Question

A stream of methane at 192 kg/hr is burned in a boiler to heat water. The boiler is fed 45.13 excess air. What is the air flow rate in kgmol/hr?

Note: Air can be treated to be 78 % N₂, 21 % O₂ and remainder percentage Argon, where all percentages are in mole percents. Average molecular weight for air of this composition is 28.970 kg/kgmol

Answer 1

First of all, write the reaction:

CH4 + 2O2 ->CO2 +2H2O

This means for 1 mole of CH4 you need 2 moles of O2 stoichiometrically.

You are feeding 192 kg/h (or 192/16 kmoles/h) of CH4.

Do everything in terms of moles.

You would need twice the number of moles of O2 to react with 192/16 kmoles/h of CH4.

Therefore, you need 2*192/16 kmoles/h of O2. This comes out to be 24 kmoles/h of O2.

Air only contains 21 % of O2. In other words, 0.21 moles of O2 are found in 1 mole of air. Therefore, 24 kmoles/h of O2 would be found in 24/0.21 kmoles/h of air. This comes out to be 114.2857 kmoles of air/h.

You have taken 45.13 % excess air. This means that you are feeding 1.4513 times the stoichiometric requirement of air. Therefore, the amount of air you are feeding is 1.4513*114.2857 kmoles of air/h. This comes out to be 165.8628 kmoles of air/h.

Hope that helps!