Question

a) Explain what is meant by the term stoichiometric combustion. b) Explain why it is often necessary to use excess air during combustion c) When pentane gas (C_5H_12) is burned with excess air, the dry products are found to contain 12% CO_2 by volume. Calculate the volume of excess air supplied and the percentage of excess air supplied.

Answer 1

stoichiometric combustion refer to combustion of fuel with stoichiometric air. For example

C5H12 is when combusted, the following reaction takes place

C5H12+ 8O2-------->5CO2+6H2O

the combustion reaction requires 8 moles of oxygen to completely combust the fuel. This is the stoichiometric requirement.

for ensuring the valuable fuel, it is prferrred to send some percentage of excess oxygen. This excess air is air supplied over and above the stoichiometric air required. This ensures complete combustion of C5H12.

The combustion efficiency increases with increased excess air -to a point where the heat loss in air is larger than heat generated by efficienct combustion.

The combustion of pentane is

C5H12+8O2---à5CO2+ 8H2O

Basis : 1 mole of pentane. Moles of oxygen required= 8 moles. air contains 21% O2 and 79%N2. Moles of air requited= 8/0.21= 38.1 moles

Let x= excess % of air supplied. Excess air supplied = 38.1*(1+0.01x)

Moles of CO2 formed= 5, moles of O2 remaining = 8*x/100 =0.08x

Moles of N2= 38.1*(1+0.01x)*0.79= 30.1*(1+0.01x)= 30.1+0.301x

Moles of products on dry basis : 5+0.381x+30.1+0.08x= 35.1+0.461x

CO2 percentage = 100*5/(35.1+0.461x)= 12

Hence 35.1+0.461x= 100*5/12= 41.67

0.461x= 41.67-35.1 , x=14.25

Moles of excess air supplied = 38.1*(1+0.01*14.25)= 43.53 moles

Assuming air is supplied at 1 atm pressure and 25deg.c (298K)

Volume of air = nRT/P= 43.53*0.0821L.atm/mole.K*298/1= 1065 L

Volume of excess air = 1065*14.25/100 L=152 L