Question

A stream of methane at 86.06 kg/hr is burned in a boiler to heat water. The boiler is fed 47.7 % excess air. Solve for air flow rate in kg/hr.

Note: Air can be treated to be 78 % N₂, 21 % O₂ and 1% Ar, where all percentages are in mole percents. The average molecular weight for air of this composition is 28.970 kg/kgmol

Answer 1

For this question, you first need to understand how much O2 is required to react with the said amount of methane, as per the reaction.

The reaction is :

CH4 + 2O2 -> CO2 + 2H2O

The reaction shows that you need 2 moles of Oxygen per mole of CH4.

You have been given 86.06 kg/h of CH4. This is equal to 5.37875 kmole of CH4 / h (86.06 divided by molecular weight of CH4, i.e. 16)

As per the reaction you need 2 * 5.37875 kmole of O2 / h. This comes out to be 10.7575 kmole of O2/h).

Air contains about 21 % O2 by moles (or volume). In other words, 0.21 kmole of O2 are found in 1 kmole of air. Therefore 10.7575 kmole of O2/h would be found in kmoles of air/h. This comes out to be about 51.2262 kmoles of air/h.

Now you have been given that you are providing 47.7 % excess AIR.

This means that the actual amount of air you're providing is 1.477 times that required. Therefore, you are providing 1.477 * 51.2262 kmoles of air/h. This comes out to be 75.6611 kmoles of air/h.

You are required to report your answer in kg/h. The molecular weight of air is 28.970 kg/kmole. Therefore, the air flow rate you are providing is 28.970*75.6611 kg/h. This comes out to be 2191.9016 kg/h of air.

I can feel that you have a problem with reactions and stoichiometry. Try to follow these steps always and it should help. Also, you could have done these calculations in volumetric flow rates as well ( apart from molar flow rates). This because, as per the ideal law, molar flow rate s are directly related to volumetric flow rates.

All the best!