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:


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:


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


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


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

  • $\begingroup$ 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 '19 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 '19 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 '19 at 19:45

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 $\omega_O + \omega_f = 1$.

| improve this answer | |
  • $\begingroup$ 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 '19 at 6:07
  • $\begingroup$ @UniqueWorldline Yes, fixed it, thanks. $\endgroup$ – Jeffrey J Weimer Jun 26 '19 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.