A solid fuel consists of 68wt carbon 8wt oxygen 124wt inert

A solid fuel consists of 68%wt carbon, 8%wt oxygen, 12.4%wt inert ash, and the rest are hydrogen, nitrogen, and sulfur. Air is added to burn the fuel, producing a dry flue gas of 10.4%mol of CO2, 1.2%mol of CO, 0.2 %mol of NO2, 0.1 %mol of SO2, and 5.7 %mol of O2. The rest is nitrogen (assume no nitrogen in the feeding air becomes NO2). If 3,000 kg air was used (air density = 1.0 kg/m 3 ), calculate

a. amount of the fuel (in kg), and

b. amount of outlet flue gas (in kmol),

c. percentage (%wt) of hydrogen in the fuel.

Solution

The air consumed is = 3000kg = 3000 m³

We know that one mole of air occupies 22.4 liters of volume = 0.0224m³

Hence no. of moles present in the air = 3000/0.0224 = 133.928 kmol

Given information states that the dry flue gas has 10.4 %mol of CO₂

It also says that the no. of moles of outward gases of CO is 1.2 %mol

No. of Oxygen molecules in the carbon composition is = 10.4 + 1.2/2 = 11 %mol

Given there is 0.1 %mol of SO₂ in the outward flue gases.

Hence, the Oxygen content in sulphur is given by 0.1 %mol.

It is given that the flue gases contains 5.7%mol of O₂

Hence the Total amount of Oxygen in the outward gases is 5.7 + 0.1 + 11 = 16.8 %mol

Assuming the moles of flue gases as b kmol

Hence Total oxygen = 0.168b kmol

Since the air input is given as 133.928 k.mol, the amount of oxygen present in air is 20%.

Hence the oxygen in air is 133.928 x 20% = 26.786 kmol

Let us assume the weight of the solid fuel as a kg

We know that 8%wt is of oxygen. Hence, no. of moles of oxygen present in solid fuel is given by 0.08a kg = 0.08a/16 k.mol = 0.005a k.mol

Hence since oxygen input = output, 0.005a + 26.786 = 0.168b

Taking account for Carbon molecules, the no. of moles of carbon content output is = 10.4b + 1.2b = 0.116b kmol

Carbon input from the solid fuel is given by 0.68a kg = 0.68a/12 = 0.0566a kmol

Since no carbon is created or destroyed, 0.0566a = 0.116b

Solving oxygen equation using the carbon data, we find b = 169.8 kmol

Hence, the moles of flue gases = 169.8 kmol

a = 348.09

Hence, weight of fuel is 348.09 kg

The output gases contain 0.2 %mol of NO₂= 0.3396 kmol = 339.6 mol

We know that molar weight of the substance is given by = 14+16+16= 46

Hence, weight of Nitrogen used is 103.35 grams

Similarly, weight of Sulphur used is = 84.9 grams

We know that solid fuel has many entities which are enumerated and they have a combined 88.4% of all weight.

Hence, Hydrogen, Sulphur and Nitrogen form 11.6% of weight = 40.378 kg

Out of this, by removing the weight of sulphur and nitrogen, we get the weight of hydrogen as 40.19 kg

Hence percentage of hydrogen in mixture is given by = 11.54 %wt