Adiabatic flame temperatures¶

Flame temperatures depend on

• stoichiometry
• products
• fuel
• oxidizer
• enthalpy

These properties are affected by

• heat losses (radiation)
• differential diffusion
• flame strain, mixing fields
• residence time
• soot formation

For convenience here

• use products of complete combustion,
• at constant pressure (1 atm),
• with reactants at 298.15 K (25 C)
• adiabatic

Two kinds of energy: internal, and sensible

• Internal is energy stored in bonds
• Sensible is thermal energy
• Combustion reactants have more energy stored in chemical bonds (high internal energy) than combustion products (low internal energy).
  • The difference is sensible energy that results in high flame temperatures.
• That is, internal energy is released as sensible energy

Enthalpy, energy, C$_p$¶

• Considering ideal gases

• Total enthalpy is the sum of species enthalpies times the amount of the species. $$H = \sum_k n_kh_k$$ $$h_k = h_{f,k}^o + \int_{T_r}^T c_{p,k}dT$$

  • $n_k$ is moles of species $k$,
  • $H$ is total enthalpy (J),
  • $h_k$ is enthalpy per mole (J/kmol).
  • $c_{p,k}$ is heat capacity of species $k$ (J/kmol*K)

• If we divide the first equation by the total number of moles we have $$ h = \sum_k x_kh_k$$

• Insert the second equation into this: $$ h = \sum_k h_{f,k}^o + \sum_k\int_{T_r}^Tc_{p,k}dT$$ $$ h = h_{f}^o + \int_{T_r}^Tc_pdT$$

• Here, $h_{f}^o = \sum_kh_{f,k}^o$ and $c_p = \sum_kc_{p,k}$

• All of this can be done on a mass basis too, just replace $x_k$ with $y_k$ and $n$ for $m$. Then $h_k$ is J/kg instead of J/kmol, and $c_p$ is J/kgK instead of J/kmolK.

  • To convert $h_k$ from per mole to per mass, divide by the molecular weight. Same for $c_p$.

• $h_{f,k}^o$ is the heat of formation of species $k$ at 1 atm and 298.15 K. This is the enthalpy of formation from elements in their standard states.

C$_p$¶

• Heat capacities are temperature dependent
  • Usually have a polynomial curve fit to represent $c_p(T)$.
  • Hence, enthalpy, entropy, etc. are temperature dependent.
  • $c_p$ are normalized by $R$, so the units on $c_p$ are whatever the units on the chosen $R$ are.
  • $c_p$ increases with temperature.

Heat of reaction¶

• $\Delta H_{rxn} = H_{prod}-H_{react}$
  • reactants and products at the SAME temperature
• $\Delta H_{rxn} < 0$ since combustion reactions are exothermic.
• $\Delta H_{comb} = -\Delta H_{rxn}$, so heats of combustion are positive.
• Usually reported per mole of fuel, or per kg of fuel (be careful).

Heating values: LHV, HHV¶

• As a basis for the amount of heat that can be extracted from combustion, the higher heating value (HHV) and lower heating value LHV are used.
• The HHV treats product $H_2O$ as a liquid.
• The LHV treats product $H_2O$ as a vapor.
• These use products of complete combustion at 298.15 K and 1 atm.
• See Turns Table B1.

T$_{ad}$ for CH$_4$¶

$$CH_4 + 2O_2 + 7.52N_2 \rightarrow CO_2 + 2H_2O + 7.52 N_2$$

$H_R(298.15\, K) = H_P(T_{ad})$

• This is one equation in one unknown.
• The reactant and product compositions are known. The reactant temperature is known, so the reactant enthalpy is known.
• Solve $T_{ad}$.

$$H_R = H_P(T_{ad})$$ $$H_P(T_{ad}) = 1\mbox{ mole}\cdot h_{CO2}(T_{ad}) + 2\cdot h_{H2O}(T_{ad}) + 7.52\cdot h_{N2}(T_{ad})$$

• Turns Table A.13 gives $h_{k}(T)$ as polynomial expressions: $$\frac{h_{k}(T)}{RT} = a_1 + \frac{a_2}{2}T + \frac{a_3}{3}T^2 + \frac{a_4}{4}T^3 + \frac{a_5}{5}T^4 + \frac{a_6}{T}$$
  • where a different set of coefficients $a$ are used for each species $k$
• We insert these expressions for $h_k(T)$ into the equation for $H_P(T_{ad})$ and solve for $T_{ad}$.

Set some data and make a thermo class for convenience¶

Calculate T$_{ad}$¶

Do the same thing using Cantera instead¶

Compute T$_{ad}$ over all $\xi$ using Cantera¶

• Species mass fractions are peicewise linear.

  • That is, the lean compositions are linear between $\xi=0$ and stoichiometric, and the rich compositions are linear between $\xi=\xi_{st}$ and $\xi=1$.
  • So, find the mass fractions at $\xi=0$, $\xi=\xi_{st}$, and $\xi=1$.

• Lean:

  • $(y-y_0)/(\xi-0) = (y_{st} - y_0)/(\xi_{st}-0)$
  • $\rightarrow y = y_0 + (y_{st} - y_0)\xi/\xi_{st}.$

• Rich:

  • $(y-y_{st})/(\xi-\xi_{st}) = (y_1 - y_{st})/(1-\xi_{st})$
  • $\rightarrow y = y_{st} + (y_1 - y_{st})(\xi-\xi_{st})/(1-\xi_{st})$

• Enthalpy is a conserved scalar, so

  • $h = h_0\cdot(1-\xi) + h_1\cdot(\xi)$