Question

Two fuel gases are mixed together prior to being burned with air in a furnace to produce a flue gas with the following composition: CO_2 7%, CO 1%, O_2 7% & N_2 85% (on a dry basis). The analyses of the fuel gases are as follows: Fuel Gas A: CH_4 80%, N_2 20% Fuel Gas B CH_4 60%, C_2H_6 20%, N_2 20% All compositions are expressed on a mole basis. Calculate a The mole ratio of Gas A to Gas B in the fuel gas mixture. b The percentage excess air used in the furnace.

Answer 1

Ans:

a) Given:

Fuel A- 80% CH₄ , 20% N₂

Fuel B- 60 % CH₄, 20% C₂H₆ , 20% N₂

Step 1: Convert all the masses into moles.

Let the total mass of both fuel gases be 100 g each.

So in fuel gas A: let the mass of CH₄ =80 g

And the mass of N₂ =20 g

Now, no. Of moles= given mass / molar mass

So no. of moles of CH₄ in fuel gas = 80 g / 16g = 5 moles

And no. Of moles of N₂ in fuel gas A= 20g/ 28g = 0.714 moles

Total no. Of moles of fuel gas A= 5.714

In fuel gas B:

Let the mass of CH₄= 60g,

Mass of C₂H₆= 20g

& mass of N₂= 20 g

Now, no. Of moles of CH₄ in fuel gas B= 60g/16g = 3.75 mol

No. Of moles of C₂H₆ in fuel gas B= 20g / 30g = 0.666 moles

& no. Of moles of N₂= 20g/ 28g = 0.714 moles

Total no. Of moles of fuel gas B= 5.13 moles

Now, mole ratio of gas A to gas B= 5.714/ 5.13 =1.11

B) Excess air: The amount of excess air within the system can be determined by analyzing the amount of O₂ in the fuel gas.

Thus the CO level in the fuel gas is assumed to be very low and incomplete combustion is neglected, hence the formula used to calculate excess air on a dry basis is:

Excess air %=( 0.895× A%) / (0.21 - A%)

A% is the percentage of Oxygen in fuel gas mixture.

A% = 7% =0.07

Therefore, Excess air %= 0.895 × 0.07 /(0.21-0.07)

= 0.4475

= 44.75%