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.

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).
  $$\frac{\mbox{kmol}_k}{\mbox{kmol}} \cdot \frac{\mbox{kg}_k}{\mbox{kmol}_k} \cdot \frac{\mbox{kmol}}{\mbox{kg}} = \frac{\mbox{kg}_k}{\mbox{kg}}.$$ $$ x_k \phantom{x}\cdot\phantom{xx} M_k \phantom{xx}\cdot\phantom{xx} \frac{1}{M} = \phantom{xx}y_k.$$

Stoichiometry¶

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

Air-to-fuel ratio¶

$$ \frac{A}{F} = \frac{m_{air}}{m_{fuel}}.$$

• Mass Basis
• $0 \le A/F \le \infty$

Equivalence ratio¶

$$\Phi = \frac{F/A}{(F/A)_{stoic}}$$

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

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 $\xi$ and $\phi$¶

$$\xi = \frac{F}{F+A} = \frac{F/A}{F/A + 1} = \frac{\Phi(F/A)_{st}}{\Phi(F/A)_{st}+1} \rightarrow \Phi = \frac{\xi}{1-\xi}\frac{1}{(F/A)_{st}}$$

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 1.

Products of complete combustion (PCC)¶

• Given a set of stream definitions (species compositions $y_{k,0}$ for $\xi=0$ and $y_{k,1}$ for $\xi=1$), we can compute the stoichiometric mixture fraction $\xi_{st}$.
• At the stoichiometric point, the only species are $N_2$, $H_2O$, and $CO_2$.
  • We presume that we have just enough $O_2$ (including considerations of fuel oxygen) present to convert all the $H$ atoms to $H_2O$, and all the $C$ atoms to $CO_2$.
  • Any fuel nitrogen is assumed to form $N_2$ and takes no oxygen, but this can be changed if want.
• Then the stoichiometric product composition is set directly from the stoichiometric elemental composition.
• Elements are conserved, so stoichiometric elemental mass fractions are given by

\begin{align} y_C &= (1-\xi_{st})y_{C,0} + (\xi_{st})y_{C,1} \\ y_H &= (1-\xi_{st})y_{H,0} + (\xi_{st})y_{H,1} \\ y_O &= (1-\xi_{st})y_{O,0} + (\xi_{st})y_{O,1} \\ y_N &= (1-\xi_{st})y_{N,0} + (\xi_{st})y_{N,1} \end{align}

• If we take one kg of mass, then the moles of elements are given by:

\begin{align} n_C &= \frac{y_C}{M_C}\\ n_H &= \frac{y_H}{M_H}\\ n_O &= \frac{y_O}{M_O}\\ n_N &= \frac{y_N}{M_N} \end{align}

• Then, the stoichiometric moles of product species are

\begin{align} n_{CO2} &= n_C, \\ n_{H2O} &= n_H/2, \\ n_{N2} &= n_N/2, \end{align}

and from this we can compute the stoichiometric mole fractions and corresponding mass fractions.

• It turns out that, for PCC, all species mass fractions are linear in $\xi$ in the lean region between $0\le\xi\le\xi_{st}$ and linear in $\xi$ the rich region between $\xi_{st}\le\xi\le 1$.
  • The lines are determined in the two regions using values at two points: $y_{k,0}$ and $y_{k,st}$ in the lean region; and $y_{k,st}$, and $y_{k,1}$ in the rich region. That is,

\begin{align} 0\le\xi\le\xi_{st}:\phantom{xxxxx} & y_k = y_{k,0} + (y_{k,st} - y_{k,0})\frac{\xi}{\xi_{st}} \\ \xi_{st}\le\xi\le 1:\phantom{xxxxx} & y_k = y_{k,st} + (y_{k,1}-y_{k,st})\frac{\xi-\xi_{st}}{1-\xi_{st}} \end{align}

Show the linearity¶

Show that $y_k(\xi)$ is linear in $\xi$ in the lean and rich regions.

• Consider streams composed as:
  • $\xi=0$: $C_xH_yO_zN_w$.
  • $\xi=1$: $C_\alpha H_\beta O_\gamma N_\delta$.
• Assume a basis of one mass unit. Then $y_k=m_k$.
• Nitrogen doesn't interact with any other species so $2N\rightarrow N_2$, and it's mass fraction is obviously linear in $\xi$ given the above $\psi$ relations.

Lean

• The only possible species are $CO_2$, $H_2O$, $O_2$, and $N_2$.
• The moles of element $k$ are given by

$$n_k = \frac{y_{k,1}}{M_k}(\xi) + \frac{y_{k,0}}{M_k}(1-\xi),$$

which are linear in $\xi$.

• All $C$ forms $CO_2$. All $H$ forms $H_2O$.
  • $n_{CO2}=n_C$; $y_{CO2}=m_{CO2}=n_{C}M_{CO2}$, hence $y_{CO2}$ is linear in $\xi$.
  • $n_{H2O}=n_H/2$; $y_{H2O}=m_{H2O}=n_{H}M_{H2O}/2$, hence $y_{H2O}$ is linear in $\xi$.
  • $n_{O2} = n_O/2 - n_{CO2} - n_{H2O}/2$; $y_{O2}=m_{O2}=n_{O2}M_{O2}$, hence $y_{O2}$ is linear in $\xi$ since its component parts are.

Rich

• The only possible species are $CO_2$, $H_2O$, $N_2$, and $C_\alpha H_\beta O_\gamma N_\delta$.
• The moles of element $k$ are again given by

$$n_k = \frac{y_{k,1}}{M_k}(\xi) + \frac{y_{k,0}}{M_k}(1-\xi),$$

which are linear in $\xi$.

• That is $n_O$, $n_C$, and $n_H$ are each linear in $\xi$. A balance on each gives:

\begin{align} n_C &= n_{CO2} + \alpha n_{C_\alpha H_\beta O_\gamma N_\delta}, \\ n_H &= 2n_{H2O} + \beta n_{C_\alpha H_\beta O_\gamma N_\delta}, \\ n_O &= 2n_{CO2} + n_{H2O} + \gamma n_{C_\alpha H_\beta O_\gamma N_\delta} \end{align}

• This is a linear system, so solving for $n_{CO2}$, $n_{H2O}$ and $n_{C_\alpha H_\beta O_\gamma N_\delta}$ will give each as linear combinations of $n_C$, $n_H$, and $n_O$, which are in turn linear in $\xi$.
• We can then multiply these by the respective species molecular weights to get masses, hence mass fractions.

Example¶

• See the Python streams.py class that includes a number of convenience functions.
  • NOTE: save the file as is, don't copy and paste from it.