Question

A furnace is burning natural gas (90% by mol methane and the balance ethane) at 20 degrees C and 1 atm with 25% excess air at 150 degrees C and 1.1 atm. The product gases are at 600 degrees C and 1 atm. 95% of each fuel gas forms CO2 while the balance forms CO. a) What is the ratio of air supply volume to fuel volume? b) What is the composition of the product gas mixture on a mol% basis? c) What is the ratio of exhaust gas volume to fuel volume?

Answer 1

The natural gas burning equation as per moles present is given as

0.9 CH4 + 0.1 C2H4 + 2.025 O₂ 0.95 CO₂ + 2 H₂O + 0.15 CO

so you will need 2.025 moles of O₂ per mole of natural gas burnt

Air has 21 mole% oxygen and 79mol% nitrogen

2.025 moles of oxygen is present in (2.025 x 1)/0.21 = 9.64 moles of air

PV=nRT

1.1 x V = 1 x 0.082057 x 293

V = 21.86 L of natural gas requires

1.1 x V = 9.64 x 0.082057 x 293

V = 210.7 L of air. Air is passed at 25% excess so actual air passed is 210.7 + 210.7 x 0.25 = 263.37 L

So the ratio of air to fuel in volume is 263.37 L to 21.86 L of natural gas

The output gas contains CO₂, CO, H₂O and N₂ and unburnt O₂

Since the out put from burning has 0.95 CO₂ + 2 H₂O + 0.15 CO these came from 2.025 mole of O₂

No we have taken 25% excess of air which means 25% excess of O2 which will be

2.025 + 2.025 x 0.25 = 2.531 moles of which 2.025 will burn and 0.506 moles will remain unburnt

the amount of nitrogen corresponding to 2.531 moles of oxygen will be (79/21) x 2.531 = 9.52 mole of nitrogen

SO the output gas will have 0.95 mole CO₂ + 2 mole H₂O + 0.15 mole CO + 0.506 mole O₂ and 9.52 mole N₂

Total number of moles of gas in exhaust is 13.12 moles

PV=nRT

1 atm x V = 13.12 x 0.082057 x 873

V = 874 litres of exhaust gas

Fuel volume is 21.86 L of natural gas so ratio is 874/21.86 = 39.98