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I am trying to calculate mass fractions of a Hydrogen Oxygen gas mixture as a function of equivalence ratio before ignition. My work is the following:

Equivalence Ratio: $$\phi=\frac{FA}{FA_s}=\frac{\frac{m_f}{m_{air}}}{\frac{m_{fs}}{m_{air,s}}}=\frac{\frac{n_f}{n_{air}}}{\frac{n_{fs}}{n_{air,s}}}$$

Where $FA$ is the fuel to air ratio and a subscript $s$ stands for the stoichiometric reaction.

The balanced combustion reaction equation (assuming a perfectly stoichiometric mixture) is:

$$2H_2+1O_2\rightarrow2H_2O$$

Thus, $n_{fs} = 2$ and $n_{air,s}=1$. Now, when I write the mass fractions as a function of equivalence ratio, my work is:

$$y_{O_2}=\frac{m_{air}}{m_f+m_{air}}=\frac{1}{FA+1}=\frac{1}{\phi\frac{n_{fs}}{n_{air,s}}+1}=\frac{n_{air,s}}{\phi n_{fs}+n_{air,s}}=\frac{1}{2\phi+1}$$

By a similar calculation:

$$y_{H_2}=\frac{n_{fs}\phi}{n_{fs}\phi+n_{air,s}}=\frac{2\phi}{2\phi+1}$$

A quick check that should be true reveals that $y_{H_2}+y_{O_2}=1$

The reason I am asking about this calculation is that I did this calculation for a previous project and the results the class said was correct was the reverse (i.e. the expression with $\phi$ in the numerator was for $y_{O_2}$ not $y_{H_2}$. Is there something I did wrong/did I calculate them the wrong way round? The code I am basing my previous work off of (written in MATLAB) is:

        yO =  32/(4*phi+32);   
        yH =  4*phi/(4*phi+32);

The 32 crops up because the molar mass of $O_2$ was in the expression for some reason and I have not been able to figure out exactly how the above code expressions were derived. I just know they gave results that were marked correct for the class I did this for.

EDIT: I have figured out the source of my confusion. The statement

$$\phi=\frac{FA}{FA_s}=\frac{\frac{m_f}{m_{air}}}{\frac{m_{fs}}{m_{air,s}}}=\frac{\frac{n_f}{n_{air}}}{\frac{n_{fs}}{n_{air,s}}}$$

is true by itself, but this does not mean that $FA_s=\frac{n_{fs}}{n_{air,s}}$! I made this assumption in my head after I canceled out the molar masses used to transform the masses into moles.

The correct expressions are: $$y_{O_2} = \frac{n_{o,s}M_O}{\phi n_{f,s}{M_f}+n_{o,s}M_O}$$

and

$$y_{H_2} = \frac{\phi n_{f,s}M_f}{\phi n_{f,s}{M_f}+n_{O,s}M_O}$$

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  • $\begingroup$ I am puzzled why $n_{air,s} = 1$. Air is not pure oxygen. Do you mean "mixture" instead of "air" for the gas feed? $\endgroup$ – Jeffrey J Weimer Jun 24 at 19:35
  • $\begingroup$ That’s a notational issue on my part. Since I’m assuming I have a combustion chamber that only contains hydrogen and oxygen, the moles of oxygen = the moles of air for the paradigm of calculating $\phi$. In this scenario, we are calculating the equivalence ratio with every reference to “air” representing “pure oxygen”. $\endgroup$ – Unique Worldline Jun 24 at 19:40
  • $\begingroup$ To answer your question more directly, any reference to "air" refers to "oxygen". Air is indeed not pure oxygen, but in this combustion reaction, there is no nitrogen or argon. $\endgroup$ – Unique Worldline Jun 24 at 19:45
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The derivation has a mistake in setting $FA = \phi n_{fs}/n_{O2s}$. Here is the correction. For simplicity, I use $F$ rather than $FA$, $m_O$ and $n_O$ as the mass and moles, respectively, for O$_2$, and a $\star$ for stoichiometric. I use the term $\omega$ for mass fraction and $M$ (capital M) for molar mass.

$$ \phi \equiv \frac{F}{F_s} = \frac{m_f/m_O}{m^\star_f/m^\star_O} $$

$$ \omega_O \equiv \frac{m_O}{m_f + m_O} = \frac{1}{F + 1} $$

$$ F = \phi\ F_s = \phi\ \frac{m^\star_f}{m^\star_O} = \phi\ \frac{n^\star_f\ M_f}{n^\star_O\ M_O}$$

For the given reaction, $n^\star_f = 2$ and $n^\star_O = 1$. Therefore

$$ \omega_O = \frac{1}{2\phi M_{f/O} + 1} $$

where $M_{f/O}$ is the ratio of molar mass of fuel (H$_2$) to oxygen O$_2$.

The expression for $\omega_f$ is found from a mass balance $\omega_O + \omega_f = 1$.

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  • $\begingroup$ I agree with your work except for one small point. isn't the ratio of $n_f^{\star}/n_O^{\star} = 2$ not 1/2? $\endgroup$ – Unique Worldline Jun 26 at 6:07
  • $\begingroup$ @UniqueWorldline Yes, fixed it, thanks. $\endgroup$ – Jeffrey J Weimer Jun 26 at 12:03

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