Stoichiometric Combustion with Air

Stoichiometry with Pure \ce{O2}

For a generic fuel $\mathrm{C}{}_a\mathrm{H}{}_b\mathrm{O}{}_d$,

$$\mathrm{C}{}_a\mathrm{H}{}_b\mathrm{O}{}_d+{\nu }_O\mathrm{O}{}_2\to a\mathrm{CO}{}_2+\frac{b}{2}\mathrm{H}{}_2\mathrm{O},$$

so the stoichiometric $\mathrm{O}{}_2$ moles per mole of fuel are

$$\boxed{{\nu }_O=a+\frac{b}{4}-\frac{d}{2}}.$$

Mass-based stoichiometric ratio (per mol fuel):

$${\left(\frac{O}{F}\right)}_{\mathrm{sto}}=\frac{{\nu }_O{W}_{\mathrm{O}{}_2}}{1\cdot W_{\text{fuel}}}.$$

Replace \ce{O2} with air

Assume dry air $\approx$ $21\%\mathrm{O}{}_2$, $79\%\mathrm{N}{}_2$ by moles (i.e., $\mathrm{air}=\mathrm{O}{}_2+3.76\mathrm{N}{}_2$).

Then the stoichiometric reaction becomes

$$\mathrm{C}{}_a\mathrm{H}{}_b\mathrm{O}{}_d+{\nu }_O(\mathrm{O}{}_2+3.76\mathrm{N}{}_2)\to a\mathrm{CO}{}_2+\frac{b}{2}\mathrm{H}{}_2\mathrm{O}+3.76{\nu }_O\mathrm{N}{}_2.$$

Example (methane):

$$\mathrm{CH}{}_4+2(\mathrm{O}{}_2+3.76\mathrm{N}{}_2)\to \mathrm{CO}{}_2+2\mathrm{H}{}_2\mathrm{O}+7.52\mathrm{N}{}_2.$$

Percent Stoichiometric & Percent Excess Oxidizer

Let $r\equiv \frac{(O/F{)}_{\text{actual}}}{(O/F{)}_{\mathrm{stoich}}}=\frac{1}{\varphi }$.

• % stoichiometric oxidizer (how much of the stoich $\mathrm{O}{}_2$ you actually have):

$$\boxed{\%\text{stoich O}=100r=\frac{100}{\varphi }}.$$

• % excess oxidizer (beyond stoichiometric):

$$\boxed{\%\text{excess O}=100(r-1)=100\left(\frac{1}{\varphi }-1\right)}.$$

(Conversely, ”% excess fuel” $=100(\varphi -1)$, when using a fuel-based metric.)