Question

ChE question

A fuel oil is analyzed and found to contain 87 wt% carbon, 11 wt% hydrogen, 1.6 wt% sulfur and the remainder incombustible matter. The oil is burned with 20% excess air based on complete combustion of carbon to CO₂, hydrogen to H₂O and sulfur to SO₂. The oil is burned completely but 5% of the carbon forms CO. Calculate the molar composition of the stack gas.

Answer 1

Basis: 100 kg oil.

C=87 kg

C forming CO2=0.95*87

=82.65 kg

=6.8875 kmoles

C forming CO=0.05 *87

= 4.35 kg

= 0.3625

S=1.6 kg=0.05 moles

H=11 kg = 5.5 moles

1) C+O2=CO2

O2 needed=6.8875 kmoles

2)C+1/2O2=CO

O2 needed= 0.3625 *0.5 = 0.18125 moles

3) S+O2=SO2

O2 needed=0.05 moles

4)H2+1/2O2=H2O

O2 needed= 0.5 x 5.5 moles = 2.75 moles

Total O2 needed=9.86875 moles

Total O2 used(including excess)=1.2*9.86875

=11.8425 moles

Air used=100/21*11.8425=56.393 moles

Oxygen remaining in stack gas=11.8425 -9.86875 moles=1.97375 moles

Nitrogen in stack gas=56.3046*79/100=44.55 moles

SO2 in stack gas=0.05 moles

CO2 in stack gas=6.8875 kmoles

CO in stack gas=0.18125 moles

Steam in stack gas=5.5 moles

Total stack gas=59.1425 moles