Question

QUESTION 1 150 Marks] a) A Coal from Indonesia which has an ultimate analysis (by mass) as 67.40 % C, 5.31 % Нг, 15.11 % 02, 1.44 % N2, 2.36 % S, and 8.38 % ash (non-combustibles) The coal is burned with 40% excess air. Determine: 1. The total mole number of the coal Marks] 2. The mole fractions of each component of the [7 Marks] 3. Stoichiometric combustion equation [12 coal Marks] 4. The mass of air required per unit mass of [2 Marks] 5. The total mole number of the product gas [2 coal burned Marks] 6. The apparent molecular weight (molar mass) [4 Marks] 7. The Stoichiometric Air-Fuel ratio by mass [6 Marks] of the product gas

Answer 1

Let us assume there is 100 kg of the particular coal:

The values in the parenthesis are the molar mass of the elements in kg/kmol

Since no molar mass of any standard ash is known we are excluding that from our calculations

Constituents	%(w/w)	Wt. of the element per 100kg	No. of kmol.
Carbon(12)	67.4	67.4	67.4/12 = 5.617
Hydrogen(2)	5.31	5.31	5.31/2 = 2.655
Oxygen(32)	15.11	15.11	15.11/32 = 0.472
Nitrogen(28)	1.44	1.44	1.44/28 = 0.051
Sulfur(32)	2.36	2.36	2.36/32 = 0.074

1) So, the total mole number of coal is the sum of the no. of moles = 5.617 + 2.655 + 0.472 + 0.051 + 0.074 = 8.869 kmol/100 kg of coal. So if we take 100 g of coal it will contain 8.869 moles of different constituents.

2) Mole Fraction of component 1 in a mixture is given by the following expression. Mole fraction is unit-less.

$\chi_{1} = \frac{n_{1}}{n_{1}+n_{2}+n_{3}+n_{4}....+n_{n}}$

We already know the total number of moles, i.e 8.869

$\chi_{C} = \frac{5.617}{8.869} = 0.633$ (Mole fraction of Carbon)

$\chi_{H_{2}} = \frac{2.655}{8.869} = 0.255$ (Mole fraction of Hydrogen)

$\chi_{O_{2}} = \frac{0.472}{8.869} = 0.053$ (Mole fraction of Oxygen)

$\chi_{N_{2}} = \frac{0.051}{8.869} = 0.0058$ (Mole fraction of Nitrogen)

$\chi_{S} = \frac{0.074}{8.869} = 0.0083$ (Mole fraction of Sulfur)

3)

This is the combustion reaction that has not been balanced. The part in the parenthesis refers to the air.

We should keep in mind that air contains 21% Oxygen and 79% Nitrogen.

Other key takeaways from the unbalanced equation are that Nitrogen doesn't produce a combustion product owing to its triple bond. Also if oxygen is present in the coal sample, it can also aid the combustion process.

We can clearly see that 2.5 Oxygen molecules are required to balance the reaction.

That means 2.5 moles of oxygen. From the data of air we can compute that if 2.5 moles of oxygen molecules is required. moles of air = 2.5 * (100/21) = 11.905 moles of air.

Amount of Nitrogen in that air = (79/100) * 11.905 = 9.405 moles of Nitrogen.

After balancing the number of Oxygen atoms, the reaction will become:

The coal sample already contains some amount of oxygen i.e = 0.472 moles.

Also the mixture already contains, 0.051 moles of Nitrogen.

2.655 moles of Hydrogen, 0.074 moles of sulfur and 5.617 moles of Carbon.

No. of moles of oxygen molecules required for complete combustion = 5.617 + (2.655/2) + 0.074 = 7.0185 moles

That means, air to be supplied = 7.0185-0.472 = 6.5465 moles

So, air required = 6.546 / .21 = 31.17 moles of air to be supplied.

Nitrogen from air 31.17*.79 = 24.627 moles of Nitrogen, total number of moles of Nitrogen = 24.627 + 0.051 = 24.678 moles.

the complete balanced equation becomes,

4) We learnt that for combustion of 8.869 mol of coal = 31.17 moles of air is to be supplied.

We know that 8.869 mol are present in 100 g of the coal sample(Take note, i have used moles here instead of kmol)

So, 8.869(100) of coal requires, 31.17 (6.546 mol of O₂ & 24.627 mol of N₂) = 209.472 g of O₂ and 689.556 g of N₂

Now, 1 g of coal will require, (209.472 + 689.556)/100 = 8.99g of theoretical air.

But the coal is burnt in 40% excess air that is 31.17 * 1.4 = 43.638 moles of air is used.

If we calculate the mass of 1 mol of air = (0.79*28 + 0.21*32) = 28.84 g/mol

So, actual air required for combustion of 1 g of of coal will require, 43.638*(28.84) / 100 = 12.585g of air.

5) Theoretically the number of mole in the product gas = 5.617 + 2.655 + 0.074 + 24.678 = 33.024 mol

But since excess air is used, the total no. of mol become, 33.024 + 0.40*(31.17) = 45.492 mol

6) Molar mass of CO₂ = 44 g/mol, for H₂O = 18 g/mol, for SO₂ = 64 g/mol, for N₂ = 28 g/mol

Apparent molecular weight of the product gas = (5.617*44) + (2.655*18) + (0.074*64) + (24.678*28) + (0.40*31.17*28.84) = 1350 g for 100g of Coal; 1350 kg per 100kg of coal.

7) Stoichiometrically, 100g of coal requires 31.17 moles of air = (31.17*28.84)g = 898.943g.

So the air-fuel ratio(by mass) = 898.943/100 = 8.989