Reference no: EM13943771
Combustion problems
Question -10
A hydrocarbon liquid fuel (containing only C and H2) is burnt in air which produces a flue gas which analyses ( orsat analysis ) on dry basis as CO2 =10.5% , O2 = 5% and N2 =84.5%.
Find out the
A, gravimetric composition ( weight%) of the fuel.
B -% excess air used for the combustion
ANSWER -A- Combustion reactions are-
C+02 =CO2 and H2 +1/2 O2 = H2O
Assuming 100 kg mole of dry flue gases -
O2 present in flue gas =10.5 (from CO2 )+ 5 ( as free oxygen) =15.5 kg mole of O2
N2 presents in 100 kg mole of flue gas =84.5 kg mole
So , actual air used = 84.5*100/79 =107 kg mole /100 kg mole of flue gas
02 present in the air supplied =107 *21/100 =22.5 kgmole /100 kg mole of the flue gas.
So , the amount of oxygen present in water , H2O , produced by the combustion of hydrogen ( H2) = 22.5 - 15.5 = 7 kg mole
Hence , the amount of H2 present =7*2 =14 kg mole
As H2 +1/2 O2 =H2O
= 14*2 = 28 kg /kg mole of dry flue gases
And , the amount of carbon present =10.5 kg mole
C+O2 = CO2
CO2% in flue gas =10.5% = 10.5*12 = 126 kg/kg mole of dry flue gases
Hence , gravimetric composition of fuel ( by weight ) is C = 126 kg and H = 28 kg
% of carbon in fuel = ( 126/126 +28 )*100 =81.5%
% of hydrogen in fuel = (28/126 +28)*100 =18.2%
Part - b - in
In 100 kg mole flue gas , CO2 =10.5 kg mole , i.e carbon is also 10.5 kg mole
C+ O2 = CO2
And hydrogen = 14 kg mole
Amount of oxygen required to burn 10.5 kg mole carbon = 10.5 kg mole
And , amount of oxygen required to burn 14 kg mole hydrogen =14/2 = 7 kg mole
So, total oxygen required to burn the fuel = 10.5 +7 =17.5 kg mole /100 kg mole of dry flue gas ,but actual oxygen supplied = 22.5 kg mole /100 kg moe of dry flue gas % excess O2 = % EXCESS AIR = (22.5 -17.5/17.5 )*100 = 28.6 %
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