"19 Putting these into a table to summarise: O required Flue gas 2 Material In fuel/air (moles) (Wet) moles Moles g (moles× × molecular mass) C 11 132 (11× 12) 11 11.0 (as CO ) 2 H 7.6 15.2 (7.6× 2) 3.8 7.6 (as H O) 2 2 O 21.8 697.6 (21.8× 32) 7.0 2 N 82 2296 (82× 28) 82 2 We cannot determine an air:fuel ratio by volume in this case because we do not know how the carbon and hydrogen in the fuel are combined or even if they are combined.We suggested earlier that they could be combined as propyne, a gas, or as cyclohexadiene, a liquid. However, we can do it on a mass basis. We have a total mass of fuel of 132+15.2=147.2 g and a total mass of air of 2296+697.6=2993.6 g. 2993.6 × 1 For 1 kg of fuel the amount of air required would be= 20.34 kg 147.2 giving an air:fuel ratio by mass of 20.34:1. When we know the compound in the fuel is a gas we usually quote air:fuel ratios in volume terms and for solids and liquids we quote air:fuel ratios by mass. Teesside University Open Learning © Teesside University 2011 (Engineering)20 Try the followingexample for yourself. A gaseous fuel consists of a compound containing only carbon and hydrogen.After it is completely burnt, the analysis of the products of combustion (on a dry basis) is CO – 9%, O – 6%, N – 85% 2 2 2 Calculate the composition of the fuel and the % excess air used. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 (Engineering)21 In 100 moles of flue gas there are 9 moles of carbon dioxide, 6 moles of oxygen and 85 moles of nitrogen. The nitrogen takes no part in the combustion process and simply passes through the process unchanged so 85 moles of nitrogen are present in the initial air used to burn the fuel.Nitrogen forms 79% of the air, so85 × 10 00 the amount of air entering the process = = 107.6 moles 79 The oxygen entering must therefore be the difference between the moles of air and moles of nitrogen=107.6–85=22.6 moles (21% of 107.6). Oxygen leaves the process in carbon dioxide (9 moles) and as unreacted oxygen (6 moles) and in water vapour, not yet accounted for. But we know that 22.6 moles oxygen are entering and we have accounted for 15, thus the amount of water formed must account for the difference, i.e. 7.6 moles of oxygen. According to the reaction equation 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 7.6 moles oxygen will require 15.2 moles hydrogen. Thus the fuel contained 15.2 moles hydrogen. All the carbon dioxide must have come from the carbon in the fuel and since the equation isCO+?CO (s) 2 (g) 2 (g) 1 mole 1 mole 1 mole the 9 moles of carbon dioxidein the flue gas must have come from 9 moles of carbon in the feed. Thus the fuel contains 9 moles of carbon. Teesside University Open Learning © Teesside University 2011 (Engineering)22 On a mole basis the fuel thus consists of 15.2 moles H and 9 moles C. 2 As a mole percentage 15.2 H = ×= 100 62.% 8 2 15.2 + 9 C = 37.2% (This is approximately C H and I can tell you that this indicates that this is a fuel mix as 9 30 no single substance can have that amount of hydrogen associated with the carbon.It actually approximates to a 50% mix of methane CH and ethane C H but this is beyond the 4 2 6 scope ofthis course!!) On a relative mass basis (using mass=moles× molecular mass) 15.2 moles H (molecular mass of 2) = 30.4 g 2 9 moles C (atomic mass 12)=× 9 12= 108 g As a mass percentage 30.4 3040 % hydrogen = ×= 100 = 21.97% % 30.4 + 108 138.4 % carbon = 78.03% The excess air (which is taken as the excess oxygen)=6 moles as this is present in the flue gas.The amount actually used (and therefore the theoretical amount as there are no other materials produced and all the fuel is burnt) to form the carbon dioxide and water =9+7.6=16.6 molesactual – theoretical exces air (oxygen) = × 100 Thus theoretical t 6 =× 100= 36.% 14 16.6 Teesside University Open Learning © Teesside University 2011 (Engineering)23 Putting these into a table to summarise: O required Flue gas 2 Material In fuel/air (moles) (Wet) moles Moles g (moles× × molecular mass) C 9 108 (9× 12) 9 9.0 (as CO ) 2 H 15.2 30.4 (15.2× 2) 7.6 15.2 (as H O) 2 2 O 22.6 723.2 (22.6× 32) 6.0 2 N 85 2380 (85× 28) 85 2 The fuel to air ratio by mass in this case is: Total fuel=138.4 g and total air=3103.2 g 3103.2 air:fuel by mass== 22.: 421 138.4 We have now completed the lesson.Test your understanding by attempting the Self-Assessment Questions which follow. Teesside University Open Learning © Teesside University 2011 (Engineering)"