Unformatted text preview: il in the final extract, if all leaching takes place in the leaching
stage.
(b) The % recovery of oil in the final extract, if leaching takes 3 of the total of 4 stages,
with the amount of leaching divided equally among the three stages.
Analysis: Because the underflow basis is on total liquid (not just the hexane), use mass fractions
to express the oil composition in the overflow and in the liquid in the underflow.
Let: yi = mass fraction of oil in the overflow from Stage i
xi = mass fraction of oil in the underflow from Stage i
Vi = kg/h of liquid in the overflow from Stage i
Li = kg/h of liquid in the underflow from Stage i
S = kg/h of fresh nhexane entering the last stage
(a) This part can be solved from oil mass balances around each stage or by applying (168) of
the McCabeThiele algebraic method. In either method: (1) the flowsheet of Figure 16.7 applies
with a leaching stage, L, and N = 3 washing stages; and (2) the leaching stage is computed
separately. Solve with (168).
For both methods, 40,000 kg/h of insoluble solids appear in each underflow.
Therefore, all L = 0.8( 40,000) = 32,000 kg/h.
The total final underflow = 40,000 + 32,000 = 72,000 kg/h
The total flow rate of flaked soybeans = 50,000 kg/h
The flow rate of entering nhexane solvent = 50,000 kg/h
Therefore, by overall total mass balance, VL = 50,000 + 50,000 – 72,000 = 28,000 kg/h
However, since all L = 32,000 kg/h, total mass balances around each washing stage give:
V = 50,000 kg/h for all stages except the overflow leaving the leaching stage.
Oil mass balance around the leaching stage:
y1V1 + 10,000 = yLVL + xLLL
or, since xL = yL, y1(50,000) + 10,000 = yL(28,000) + yL(32,000) = yL(60,000)
Because there are two unknowns, we cannot compute yL. We need more equations.
A total oil mass balance around all stages gives:
10,000 + 0 = x3(32,000) + yL(28,000) (1)
(2) Exercise 16.5 (continued)
From (168),
log
N= x3 − 0
yL − y1
or, 3 =
32, 000
log
50, 000
log or rearranging, x3 − y N +1
yL − y1
L
log
V x3
3log 0.64
= 10 ( ) = 0.262
yL − y1 (3) Equations (1) to (3) are three linear equations in three unknowns: y1, yL, and x3
Solving with Polymath,
y1 = 0.1794, yL = 0.3162, x3 = 0.03585
Oil in the final extract = yLVL = 0.3162(28,000) = 8,854 kg/h
The % recovery of oil = 8,854/10,000 = 0.8854 or 88.54%
(b) We cannot use (168) for this part because the leaching is divided equally among the
leaching stage and the first 2 washing stages. Until the oil is leached, it is part of the solid.
However, because the underflow from each stage is given as 0.8 kg liquid/kg of oilfree
soybeans, values of V and L remain the same as in Part (a) and xi = yi still applies.
Write oil mass balances for each stage, using the nomenclature of Figure 16.7, noting that the
following flow rates apply for unleached oil in the underflows:
10,000 – (1/3)(10,000) = 6, 667 kg/h to Stage 1
6,667 – (1/3)(10,000) = 3,333 kg/h to Stage 2
3,333 – (1/3)(10,000) = 0 kg/h to Stage 3
Stage L:
y1V1 + 10,000 = yLVL + xLLL + 6,667
(4)
or,
y1(50,000) + 10,000 = yL(28,000) + yL(32,000) + 6,667
Stage 1:
y2V2 + xLLL + 6,667 = y1V1 + x1L1 + 3,333
or,
y2(50,000) + yL(32,000) + 6,667 = y1(50,000) + y1(32,000) + 3,333
(5)
Stage 2:
y3V3 + x1L1 + 3,333 = y2V2 + x2L2
or, y3(50,000) + y1(32,000) + 3,333 = y2(50,000) + y2(32,000)
(6)
Stage 3:
0 + x2L2 = y3V3 + x3L3
or, y2(32,000) = y3(50,000) + y3(32,000)
(7)
Solving linear Equations (4) to (7) with Polymath for the fou...
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 Spring '11
 Levicky
 The Land

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