Determining the Air–Fuel Ratio for Complete Combustion of Octane
Determine the air–fuel ratio on both a molar and mass basis for the complete combustion of octane, C _{8} H _{18}, with (a) the theoretical amount of air, (b) 150% theoretical air (50% excess air).
Known Octane, C _{8} H _{18}, is burned completely with (a) the theoretical amount of air, (b) 150% theoretical air.
Find Determine the air–fuel ratio on a molar and a mass basis.
Engineering Model
1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen.
2. The nitrogen is inert.
3. Combustion is complete.
Analysis
a. For complete combustion of C _{8} H _{18} with the theoretical amount of air, the products contain carbon dioxide, water, and nitrogen only. That is,
C _{8} H _{18}+a\left( O _{2}+3.76 N _{2}\right) \rightarrow b CO _{2}+c H _{2} O +d N _{2}
Applying the conservation of mass principle to the carbon, hydrogen, oxygen, and nitrogen, respectively, gives
C: b = 8
H: 2c = 18
O: 2b + c = 2a
N: d = 3.76a
Solving these equations, a = 12.5, b = 8, c = 9, d = 47. The balanced chemical equation is
C _{8} H _{18}+12.5\left( O _{2}+3.76 N _{2}\right) \rightarrow 8 CO _{2}+9 H _{2} O +47 N _{2}
The air–fuel ratio on a molar basis is
\overline{A F}=\frac{12.5+12.5(3.76)}{1}=\frac{12.5(4.76)}{1}=59.5 \frac{ kmol (\text { air })}{ kmol (\text { fuel })}
The air–fuel ratio expressed on a mass basis is
A F=\left[59.5 \frac{ kmol ( air )}{ kmol (\text { fuel })}\right]\left[\frac{28.97 \frac{ kg ( air )}{ kmol ( air )}}{114.22 \frac{ kg ( fuel )}{ kmol ( fuel )}}\right]=15.1 \frac{ kg ( air )}{ kg (\text { fuel })}
b. For 150% theoretical air, the chemical equation for complete combustion takes the form
1 C _{8} H _{18}+1.5(12.5)\left( O _{2}+3.76 N _{2}\right) \rightarrow b CO _{2}+c H _{2} O +d N _{2}+e O _{2}
Applying conservation of mass
C: b = 8
H: 2c = 18
O: 2 b+c+2 e=(1.5)(12.5)(2)
N: d=(1.5)(12.5)(3.76)
Solving this set of equations, b = 8, c = 9, d = 70.5, e = 6.25, giving a balanced chemical equation
C _{8} H _{18}+18.75\left( O _{2}+3.76 N _{2}\right) \rightarrow 8 CO _{2}+9 H _{2} O +70.5 N _{2}+6.25 O _{2}
The air–fuel ratio on a molar basis is
\overline{A F}=\frac{18.75(4.76)}{1}=89.25 \frac{ kmol (\text { air })}{ kmol (\text { fuel })}
On a mass basis, the air–fuel ratio is 22.6 kg (air)/kg (fuel), as can be verified.
1 When complete combustion occurs with excess air, oxygen appears in the products, in addition to carbon dioxide, water, and nitrogen.
Skills Developed
Ability to…
• balance a chemical reaction equation for complete combustion with theoretical air and with excess air.
• apply definitions of air–fuel ratio on mass and molar bases.
Quick Quiz
For the condition in part (b), determine the equivalence ratio. Ans. 0.67.