Library / Aerospace / AER-49

Aerospace

Combustor fuel-to-air mass ratio

Mass of fuel burned per mass of air from an energy balance, in the constant-specific-heat model.

Ref: AER-49
Rev: v1
Field: Aerospace
Result: f

f = \frac{c_{p,g} T_{t4} - c_{p,a} T_{t3}}{η_{b} Q_{LHV} - c_{p,g} T_{t4}}

An energy balance on the combustor sets the fuel needed to raise the gas from Tt3 to Tt4: f·ηb·QLHV + cpa·Tt3 = (1 + f)·cpg·Tt4, which gives f = (cpg Tt4 − cpa Tt3)/(ηb QLHV − cpg Tt4). The specific heats are chosen so that the sensible enthalpies refer to a common datum. For a turbine inlet temperature of 1400 K and a compressor exit of 600 K, f is about 2.3 %.

Mass of fuel burned per mass of air from an energy balance, in the constant-specific-heat model. cpa and cpg are chosen consistently so that sensible enthalpies refer to a common datum. T3 and T4 are total temperatures, ηb is combustion efficiency and Q the fuel lower heating value. Fuel sensible enthalpy and bleed are ignored.

Calculator

Solve for

fFuel-to-air ratioComputed

c_p,gSpecific heat of the hot gas

J/(kg·K)

1,0501,250

T_t4Combustor exit total temperature

9002,000

c_p,aSpecific heat of the compressed air

J/(kg·K)

9501,100

T_t3Combustor inlet total temperature

400900

η_bCombustion efficiency

0.91

Q_LHVFuel lower heating value

J/kg

4e75e7

Fuel energy per kg of air, f·ηb·Q975,700 J/kg
Total temperature rise, Tt4 − Tt3800 K
Fuel flow for 20 kg/s of air0.4631 kg/s

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Fuel-to-air ratio () current value 0.02315

T_t4Combustor exit total temperatureKey driver

0.009480.0401

T_t3Combustor inlet total temperature

0.01580.0281

c_p,gSpecific heat of the hot gas

0.02140.0285

Q_LHVFuel lower heating value

0.01980.025

η_bCombustion efficiency

0.02270.0253

c_p,aSpecific heat of the compressed air

0.02170.0239

The combustor burns 0.02315 kg of fuel per kg of air (2.315 %) to raise the gas by 800 K, so 0.4631 kg/s of fuel for a 20 kg/s core.
The air is far leaner than stoichiometric (about 0.068), which is why jet engines carry large excess air for cooling and dilution.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing ()	Local step	Change in result
T_t4 Combustor exit total temperature	900 to 2,000 K	0.009478	0.04006	0.03058	+10%	+16.8 %
T_t3 Combustor inlet total temperature	400 to 900 K	0.01576	0.02808	0.01232	+10%	−6.38 %
c_p,g Specific heat of the hot gas	1,050 to 1,250 J/(kg·K)	0.02139	0.02847	0.007081	+10%	+16.8 %
Q_LHV Fuel lower heating value	4e7 to 5e7 J/kg	0.01981	0.02496	0.005154	+10%	−9.4 %
η_b Combustion efficiency	0.9 to 1	0.02267	0.0253	0.002624	+10%	−9.4 %
c_p,a Specific heat of the compressed air	950 to 1,100 J/(kg·K)	0.02167	0.02389	0.002217	+10%	−6.38 %

Sweep of Combustor exit total temperature

Combustor exit total temperature (K)	Fuel-to-air ratio
900	0.009478
1,038	0.0132
1,175	0.01695
1,313	0.02073
1,450	0.02454
1,588	0.02838
1,725	0.03224
1,863	0.03614
2,000	0.04006

Sweep one variable

Vary

Fuel-to-air ratio ()Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Fuel-to-air ratio

Result ·

What the equation gives you. Everything else feeds it.

c_p,g

Specific heat of the hot gas

J/(kg·K) · typical 1,050 to 1,250

The specific heat of the hot gas, in J/(kg·K). Raises the fuel needed to heat the gas. Raising it by 10 % (to 1,210) moves the result by +16.8 %, from 0.02315 to 0.02705.

T_t4

Combustor exit total temperature

K · typical 900 to 2,000

The combustor exit total temperature, in K. Raises the fuel-air ratio, steeply at high temperatures. Raising it by 10 % (to 1,540) moves the result by +16.8 %, from 0.02315 to 0.02705.

c_p,a

Specific heat of the compressed air

J/(kg·K) · typical 950 to 1,100

The specific heat of the compressed air, in J/(kg·K). A higher value lowers the fuel needed, since the air carries more sensible enthalpy in. Raising it by 10 % (to 1,100) moves the result by −6.38 %, from 0.02315 to 0.02167.

T_t3

Combustor inlet total temperature

K · typical 400 to 900

The combustor inlet total temperature, in K. A hotter inlet air lowers the fuel needed. Raising it by 10 % (to 660) moves the result by −6.38 %, from 0.02315 to 0.02167.

η_b

Combustion efficiency

· typical 0.9 to 1

The combustion efficiency. A lower efficiency raises the fuel needed in inverse proportion. Raising it by 10 % (to 1.078) moves the result by −9.4 %, from 0.02315 to 0.02098.

Q_LHV

Fuel lower heating value

J/kg · typical 4e7 to 5e7

The fuel lower heating value, in J/kg. A higher heating value lowers the fuel needed. Raising it by 10 % (to 4.73×10⁷) moves the result by −9.4 %, from 0.02315 to 0.02098.

Worked example with the values above

Heat to be added: cpg Tt4 − cpa Tt3 = 1,540,000 − 600,000 = 940,000 J/kg.
Available: ηb Q − cpg Tt4 = 4.214×10⁷ − 1,540,000 = 4.06×10⁷ J/kg.
f = 940,000 / 4.06×10⁷ = 0.0231527 kg fuel/kg air.

Assumptions and limits

Constant specific heats in each stream, adiabatic components and total-to-total efficiencies.
Steady operation at one operating point, with no bleed and no leakage unless the page says so.

Common mistakes

Mixing total and static states.
Using a total-to-static efficiency where a total-to-total one is needed.
Forgetting the fuel added to the turbine mass flow.

Statistics and uncertainty

Constant-cp cycle calculations are rough: cp rises from 1005 J/(kg·K) in cold air to about 1150 J/(kg·K) in hot gas, which shifts results by 5 % to 10 %. Component efficiencies from maps are good to 1 % to 2 % at design and worse off design.

Where it is used

Matching compressors, combustors and turbines and finding fuel flow and power.
Sizing engines and checking cycle designs.

Maths you need

Isentropic efficiency: ideal work divided by actual work (or the reverse for a turbine)
Total temperature: the temperature the gas would reach at rest
Brayton cycle: compression, heat addition, expansion

Related formulas

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