Defines adiabatic flame temperature, describes how to calculate it, and discusses issues that affect the accuracy of the calculation.

We suggest that after watching this screencast, you list the important points as a way to increase retention.

Demonstrates how to use an Excel spreadsheet to calculate the adiabatic flame temperature for methane oxidation in air. Dissociation reactions are ignored.

We suggest that after watching this screencast and using the spreadsheet, you list the important points as a way to increase retention.

##### Important Equations:

Heat capacity

$C_{P,i} = A_i +B_iT + C_iT^2 +D_iT^3 + \frac{E_i}{T^2}$

where $$T$$ is in kelvin, $$C_{P,i}$$ = heat capacity of component $$i$$ (J/(molK), and $$A_i, B_i, C_i, D_i, and E_i$$ are constants for component $$i$$; these constants are only valid over a specified temperature range.

Heat of reaction

For the gas-phase reaction: $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$

$\Delta H_{rxn} = \sum\nu_i\Delta H_{f,i} = 2 \Delta H_{f,H2O} + \Delta H_{f,CO2} – \Delta H_{f,CH4} – 2\Delta H_{f,O2}$

where $$\Delta H_{rxn} =$$ heat of reaction at 298 K,

$$\nu =$$ stoichiometric coefficient, and

$$\Delta H_{f,i} =$$ heat of formation (enthalpy of formation) for species $$i$$ at 298 K

Energy balance

$\Delta H = Q = n_{fuel} \Delta H_{rxn} + (n_{O_2}H_{O_2} + n_{N_2}H_{N_2} + n_{CO_2}H_{CO_2} + n_{H_2O}H_{H_2O})_{out} – (n_{O_2}H_{O_2} + n_{N_2}H_{N_2})_{in} – (n_{fuel}H_{fuel})_{in}$

$\Delta H = n_{fuel}\Delta H_{rxn} + \sum (n_i^{out} \int_{298}^T C_PdT) – \sum (n_i^{in} \int_{298}^{T_in}C_PdT)$

where $$n_i^{out}$$is the moles of component $$i$$ leaving the furnace,

$$T$$ is the adiabatic flame temperature,

$$n_i^{in}$$ is the moles of component $$i$$ fed to the furnace, and

$$T_in$$ is the feed temperature for each component fed to the furnace.

Using NIST webbook to calculate enthalpies

The adiabatic flame temperature spreadsheet uses the equations for enthalpy in the NIST Chemistry Webbook (webbook.nist.gov). These equations result from integrating the heat capacity as a function of temperature, applying the limites, and then grouping terms into constants $$F$$ and $$H$$ (where $$H$$ is not enthalpy) so that enthalpy at a given temperature is found using this equation:

$H_T – H_{298} = A_iT + \frac {B_iT^2}{2} + \frac {C_iT^3}{3} + \frac {D_iT^4}{4} -\frac{E_i}{T} +F -H$

where $$H_T$$ is enthalpy (kJ/mol) at temperature T(K)

and $$H_{298}$$ is enthalpy (kJ/mol) at 298 K.