 tons/h of solution. The solution contains 40 – 0.4 = 39.6 tons/h of water.
By overall water balance, the fresh wash water flow rate = 111.1 + 39.6 = 150.7 tons/h
Summary of overall mass balance in tons/h:
Component
Feed
Wash Water
Final Underflow
Final Extract
Na2CO3
20.0
0.4
19.6
Insolubles
80.0
80.0
Water
150.7
39.6
111.1
Total
100.0
150.7
120.0
130.7
Because the underflow basis is on total liquid (not just the water), use mass fractions to express
the Na2CO3 composition in the overflow and in the liquid in the underflow.
Let:
yi = mass fraction of Na2CO3 in the overflow from Stage i
xi = mass fraction of Na2CO3 in the underflow from Stage i
Vi = tons/h of liquid in the overflow from Stage i
Li = tons/h of liquid in the underflow from Stage i
All flow rates of L = 40 tons/h and by total mass balance around each washing stage, all
V = 150.7 tons/h. Also, yi = xi. Also for the last washing stage, xN = 0.4/40 = 0.01
Calculate the leaching stage by a Na2CO3 mass balance, referring to Figure 16.7:
20 + y1V1 = 19.6 + xL (40)
or, 20 + y1(150.7) = 19.6 + 0.15(40)
Solving, y1 = 0.0372
x − y N +1
0.01 − 0
log N
log
yL − y1
0.150 − 0.0372
From (168),
N=
=
= 1.8 or 2 washing stages
L
40
log
log
V
150.7 Exercise 16.7
Subject: Production of pure, insoluble TiO2 by leaching and washing soluble impurities with
water in a continuous, countercurrent system.
Given: Feed contains 50 wt% TiO2, 20 wt% soluble salts, and 30 wt% water. Final product
after filtering and drying is 200,000 kg/h of 99.9 wt% TiO2. Pure wash water flow rate = feed
flow rate on a massflow basis.
Assumptions: One ideal leaching stage. No solids in the overflows. Liquid in the underflow
and liquid in the overflow leaving a stage have the same composition.
Find: (a) Number of washing stages if the underflow from each stage = 0.4 kg solution/kg TiO2.
(b) Number of washing stages if the underflow from each stage in terms of retention of
solution to solid TiO2 in un...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details