SME
Separation Processes
Extraction
Use this for “KD = 3 and the solvent is 0.8 times the feed: what fraction of the solute is left after one stage?”
Diagram
Each
stage:
E
=
4
×
0.5
1
=
2,
so
0.3333
of
the
solute
stays.
After 1: 0.3333.
This is an example. Type your own values to start a new problem.
E
=
KD
×
S/F
fleft
=
1
1
+
fₑₓₜ
Fraction extracted
calculated
Formulas
E
=
KD
×
(S/F)
2 = 4 × 0.5
fleft
=
1
1 +
E
0.3333
=
1
1 + 2
fₑₓₜ
=
E
1 +
E
0.6667
=
2
1 + 2
fₑₓₜ = 1 − fleft
0.6667 = 1 − 0.3333
Assumptions
•
The feed liquid and the solvent don’t mix; the solute is dilute.
•
One equilibrium stage: the extract and raffinate leave in equilibrium, y = KDx.
•
S ÷ F compares solute-free solvent with solute-free feed liquid.
Step-by-step
WE KNOW
KD = 4, S/F = 0.5
FIND
extraction factor (E), fraction left (fleft), fraction extracted (fₑₓₜ)
Step 1 · Find extraction factor (E)
Use E = KD × (S/F)
The solvent holds KD times the concentration, and there is S/F as much of it.
E
=
4
×
0.5
E
=
2
Step 2 · Find fraction left (fleft)
Use fleft = 1 ÷ (1 + E)
For
each
part
of
solute
left
in
the
feed,
E
parts
go
to
the
solvent.
fleft
=
1
1 + 2
fleft
=
1
3
fleft = 0.3333
Step 3 · Find fraction extracted (fₑₓₜ)
Use fₑₓₜ = E ÷ (1 + E)
For
each
part
of
solute
left
in
the
feed,
E
parts
go
to
the
solvent:
E
of
the
1
+
E
parts.
fₑₓₜ
=
2
1 + 2
fₑₓₜ
=
2
3
fₑₓₜ = 0.6667
Check
Put the numbers back in. Both sides should match.
2 = 4 × 0.5 ✓
0.3333
=
1
1 + 2
✓
0.6667
=
2
1 + 2
✓
0.6667 = 1 − 0.3333 ✓
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