Adiabatic flame temperature

Common flame temperatures

Assuming initial atmospheric conditions (1 bar and 20 °C), the following table lists the flame temperature for various fuels under constant pressure conditions. The temperatures mentioned here are for a stoichiometric fuel-oxidizer mixture (i.e. equivalence ratio φ = 1).

Note that these are theoretical, not actual, flame temperatures produced by a flame that loses no heat. The closest will be the hottest part of a flame, where the combustion reaction is most efficient. This also assumes complete combustion (e.g. perfectly balanced, non-smoky, usually bluish flame). Several values in the table significantly disagree with the literature or predictions by online calculators.

Thermodynamics

From the first law of thermodynamics for a closed reacting system we have

${\displaystyle {}_R{Q}_P-{}_R{W}_P=U_P-U_R}$

where, ${\displaystyle {}_R{Q}_P}$ and ${\displaystyle {}_R{W}_P}$ are the heat and work transferred from the system to the surroundings during the process, respectively, and ${\displaystyle U_R}$ and ${\displaystyle U_P}$ are the internal energy of the reactants and products, respectively. In the constant volume adiabatic flame temperature case, the volume of the system is held constant and hence there is no work occurring:

${\displaystyle {}_R{W}_P=\underset{R}{\overset{P}{\int }}pdV=0}$

There is also no heat transfer because the process is defined to be adiabatic: ${\displaystyle {}_R{Q}_P=0}$. As a result, the internal energy of the products is equal to the internal energy of the reactants: ${\displaystyle U_P=U_R}$. Because this is a closed system, the mass of the products and reactants is constant and the first law can be written on a mass basis,

${\displaystyle U_P=U_R\Rightarrow m_P{u}_P=m_R{u}_R\Rightarrow u_P=u_R}$.

In the case of the constant pressure adiabatic flame temperature, the pressure of the system is held constant, which results in the following equation for the work:

${\displaystyle {}_R{W}_P=\underset{R}{\overset{P}{\int }}pdV=p\left(V_P-V_R\right)}$

Again there is no heat transfer occurring because the process is defined to be adiabatic: ${\displaystyle {}_R{Q}_P=0}$. From the first law, we find that,

${\displaystyle -p\left(V_P-V_R\right)=U_P-U_R\Rightarrow U_P+p{V}_P=U_R+p{V}_R}$

Recalling the definition of enthalpy we obtain ${\displaystyle H_P=H_R}$. Because this is a closed system, the mass of the products and reactants is the same and the first law can be written on a mass basis:

${\displaystyle H_P=H_R\Rightarrow m_P{h}_P=m_R{h}_R\Rightarrow h_P=h_R}$.

We see that the adiabatic flame temperature of the constant pressure process is lower than that of the constant volume process. This is because some of the energy released during combustion goes, as work, into changing the volume of the control system.

If we make the assumption that combustion goes to completion (i.e. forming only CO
₂ and H
₂O), we can calculate the adiabatic flame temperature by hand either at stoichiometric conditions or lean of stoichiometry (excess air). This is because there are enough variables and molar equations to balance the left and right hand sides,

${\displaystyle {\mathrm{C}}_{\alpha }{\mathrm{H}}_{\beta }{\mathrm{O}}_{\gamma }{\mathrm{N}}_{\delta }+\left(a{\mathrm{O}}_2+b{\mathrm{N}}_2\right)\to {\nu }_1{\mathrm{C}\mathrm{O}}_2+{\nu }_2{\mathrm{H}}_2\mathrm{O}+{\nu }_3{\mathrm{N}}_2+{\nu }_4{\mathrm{O}}_2}$

Rich of stoichiometry there are not enough variables because combustion cannot go to completion with at least CO and H
₂ needed for the molar balance (these are the most common products of incomplete combustion),

${\displaystyle {\mathrm{C}}_{\alpha }{\mathrm{H}}_{\beta }{\mathrm{O}}_{\gamma }{\mathrm{N}}_{\delta }+\left(a{\mathrm{O}}_2+b{\mathrm{N}}_2\right)\to {\nu }_1{\mathrm{C}\mathrm{O}}_2+{\nu }_2{\mathrm{H}}_2\mathrm{O}+{\nu }_3{\mathrm{N}}_2+{\nu }_5\mathrm{C}\mathrm{O}+{\nu }_6{\mathrm{H}}_2}$

However, if we include the water gas shift reaction,

${\displaystyle {\mathrm{C}\mathrm{O}}_2+H_2\iff \mathrm{C}\mathrm{O}+{\mathrm{H}}_2\mathrm{O}}$

and use the equilibrium constant for this reaction, we will have enough variables to complete the calculation.

Different fuels with different levels of energy and molar constituents will have different adiabatic flame temperatures.

We can see by the following figure why nitromethane (CH₃NO₂) is often used as a power boost for cars. Since each molecule of nitromethane contains an oxidant with relatively high-energy bonds between nitrogen and oxygen, it can burn much hotter than hydrocarbons or oxygen-containing methanol. This is analogous to adding pure oxygen, which also raises the adiabatic flame temperature. This in turn allows it to build up more pressure during a constant volume process. The higher the pressure, the more force upon the piston creating more work and more power in the engine. It stays relatively hot rich of stoichiometry because it contains its own oxidant. However, continual running of an engine on nitromethane will eventually melt the piston and/or cylinder because of this higher temperature.

In real world applications, complete combustion does not typically occur. Chemistry dictates that dissociation and kinetics will change the composition of the products. There are a number of programs available that can calculate the adiabatic flame temperature taking into account dissociation through equilibrium constants (Stanjan, NASA CEA, AFTP). The following figure illustrates that the effects of dissociation tend to lower the adiabatic flame temperature. This result can be explained through Le Chatelier's principle.