SME
Propulsion · Problem type
Fuel–air ratio of a combustor
Use this for “Air enters the burner at 600 K. What fuel–air ratio heats it to 1400 K with 43,000 kJ/kg fuel, and what is φ?”
Diagram
Total in: 1 + 0.02223 = 1.022; out: 1.022 kg per kg of air.
Energy in 1,646 = energy out 1,646 kJ per kg of air.
Stoichiometric fuel–air ratio fₛₜ = 0.068
Equivalence ratio φ = 0.3269
This is an example. Type your own values to start a new problem.
f
=
cₚ (kJ/(kg·K))
×
(
T₀₄ (K)
−
T₀₃ (K)
)
÷
(
QR (kJ/kg)
−
cₚ (kJ/(kg·K))
×
T₀₄ (K)
)
φ
=
f
÷
fₛₜ
Per kilogram of air, the fuel’s fQR heats the air from T₀₃ and 1 + f kg of gas leaves at T₀₄: cₚT₀₃ + fQR = (1 + f)cₚT₀₄. Solving it for f gives the first equation.
Formulas
f
=
cₚ
×
T₀₄
−
T₀₃
QR − cₚ ×
T₀₄
0.02223
=
1.15
×
1,400 − 600
43,000 − 1.15 × 1,400
φ
=
f
fₛₜ
0.3269
=
0.02223
0.068
Assumptions
•
Per kilogram of air: f kg of fuel releases fQR, and 1 + f kg of gas leaves.
•
One cₚ for the air and the gas; total temperatures; no heat lost through the walls.
•
φ compares f with the stoichiometric ratio fₛₜ of the fuel.
Step-by-step
WE KNOW
T₀₃ = 600 K, T₀₄ = 1,400 K, cₚ = 1.15 kJ/(kg·K), QR = 43,000 kJ/kg, fₛₜ = 0.068
FIND
fuel–air ratio (f), equivalence ratio (φ)
Step 1 · Find fuel–air ratio (f)
Use f = cₚ × (T₀₄ − T₀₃) ÷ (QR − cₚ × T₀₄)
Energy
in,
cₚT₀₃
+
fQR,
equals
energy
out,
(1
+
f
)cₚT₀₄;
solve
for
f
.
f
=
1.15
×
1,400 − 600
43,000 − 1.15 × 1,400
f
=
1.15
×
800
43,000 − 1,610
f
=
1.15
×
800
41,390
f
=
920
41,390
f
=
0.02223
Step 2 · Find equivalence ratio (φ)
Use φ = f ÷ fₛₜ
The fuel used, as a share of what would burn all the oxygen.
φ
=
0.02223
0.068
φ
=
0.3269
Check
Put the numbers back in. Both sides should match.
0.02223
=
1.15
×
1,400 − 600
43,000 − 1.15 × 1,400
✓
0.3269
=
0.02223
0.068
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.