SME
Chemical Engineering Thermodynamics · Problem type
Binary flash
Use this for “A feed of 60% benzene and 40% toluene flashes at 90 °C and 760 mmHg (P₁sat = 1021 mmHg, P₂sat = 406.7 mmHg). Find x₁, y₁ and V ÷ F.”
Diagram
Pxy diagram of 1–2: the bubble curve runs from P₂ˢᵃᵗ = 406.7 to P₁ˢᵃᵗ = 1,021 mmHg, the dew curve under it; between them liquid and vapor coexist.
At
P
=
760
mmHg:
x₁
=
P − P₂ˢᵃᵗ
P₁ˢᵃᵗ − P₂ˢᵃᵗ
=
0.575,
y₁
=
0.773.
Lever
rule:
V
F
=
z₁ − x₁
y₁ − x₁
=
0.1259,
the
feed’s
distance
from
the
liquid
end
over
the
tie
line’s
length.
This is an example. Type your own values to start a new problem.
K₁
=
P₁ˢᵃᵗ (mmHg)
P (mmHg)
K₂
=
P₂ˢᵃᵗ (mmHg)
P (mmHg)
V ÷ F
=
−
−
Each yᵢ = Kᵢxᵢ, and x₁ + x₂ = y₁ + y₂ = 1, so x₁ = (1 − K₂) ÷ (K₁ − K₂) and y₁ = K₁x₁. V ÷ F is then the lever rule on the tie line from x₁ to y₁.
Formulas
K₁
=
P₁
ˢᵃᵗ
P
1.343
=
1,021
760
K₂
=
P₂
ˢᵃᵗ
P
0.5351
=
406.7
760
x₁
=
1 −
K₂
K₁
−
K₂
0.5751
=
1 − 0.5351
1.343 − 0.5351
y₁
=
K₁
×
x₁
0.7726 = 1.343 × 0.5751
V
F
=
z₁
−
x₁
y₁
−
x₁
0.1259
=
0.6 − 0.5751
0.7726 − 0.5751
Assumptions
•
Ideal liquid (Raoult’s law) and ideal vapor; the drum’s liquid and vapor are at equilibrium.
•
Both vapor pressures are at the drum’s temperature; component 1 is the more volatile one.
•
The feed lies between the bubble and dew pressures, so V ÷ F is between 0 and 1.
Step-by-step
WE KNOW
P₁ˢᵃᵗ = 1,021 mmHg, P₂ˢᵃᵗ = 406.7 mmHg, P = 760 mmHg, z₁ = 0.6
FIND
K-value of 1 (K₁), K-value of 2 (K₂), liquid mole fraction of 1 (x₁), vapor mole fraction of 1 (y₁), fraction vaporized (V ÷ F)
Step 1 · Find K-value of 1 (K₁)
Use K₁ = P₁ˢᵃᵗ ÷ P
Raoult’s
law:
y
x
is
the
vapor
pressure
over
the
total
pressure.
K₁
=
1,021
760
K₁
=
1.34342
Step 2 · Find K-value of 2 (K₂)
Use K₂ = P₂ˢᵃᵗ ÷ P
Raoult’s
law:
y
x
is
the
vapor
pressure
over
the
total
pressure.
K₂
=
406.7
760
K₂
=
0.535132
Step 3 · Find liquid mole fraction of 1 (x₁)
Use x₁ = (1 − K₂) ÷ (K₁ − K₂)
The
vapor’s
fractions
K₁
x₁
and
K₂x₂
add
to
1;
solve
that
for
x₁
.
x₁
=
1 − 0.535132
1.34342 − 0.535132
x₁
=
0.4649
0.8083
x₁
=
0.575126
Step 4 · Find vapor mole fraction of 1 (y₁)
Use y₁ = K₁ × x₁
The
vapor
in
equilibrium
holds
K₁
times
the
liquid’s
share
of
1.
y₁
=
1.34342
×
0.575126
y₁
=
0.77264
Step 5 · Find fraction vaporized (V ÷ F)
Use V ÷ F = (z₁ − x₁) ÷ (y₁ − x₁)
Lever rule: the feed’s distance from the liquid end of the tie line over the tie line’s length.
V
F
=
0.6 − 0.575126
0.77264 − 0.575126
V
F
=
0.024874
0.19751
V
F
=
0.1259
Check
Put the numbers back in. Both sides should match.
1.34342
=
1,021
760
✓
0.535132
=
406.7
760
✓
0.575126
=
1 − 0.535132
1.34342 − 0.535132
✓
0.77264 = 1.34342 × 0.575126 ✓
0.1259
=
0.6 − 0.575126
0.77264 − 0.575126
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Vapor–liquid equilibrium
Main lesson
Dew pressure
Use this for “A vapor of 50% benzene and 50% toluene is at 90 °C, where P₁sat = 1021 mmHg and P₂sat = 406.7 mmHg. Find its dew pressure.”
Margules activity coefficients
Use this for “A binary follows the one-parameter Margules equation with A = 0.8. Find γ₁ and γ₂ at x₁ = 0.3.”