This preview shows page 1. Sign up to view the full content.
Unformatted text preview: .
Final Overflow
10 wt% Na2S 1 120,000 kg/d H2O
40,000 kg/d BaS
solid Na2CO3 Washing Water 2 3 4 Each underflow is two parts water/part BaCO3. (b) Na2CO3 requirement and BaCO3 and Na2S produced by stoichiometry:
Molecular Weight: BaS (aq) + Na2CO3 (s) = BaCO3 (s) + Na2S (aq)
169.42
106.00
197.37
78.05 5 Final Underflow Exercise 16.2 (continued)
Flow rate of Na2CO3 required = 40,000(106.00/169.42) = 25,027 kg/d
Flow rate of BaCO3 produced = 40,000(197.37/169.42) = 46,599 kg/d
Flow rate of Na2S produced = 40,000(78.05/169.42) = 18,428 kg/d
Determination of fresh water needed washing:
Final underflow contains all of the solid BaCO3 = 46,599 kg/d
Water in the final underflow = 2(46,599) = 93,198 kg/d
Since the precipitation reaction is complete and under stoichiometric conditions, the final
overflow contains only water and Na2S.
Need to write mass balances for Na2S for each stage. Best to write these in terms of mass
ratios of Na2S because underflow ratios are given as parts of H2O (rather than parts solution) per
part of solid BaCO3.
Yi = mass ratio of Na2S to water in the overflow from Stage i
Xi = mass ratio of Na2S to water in the underflow from Stage i
Vi = kg/d of water in the overflow from Stage i.
Li = kg/d of water in the underflow from Stage i.
S = kg/d of wash water entering Stage 5.
Then: Y1 = 0.1/0.9 = 0.1111 = X1 and all Yi = Xi .
Since the flow rate of water in the underflows is constant,
L1 = L2 = L3 = L4 = L5 = 93,198 kg/d of water
Overall water balance:
120,000 + S = V1 + 93,198 or S = V1 – 26,802
A water balance around each stage results in,
V2 = V3 = V4 = V5 = V1 – 26,802 = S
Mass balances for Na2S around each stage:
Stage 1:
18,428 + Y2V2 = X1L1 + Y1V1
or
18,428 + Y2S = 0.1111(93,148) + 0.1111(S + 26,802)
(1)
Stage 2:
X1L1 + Y3V3 = X2L2 + Y2V2
or
0.1111(93,198) + Y3S = Y2(93,198) + Y2S
(2)
Stage 3:
X2L2 + Y4V4 = X3L3 + Y3V3
or
Y2(93,198) + Y4S = Y3(93,198) + Y3S
(3)
Stage 4:
X3L3 + Y5V5 = X4L4 + Y4V4
or
Y3(93,198) + Y5S = Y4(93,198) + Y4S
(4)
Stage 5:
X4L4 = X5L5 + Y5V5
or
Y4(93,198) = Y5(93,198) + Y5S
(5)
We have five linear equations in five variables: S, Y2, Y3, Y4, and Y5
Solving with Polymath, S = 130,600 kg/d and Yi values from Stage 2 to 5 are
0.07207, 0.04421, 0.02433, and 0.01013.
(c) Since Yi = Xi and Wt% Na2S in an underflow = [Xi/(1 + Xi)](100%) =[Yi/(1 +
Yi)](100%),
the Wt% values from Stage 2 to 5 are: 6.72%, 4.23%, 2.37%, and 1.00%.
The final dried underflow contains 0.01013(93,198) = 944 kg/d of Na2S. Therefore,
purity of the final dried BaCO3 is 46,599/(944 + 46,599) = 0.980 or 98.0%.
Let: Exercise 16.3
Subject: Precipitation of CaCO3 from the reaction of solid CaO and an aqueous solution of
Na2CO3, followed by two stages of washing.
Given: A process consisting of a precipitation vessel followed by a twostage, continuous,
countercurrent washing system using water. Slurry of 100,000 lb/h from the precipitation tank is
5 wt% solid CaCO3, 0.1 wt% NaOH, and 94.9 wt% water. Wash is 20,000 lb/h of water.
Underflow from each of the two washing stages is 20 wt% solids.
Assumptions: No solids in the overflows. Liquid in the underflow and liquid in the overflow
leaving a sta...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details