SME
Separation Processes · Problem type
Crosscurrent stages: one wash or several
Use this for “KD = 2 and S/F = 1.5 in all: what is left after three washes of 0.5 each, and after one wash with the lot?”
Diagram
Each
stage:
E
=
4
×
0.5
3
=
0.6667,
so
0.6
of
the
solute
stays.
After 3: 0.216.
This is an example. Type your own values to start a new problem.
E
=
KD
×
S/F
n
fleft
=
1
1
+
f₁
=
1
1
+
×
Formulas
E
=
KD
×
S/F
n
0.6667
=
4
×
0.5
3
fleft
=
(
1
1 +
E
)
n
0.216
=
(
1
1 + 0.6667
)³
f₁
=
1
1 + KD × (S/F)
0.3333
=
1
1 + 4 × 0.5
Assumptions
•
The total solvent is split into n equal parts, fresh solvent into each stage.
•
The liquids don’t mix; each stage reaches equilibrium with the same KD.
•
The solute is dilute, so KD and the flows stay the same stage to stage.
Step-by-step
WE KNOW
KD = 4, S/F = 0.5, n = 3
FIND
extraction factor per stage (E), fraction left after n stages (fleft), fraction left after one big stage (f₁)
Step 1 · Find extraction factor per stage (E)
Use E = KD × (S/F) ÷ n
Each
stage
gets
one
n
-th
of
the
solvent,
so
its
factor
is
one
n
-th
of
KDS
F
.
E
=
4
×
0.5
3
E
=
2
3
E
=
0.6667
Step 2 · Find fraction left after n stages (fleft)
Use fleft = (1 ÷ (1 + E))ⁿ
Each
stage
keeps
the
same
share
of
what
reaches
it;
n
stages
multiply.
fleft
=
(
1
1 + 0.6667
)³
fleft
=
(
1
1.6667
)
3
fleft
=
0.6
3
fleft = 0.216
Step 3 · Find fraction left after one big stage (f₁)
Use f₁ = 1 ÷ (1 + KD × (S/F))
All
the
solvent
in
one
stage,
for
comparison:
its
factor
is
the
whole
KDS
F
.
f₁
=
1
1 + 4 × 0.5
f₁
=
1
1 + 2
f₁
=
1
3
f₁ = 0.3333
Check
Put the numbers back in. Both sides should match.
0.6667
=
4
×
0.5
3
✓
0.216
=
(
1
1 + 0.6667
)³
✓
0.3333
=
1
1 + 4 × 0.5
✓
Topic check
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