Mixture Ratios

Mixture Ratios (mass vs. mole)

Let oxidizer (ox) and fuel (f) be delivered separately.

• Mass-based oxidizer/fuel ratio:

$$\left(\frac{O}{F}\right)\equiv \frac{m_{\mathrm{ox}}}{m_{\mathrm{fuel}}}.$$

• Mole-based oxidizer/fuel ratio:

$$\left(\frac{N_{\mathrm{ox}}}{N_{\mathrm{fuel}}}\right)=\frac{\text{moles ox}}{\text{moles fuel}}.$$

They are not numerically equal unless $W_{\mathrm{ox}}=W_{\mathrm{fuel}}$; in general

$$\left(\frac{O}{F}\right)=\left(\frac{N_{\mathrm{ox}}}{N_{\mathrm{fuel}}}\right)\frac{W_{\mathrm{ox}}}{W_{\mathrm{fuel}}},\left(\frac{N_{\mathrm{ox}}}{N_{\mathrm{fuel}}}\right)=\left(\frac{O}{F}\right)\frac{W_{\mathrm{fuel}}}{W_{\mathrm{ox}}}.$$

Equivalence Ratio $\varphi$

Define equivalence ratio (fuel-based):

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

• Lean mixture: $\varphi <1$ (fuel-poor / oxidizer-rich)
• Rich mixture: $\varphi >1$ (fuel-rich)
• Stoichiometric: $\varphi =1$

Why use $\varphi$? It’s basis-independent (same numerical value whether ratios are mass-based or mole-based) and avoids confusion across specs.