Methane burns with stoichiometric
quantity of air. The AFR ratio by weight is
(a)13. 41, (b) 16.79, (c)
12.51, (d) 17.39
Solution: The reaction
involved in the combustion of methane is
CH4 + 2O2
à CO2 +2H2O
Molar mass of methane
is 16g
Molar mass of 2O2 is
64g
It means, for combustion
16g methane needs 64g O2 OR
1g of methane needs
64/16=4g of O2
Therefore 1kg of
methane needs 4kg of O2
Now as we know 23kg of
O2 by mass is present in 100kg of air.
Therefore mass of air
needed will be (100/23)*4=17.39kg of air
So AFR= Mass of
air/Mass of fuel =17.39/1=17.39
So the option (d) is
correct
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