A liquid fuel containing 88 wt% C and 12% H is burned with air, comprising 21vol

Question

A liquid fuel containing 88 wt% C and 12% H is burned with air, comprising 21vol% O2 and 79% N2, in a combustor. The combustion product is called "flue gas", and the flue gas leaving the combustor is then passed through a desiccant column, to completely removed water from the flue gas. The resulting dry flue gas is then passed through a gas analyser, to analyse for the composition of the dry flue gas, and the analytical results are as follows: CO2 13.4% by vol., O2 3.6%, and N2 83.0%. If the amount of fuel fed into the combustor is 100 g, calculate dry the amount of the flue gas, in g, and determine the percentage excess of O_2.

Explanation / Answer

Fuel contains   88 gm of C and 12 gm of H. The combustion reactions are

C+ O2----àCO2 and H2+0.5O2--àH2O

Moles= mass/atomic mass

Moles : C= 88/12=7.33, H2= 12/2= 6

Moles of O2 required for combustion of C= 7.33 moles of oxygen required for combustion of H2= 6/3=3

Total moles of O2= 7.33+3= 10.33

Air contains 21% O2. Moles of air = 10.33/0.21= 49.2 moles

Let x= percentage of excess air, moles of air supplied= 49.2*(1+x/100)= 49.2*(1+0.01x)

Products contains ( moles ) :CO2= 7.33, H2O= 6, N2= 49.2*0.79*(1+0.01x)= 39*(1+0.01x)

Moles of O2= moles of oxygen supplied-consumed = 49.2*0.21*(1+0.01x)- 10.33 =0.10332x

Water is removed through desiccation.

Now the products contains CO2= 7.33, N2= 39*(1+0.01x), O2= 0.10332x

Total moles of product = 7.33+39*(1+0.01x)+0.10332x= 46.33 +0.49332x

Percentage of CO2= 100* 7.33/(46.39+0.49332x)= 13.4

1/(46.39+0.49332x)= 13.4/ (100*7.33)

46.39+0.49332x= 100*7.33/13.4=54.7

0.49332x= 54.7-46.39= 8.31, x= 16.84%

Moles of products = CO2= 7.33, N2= 39*(1+0.1684)=45.6 , O2= 0.10332*16.48=1.70

Mass of dry flue gas= 7.33*44+ 45.6*28+1.7*32= 1653.72 gm

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