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**The following article will guide you about: How much Air is Required for Complete Combustion?**

**Stoichiometric Air-Fuel Ratio:**** **

The stoichiometric air-fuel ratio can be defined as ratio of amount air required for complete combustion of 1 kg of fuel. It is also called as chemically correct air-fuel ratio.

If the combustion is complete then and then only maximum heat is available from a given fuel. The theoretically exact amount of oxygen required can be calculated with the help of equations or with the help of the formula derived from the above equations and it will give us directly the theoretically required oxygen if we know the ultimate analysis of the fuel.

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The oxygen for the combustion of a fuel is to be obtained from the atmospheric air although in some cases a certain amount of oxygen is a constituent of the fuel. Air is a mixture of oxygen, nitrogen, a small amount of carbon dioxide and small traces of rare gases such as neon, argon, krypton, etc.

For all practical purposes we assume that air is made up of 23% by weight of oxygen, the remaining 77% being nitrogen. If considered by volume, air consists of 21% of oxygen and 79% of nitrogen. When we know the amount of oxygen necessary for the combustion of a fuel, we can determine the amount of air necessary for the complete combustion of one kg of fuel.

Let us consider 1 kg of a fuel, the ultimate analysis of which shows that carbon is C kg, hydrogen H kg, oxygen O kg and sulphur S kg.

1 kg of carbon requires 8/3 kg of oxygen for its complete combustion; therefore C kg of carbon will require C x 8/3 kg of oxygen which is equivalent to 2.66C kg of oxygen (considering upto two places of decimal).

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1 kg of hydrogen requires 8 kg of oxygen; therefore H kg of hydrogen requires 8H kg of oxygen.

1 kg of sulphur requires 1 kg of oxygen; therefore S kg of sulphur requires S kg of oxygen.

**Therefore the quantity of oxygen required for combustion of 1 kg of the fuel is:**

(2.66C + 8H + S) kg.

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If the fuel contains O kg of oxygen; then it is taken into account

∴ Oxygen required from air for the complete combustion of fuel will be (2.66C + 8H + S – O) which can be written as 2.66C + 8 (H – O/8) + S, the term in the bracket being known as the available hydrogen. Thus, we get the formula for determination of minimum quantity of oxygen for complete combustion of a solid or liquid fuel whose ultimate analysis is known.

**The formula can be written as under: **

#### Determination of Excess Air:

We have calculated the minimum quantity of air required for complete combustion of one kg of fuel. In actual practice we supply air more than the theoretical minimum amount in order to ensure the complete combustion of the fuel because all of the air supplied does not come into intimate contact with the particles of the fuel.

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A large amount of excess air has a cooling effect on the process of combustion and represent a loss, and in order to avoid this cooling effect air is pre-heated before it enters the furnace of a boiler.

With natural draught system the excess air is more in comparison with the artificial draught system. The total quantity of air supplied varies with the quantity of the fuel, rate of combustion, system of firing and the draught intensity. The excess air may approach 100 per cent but the modern practice is to use 25 to 50 per cent.

The excess air is indicated by CO_{2}% in flue gases. When the boiler plant of Lancashire boiler type employs hand firing with natural draught system 10% to 12% CO_{2 }in flue gases would be considered good practice. With mechanical strokers and artificial draught it would be quite reasonable to expect 12% to 15% CO_{2}.

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In case of internal combustion engine all the air taken in during suction stroke will not come in contact with the fuel particles; as a result excess air is supplied and it should be reduced to minimum to get more specific output. If the supply of air to an engine cylinder is less than 15% excess, the mixture of air and fuel in the cylinder is known as rich, If the supply of air is more than 30% in excess of the theoretical minimum, the mixture is known as a weak mixture.

**Example: **

**Determine the theoretical mass of air required for the complete combustion of 1 kg of coal whose analysis by mass is given as under: **

Carbon — 83%, hydrogen – 5%; oxygen – 2%;

Sulphur – 0.2%; remainder being incombustible.

**Solution: **

**As we know in the ultimate analysis of fuel we determine the mass of various combustible elements in the fuel. In 1 kg of fuel: **

Carbon is 0.83 kg, hydrogen 0.05 kg, oxygen 0.02 kg and sulphur 0.002 kg.

**Minimum quantity of air required for complete combustion of 1 kg of fuel is given by: **

**Determination of the Flue Gas Analysis by Mass and by Volume:**** **

When the combustion of fuel takes place, the products of combustion will be carbon dioxide, water vapour, sulphur dioxide and nitrogen of the air. As an excess air is supplied, oxygen also will be the one of the constituents of flue gas analysis.

In order to determine the analysis of flue gas by mass, determine the mass of each constituent separately and express each of them as percentage of the total mass. When the mass analysis of flue gas is known, it can be converted into volumetric analysis by the help of Avogadro’s hypothesis (molecular weights of all gases at equal volumes).

**Determination of Air Supplied from Volumetric Analysis of Flue Gases:**** **

The amount of air supplied to a boiler furnace is very great and so it cannot be measured directly. Its indirect measurement can be done if we know the volumetric analysis of dry flue gas. When we consider the volumetric analysis of flue gas we consider carbon monoxide, carbon dioxide, oxygen and nitrogen because the quantity of sulphur dioxide is negligible and water vapour is condensed.

Let us consider the fuel which contains C% by weight of carbon. Analysis of dry flue gas of the combustion of this fuel shows-

**We know that the molecular weights of gases are the relative weight of equal volumes of them. It follows that relative weight of the given relative volume of gases are shown as under: **

**Determination of Air Leakage in Boiler Flues:**** **

The pressure within the boiler furnace and various flues is less than atmospheric with induced and natural draught; so atmospheric air is likely to leak in the various flues through cracks. By taking the flue gas analysis at two points we can determine the leakage of air between these two test points. The amount of air leakage can be determined with the help of formula.

#### Determination of the Quantity of Air Supplied Per kg of Fuel:

**Determination of the quantity of air supplied per kg of fuel from the analysis of flue gases when given by mass:**

**Let C be the % of carbon in one kg of fuel and the gas by weight is given below: **

**Chemically Correct Air Fuel Ratio:**** **

The minimum quantity of air required for complete combustion of one m_{3} of gaseous fuel which is calculated as below-

**I. Combustion of Hydrogen: **

**II. Combustion of Carbon Monoxide: **

**III. Combustion of Marsh Gas:**

By applying the formula derived above, we can determine the minimum quality of air required for complete combustion of one m^{3} of gaseous fuel. If we know the percentage of excess air supplied we can find out the actual quality of air supplied for combustion of gaseous fuel.

**Excess Air Coefficient:**** **

It is the amount of air supplied over and above the minimum quantity of air required for complete combustion. This can be expressed as % of excess of required air or by a coefficient known as excess air coefficient α.

**Determination of Flue Gas Analysis by Volume and by Mass in Case of Gaseous Fuel:**** **

When the volumetric analysis of gaseous fuel is known, we can easily determine the volumetric analysis of dry flue gas. When volumetric analysis is known the analysis by mass can be obtained by means of the method. An illustration is taken to explain the method of determination of volumetric analysis of flue gas.

**A producer gas has the following percentage analysis by volume: **

H_{2} – 15; CH_{4} – 2; CO – 20; CO_{2} – 6; O_{2} = 3; N_{2} – 54.

50% excess air is supplied for combustion. Calculate the volume of air supplied per m^{3} of gas and the analysis by volume of the dry products of combustion.

Quantity of oxygen required for complete combustion of hydrogen

= 0.5 x 0.15 = 0.075 m^{3}.

Quantity of oxygen required for complete combustion of methane

Quantity of oxygen required for complete combustion of carbon monoxide

**Determination of Quantity of Air Supplied Per M**^{3} of Gas:

^{3}

For determining the quantity of air supplied, we require the analysis of the gas and also that of the exhaust gases. From volumetric analysis of gas we can determine the minimum quantity of air required, the total volume of products of combustion and water vapour formed. When the steam has been condensed, the volume of the dry products of combustion can be calculated. When the exhaust gases are analysed the steam will be condensed.

Let V_{1} m^{3} be the amount of air supplied per m^{3} of gas in excess of that required for complete combustion and V be the volume of dry products of combustion when minimum quantity of air is supplied.

∴ Total volume of actual products of combustion = (V_{1} + V) m^{3}.

Let O be the percentage of oxygen (by volume) present in the exhaust gases, then the excess quantity of air which will contain this volume of oxygen will be O/21 m^{3}.

When the excess air is known and minimum quantity of air required is known, we can determine the quantity of air supplied per m^{3} of gas.