Propane flame

How hot does a propane flame in air get, and how much CO and NO does it make?

You ask

How hot does a propane flame in air get, and how much CO and NO does it make?

Wobbl runs

Terminal
$ wobbl flame propane --phi 1 --pressure "1 atm" --temperature "25 C"

Checks

  • passelement balance
  • passenergy balance
  • passthermo data
  • passno dissociation bound
  • passflammability
Flame temperature across mixtures. Adiabatic flame temperature 2266 K (1992 C) for propane in air at an equivalence ratio of 1, with 1.25% CO and 0.234% NO at equilibrium.

OKAdiabatic flame temperature 2266 K (1992 C) for propane in air at an equivalence ratio of 1, with 1.25% CO and 0.234% NO at equilibrium.

Flame temperature
2266 K
Equivalence ratio
1
CO
1.25%
Confidence
high
summary.txt
summary.txt
OK. Adiabatic flame temperature 2266 K (1992 C) for propane in air at an equivalence ratio of 1, with 1.25% CO and 0.234% NO at equilibrium. Confidence: high. passed element balance, energy balance, thermo data, no dissociation boundflame temperature: 2266 Kequivalence ratio: 1CO: 1.25%NO: 0.234% Checks:  [pass] element balance: every element of the reactants is in the products to 6e-16  [pass] energy balance: the products at 2266 K hold the reactants' enthalpy to 2e-09 of cp T  [pass] thermo data: GRI-Mech 3.0's own thermodynamic data give 2266 K for the same flame, 0.0694 K from NASA's (0.0031% apart)  [pass] no dissociation bound: burning completely to CO2 and H2O with nothing breaking apart would reach 2392 K, and dissociation takes 126 K off it  [pass] flammability: propane burns in air from 2.1% to 9.5% by volume, an equivalence ratio of 0.51 to 2.50, and this mixture is 4.03% fuel Assumptions:  - Propane (C3H8) and air as O2 + 3.76 N2 at 298 K and 101 kPa, fully mixed.  - Adiabatic and at constant pressure, with the products in chemical equilibrium on NASA's thermodynamic data (McBride, Gordon and Reno, NASA TM-4513). A real flame loses heat by radiation and to walls, so its peak runs below this temperature.  - Equilibrium NO is an upper bound on what a flame makes, since forming NO takes longer than most flames last. Images: sweep: images/sweep.pngReport: report.html

Open the full report

About this example

This is propane mixed with exactly the air it needs, an equivalence ratio of 1, at 1 atm and 25 C, as in a torch or a premixed burner. Wobbl burns it to chemical equilibrium on NASA’s thermodynamic data with Cantera.

Result Value
Adiabatic flame temperature 2266 K (1992 C)
Without dissociation 2392 K, so breaking CO2 and H2O apart at the flame takes 126 K off
GRI-Mech 3.0’s own data 2266 K, 0.07 K from NASA’s
Equilibrium CO and NO at the flame 1.25% and 0.234%
Flammability 4.03% propane, inside the 2.1% to 9.5% it burns over in air
Flame temperature against equivalence ratio

The flame is hottest just rich of stoichiometric, near an equivalence ratio of 1.05, and cools steeply on either side, which is why a burner runs lean to keep its walls and its NO down. The temperature is an upper bound for a real flame, which loses heat by radiation and to the burner, and the equilibrium NO is more than a flame makes in the few milliseconds the gas spends in it. For a flame that has to stay under a temperature, wobbl flame propane --limit "1800 C" finds the leaner mixture that meets it.

Run it yourself

Terminal
$ git clone https://github.com/sohan-bhat/wobbl$ cd wobbl/examples/propane-flame && ./run.sh

The part, the script and every output are in examples/propane-flame on GitHub.