Combustion Stoichiometry

• Combustion occurs between Fuel and Air (or oxidizer)
• Need to get the mixing ratio / stoichiometry right.

Air properties

• Use $x$ for mole fraction, and $y$ for mass fraction.

• Lump Ar with N$_2$:

• $n_{N2}$/$n_{O2}$ = 3.76
• $n_{air}$/$n_{O2}$ = 4.76

Mean molecular weight

• M = 28.96 kg/kmol
• M = 29 kg/kmol

• Make sure these last two equations are intuitive
  • $M$ (=) kg/kmol.
  • $M = m/n.$
  • $M = \sum_k x_k M_k,$
    • Take a 1 kmol basis. Then $n=1$ and $n_k=x_k$.
    • $m = \sum_km_k = \sum_kn_kM_k = \sum_kx_kM_k.$
    • So, $M = m/n = \sum_kx_kM_k.$
  • $M = 1/(\sum_k y_k/M_k)$.
    • Take a 1 kg basis. Then $m=1$ and $m_k=y_k$.
    • $n = \sum_kn_k = \sum_km_k/M_k = \sum_ky_k/M_k.$
    • So, $M=m/n = 1/(\sum_k y_k/M_k)$.

Convert x to y

• Intuitively
  • $x_k$ to $y_k$: (kmol$_k$/kmol) to (kg$_k$/kg). $$\frac{\text{kmol}_k}{\text{kmol}} \cdot \frac{\text{kg}_k}{\text{kmol}_k} \cdot \frac{\text{kmol}}{\text{kg}} = \frac{\text{kg}_k}{\text{kg}}.$$ $$ x_k \phantom{x}\cdot\phantom{xx} M_k \phantom{xx}\cdot\phantom{xx} \frac{1}{M} = \phantom{xx}y_k.$$

Stoichiometry

• Rich (too much fuel)
• Lean (too much air)
• Stoichiometric (just right)

Air-to-fuel ratio $A/F$

• Mass Basis
• $0 \le A/F \le \infty$

Equivalence ratio $\Phi$

• Mass or mole basis (same).
• $0 \le \Phi \le \infty$
• $\Phi=0$ is lean, $\Phi>1$ is rich.
• $\Phi$ is common in practical applications.

Excess air $E$

The fraction excess air $E$ is giveen by

Mixture fraction

• Notation, $\xi$, or $f$, or $Z$ is common.
• $\xi$ is the mass fraction of local material originating in the fuel ($\xi=1$) stream.
• $0\le \xi\le 1$
• $\xi=0$ is pure air, $\xi=1$ is pure fuel.
• $\xi$ is common in modeling.
• $\xi$ can be defined in terms of any two distinct streams, not just fuel and air.

Convert between mixture fraction and equivalence ratio

But,

Mixing

• $\xi$ defines a linear state of mixing between two streams.
  • This is true for conserved quantities (termed conserved scalars), like elemental masses, or for species in the absence of reaction.
• Conserved quantities are elemental masses, enthalpy (in the absence of heat losses), and linear combinations of these.
• Consider elemental carbon. Denote FS as “fuel stuff” (meaning mass that originated in the $\xi=1$ stream), and AS as “air stuff” (meaning mass that originated in the $\xi=0$ stream).

or

• In general, simple mixing (no reaction) of some scalar $\psi$ is given by $$\psi = (1-\xi)\psi_{\xi=0} + (\xi)\psi_{\xi=1}.$$
• We can rearrange the above expression to give: $$\xi = \frac{\psi-\psi_{\xi=0}}{\psi_{\xi=1}-\psi_{\xi=0}}.$$
• This shows that $\xi$ is a centered and scaled version of $\psi$ that varies between 0 and

Balanced reactions

Methane combustion

• Normally write 1 mole of fuel for convenience.
• 1 mole O$_2$ gives 4.75 moles Air = O$_2$ + 3.76 N$_2$.
• Mole basis
• Products
  • Carbon $\rightarrow$ CO$_2$
  • Hydrogen $\rightarrow$ H$_2$O

• Solve for $\alpha$, $\beta$, $\gamma$, using elemental balnces on $C$, $H$, and $O$.
  • C $\rightarrow$ $1 = \beta$ $\rightarrow$ $\beta = 1$
  • H $\rightarrow$ $4 = 2\gamma$ $\rightarrow$ $\gamma = 2$
  • O $\rightarrow$ $2\alpha = 2\beta + \gamma$ $\rightarrow$ $\alpha = 2$.

• Note, no change in moles for methane combustion.
• Note the high $m_f/m_a = M_{CH4}/(9.52M_{air}) = 16/(9.52*29) = 0.058$
• Note: A/F = 17.2.

Generalize the fuel

• (x and y are number of carbons and hydrogens here, not mole and mass fractions).
• C, H balances $\rightarrow$ $\beta = x$, $\gamma = y/2$.
• O balance $\rightarrow$ $2\alpha = 2\beta + \gamma$ $\rightarrow$ $\alpha = x + y/4$

• No changee in moles from reactants to products when $y=4$.

Lean

Rich

Mass basis

It is common to write the stoichiometry on a mass basis:

where $F$, $A$, and $P$ are fuel, air (or oxidizer) and products, with masses of $1$, $r$, and $1+r$, respectively. Here, fuel, air, and products can be composed of several species, i.e., air is a mixture of oxygen and nitrogen, and products include water, carbon dioxide, and nitrogen.

Rich and lean flames can both be written in terms of equivalence ratio:

This can be re-expressed for both rich and lean as

Complete reaction in terms of mixture fraction

The stoichiometric mixture fraction is given by

The mixture fraction is related to equivalence ratio and excess air by

The mixture fraction is the fraction of mass originating in the fuel stream, so in a given mixture, $\xi$ is the mass fraction of fuel plus the fraction of products originating in the fuel, which is $\xi = Y_F + Y_P/(1+r)$, or

Similarly, $1-\xi$ is the mass fraction originating in the air (oxidizer) stream, $1-\xi = Y_A + Y_Pr/(1-r)$, or

These are written in terms of the product mass fraction $Y_P$.

For lean reaction, $E>1$ and

Using the above relation for $E$ in terms of $\xi$ gives

For rich reaction, $\Phi>1$ and

Using the above relation for $\Phi$ in terms of $\xi$ gives

Note that $Y_P$ is linear in $\xi$ on either side of $\xi_{st}$, hence $Y_F$ and $Y_A$ are linear on either side of $\xi_{st}$. $Y_P$ can be written for either rich or lean as

Example

Plot products of complete combustion for methane-air reaction for all mixture fractions. The stoichiometric reaction is $CH_4 + 2O_2 + 7.52N_2 \rightarrow CO_2 + 2H_2O + 7.52N_2.$ This gives $r = (2M_{O2} + 7.52M_{N2})/M_{CH4} = 17.2$, and $\xi_{st}=0.055$. Care must be taken in defining the constituents of $Y_F$, $Y_A$, and $Y_P$. For instance, nitrogen is present in both products and air species, so, $Y_{N2} = Y_{N2,A}Y_A + Y_{N2,P}Y_P.$ Similar treatment would be required if the fuel included diluent species, such as $CO_2$ or $N_2$.

import numpy as np
import matplotlib.pyplot as plt

nxi = 1000
xi = np.linspace(0,1,nxi)
xist = 0.055

Yp = xi/xist * np.minimum(np.ones(nxi), xist/(xi+1E-8)*(1-xi)/(1-xist+1E-8))
Yf = xi-Yp*xist
Ya = 1-xi-Yp*(1-xist)

Mo2 = 32
Mn2 = 28
Mco2 = 44
Mh2o = 18
Yo2 = Ya*Mo2/(Mo2+3.76*Mn2)
Yn2 = Ya*3.76*Mn2/(Mo2+3.76*Mn2) + Yp*7.52*Mn2/(Mco2+2*Mh2o+7.52*Mn2)
Yco2 = Yp*Mco2/(Mco2+2*Mh2o+7.52*Mn2)
Yh2o = Yp*2*Mh2o/(Mco2+2*Mh2o+7.52*Mn2)
Ych4 = Yf

plt.plot(xi,Yo2, label='O2')
plt.plot(xi,Yn2, label='N2')
plt.plot(xi,Yco2, label='CO2')
plt.plot(xi,Yh2o, label='H2O')
plt.plot(xi,Ych4, label='CH4')
plt.legend(frameon=False)
plt.xlabel(r'$\xi$')
plt.ylabel(r'$Y_i$');