Question

Two kmol of ethane (C₂H₆) is burned with an unknown amount of air during a combustion process. An analysis of the combustion products reveals that the combustion is complete, and there are 4 kmol of free O₂ in the products. Determine (a) the air–fuel ratio and (b) the percentage of theoretical air used during this process (c) The equivalent ratio

Answer 1

Answer : for ethane

C2H6 + xO2 = 2CO2 + 3H2O + 4O2

By balancing oxygen, we get x = 7.5

1 mol of Air consists of 0.21 moles of oxygen and 0.79 moles of nitrogen.

Nitrogen quantity = 7.5 * 0.79 / 0.21 = 28.214 kmol

Total air quantity = 7.5 + 28.214 = 35.714 kmol

1 mole of oxygen = 32 grams of oxygen

1 kmol of air = 29 kg air

1 kmol of C2H6 = 30 kg C2H6

1.Air-fuel ratio = (35.714 * 29) / 30 = 34.52

2.For theoretical air

C2H6 + 3.5O2 = 2CO2 + 3H2O

Nitrogen quantity = 3.5 * 79 / 21 = 13.17 kmol

Air quantity = 3.5 + 13.17 = 16.67 kmol

% Theoretical air = (35.714 / 16.67) * 100 = 214.3%