Adiabatic flame temperature

Remarks

Example: assume complete combustion, a heat of reaction $Q > 0$ per unit mass of mixture at $T_i$, and a constant specific heat $c_p$ of the products. Then $H_p(T, p) = H_r(T_i, p) - Q + c_p (T - T_i)$, and the defining equation gives $T_{\mathrm{ad}} = T_i + Q / c_p$. Non-example: if the mixture loses an amount of heat $q > 0$ to the walls, the final temperature $T_f$ satisfies $H_p(T_f, p) = H_r(T_i, p) - q$. The process is not adiabatic, and $T_f < T_{\mathrm{ad}}$ because $H_p$ increases with $T$. The constant-volume value is higher than the constant-pressure value, since at constant volume no energy is spent on expansion work. With chemical equilibrium, dissociation of the products at high temperature absorbs energy and lowers $T_{\mathrm{ad}}$ below the complete-combustion value. For a stoichiometric methane–air mixture at room conditions, the constant-pressure value is about $2200$ K.