"29 On a relative mass basis (using mass in kg=kilomoles× molecular mass) 19.2 kmol H (molecular mass of 2)=38.4 kg 2 9.6 kmol carbon (atomic mass 12)=9.6× 12=115.2 kg As a mass percentage 38.4 %%hydrogen = ×= 100 25 38.4 + 115.2 % carbon = 75% 5 There are twice as many kmol of H compared to carbon in the fuel 2 so we could say we have 1 C and 2× H , i.e CH .This would also 2 4 be confirmed on a mass basis (C=12, H =4,which gives a total 4 of 16). 12 =× 100= 75% Thus % C in CH 4 16 The fuel could be methane.If we assume this is the case, then the amount of methane is 9.6 kmol, the same as the carbon. (ii) The air:fuel ratio by volume for gases is same as by mole if at the same temperature and pressure. So there are 109.6 kmol air :9.6 kmol fuel=11.42:1. Teesside University Open Learning © Teesside University 2011 (Engineering)30 (iii) The excess air (which could be taken as the excess oxygen) = 3.8 kmol as this is present in the flue gas.The amount actually used (and therefore the theoretical amount since complete combustion) to form the carbon dioxide and water = 9.6 + 9.6 = 19.2 kmoland we calculated that 23.0 kmol were actually used in the process (which confirms the excess as 3.8 kmol). actual – theoretical Thus excess air (oxygen) =× 100 theoretical 23.– 0 19.2 2 = ×= 100 19.% 8 19.2 Putting these into a table to summarise: O required Flue gas (Wet) 2 Material In fuel/air (kmol) Moles kg (kmol× × molecular mass) (kmol) % C 9.6 115.2 9.6 9.6 (as CO ) 8.0 2 H 19.2 38.4 9.6 19.2 (as H O) 16.1 2 2 O 23.0 736 3.8 3.2 2 N 86.6 2424.8 86.6 72.7 2 Total 119.2 100.0 Teesside University Open Learning © Teesside University 2011 (Engineering)31 Total mass of fuel=+ 115.2 38.4= 153.6 kg (iv) Totalmass of air=+ 736 2424.8= 3160.8 kg 3160.8 The air:fu uel ratio by mass== 20.: 261 153.6 (v) The composition of the wet flue gas by volume (kmol) is shown in the table.It is found by dividing the kmol of each component by the total number of kmol and multiplying by 100. Teesside University Open Learning © Teesside University 2011 (Engineering)32 ________________________________________________________________________________________ SUMMARY ________________________________________________________________________________________ In this lesson we looked at how to balance equations for the complete combustion of selected fuels which contain carbon and hydrogen only.In these circumstances, the only products of combustion are carbon dioxide and water vapour.The balance consists of ensuring that the numbers of carbon, hydrogen and oxygen atoms on each side of the equation are the same by putting numbers in front of the oxygen reactant, the carbon dioxide and/or water as necessary. The balanced equation enables either the mass or volume (for gaseous fuels) of oxygen, which in theory will completely combust a given amount of fuel (theoretical oxygen requirement), to be determined.In most practical cases, however, this amount is insufficient to enable complete combustion as the oxygen is usually present in air (only 21% of whichis oxygen) and this makes mixing of fuel and air more difficult.The extra air supplied is known as excess air.This, when expressed as a percentage, could be either based on oxygen or air and is given by: actual – theoretical % excess air (oxygen) = × 100 theo oretical Calculations can be done to determine the amount of air required to burn a 3 fixed amount of fuel (1 m or 1 kg) and an air:fuel ratio determined either by volume or by mass.Usually air:fuel ratios are quoted for gases by volume and for solid and liquid fuels by mass. Teesside University Open Learning © Teesside University 2011 (Engineering)33 It is possible, by carrying out a materials balance based on flue gas (products of combustion) analysis, to determine: • the composition of fuels containing carbon and hydrogen only • the % excess air • air:fuel ratios by mass (and possibly by volume). Teesside University Open Learning © Teesside University 2011 (Engineering)"