A natural gas has the following molar analysis CH4 8162 C2H6

A natural gas has the following molar analysis, CH_4: 81.62%; C_2H_6: 4.41%; C_3H_8: 1.85%; C_4H_10: 1.62%; N_2: 10.50%. The gas is burned with dry air, giving products having a molar analysis on a dry basis: CO_2: 8.8%; CO: 0.2%; O_2: 6%; N_2: 85 %. Determine the air fuel ratio on a molar basis.

Solution

For this answer following assumptions are made:

1. .Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert.

2. . The fuel mixture can be modeled as an ideal gas.

The solution can be conducted on the basis of an assumed amount of fuel mixture or on the basis of an assumed amount of dry products. Let us illustrate the first procedure, basing the solution on 1 kmol of fuel mixture. The chemical equation then takes the form:

(0.8162 CH₄ + 0.0441 C₂H₆ + 0.0185 C₃H₈ + 0.0162 C₄H₁₀ + 0.105 N₂) + a(O₂ + 3.76 N₂) = b(0.088 C0₂ + 0.002 CO +0.06 O₂ + 0.85 N₂) + cH₂O

The products consist of b kmol of dry products and c kmol of water vapor, each per kmol of fuel mixture. Applying conservation of mass to carbon

b(0.088+0.002) = 0.8162 + 0.0441 (2) + 0.0185 (3) + 0.0162 (4)

Therefore b= 11.39

Conservation of mass for hydrogen results in

2c = 0.8162(4) + 0.0441 (6) + 0.0185 (8) + 0.0162 (10)

c = 1.92

The unknown coefficient a can be found from either an oxygen balance or a nitrogen balance. Applying conservation of mass to oxygen

2a = 11.39 [0.088 (2) + 0.002 + 0.06 (2) ] + 1.92

a = 2.66

Therefore balanced eqn:

(0.8162 CH₄ + 0.0441 C₂H₆ + 0.0185 C₃H₈ + 0.0162 C₄H₁₀ + 0.105 N₂) + 2.66(O₂ + 3.76 N₂) = 11.39(0.088 C0₂ + 0.002 CO +0.06 O₂ + 0.85 N₂) + 1.92H₂O

The air–fuel ratio on a molar basis is = 2.66 (4.76) / 1 = 12.66