Answer:
57.33%, 38.3%
Explanation:
The stack gas leaves the furnace with
5.45 mole% CO2, 2.14% CO, 6.44% O2, 12.91% H2O
Basis: 100 moles of stack gas
The molecular formula of methane is CH4 and ethane is C2H6,
Assume x = number of moles of methane
y = number of moles of ethane
Hence we have the carbon balance
x + 2y = 5.45 + 2.14 = 7.59
And we have the hydrogen balance
4x + 6y = 2(12.91) = 25.82
Simultaneously solving the two equations, we obtain
x = 3.05 and y = 2.27
Hence, the molar percentage of methane in the fuel = 3.05 / (3.05+2.27) * 100% = 57.33%
Similarly, molar percentage of ethane is 42.67%
The oxygen entering = (100 - 5.45 - 2.14 - 6.44 - 12.91)*(21/79) = 19.42
And oxygen required = 2(3.05) + 3.5(2.27) = 14.045
Hence, excess air feed = 100 * (19.42 - 14.045) / 14.045 = 38.3%