Known A natural gas with a specified molar analysis burns with dry air, giving products having a known molar analysis on a dry basis.
Find Determine the air–fuel ratio on a molar basis, the amount of products in kmol that would be formed from 100 m ^{3} of natural gas at 300 K and 1 bar, and the percent of theoretical air.
Engineering Model
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.
Analysis
a. 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
\begin{aligned}&\left(0.8062 CH _{4}+0.0541 C _{2} H _{6}+0.0187 C _{3} H _{8}+0.0160 C _{4} H _{10}\right. \\&\left.\quad+0.1050 N _{2}\right)+a\left( O _{2}+3.76 N _{2}\right) \rightarrow b\left(0.078 CO _{2}\right. \\&\left.\quad+0.002 CO +0.07 O _{2}+0.85 N _{2}\right)+c H _{2} O\end{aligned}
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(0.078+0.002)=0.8062+2(0.0541)+3(0.0187)+4(0.0160)
Solving gives b = 12.931. Conservation of mass for hydrogen results in
2c=4(0.8062)+6(0.0541)+8(0.0187)+10(0.0160)
which gives c = 1.93. The unknown coefficient a can be found from either an oxygen balance or a nitrogen balance. Applying conservation of mass to oxygen
12.931[2(0.078)+0.002+2(0.07)]+1.93=2 a
1 giving a = 2.892.
The balanced chemical equation is then
\begin{aligned}&\left(0.8062 CH _{4}+0.0541 C _{2} H _{6}+0.0187 C _{3} H _{8}+0.0160 C _{4} H _{10}\right. \\&\left.\quad+0.1050 N _{2}\right)+2.892\left( O _{2}+3.76 N _{2}\right) \rightarrow 12.931\left(0.078 CO _{2}\right. \\&\left.\quad+0.002 CO +0.07 O _{2}+0.85 N _{2}\right)+1.93 H _{2} O\end{aligned}
The air–fuel ratio on a molar basis is
\overline{A F}=\frac{(2.892)(4.76)}{1}=13.77 \frac{ kmol (\text { air })}{ kmol (\text { fuel })}
b. By inspection of the chemical reaction equation, the total amount of products is b + c = 12.931 + 1.93 = 14.861 kmol of products per kmol of fuel. The amount of fuel in kmol, n_{ F }, present in 100 m ^{3} of fuel mixture at 300 K and 1 bar can be determined from the ideal gas equation of state as
n_{ F }=\frac{p V}{\bar{R} T}
=\frac{\left(10^{5} N / m ^{2}\right)\left(100 m ^{3}\right)}{(8314 N \cdot m / kmol \cdot K )(300 K )}=4.01 kmol \text { (fuel) }
Accordingly, the amount of product mixture that would be formed from 100 m ^{3} of fuel mixture is (14.861)(4.01) = 59.59 kmol of product gas.
c. The balanced chemical equation for the complete combustion of the fuel mixture with the theoretical amount of air is
\begin{gathered}\left(0.8062 CH _{4}+0.0541 C _{2} H _{6}+0.0187 C _{3} H _{8}+0.0160 C _{4} H _{10}+0.1050 N _{2}\right) \\+2\left( O _{2}+3.76 N _{2}\right) \rightarrow 1.0345 CO _{2}+1.93 H _{2} O +7.625 N _{2}\end{gathered}
The theoretical air–fuel ratio on a molar basis is
(\overline{A F})_{\text {theo }}=\frac{2(4.76)}{1}=9.52 \frac{\text { kmol (air) }}{\text { kmol (fuel) }}
The percent theoretical air is then
\% \text { theoretical air }=\frac{13.77 kmol (\text { air }) / kmol \text { (fuel) }}{9.52 kmol \text { (air)/kmol (fuel) }}=1.45(145 \%)
1 A check on both the accuracy of the given molar analyses and the calculations conducted to determine the unknown coefficients is obtained by applying conservation of mass to nitrogen. The amount of nitrogen in the reactants is
0.105+(3.76)(2.892)=10.98 kmol / kmol \text { of fuel }
The amount of nitrogen in the products is (0.85)(12.931) = 10.99 kmol/kmol of fuel. The difference can be attributed to round-off.
Skills Developed
Ability to…
• balance a chemical reaction equation for incomplete combustion of a fuel mixture given the analysis of dry products of combustion.
• apply the definition of air–fuel ratio on a molar basis as well as percent theoretical air.
Quick Quiz
Determine the mole fractions of the products of combustion. Ans. y_{ co _{2}}=0.0679, y_{ co }=0.0017, y_{0_{2}}=0.0609, y_{ N _{2}}=0.7396, y_{ H _{2} O }=0.1299.