FUEL AND COMBUSTION

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Prob: A hydrocarbon fuel gave a flue gas
CO 2 % 13.53, O2 % 3.56 , rest nitrogen 82.91 %
a)determine composition of fuel by weight
b)the excess air percentage
c) volume of air supplied per kg. of fuel




soln:
Combustion reaction of Carbon (C)
C + O2 = CO2

This indicates that moles of carbon (C) = moles of CO2=moles of Oxygen (O)

Assuming let dry flue gas is 100 KMOL , so CO2= 13.53 KMol
Thus amount of O 2 in flue gas
= 13.53 ( from CO2) + 3.56( from free O2 in flue gas) = 17.09 K Mol
Given amount of N2 in the flue gas = 82.91 kmol
we know that in 100 kmol 79 Kmol N2 is present

Thus amount of air supplied for combustion = (100/79) X82.91= 104.95 Kmol
Amount of O2 supplied= 104.95x21/100= 22.039= 22.04 kmol
Amount of O2 that has combined with H2= 22.04-17.09= 4.95 kmol
the combustion equation for H2 is H2 +1/2 O2= H2O
So amount of H2 burnt= 2X 4.95= 9.9 Kmol
So amount of H2 burnt= 9.9X2= 19.8 Kg
As 100 kmol of flue gas containing 13.53 kmol of CO2 or 13.53 kmol of C
So of Carbon in the fuel= 13.53 x 12 = 162.36 kg.

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