Calculator Inputs
Example Data Table
The example below uses methane with room-temperature reactants, 10% excess air, 21% oxygen in the oxidizer, 99% combustion efficiency, and 2% heat loss.
| Fuel | Fuel Flow (kmol/h) | Fuel Temp (°C) | Oxidizer Temp (°C) | Excess Air (%) | O₂ Fraction (%) | Efficiency (%) | Heat Loss (%) | Tad (K) | Tad (°C) | Heat Rate (kW) |
|---|---|---|---|---|---|---|---|---|---|---|
| Methane (CH4) | 1.00 | 25.00 | 25.00 | 10.00 | 21.00 | 99.00 | 2.00 | 2,225.16 | 1,952.01 | 216.22 |
Formula Used
νO₂,st = x + y/4 − z/2 for fuel CxHyOz
λ = 1 + (Excess Air / 100), and φ = 1 / λ
Qavailable = LHV × η × (1 − Heat Loss) + Hfuel,sensible + Hoxidizer,sensible
Σ nproducts ∫ Cp,i(T) dT = Qavailable
This implementation uses a complete-combustion model with temperature-dependent product heat capacities and solves the final temperature numerically by bisection. It is suitable for fast engineering estimates, screening studies, and classroom calculations. It does not model chemical dissociation, radiation exchange, pressure dependence, soot formation, or finite-rate kinetics.
How to Use This Calculator
- Select the fuel from the list.
- Enter the fuel flow rate in kmol/h.
- Enter the fuel and oxidizer inlet temperatures.
- Set the excess air percentage and oxidizer oxygen fraction.
- Enter combustion efficiency and expected heat loss.
- Click the calculate button.
- Review the results shown above the form.
- Use the CSV or PDF buttons for reporting.
8 FAQs
1. What is adiabatic flame temperature?
It is the theoretical product temperature reached when combustion occurs without heat transfer to the surroundings. It represents an upper-limit estimate for many burners and reactors.
2. Why does excess air reduce flame temperature?
Extra air adds more mass that must be heated. That additional nitrogen and unused oxygen absorb energy, so the same chemical heat is spread across more product moles.
3. Why do hotter reactants raise flame temperature?
Preheated fuel and oxidizer carry sensible enthalpy into the reaction zone. That extra energy adds to the combustion heat, increasing the calculated final temperature.
4. Does this model include dissociation?
No. The calculator assumes complete combustion products and neglects high-temperature equilibrium dissociation. Real peak temperatures at very high temperatures can therefore be lower than the estimate.
5. What is the role of combustion efficiency?
Combustion efficiency reduces the usable chemical energy. Lower efficiency means less heat is released into the products, which lowers the predicted flame temperature.
6. Why is oxygen fraction an input?
It allows comparison between normal air, oxygen-enriched oxidizers, and other oxidizer mixtures. Higher oxygen fraction reduces inert dilution and usually raises flame temperature.
7. Can I use this for burner design?
Yes, for preliminary sizing and comparison. For final design, validate with detailed equilibrium tools, radiation analysis, residence-time effects, material limits, and safety margins.
8. Why might plant temperatures differ from this result?
Real systems lose heat, mix imperfectly, dissociate at high temperature, and experience wall cooling, incomplete combustion, humidity effects, and measurement uncertainty.