How much air is needed to burn 1m3 of gas containing 90% methane, 5% ethane, 3% propane, 2% nitrogen?
Given:
V (gas) = 1m3
+ O2
ϕ (CH4) = 90% or 0.9
ϕ (C2H6) = 5% or 0.05
ϕ (C3H8) = 3% or 0.03
ϕ (N2) = 2%
V (air) -?
Decision:
1) Calculate the amount of oxygen consumed for the combustion of methane:
V (CH4) = 1m3 * 0.9 = 0.9m3
0.9 m3 x m3
CH4 + 2 O2 = CO2 + 2 H2O
1 volume 2 volumes
X = 0.9 * 2/1 = 1.8 m3 (O2)
2) Calculate the volume of oxygen consumed for ethane combustion:
V (C2H6) = 1m3 * 0.05 = 0.05 m3
0.05 m3 x m3
2 С2Н6 + 7 О2 = 4 СО2 + 6 Н2О
2 vol. 7 vol.
X = 0.05 * 7/2 = 0.175 m3 (O2)
3) Calculate the volume of oxygen consumed for propane combustion:
V (C3H8) = 1 m3 * 0.03 = 0.3 m3
0.03 m3 x m3
С3Н8 + 5 О2 = 3 СО2 + 4 Н2О
1 vol. 5 vol.
X = 0.03 * 5/1 = 0.15 m3 (O2)
Nitrogen does not burn.
4) Find the volume of oxygen:
V (O2) = 1.8 m3 + 0.175 m3 + 0.15 m3 = 2.125 m3
Oxygen in the air contains 21% or 0.21
V (air) = 2.125 m3 / 0.21 = 10.12 m3
Answer: 10.12 m3 (air).