- 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.
| Species | x | y |
|---|---|---|
| N$_2$ | 0.7809 | 0.75532 |
| O$_2$ | 0.2095 | 0.23144 |
| Ar | 0.0096 | 0.01324 |
- Lump Ar with N$_2$:
| Species | x | y |
|---|---|---|
| N$_2$ | 0.79 | 0.77 |
| O$_2$ | 0.21 | 0.23 |
- n${N2}$/n${O2}$ = 3.76
- n${air}$/n${O2}$ = 4.76
Mean molecular weight
- M = 28.96 kg/kmol
- M = 29 kg/kmol
$$ M = \sum_k x_k M_k,$$ $$ M = \frac{1}{\sum_k \frac{y_k}{M_k}}.$$
- 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
$$ x_kM_k = y_kM,$$
$$ y_k = \frac{x_kM_k}{M},$$ $$ x_k = \frac{y_kM}{M_k}.$$
- 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 $A/F$
$$ \frac{A}{F} = \frac{m_{air}}{m_{fuel}}.$$
- Mass Basis
- $0 \le A/F \le \infty$
Equivalence ratio $\Phi$
$$\Phi = \frac{F/A}{(F/A)_{stoic}}$$
- 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
$$E = \frac{A-A_s}{A_s} = \frac{A/F - A_s/F}{A_s/F} = \frac{A/F}{A_s/F} - 1 = \frac{1}{\Phi}-1.$$
Mixture fraction
$$\xi = \frac{m_f}{m_f + m_a}$$
- 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
$$\xi = \frac{F}{F+A} = \frac{F/A}{F/A + 1}=\frac{\Phi(F/A)_{st}}{\Phi(F/A)_{st}+1}, \rightarrow$$
$$\Phi = \frac{\xi}{1-\xi}\frac{1}{(F/A)_{st}}$$
But,
$$\xi_{st} = \frac{(F/A)_{st}}{(F/A)_{st}+1} \rightarrow (F/A)_{st} = \frac{\xi_{st}}{1-\xi_{st}}\rightarrow$$
$$\Phi = \frac{\xi(1-\xi_{st})}{\xi_{st}(1-\xi)}$$
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).
$$y_C = \frac{m_C}{m_{tot}} = \frac{\text{(mass of C from AS)}+\text{(mass of C from FS)}}{m_{tot}} = \frac{m_{AS}}{m_{tot}}\left(\frac{m_C}{m_{AS}}\right) + \frac{m_{FS}}{m_{tot}}\left(\frac{m_C}{m_{FS}}\right), $$ or $$y_C = (1-\xi)y_{C,\xi=0} + (\xi)y_{C,\xi=1}.$$
- 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
$$(CH_4) + \alpha (air) \rightarrow b(products)$$ $$CH_4 + \alpha O_2 + 3.76\alpha N_2 \rightarrow \beta CO_2 + \gamma H_2O + 3.76\alpha N_2$$
- 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 $\rightarrow$ $2\alpha = 2\beta + \gamma$ $\rightarrow$ $\alpha = 2$.
$$CH_4 + 2O_2 + 7.52N_2 \rightarrow CO_2 + 2H_2O + 7.52N_2$$
- 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
$$C_xH_y + \alpha O_2 + 3.76\alpha N_2 \rightarrow \beta CO_2 + \gamma H_2O + 3.76\alpha N_2$$
- (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 $\rightarrow$ $2\alpha = 2\beta + \gamma$ $\rightarrow$ $\alpha = x + y/4$
$$C_xH_y + \left(x+\frac{y}{4}\right)O_2 + 3.76\left(x+\frac{y}{4}\right)N_2 \rightarrow xCO_2 + \frac{y}{2}H_2O + 3.76\left(x+\frac{y}{4}\right)N_2$$
- No changee in moles from reactants to products when $y=4$.
Lean
$$C_xH_y + (1+E)\left(x+\frac{y}{4}\right)(O_2 + 3.76 N_2) \rightarrow xCO_2 + \frac{y}{2}H_2O + E\left(x+\frac{y}{4}\right)O_2 + (1+E)3.76\left(x+\frac{y}{4}\right)N_2$$
Rich
$$\Phi C_xH_y + \left(x+\frac{y}{4}\right)O_2 + 3.76\left(x+\frac{y}{4}\right)N_2 \rightarrow xCO_2 + \frac{y}{2}H_2O + 3.76\left(x+\frac{y}{4}\right)N_2 + (\Phi-1)C_xH_y$$
Mass basis
It is common to write the stoichiometry on a mass basis:
$$F + rA \rightarrow (1+r) P,$$
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:
$$\text{Rich,}\,\Phi\ge 1:\; \Phi F + rA \rightarrow (1+r)P + (\Phi-1)F,$$ $$\text{Lean,}\,\Phi\le 1:\; \Phi F + rA \rightarrow \Phi(1+r)P + (1-\Phi)rA.$$
This can be re-expressed for both rich and lean as
$$\Phi F + rA \rightarrow \min(1,\Phi)(1+r)P + \max(0,\Phi-1)F + \max(0,1-\Phi)rA.$$
Complete reaction in terms of mixture fraction
The stoichiometric mixture fraction is given by
$$\xi_{st} = \frac{1}{1+r}.$$
The mixture fraction is related to equivalence ratio and excess air by
$$\Phi = \frac{r\xi}{(1-\xi)} = \frac{\xi}{(1-\xi)}\frac{(1-\xi_{st})}{\xi_{st}},$$ $$E = \frac{1}{\Phi} - 1 = \frac{1-\xi(1+r)}{r\xi} = \frac{\xi_{st}-\xi}{\xi(1-\xi_{st})}.$$
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
$$\color{blue}Y_F = \xi - \frac{Y_P}{1+r} = \xi - \xi_{st}Y_P.$$
Similarly, $1-\xi$ is the mass fraction originating in the air (oxidizer) stream, $1-\xi = Y_A + Y_Pr/(1-r)$, or
$$\color{blue}Y_A = (1-\xi) - Y_P\frac{r}{1+r} = (1-\xi) - Y_P(1-\xi_{st}).$$
These are written in terms of the product mass fraction $Y_P$.
For lean reaction, $E>1$ and
$$F + r(1+E)A \rightarrow (1+r)P + (rE)A.$$
Using the above relation for $E$ in terms of $\xi$ gives
$$Y_P^L = \xi(1+r) = \frac{\xi}{\xi_{st}}.$$
For rich reaction, $\Phi>1$ and
$$\Phi F + rA \rightarrow (1+r)P + (\Phi-1)F.$$
Using the above relation for $\Phi$ in terms of $\xi$ gives
$$Y_P^R = \frac{(1-\xi)(1+r)}{r} = \frac{1-\xi}{1-\xi_{st}}.$$
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
$$\color{blue}Y_P = \min\left(1,\,\frac{\xi_{st}(1-\xi)}{\xi(1-\xi_{st})}\right)\frac{\xi}{\xi_{st}}.$$
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'$Y_i$');
