Separation Processes

Distillation

Use this for “A column makes 98% methanol overhead and leaves 2% in the bottoms, α = 3.2. What is the fewest number of stages?”

Diagram

00.20.40.60.8100.20.40.60.81x (liquid mole fraction of A)y (vapor)1234567xBxDEquilibrium, α = 2.5Operating
Equilibrium
curve:
y
=
αx
1 + (α − 1)x
with
α
=
2.5;
the
45°
line
is
y
=
x.
At total reflux both operating lines are the 45° line: from xD = 0.95 down to xB = 0.05 the stairs take 7, the last one partial.
Fenske:
Nₘᵢₙ
=
ln[(
xD
1 − xD
)(
1 − xB
xB
)]
÷
ln
α
=
6.427
stages,
the
reboiler
counted,
so
7
whole
stairs.
This is an example. Type your own values to start a new problem.
Nₘᵢₙ
=
ln
xD
1
−
×
1
−
xB
÷
ln(
α
)

Formulas

Nₘᵢₙ
=
ln((
xD
1 − xD
)
×
(
1 − xB
xB
))
÷
ln(
α
)
6.4269
=
ln((
0.95
1 − 0.95
)
×
(
1 − 0.05
0.05
))
÷
ln(2.5)

Assumptions

•
Constant relative volatility α over the whole column.
•
Total reflux: no product is drawn, so both operating lines are the 45° line y = x.
•
Nₘᵢₙ counts the reboiler as a stage; a real column needs more.

Step-by-step

WE KNOW
xD = 0.95, xB = 0.05, α = 2.5
FIND
fewest stages (Nₘᵢₙ)
Step 1 · Find fewest stages (Nₘᵢₙ)
Use Nₘᵢₙ = ln((xD ÷ (1 − xD)) × ((1 − xB) ÷ xB)) ÷ ln(α)
Each
stage
at
total
reflux
multiplies
the
light-to-heavy
ratio
by
α
,
so
divide
the
log
of
the
whole
separation
by
ln
α
.
Nₘᵢₙ
=
ln((
0.95
1 − 0.95
)
×
(
1 − 0.05
0.05
))
÷
ln(2.5)
Nₘᵢₙ
=
ln(19 × 19)
ln(2.5)
Nₘᵢₙ
=
ln(361)
ln(2.5)
Nₘᵢₙ
=
5.8889
0.91629
Nₘᵢₙ = 6.4269
Check
Put the numbers back in. Both sides should match.
6.4269
=
ln((
0.95
1 − 0.95
)
×
(
1 − 0.05
0.05
))
÷
ln(2.5)
✓
Topic check
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