"24 ________________________________________________________________________________________ SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________ 1. Write balanced equations for the complete combustionof the following fuels: (i) ethyne (acetylene) C H 2 2 (g) (ii) ethane C H 2 6 (g) (iii) fuel oil (take as C H ) 20 42 (l) (iv) propene C H 3 6 2. For each of the above reactions determine the theoretical amount of oxygen required for complete combustionof: (i) 1 kg of fuel 3 (ii) 1 m of fuel (gaseous fuels only) at the same temperature and pressure. Data: Atomic mass ofC=12, H=1andO=16. Teesside University Open Learning © Teesside University 2011 (Engineering)25 3. The flue gas leaving a combustion process for a gaseous fuel containing carbon and hydrogen onlyhas the following analysis on a dry basis: 86.6% N , 9.6% CO and 3.8% O 2 2 2 Calculate: (i) the composition of the fuel on a kmol and mass basis.Can you suggest a formula for the fuel? (ii) the air:fuel ratio by volume (iii) the % excess air used (iv) the air:fuel ratio by mass (v) the wet flue gas composition. (Base your answers on 100 kmol of dry flue gas.) Teesside University Open Learning © Teesside University 2011 (Engineering)26 ________________________________________________________________________________________ ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________ 1. (i) CH+? 25 . O 2CO+ H O 2 2 (g) 2 (g) 2 (g) 2 (g) (ii) CH + 35 . O?+23 CO H O 26(g) 2 2 (g) 2 (g) 2 (g) (iii) CH+? 30.C 5O 20CO + 21H O 20 42 (l) 2 (g) 2 (g) 2 (g) (iv) CH+? 45 . O 3CO+ 3H O 3 6 (g) 2 (g) 2 (g) 2( 2 g) 2. For the mass of oxygen, you need to determine the relative masses according to the formulae and stoichiometry of the equation and then adjust to 1 kg of fuel. C H+? 2.5O 2CO+ H O 2 2 (g) 2 (g) 2 (g) 2 (g) 21×+ 2 2×1=26 g g g 25 .×= 32 80 80 13kg ×= 1 .08 kg 26 33 12m .5 m C H+? 3.5O 2CO+ 3H O 2 6 (g) 2 (g) 2 (g) 2 (g) 21×+ 2 6×1= 30 335 .×= 32 112 112 13kg ×= 1 .73 kg 30 33 13m .5 m C H+? 30.5O 20CO+ 21H O 20 42 (l) 2 (g) 2 (g) 2 g 20×+ 12 42××=13 282 0.5×32= 976 976 13kg ×= 1 .46 kg 282 As fuel oil is a liquid we cannot work out the relative volumes from the stoichiometry. Teesside University Open Learning © Teesside University 2011 (Engineering)27 C H+? 4.5O 3CO+ 3H O 3 6 (g) 2 (g) 2 (g) 2 (g) 31×+ 2 6×1= 42 445 .×= 32 144 144 13kg ×= 1 .43 kg 42 33 14m .5 m One point to note from these answers is that the masses and volumes of oxygen required do not have a simple correlation.The ethyne with the lowest mass of oxygenhas the least amount of oxygen in terms of stoichiometry (2.5).However, ethane with the highest relative mass of oxygen has the next loweststoichiometric amount of oxygen (3.5).This is due to the differing relative amount of hydrogen, a lighter substance, which requires more oxygen in mass terms compared to what it needs in volume. 3. (i) In 100 kmol of flue gas there are 9.6 kmol of carbon dioxide, 3.8 kmol of oxygen and 86.6 kmol of nitrogen.86.6 kmol of nitrogen must come from the air used to burn the fuel.Nitrogen forms 79% of the air, so 86.6 the amount of air entering the process =× 100= 109.6 kmol 79 Oxygen present in this air=109.6–86.6 = 23 kmol (21% of 109.6) Oxygen leaves the process in carbon dioxide (9.6 kmol) and as unreacted oxygen (3.8 kmol), a total of 13.4 kmol, and in water vapour. Thus, the amount of oxygen used to form water must account for the difference between in and out, i.e. 23–13.4=9.6 kmol oxygen. Teesside University Open Learning © Teesside University 2011 (Engineering)28 According to the reaction equation H+? 0.5O H O 2 (g) 2 (g) 2 (g) 1 kmol 0.5 kmol 1 kmol Thus the oxygen will require 2× 9.6= 19.2 kmol hydrogen and form 9.6 kmol of water. All the carbon dioxide must have come from the carbon in the fuel and since the equation is CO+?CO (s) 2 (g) 2 (g) 1 kmol 1 kmol 1 kmol the 9.6 kmol of carbon dioxidein the flue gas must have come from 9.6 kmol of carbon in the feed. Thus the fuel contains 9.6 kmol of carbon. On a kmol basis the fuel thus consists of 19.2 kmol H and 9.6 kmol C. 2 As a mole percentage 19.2 H = ×= 100 66.% 67 2 19.. 2 + 9 6 C = 33.3% Teesside University Open Learning © Teesside University 2011 (Engineering)"