Known Methane is burned with dry air. The molar analysis of the products on a dry basis is provided.
Find Determine (a) the air–fuel ratio on both a molar and a mass basis, (b) the percent theoretical air, (c) the dew point temperature of the products, in °F, if cooled at 1 atm, (d) the amount of water vapor present if the products are cooled to 90°F at 1 atm.
Engineering Model
1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert.
2. The products form an ideal gas mixture and the dew point temperature of the mixture is conceptualized as in Sec. 12.5.4.
Analysis
1 a. The solution is conveniently conducted on the basis of 100 lbmol of dry products. The chemical equation then reads
\begin{aligned}&a CH _{4}+b\left( O _{2}+3.76 N _{2}\right) \rightarrow \\&\quad 9.7 CO _{2}+0.5 CO +2.95 O _{2}+86.85 N _{2}+c H _{2} O\end{aligned}
In addition to the assumed 100 lbmol of dry products, water must be included as a product.
Applying conservation of mass to carbon, hydrogen, and oxygen, respectively,
C: 9.7+0.5=a
H: 2 c=4 a
O: (9.7)(2)+0.5+2(2.95)+c=2 b
2 Solving this set of equations gives a = 10.2, b = 23.1, c = 20.4. The balanced chemical equation is
\begin{aligned}&10.2 CH _{4}+23.1\left( O _{2}+3.76 N _{2}\right) \rightarrow \\&9.7 CO _{2}+0.5 CO +2.95 O _{2}+86.85 N _{2}+20.4 H _{2} O\end{aligned}
On a molar basis, the air–fuel ratio is
\overline{A F}=\frac{23.1(4.76)}{10.2}=10.78 \frac{\text { lbmol (air) }}{\text { lbmol (fuel) }}
On a mass basis
A F=(10.78)\left(\frac{28.97}{16.04}\right)=19.47 \frac{ lb ( air )}{ lb (\text { fuel })}
b. The balanced chemical equation for the complete combustion of methane with the theoretical amount of air is
CH _{4}+2\left( O _{2}+3.76 N _{2}\right) \rightarrow CO _{2}+2 H _{2} O +7.52 N _{2}
The theoretical air–fuel ratio on a molar basis is
(\overline{A F})_{\text {theo }}=\frac{2(4.76)}{1}=9.52 \frac{\text { lbmol (air) }}{\text { lbmol (fuel) }}
The percent theoretical air is then found from
\% \text { theoretical air }=\frac{(\overline{A F})}{(\overline{A F})_{\text {theo }}}
=\frac{10.78 \text { lbmol (air)/lbmol (fuel) }}{9.52 \text { lbmol (air)/lbmol (fuel) }}
= 1.13 (113%)
c. The dew point temperature is determined using the methodology of Secs. 12.5.3 and 12.5.4 but with the combustion products playing the role of the moist air in that discussion. Accordingly, the partial pressure of the water vapor in the combustion products is the focus. The partial pressure p_{ v } is found from p_{ v }=y_{ v } p, \text { where } y_{ v } is the mole fraction of the water vapor in the combustion products and p is 1 atm.
Referring to the balanced chemical equation of part (a), the mole fraction of the water vapor is
y_{ v }=\frac{20.4}{100+20.4}=0.169
Thus, p_{ v }=0.169 atm =2.484 lbf / in .^{2} Interpolating in Table A-2E, the dew point temperature is 134°F.
d. If the products of combustion are cooled at 1 atm below the dew point temperature of 134°F to 90°F, some condensation of the water vapor present will occur, giving a gas phase including water vapor in equilibrium with a liquid water phase.
Expressing the balanced reaction equation of part (a) on a per mole of fuel basis, at 90°F the products will consist of 9.8 lbmol of “dry” products \left( CO _{2}, CO , O _{2}, N _{2}\right) plus 2 lbmol of water, each per lbmol of fuel. Of the water, n lbmol is water vapor and the rest is liquid. Considering the gas phase, the partial pressure of the water vapor is the saturation pressure at 90°F: 0.6988 lbf / in .^{2} The partial pressure also is the product of the water vapor mole fraction and the mixture pressure, 14.696 lbf/in.2 Collecting results
0.6988=\left(\frac{n}{n+9.8}\right) 14.696
Solving, n = 0.489 lbmol of water vapor per lbmol of fuel consumed.
1 The solution could be obtained on the basis of any assumed amount of dry products—for example, 1 lbmol. With some other assumed amount, the values of the coefficients of the balanced chemical equation would differ from those obtained in the solution, but the air–fuel ratio, the value for the percent of theoretical air, and the dew point temperature would be unchanged.
2 The three unknown coefficients, a, b, and c, are evaluated here by application of conservation of mass to carbon, hydrogen, and oxygen. As a check, note that the nitrogen also balances
N : \quad b(3.76)=86.85
This confirms the accuracy of both the given product analysis and the calculations conducted to determine the unknown coefficients.
Skills Developed
Ability to…
• balance a chemical reaction equation for incomplete combustion given the analysis of dry products of combustion.
• apply definitions of air–fuel ratio on mass and molar bases as well as percent theoretical air.
• determine the dew point temperature of combustion products.
Quick Quiz
When the combustion products are at 90°F, 1 atm, how much liquid water is present, in lbmol per lbmol of fuel consumed? Ans. 1.511.