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Chapter 14 Extraction Page 14-1
Chapter 14 Extraction
1. Mushroom flower
The general equation for the length of a column is:
For small m, we have:
To convert to KG, we use:
But the hexane flux, G, can be expressed in terms of the linear flow rate, v:
2. Anti-inflammatory recovery
a) Mass balance: L, x0 = 0
L, xn
For minimum solvent ratio, yn+1 =
*0.2
nn
yx
b)
From the Kremser equation:
3. Protein extraction
Chapter 14 Extraction Page 14-3
E - 1 y1 = 1 - E4
Since we want to extract 95% of protein, y1 = 0.05y4, thus this equation becomes
4. Staged separation of rare earths
From the Kremser equation we have:
Since the organic phosphates enter free of rare earth, y0 = 0 and we have:
The fraction which is not recovered is y1 / yN+1 so the fraction recovered for N = 3 is:
1
A
Solving numerically, we find that A = 0.742. The recovery for N = 6 is:
5.
First we perform a mass balance on the Nth stage:
Chapter 14 Extraction Page 14-4
Now we use the equilibrium relationship to relate xN and yN:
For a cascade of N stages we will have:
Solving for L, we have:
Since we have four stages, our total extractant will be:
6. Beet sugar
Using a basis of 100 kg vegetable matter and 200 kg of water fed we can draw the following
diagram using simple mass balances.
Chapter 14 Extraction Page 14-5
Overall sugar balance:
Sugar balance on stages 1 through N-1:
Using the Kremser equation:
7. Rare earth extraction by multiple-stage extractor
(a) Recover fraction of 4-stage extractor
For A, xFA /x1A = (1 + 0.8/0.733 + 0.8/0.7332 + 0.8/0.7333 + 0.8/0.7334) = 8.37
Chapter 14 Extraction Page 14-6
(b) The purity:
(c) 8-stage case
This case can be viewed in two sections. Section I is stages 1~4, as the figure shows.
Note that the notations are different from those used in the textbook. To cope with these
changes, some of the equations are rewritten below:
x*
n - xn+1 = 0.8 (P5-3)
From Eq. P5-1, we find that
Again, to cope with the notation, we use the following equations:
Chapter 14 Extraction Page 14-7
where
and from mass balance,
8.
a) From the Kremser equation we that for y0 = 0:
For rare earth “A” the recovery is:
For rare earth “B” the recovery is:
b) An overall mass balance gives:
Dividing by the feed concentration gives:
The purity is:
c) We analyze both ends of the cascade separately. For the first three stages, the Kremser
equation gives:
For the last four stages, we have:
We now use the condition of equilibrium on the fourth stage:
Chapter 14 Extraction Page 14-9
We can now write:
An overall mass balance gives:
For rare earth “A” we have:
The fraction recovered is:
Chapter 14 Extraction Page 14-10
The fraction recovered is:
d) The purity is:
