Unformatted text preview: flow rates, of overflow and underflow,
and the percentage of oil in the feed that is recovered in the overflow.
Analysis:
The flakes contain (0.19)(100,000) = 19,000 kg/h of oil and (100,000 – 19,000) = 81,000
kg/h of oilfree solids. However, all of the oil is not leached. For convenience in the
calculations, lump the unleached oil with the oilfree solids to give an effective solids.
The flow rate of unleached oil = (81,000)(0.5/99.5) = 407 kg/h.
Therefore, take the solids flow rate as (81,000 + 407) = 81,407 kg/h.
Take the oil in the feed as just the amount that is leached or (19,000 – 407) = 18,593
kg/h.
Therefore, in the feed, F,
YA = (81,407/18,593) = 4.38
XB = 1.0
The sum of the liquid solutions in the underflow and overflow include 200,000 kg/h of hexane
and 18,593 kg/h of leached oil. Therefore, for the underflow and overflow,
XB = YB = [18,593/(200,000 + 18,593)] = 0.0851
This is a case of variable solution underflow. Using data in the table that accompanies
Example 4.9, and the conversion shown there of that table to XA as a function of XB, we obtain
XA = 2.09.
Using the algebraic method, since the flow rate of solids in the underflow = 81,407 kg/h, the
flow rate of liquid in the underflow = 81,407/2.09 = 38,950 kg/h. The total flow rate of
underflow is U = 81,407 + 38,950 = 120,357 kg/h. By mass balance, the flow rate of overflow =
300,000 – 120357 = 179,643 kg/h. Exercise 4.52 (continued)
Now compute the compositions of the underflow and overflow. In the overflow, with XB
= 0.0851, the mass fractions of solute B and solvent C are, respectively, 0.0851 and (1 – 0.0851)
= 0.9149. In the underflow, using XA = 2.09 and XB = 0.0851, the mass fractions of solids, B,
and C, are, respectively, [2.09/(1 + 2.09)] = 0.676, 0.0851(1 – 0.676) = 0.0276, and (1 – 0.676 –
0.0276) = 0.2964.
From above, the oil flow rate in the feed is 19,000 kg/h. The oil flow rate in the overflow
= YBV = 0.0851(179,643) = 15,288 kg/h. Thus, the percentage of the oil in the feed that is
recovered in the overflow = 15,288/19,000 = 0.805 or 80.5%. Exercise 4.53
Subject: Leaching of Na2CO3 from a solid by water. Given: Lo = 3,750 kg/h of a solid containing 1,350 kg/h Na2CO3, contacted with So = 4,000
kg/h of water.
Assumptions: Ideal leaching stage so that compositions of overflow and underflow liquid are
equal. Underflow contains 40 wt% water on a solutefree basis.
Find: Compositions and flow rates, L1 and S1, respectively, of overflow and underflow.
Analysis: Let x = mass fraction of Na2CO3 (solute).
Lo + So = 3,750 + 4,000 = 7,750 = L1 + S1
(a) Total balance:
Na2CO3 balance: x Lo Lo = 1,350 = x L1 L1 + xS1 S1
Insoluble solids balance:
(3,750 − 1,350) = 2,400 = (1 − 0.40) 1 − xS1 S1 = 0.60 1 − xS1 S1
Equilibrium: x L1 = xS1 S1
S1 − (0.60) 1 − xS1 S1 = (1)
(2)
(3) xS1 (4) 0.4 + 0.6 xS1 Eqs. (1) to (4) can be reduced to one quadratic equation in xS1 . Solving, xS1 = 01189 ; x L1 = 0.2523 ; S1 = 4,540 kg / h ; and L1 = 3,210 kg / h
. The material balance may be summari...
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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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