Question

A furnace is burning natural gas (90% by mol methane and the balance ethane) at 20 Celsius degree and 1 atm with 25% excess air at 150 Celsius degree and 1.1 atm. The product gases are at 600 Celsius degree 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

first we will calculate the densities of fuel gases at 20⁰ C and at 1 atm. pressure.

methane =0.668 gms , ethane=1.19 gms

now we will calculate the air density at 150⁰ C and 1.1 atm. pressure, which is equals to 0.9144 gms.

now we will look at the reaction of fuel gases with air it gives CO2 and carbon monoxide Products,

then, one mole gives 95% conversion of methane to CO2, and 5&% conversion to CO.

0.11 moles of ethane gives 95% CO2 and 5% CO. (Ratio of fuel gases( metahe: ethane 1:0.11) that is (9:1).

now, for the reaction 1 mol methane requires one mole of oxygen gas and for 0.11 moles of ethane requires double (2O₂) That IS 0.22 MOLES OF OXYGEN GAS,

methane =mol wt 16* 1 mole=16 grams

ethane= mol wt.=30*0.11 moles=3.3 grams

(For reaction total oxygen require will be 1.22 moles*32=39.04 grams) and also excess 25% ,=39*25=975/100=9.75=48.79 grams

now we will calculate the ratio of air supply to fuel volume=1:6.20 (fuel:air supply)

B) Methane 1 mol*95% co2 and 5% co means , 1*95=95/100=0.95 moles co2)=mol wt =44*0.95=41.8 grams co2,

0.5*28=14 grams co,

ethane= 0.11*95=0.1045 moles co2, =0.1045*44=4.598 grams co2,

0.11*5=0.005 moles co=0.005*28=0.154 grams co2

C)Fxhaust gas volume/fuel gas volume

=total exhaust gas co2 and co(60.55 grams)/total fuel gas in gms(52.09)

=60.55/52.09

=1.16