"14 The measure of excess air is given byamount of air actually used – the eoretical amount of air () % excess air = × 100 theoretical amount t of air [Note 'air' can be replaced by 'oxygen' in this equation.] So, for our example fuel, to ensure complete combustion, it is found, in practice, that the air:fuel ratio by volume has to be 6.5:1.What is the % excess air? 3 3 3 For 1 m of fuel, actual air used=6.5 m and the theoretical air is 5.24 m so:6.5 – 5.24 () excess air = ×= 100 24% 5.24 What would happen if this amount was not used? ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 (Engineering)15 Here the combustion would not be complete.Depending on the actual amount used several situations may occur.For our sample fuel (and this is also the general case when the mixing of fuel and air is good), the pure hydrogen would preferentially burn first, followed by the hydrogen in the methane and then the carbon would finally burn.Some of the carbon may only be converted to carbon monoxide and some may even be left behind as carbon solid ('soot').Unburnt fuel would only be the case if mixing was poor and the amount of excess air was minimal. So far we have determined the theoretical amount of oxygen (or air) required from the balanced reaction equations.It is possible to find the composition of a fuel and the amount of excess air used in a combustion process by looking at the products of combustion. Let's do an example of this type of calculation.We will only be considering fuels based on carbon and hydrogen, though the method is suitable for other fuels (e.g. sulphur based or fuels containing oxygen such as methanol CH OH).We will also be dealing with complete combustion only though it 3 can be used for situations where incomplete combustion occurs. Example A gaseous fuel consists of a compound containing only carbon and hydrogen. After it was completely burnt, the analysis of the products of combustion (on a dry basis) was CO – 11%,O – 7%andN – 82% 2 2 2 Calculate the composition of the fuel and the % excess air used. Teesside University Open Learning © Teesside University 2011 (Engineering)16 Solution Firstly, what do we mean by 'on a dry basis'?This is after any water vapour has been condensed and removed from the products of combustion (often known as flue gases).The actual products, as we have seen, will include water vapour but at the temperature of analysis any water vapour present generally would have cooled and condensed out of the sample.This is not a problem as we shall see.We will be able to determine the amount of water produced as part of our calculation. To solve this problem, we need to remember that volume % is the same as the mole % for gases at the same temperature and pressure. Let's assume there are 100 moles of dry flue gas.This means that the flue gas contains 11 moles (11%) of carbon dioxide, 7 moles (7%) of oxygen and 82 moles (82%) of nitrogen.The nitrogen takes no part in the combustion process and simply passes through the process unchanged, so 82 moles of nitrogen are therefore present in the initial air used to burn the fuel.Nitrogen forms 79% of the air, so82 × 100 0 the amount of air entering the process = 79 = 103.8 moles The oxygen entering must therefore be the difference between the moles of air and moles of nitrogen=103.8–82=21.8 moles (alternatively it is 21% of the air=0.21 × 103.8=21.8 moles). Oxygen leaves the process in the flue gases ascarbon dioxide (11 moles) and as unreacted oxygen (7 moles) and in water vapour, which has not yet been accounted for. Teesside University Open Learning © Teesside University 2011 (Engineering)17 But we know that 21.8 moles oxygen are entering and we have accounted for 18, thus the amount of water formed must account for the difference, i.e. 3.8 moles of oxygen must have been used to burn the hydrogen and form water.(Oxygen in must equal oxygen out–a material balance). According to the reaction equation for the production of water from hydrogen H+? 0.5O H O 2 (g) 2 (g) 2 (g) 1 mole 0.5 mole 1 mole 0.5 mole of oxygen requires 1 mole of hydrogen, so 3.8 moles oxygen will use 7.6 moles hydrogen.Thus the fuel contained 7.6 moles hydrogen based on 100 moles of dry flue gas. All the carbon dioxide must have come from the combustion of carbon in the fuel and since the equation for this is CO+?CO (s) 2 (g) 2 (g) 1 mole 1 mole 1 mole the 11 moles of carbon dioxidein the flue gas must have come from 11 moles of carbon in the feed. Thus the fuel contains 11 moles of carbon based on 100 moles of dry flue gas. On a mole basis, the fuel thus consists of 7.6 moles H and 11 moles C, or as a 2 mole percentage: 76 . 760 H = ×= 100 = 40.% 9 2 76.. + 11 18 6 C = 59.1% Teesside University Open Learning © Teesside University 2011 (Engineering)18 [This is approximately 60/40 on a mole basis, which means 6 atoms of C for every 4 molecules of H (8 atoms of H).If this was a single fuel and not a mix 2 we could say the formula of this fuel could be C H which is the formula for 6 8 cyclohexadiene, or C H which is propyne.] 3 4 On a relative mass basis (using mass=moles× molecular mass) 7.6 moles H (molecular mass of 2) = 15.2 g 2 11 moles carbon (atomic mass 12)=× 11 12= 132 g So, by mass, the fuel composition is15.2 % hydrogen = × 100 15.2 + 132 1520 == 10.% 3 147.2 % % carbon = 89.7% Since we have the actual amount of oxygen presentin the process (21.8 moles) and the amount actually used (the theoretical amount) to form the carbon dioxide and water in the process=11+3.8=14.8, actual – theoretical thus excess oxygen = × × 100 theoretical 21.– 8 14.8 =× 100 14.8 = 47.% 3 As a check:actual–theoretical=7 moles, which is the amount of oxygen given as present in the flue gas, confirming a balance. Teesside University Open Learning © Teesside University 2011 (Engineering)"