Calculation of Excess Air

A bituminous coal has the following ultimate analysis: C = 71.6

N = 1.3

A = 9.1

H = 4.8

S = 3.4

W = 3.5

O = 6.3 Calculate the total air required r equired for complete combustion per pound of coal. The excess air is assumed 20%.

The theoretical air requirement is calculated as follows: Wta = 11.53C + 34.36 (H –  (H  – O/8) O/8) + 4.32S Or, Wta = 11.53*0.716 + 34.36*(0.048-0.063/8) + 4.32*0.034 Or, Wta = 9.78 lb/lb of coal.  Now, calculate the total air: Excess air percentage = (Wa – W  – Wta)/ Wta *100 Or, 0.2 = (Wa – 9.78)/9.78  – 9.78)/9.78 *100 Or, Wa = 11.74 lb/lb of coal.

CONVERSION OF VOLUMETRIC ANALYSIS BY MASS AND VICE-VERSA

If the volumetric analysis of any fuel is given it can be reduced to mass analysis by applying “Avogadro’s Law”, thus: 1. Multiply the volume of each constituent by its own molecular mass. This will give the  proportional mass of the constituents. 2. Add up these masses and then divide each mass by the total and express it as a  percentage. This will give the percentage analysis by mass. For converting mass analysis into volume analysis the following procedure may be adopted: a. Divide the volume of each constituent by its own molecular mass. This will give the  proportional volumes of the constituents.  b. Add up all these volumes and divide each by the total and express it as a percentage. This will give the analysis by volume.

MASS OF AIR REQUIRED FOR COMPLETE COMBUSTION OF FUEL It is clear that an adequate supply of oxygen is essential for the complete combustion of fuel and hence to obtain maximum amount of heat from it. Theoretically the exact amount of oxygen required for complete combustion of one kg of fuel can be determined from the analysis of the fuel. Firstly, the amount of oxygen required for  each of the constituents of the fuel is calculated separately with the help of chemical equations. Then by adding these requirements the total amount of oxygen required is obtained. The oxygen for the combustion of fuel has to be obtained from atmosphere air, which consists of oxygen, nitrogen, a small amount of carbon dioxide and small traces of rare gases like neon, argon and kryspton etc. But for all engineering calculations, composition of  air is taken as follows.  Nitrogen = 77 %, Oxygen = 23 % …………………..by mass.  Nitrogen = 79 %, Oxygen = 21 % …………………...by volume. So, if we know the amount of oxygen required for the combustion of certain fuel we can determine the amount of air necessary for the combustion of one kg of the fuel. The theoretical or stoichiometric quantity of air is that quantity which is required for  complete combustion of 1 kg of fuel without any oxygen appearing in the products of  combustion.  NOTE: If some oxygen is already present in the fuel then it must be deducted from the calculated amount of oxygen required for the combustion of the constituents.

EXCESS AIR  In actual practice, the amount of air supplied is always more than the theoretical min. one, so as to ensure complete and rapid combustion of the fuel. This is because of the fact that all this air does not come into intimate contact with the fuel particles, and if just a part of minimum amount of air as theoretically required were supplied, the fuel might have remained un-burnt; on the other hand, however, a large amount of excess air has c ooling effect on the process of  combustion and thus results in loss of heat energy. In order to overcome this effect air is preheated before supplying it to the furnace of a boiler. The amount of excess air supplied varies with the type of fuel and the firing conditions. It may approach a value of 100% but the modern practice is to use 25% to 50% excess air.

MASS OF CARBON IN FLUE GASES The mass of carbon contained in one kg of flue or exhaust gas can be calculated from the amounts of co2 and co contained in it. In equation (i) of combustion of solid fuels it was shown that 1 kg of carbon produces 11/3 kg of CO2 when completely burnt. Hence 1 kg of CO2 will contain 3/11 kg of carbon. Also from equation (ii) it can be seen that 1 kg of carbon produces 7/3 kg of CO hence 1 kg CO contains 3/7 kg of carbon. Therefore mass of carbon per kg of fuel = 3/11 CO2 + 3/7 CO Where, CO2 and CO is the quantities of carbon dioxide and carbon monoxide present in 1 kg of flue or exhaust gas.

MASS OF FLUE GAS PER KG OF FUEL BURNT The mass of flue gas or exhaust gas is always more than that of fuel burnt on account of the air supplied. The actual mass of dry flue gases can be obtained by comparing the mass of  carbon present in the flue gases with the mass of carbon in the fuel, since there is no loss of  carbon during the combustion process. As the analysis of the exhaust gas is volumetric, so it must first be reduced to mass analysis  by the method as explain earlier. The total mass of carbon in 1 kg of the flue gas is = 3/11 CO2 + 3/7 CO The mass of flue gas/kg of fuel burnt = (Mass of carbon in one kg of fuel) / (Mass of carbon in one kg of flue gas)

THE MASS OF EXCESS AIR SUPPLIED The mass of excess air supplied can be determined from the mass of oxygen remained unused after following for combustion of carbon monoxide to carbon dioxide. The mass of oxygen must then be converted to the mass per kg of fuel burnt. Mass of excess oxygen per kg of fuel = mass of excess oxygen per kg of flue gas x mass of flue gas per kg of fuel. Since 23 parts of oxygen are contained in 100 parts of air, therefore mass of excess air  supplied = mass of excess oxygen x (100/23) Total mass of air supplied = mass of necessary air + mass of excess air.

CALORIFIC VALUE OF FUELS The calorific value of a fuel is defined as the no. of heat units produced by complete combustion of a unit mass or unit volume of the fuel. It is expressed as KJ per kg or KJ per cu metre. HIGHER OR GROSS CALORIFIC VALUE

This is the total heat evolved by the complete combustion of one kg or one cubic metre of  fuel. In this case the products of combustion are cooled down to room temperature (usually taken as 288 K) at which the fuel and air were supplied. In doing so, all of the water vapours formed, during combustion, are condensed and thus the heat which was used for its evaporation is restored. Also sensible heat carried by flue gases is recovered. So, this type of  test will give, the total heat produced during combustion which is known as higher or gross calorific value of the fuel. LOWER CALORIFIC VALUE

In actual practice, whether the fuel is burnt in the furnace of a boiler, the fl ue gases leave at a considerably in excess of 373 K and so the gases c arry with them sensible heat in dry flue gases and the total heat in steam formed as a result of the combustion of the hydrogen of the fuel. In such case the heat actually obtained per kg or cubic metre of fuel burnt will be less than the gross value and is known as net or lower calorific value (LCV). If higher calorific value of a fuel is known, the lower calorific value is obtained by subtracting from it, the amount of heat carried by the products of combustion especially by steam which may be taken as 2472 KJ/kg of water vapour formed by combustion of 1 kg of  fuel. Thus lower calorific value may be given by L.C.V. = (H.C.V. – 2472 m)

KJ/kg

Where, “m” is the mass of water vapour present in the products of combustion due to combustion of kg of fuel.

ESTIMATION OF CALORIFIC VALUE FROM CHEMICAL ANALYSIS

The calorific value of a solid or liquid fuel may be obtained approximately from a chemical analysis of a dried sample. The analysis usually obtained gives the masses in one kg of dry sample of carbon, hydrogen, sulphur, nitrogen, ash and oxygen. Out of these elements only the first three contribute to calorific value of the fuel. Let C, H, O and S be the percentage of carbon, hydrogen, oxygen and sulphur, contents of the fuel respectively. Let us