Question

Given the measured oxygen concentration in the exhaust gas of a power plant is 6.78%, and assuming natural gas only consists of methane (CH₄) calculate the excess air ratio the plant is operating at.

Answer 1

the combustion of methane is represented as CH4 (g)+ 2O2(g)--->CO2(g)+2H2O (g)

basis : 1 mole of CH4, moles of O2 required= 2 moles, since air contains 21% O2 and 79% N2, moles of air to be supplied = 2/0.21= 9.5, let x= % excess air, air supplied= 9.5+x/100 =9.5+0.01x

so the prodcuts of combustion : CO2 =1 mole, H2P=2 moles, N2= (9.5+0.01x)*0.79 and O2= 0.01x*0.21= 0.0021x

total moles of products= 1+2+7.5 +0.0079x+0.0021x = 10.5+0.01x

percentage of O2 in the exhaust =100* 0.0021x/(10.5+0.01x)= 6.78

0.0021x/(10.5+0.01x)= 6.78/100 = 0.0678

0.0021x= 0.0678((10.5+0.01x)

0.0021x= 0.7119 +0.000678

x= 500.63

excess air ratio = 500.63/1 moles of air/moles of methane