Question

Consider a jet diffusion flame in which the fuel is methanol vapor (CH₃OH) and the oxidizer is air. The species existing within the flame are CH₃OH, CO, CO₂, O₂, H₂, H₂O, N₂, and OH.

A. Determine the numerical value of the stoichiometric mixture fraction.

B. Write out an expression for the mixture fraction, f, in terms of the flame-species mass fractions, Y_i. Assume all pairs of binary diffusion coefficients are equal, i.e., no differential diffusion.

Answer 1

If we look up the atomic weights of the atoms that make up octane and oxygen, we get the following numbers:

Carbon (C) = 12.01

Oxygen (O) = 16

Hydrogen (H) = 1.008

• So 1 molecule of methane has a molecular weight of: 1 * 12.01 + 4 * 1.008 = 16.042

• One oxygen molecule weighs: 2 * 16 = 32

• The oxygen-fuel mass ratio is then: 2 * 32 / 1 * 16.042 = 64 / 16.042

• So we need 3.99 kg of oxygen for every 1 kg of fuel

• Since 23.2 mass-percent of air is actually oxygen, we need : 3.99 * 100/23.2 = 17.2 kg air for every 1 kg of methane.

So the stoichiometric air-fuel ratio of methanol is 17.2.