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Unformatted text preview: 40 kg/h. Therefore, xN = 200/59,940 = 0.00334 is the target.
Feed to the system is 199,800 kg/h TiO2, (20/50)199,800 = 79,920 kg/h of solutes, and
(30/50)199,800 = 119,880 kg/h of water.
Fresh water flow rate = total feed flow rate = 199,800 + 79,920 + 119,880 = 399,600 kg/h
Therefore, by overall mass balances,
Water in final extract = 119,880 + 399,600 – 59,740 = 439,660 kg/h
Solutes in final extract = 79,920  200 = 79,720 kg/h
Summary of overall mass balance in kg/h:
Component
Feed
Wash Water
Final Underflow
Final Extract
TiO2
199,800
199,800
Solutes
79,920
200
79,720
Water
119,880
399,600
59,740
459,740
Total
399,600
399,600
259,740
539,460
Refer to Figure16.7. Calculate the leaching stage. xL = yL = 79,720/539,460 = 0.148.
A solute mass balance gives:
79,920 + y1V1 = 79,720 + xL LL
or,
y1V1 = 200 + 0.148LL
(2) Exercise 16.7 (continued)
A solution mass balance gives:
V1 + 79,920 + 119,800 = LL + 79,720 + 459,460
or,
V1 = LL + 339,460
We have 2 equations in 3 unknowns. However, from (1),
RL = LL/199,800 = 0.2 xL + 0.3 = 0.2(0.148) + 0.3 = 0.3296
or, LL = 65,850 kg/h
From (3), V1 = 65,850 + 339,460 = 405,310 kg/h
From (2), y1 = [200 + 0.148(65,850)]/405,310 = 0.0236 = x1 (3) Now compute Stage 1:
From (1), R1 = L1/199,800 = 0.2x1 + 0.3 = 0.2(0.0236) + 0.3 = 0.3047
Therefore, L1 = 60,880 kg/h
From a liquid flow balance around Stage 1, LL + V2 = L1 + V1
or, V2 = 60,880 + 405,310 – 65,850 = 400,340 kg/h
From a solute balance around Stage 1, xLLL + y2V2 = x1L1 + y1V1
or, y2 = [0.0236(60,880) + 0.0236(405,310) – 0.148(65,850)]/400,340 = 0.00314
Therefore, x2 = y2 = 0.00314
But, x2 is already below the target of 0.00334. Therefore, 2 washing stages is sufficient together
with the leaching stage or a total of 3 stages. Exercise 16.8
Subject: Rate of leaching of solute from a flake when the internal resistance to mass transfer is
negligible, such that mass transfer is controlled by the external resistance of the solvent.
Find: Derive Equation (1620)
Analysis: For diffusion from either side of a flake when the rate of mass transfer is controlled
by the external resistance, (1613), gives for masstransfer from the surface of the solid to the
solvent surrounding the solid,
ni = kc A ( Yi ) s − (Yi )b
(1)
where, ni = rate of mass transfer of solute from the flake to the solvent, mass/time
kc = external masstransfer coefficient on a concentration driving force basis,
A = area for mass transfer at the surface of the flake
(Yi )s = concentration of solute in the solvent at the surface of the flake,
mass solute/volume of solvent
(Yi )b = concentration of solute in the bulk of the solvent, mass solute/volume of solvent
Also, let,
( X i ) = concentration of solute in the solid, mass solute/volume of flake ( X i )o = initial concentration of solute in the solid, mass solute/volume of flake
Because the masstransfer resistance of the solid is negligible, values of ( X i ) stay uniform throughout the solid, but decrease with time, t.
Let the equilibrium ratio for the solute concentration between t...
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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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