If you burn methane with 50% excess air, what would be the flue gas composition?
Stoichiometric Equation for combustion of alkanes in air
here
Here the case is of methane
a=1
b=4/2=2
z=1+2/2=2
c=3.76*2=7.52
If there is no excess air
then combustion equation becomes
If there is excess 50% air
then air amount should multiplied with 1.5 (since total air 150%)
ie
air becomes
then the equation will be
There fore fuel gas composition will be as shown above
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