Question

A fired heater uses natural gas as fuel which is burned with 25% excess air.

90% of the fuel burns to give CO2, 4% burns to give CO, and the rest is unburned.

(a) Calculate the molar composition of the feed (fuel + air).

(b) Evaluate the composition of the flue gases on dry basis.

Hints: Air composition may be assumed as 21% oxygen and 79% nitrogen on molar basis..

Natural gas may be assumed as 100% methane

Answer 1

Solution.

a) There are two combustion reactions:

(first reactcion)

(second reaction)

Let's suppose we have 1 mol of methane. In this case the first reaction get a multiplication factor 0.9, the second 0.04 and 0.06 moles of methane rest unburned.

The first reaction requires 0.9*2 = 1.8 mol of oxygen;

The second reaction requires 0.04*3/2 = 0.06 mol of oxygen;

The total amount of oxygen required is 1.8+0.06 = 1.86 mol. It corresponds to 1.86*0.79/0.21 = 7 mol of nigrogen to compose air. Hovewer, we have 25% excess air, so the amount of oxygen is 1.86+1.86*0.25 = 2.325 mol, and the amount of nitrogen is 7+7*0.25 = 8.75 mol.

The total number of moles is: 1 + 2.325 + 8.75 = 12.075 mol. The molar composition of the feed is (molar fractions):

b) Assuming again burning of 1 mol of methane, the first reaction gives 0.9 moles of carbon dioxide, the second reaction gives 0.04 moles of carbon monoxide and 0.06 moles of methane remains unchanged. As we used an excessive amount of air, this excess will appear in the flue gases in amounts 1.86*0.25 = 0.465 moles of oxygen and 7*0.25 = 1.75 moles of nitrogen. Water steam should not be taken into account as the calculations are performed "on dry basis". The total number of moles is: 0.9 + 0.04 + 0.06 + 0.465 + 1.75 = 3.125 mol.

The molar composition of the flue gases is (molar fractions):