SME
Process Design
Process safety
Use this for “A fuel gas is 80% methane (LFL 5.0%), 15% ethane (3.0%) and 5% propane (2.1%), with a mixture UFL of 14.1%. Find its LFL. Is 6% in air flammable?”
Diagram
4.277 ≤ c ≤ 14.1: 4.277 ≤ c and c ≤ 14.1, both true at once: the numbers between.
Both circles are closed: 4.277 and 14.1 are taken in.
Test 6 %: 6 ≥ 4.277 is true, 6 ≤ 14.1 is true: both true, a solution.
This is an example. Type your own values to start a new problem.
y₁
+
y₂
+
y₃
=
1
LFLₘᵢₓ (%)
=
1
÷
(
y₁
÷
LFL₁ (%)
+
y₂
÷
LFL₂ (%)
+
y₃
÷
LFL₃ (%)
)
UFLₘᵢₓ
Mixture UFL
example · %
c
Fuel in air
example · %
state
Mixture in air
worked out (not typed)
Formulas
y₁
+
y₂
+
y₃
=
1
0.8 + 0.15 + 0.05 = 1
LFLₘᵢₓ
=
1
y₁
÷ LFL₁ +
y₂
÷ LFL₂ +
y₃
÷ LFL₃
4.277
=
1
0.8 ÷ 5 + 0.15 ÷ 3 + 0.05 ÷ 2.1
state
=
too
lean
if
c
<
LFLₘᵢₓ,
flammable
if
c
≤
UFLₘᵢₓ,
else
too
rich
6 ≤ 14.1, so flammable
Assumptions
•
Le Chatelier’s rule, LFLₘᵢₓ = 1 ÷ Σ(yᵢ ÷ LFLᵢ), works best for similar fuels such as light hydrocarbons.
•
Each yᵢ is that fuel’s mole fraction in the fuel alone, so they sum to 1; limits are vol % in air.
•
Air at 25 °C and 1 atm; UFLₘᵢₓ is typed, found the same way from each fuel’s upper limit.
Step-by-step
WE KNOW
y₁ = 0.8, y₂ = 0.15, LFL₁ = 5%, LFL₂ = 3%, LFL₃ = 2.1%, UFLₘᵢₓ = 14.1%, c = 6%
FIND
fuel 3 share (y₃), mixture LFL (LFLₘᵢₓ), mixture in air (state)
Step 1 · Find fuel 3 share (y₃)
Use y₁ + y₂ + y₃ = 1
The shares of the fuel add to 1, so the third is what the first two leave.
y₃
=
1
−
y₁
−
y₂
y₃
=
1
−
0.8
−
0.15
y₃
=
0.05
Step 2 · Find mixture LFL (LFLₘᵢₓ)
Use LFLₘᵢₓ = 1 ÷ (y₁ ÷ LFL₁ + y₂ ÷ LFL₂ + y₃ ÷ LFL₃)
Le
Chatelier:
at
the
mixture’s
lower
limit,
each
fuel’s
own
vol
%
in
air
over
its
own
LFL
adds
up
to
1,
which
gives
1
Σ(yᵢ ÷ LFLᵢ)
.
LFLₘᵢₓ
=
1
0.8 ÷ 5 + 0.15 ÷ 3 + 0.05 ÷ 2.1
LFLₘᵢₓ
=
1
0.16 + 0.05 + 0.02381
LFLₘᵢₓ
=
1
0.2338
LFLₘᵢₓ = 4.277%
Step 3 · Find mixture in air (state)
Use state = too lean if c < LFLₘᵢₓ, flammable if c ≤ UFLₘᵢₓ, else too rich
Between the limits, one spark can set off flame that spreads through the whole mixture.
6 ≤ 14.1: flammable
state = too lean if 6 < 4.277, flammable if 6 ≤ 14.1, else too rich
state = flammable
Check
Put the numbers back in. Both sides should match.
0.8 + 0.15 + 0.05 = 1 ✓
4.277
=
1
0.8 ÷ 5 + 0.15 ÷ 3 + 0.05 ÷ 2.1
✓
6 ≤ 14.1, so flammable ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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