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The combustion heat initially shows up as sensible heat in the combustion products. The adiabatic flame temperature (AFT) is the highest-attainable flame temperature before any appreciable heat has been transferred to the charge. Heat is transferred from the combustion gas to the charge mainly by radiation, which is proportional to the 4th power of temperature.
Clearly, the higher the AFT, the more effective the flame is in transferring heat to the furnace contents. The calculated AFT is never attained in practice but is closely approached when combustion is rapid, and the flame does not directly impinge on the furnace contents or walls. The AFT is mainly used to compare changes in thermal intensity for different combustion settings (i.e. type of fuel, % excess air, oxygen enrichment, etc.). Here we show how to make an AFT calculation for methane (CH4) burned with 40% excess air.
Figure 1 shows the three steps for calculating an AFT. Please download workbook AFT-Calc.xlsx, which has details of this example and handles several fuels at variable “% excess air.” The heat effects of steps 1 and 2 are easily calculated using techniques from earlier columns.[4,5] However, a different procedure is required for the heat effect of step 3. To illustrate, we show the AFT calculation when air enters at 38°C (100˚F) and methane enters at 20˚C (68˚F).
Table 1 displays the relevant enthalpy data obtained from FREED. Table 2 displays the amounts of reactants and products. When reactants enter at temperatures between 0˚ and 100˚C, we can use their Cp instead of quadratic enthalpy equations. For normal air, Cp = 29.16 J/g-mole-deg, and for CH4, Cp = 36.0 J/g-mole-deg.
Equations 1a and 1b show the heat effects for step 1, and Equation 2 is the DH˚comb for methane (step 2). The algebraic sum of these terms must equal the value of step 3 (+807,200 J), which is the HAFT – H25 of the combustion gas.
Cool 13.33 moles of air from 38-25˚C: H25 – H38 = 13.33(29.16)(25 – 38) = –5,050 J [1a]
Heat one mole of CH4 from 20˚-25˚C: H25 – H20 = 36(25 – 20) = +180 J [1b]
Fuel combustion: CH4 + 2O2 ’ CO2 + 2H2O(g); DH˚comb at 25°C = –802,330 J. 
The heat-content equation for each Table 1 species was multiplied by the species amount (Table 2). Equation 3 shows the sum of the combined terms set to zero. The quadratic formula was used to calculate the AFT.
0.02430(T2) + 488.25T – 847,800 = 0; AFT = 1608˚C 
The workbook at www.industrialheating.com/AFT-Calc.xlsx does all this arithmetic and more, and you can customize it for your situation.
Theoretical vs. Actual (Equilibrium) AFT
The AFT calculated in Equation 3 is called the theoretical AFT because it assumes that the combustion products are entirely CO2 and H2O. Above about 1650˚C, however, there is significant equilibrium disproportionation of CO2 and H2O to other gases, plus the formation of NO(g). The workbook shows the difference between the theoretical and actual AFT. This will be the subject of a later article. IH
1. Arthur Morris, “Available Combustion Heat,” Industrial Heating, May 2013
2. Wikipedia contributors, “Adiabatic flame temperature,” Wikipedia, the Free Encyclopedia, March 2013 http://en.wikipedia.org/wiki/Adiabatic_flame_temperature
3. Wikipedia contributors, “Thermal radiation,” Wikipedia, the Free Encyclopedia, March 2013. http://en.wikipedia.org/wiki/Thermal_radiation.
4. Arthur Morris, “Calculating the Heat of Combustion of Natural Gas,” Industrial Heating, September 2012
5. Arthur Morris, “Making a Heat Balance,” Industrial Heating, December 2012
7. Wikipedia contributors, “Quadratic equation,” Wikipedia, the Free Encyclopedia, March 2013. http://en.wikipedia.org/wiki/Quadratic_equation.