SME
Propulsion
Combustion
Use this for “Methane, CH₄, burns in air. What are the stoichiometric fuel–air and air–fuel ratios? At f = 0.04, what is φ?”
Diagram
Per
kmol
of
fuel:
O₂
for
complete
burning
is
x
+
y
4
=
1
+
4
4
=
2
kmol.
At
φ
=
0.6876
the
air
is
1
φ
=
1.454
times
that:
2.909
kmol
of
O₂
in,
0.9085
left
over.
Air brings 3.76 N₂ with each O₂: 10.94 kmol of N₂, 13.84 kmol of air in all; the N₂ passes through.
Atoms: C 1 = 1, H 4 = 2 × 2, O 2 × 2.909 = 2 × 1 + 2 + 2 × 0.9085.
Air for 1 kmol of fuel nₐᵢᵣ = 9.52 kmol
Air mass per kmol of fuel mₐᵢᵣ = 275.8 kg/kmol
Stoichiometric fuel–air ratio fₛₜ = 0.05817
Stoichiometric air–fuel ratio AFRₛₜ = 17.19
Fuel–air ratio f = 0.04
Equivalence ratio φ = 0.6876
This is an example. Type your own values to start a new problem.
x
Carbon atoms
example
y
Hydrogen atoms
example
nₐᵢᵣ
Air for 1 kmol of fuel
calculated · kmol
mₐᵢᵣ
Air mass per kmol of fuel
calculated · kg/kmol
Mfuel
Fuel molar mass
calculated
fₛₜ
Stoichiometric fuel–air ratio
calculated
AFRₛₜ
Stoichiometric air–fuel ratio
calculated
f
Fuel–air ratio
example
φ
Equivalence ratio
calculated
Formulas
nₐᵢᵣ
=
(
x
+
y
4
)
×
4.76
9.52
=
(1
+
4
4
)
×
4.76
mₐᵢᵣ = nₐᵢᵣ × 28.97
275.8 = 9.52 × 28.97
Mfuel
=
12.011
×
x
+
1.008
×
y
16.04 = 12.011 × 1 + 1.008 × 4
fₛₜ
=
Mfuel
mₐᵢᵣ
0.05817
=
16.04
275.8
AFRₛₜ
=
mₐᵢᵣ
Mfuel
17.19
=
275.8
16.04
φ
=
f
fₛₜ
0.6876
=
0.04
0.05817
Assumptions
•
Air is 21% O₂ and 79% N₂: 3.76 N₂ ride with each O₂, 4.76 kmol of air.
•
Complete combustion: all carbon to CO₂, all hydrogen to H₂O; the N₂ passes through.
•
Molar masses: air 28.97, carbon 12.011, hydrogen 1.008 kg/kmol.
•
φ below 1 is lean (O₂ left over), above 1 rich.
Step-by-step
WE KNOW
x = 1, y = 4, f = 0.04
FIND
air for 1 kmol of fuel (nₐᵢᵣ), air mass per kmol of fuel (mₐᵢᵣ), fuel molar mass (Mfuel), stoichiometric fuel–air ratio (fₛₜ), stoichiometric air–fuel ratio (AFRₛₜ), equivalence ratio (φ)
Step 1 · Find air for 1 kmol of fuel (nₐᵢᵣ)
Use nₐᵢᵣ = (x + y ÷ 4) × 4.76
Each carbon takes one O₂ and each four hydrogens one more; each O₂ comes in 4.76 kmol of air.
nₐᵢᵣ
=
(1
+
4
4
)
×
4.76
nₐᵢᵣ = (1 + 1) × 4.76
nₐᵢᵣ = 2 × 4.76
nₐᵢᵣ = 9.52 kmol
Step 2 · Find air mass per kmol of fuel (mₐᵢᵣ)
Use mₐᵢᵣ = nₐᵢᵣ × 28.97
Multiply the kilomoles of air by its molar mass, 28.97 kg/kmol.
mₐᵢᵣ = 9.52 × 28.97
mₐᵢᵣ = 275.79 kg/kmol
Step 3 · Find fuel molar mass (Mfuel)
Use Mfuel = 12.011 × x + 1.008 × y
Add up the atoms: carbon 12.011 and hydrogen 1.008 kg/kmol each.
Mfuel = 12.011 × 1 + 1.008 × 4
Mfuel = 12.011 + 4.032
Mfuel = 16.043 kg/kmol
Step 4 · Find stoichiometric fuel–air ratio (fₛₜ)
Use fₛₜ = Mfuel ÷ mₐᵢᵣ
One kmol of fuel over the air that burns it exactly, both in kilograms.
fₛₜ
=
16.043
275.79
fₛₜ = 0.05817
Step 5 · Find stoichiometric air–fuel ratio (AFRₛₜ)
Use AFRₛₜ = mₐᵢᵣ ÷ Mfuel
The same masses the other way up: kilograms of air for each kilogram of fuel.
AFRₛₜ
=
275.79
16.043
AFRₛₜ = 17.19
Step 6 · Find equivalence ratio (φ)
Use φ = f ÷ fₛₜ
The fuel actually burned, as a share of what would use up all the oxygen.
φ
=
0.04
0.05817
φ
=
0.6876
Check
Put the numbers back in. Both sides should match.
9.52
=
(1
+
4
4
)
×
4.76
✓
275.79 = 9.52 × 28.97 ✓
16.043 = 12.011 × 1 + 1.008 × 4 ✓
0.05817
=
16.043
275.79
✓
17.19
=
275.79
16.043
✓
0.6876
=
0.04
0.05817
✓
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Fuel–air ratio of a combustor
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