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Combustion Chamber Essay

Combustion chambers or engines are thermal units where the chemical energy of liquid or gaseous fuel is converted to heat energy or any other related form of energy depending on the purpose for which it is needed (Bonnick 2008). The energy of the fuel (mainly hydrocarbons) is released by igniting the fuel in the presence of air (oxygen). A typical reaction in the chamber is given as CxHy + O2(from air) + N2(from air) + H2(from air) ? CO2 + H2O + other gases(e. g. CO, O2, H2, N2)….. (1) The equation above indicates that, energy is imputed from the fuel and is released out through the constituents of the combustion products.

Generally speaking, the combustion products or flue gases are primarily carbon dioxide and water vapour. However, incomplete combustion (due to inadequate supply of oxygen) may sometime lead to the formation of carbon monoxide (CO) as part of the flue constituents. Alternatively, the presence of more than the required amount of oxygen gives rise to a complete combustion (CO2) and any excess oxygen form part of the flue gases. Besides this, other inert gases in air apart from oxygen (mainly nitrogen and hydrogen) are present in the by products (Bonnick 2008).

For the experiment being reported, methane is used as the fuel. So, equation 1 above is likely to be of the form: CH4 + O2(from air) + N2(from air) + H2(from air)? CO2 + H2O + other gases(e. g. CO, O2, H2, N2)….. (2) However, if the masses of hydrogen, carbon monoxide and oxygen (in by products) and the heat energy contained in them are assumed to be negligible, then the equation reduces to: CH4 + O2(from air) + N2(from air) ? CO2 + H2O + N2……… (3) As expected, the laws of conservation of mass and energy hold for the equation (Bonnick 2008).

Therefore, a input-output heat balance is expected in the combustion unit. Input heat is supplied by the calorific contents of fuel and the air masses. For the fuel this is obtained as: Input energy from fuel = (mass flow rate of fuel) x (gross calorific value of the fuel)…. (4) The calorific value is provided. Also, Input energy from air = (mass flow rate of air) x (specific heat capacity of air) x (? T)…(5) ?T = (inlet air temperature – Tref )…… (6) Tref refers to a reference temperature which is usually taken as 25°C. The output heat are accounted for as:

(i) sensible heat loss due to the wet flue gases (that is, water vapour is inclusive). This is calculated as: sensible heat loss = mass x cp x (T – Tref) ……………(7) The mass is obtained using the principles of conservation of mass. The mass of the reacting materials (gas and air) should equal the mass of the wet flue gases (the total by products). T refers to the exit temperature of the flue gases and cp denotes the mean of the specific heat capacities of the flue gases (usually provided). (ii) heat loss due to the vaporization of water formed in the by products

= ((mass of the water vapour) x (specific heat capacity of water) x (T -Tref)) + ((mass of the water vapour) x (specific latent heat of vaporization of water))…(8) The water vapour mass is determined from the product of the stoichiometric ratio of the gas/water vapour in the combustion equation (= 16/36) and the mass of the gas (iii) heat lost to the unit cooling/working medium (water) = (mass of the medium) x (specific heat capacity of water) x (T1 – T2)…(9) T1 and T2 are the exit and inlet temperatures of the cooling medium respectively (iv) heat loss due to the unit surface cooling.

This is usually pre-estimated. (v) heat loss due to an incomplete combustion of fuel giving rise to carbon monoxide and hydrogen (can be taken as negligible depending on the prevailing conditions). (vi) other forms of losses depending on prevailing conditions Results The results obtained are as presented in the table below. See sample of calculations in the appendix. Table 1: HCC Observations – the results of the laboratory work Gas/ Air Flame Test 1 Test 2 Test 3 Test 4 Other Air Setting 2 2. 5 3 3. 5 Other Air Inlet temp oC 23 23 23 23 2. 3 Flue Gas Temp oC 816 774 745 772 730

Water Inlet Temp (T1) oC 14 13 13 14 13 Water Outlet Temp (T2) oC 50 49 48 48 43 Water Flow Rate g/s 182 183 181 180 183 Gas flow rate m3/hr 6. 00 6. 00 5. 81 5. 81 6. 32 Gas Flow Rate kg/hr 4. 08 4. 08 3. 95 3. 95 4. 29 Air Flow rate kg/hr 70. 18 70. 18 67. 91 67. 91 73. 86 Air factor (n=1+e) 0. 90 1. 19 1. 38 1. 43 0. 90 Efficiency 73. 7 68. 4 66. 5 66. 1 75. 0 % oxygen (dry) 0. 2 3. 4 5. 8 6. 3 0. 0 % CO2 (dry) 11. 8 9. 9 8. 6 8. 3 11. 8 % Excess air (100e) -10. 0 19. 4 37. 5 43. 1 -10. 0 Analysis and Discussion of Result The result are further analysed as presented in the table 2.

1 following. The table shows the heat balance of the combustion unit. The differences between the input and output values could be accounted for as other heat losses that were neglected. Besides this, table 2. 2 shows the corresponding percentages for all the heat inputs and outputs. Table 2. 1: Analysis of Results Gas/ Air Flame Test 1 Test 2 Test 3 Test 4 Other Air Setting 2 2. 5 3 3. 5 Other Heat input from fuel (kJ/hr) 212160 212160 205300 205300 223298 Heat input from combustion air (kJ/hr) 57404 54457 50727 52560 54066 Total heat input (kJ/hr) 269564 266617 256027 257860 277364

% sensible heat lost 23. 521 26. 551 29. 384 31. 584 20. 970 Sensible heat losses (kJ/hr) 63405 70788 75230 81443 58164 Latent heat losses due to H20 (kJ/hr) 4089. 1 4089. 1 3956. 9 3956. 9 4303. 7 Heat supplied/lost to the working fluid (water) (kJ/hr) 27446 27597 26537 25637 22998 Surface losses from system (kJ/hr) 1700 1700 1700 1700 1700 Other losses (kJ/hr) 172924 162442 148602 145123 190199 Table 2. 2 shows that there is no significant difference in the percentages of each of the heat losses as the air setting changes from 2 to 3. 5 (in steps of 0. 5).

For instance, comparing between air settings of 2. 0 and 2. 5 shows that the percentages of the heat losses are almost the same in all and the same trend is observed for others. The same also goes for the input percentages (or heat gained). Invariably, it can be established that though, the air settings may differ, the percentages are almost constant. This is mainly due to the value of the fuel/air weight ratio remaining at a constant value through out the experiment. In addition, the gas flow rate remains almost the same as well. A change in these values will result in variations in the percentages.

Moreover, the efficiencies of the combustion unit as a heating device is determined from the percentages of heat loss to the cooling medium. This, in other words, refers to the amount of heat used to raise the temperature of the cooling medium compared to the amount of heat available (input). Tables 2. 1 and 2. 2 also shows that these can be taken as fairly constant, only ranging between 8% and 10% for the different air setting values. The reason for this also lie in the constancy of the fuel/air ratio at 17. 2 Table 2. 2: Corresponding Percentages of Heat inputs and Outputs Gas/ Air Flame

Test 1 Test 2 Test 3 Test 4 Other Air Setting 2 2. 5 3 3. 5 Other Heat input from fuel (kJ/hr) 78. 70 79. 57 80. 19 79. 62 80. 51 Heat input from combustion air (kJ/hr) 21. 30 20. 43 19. 81 20. 38 19. 49 Total heat input (kJ/hr) 100. 00 100. 00 100. 00 100. 00 100. 00 Sensible heat losses (kJ/hr) 23. 52 26. 55 29. 38 31. 58 20. 97 Latent heat losses due to H20 (kJ/hr) 1. 52 1. 53 1. 55 1. 53 1. 55 Heat supplied/lost to the working fluid (water) (kJ/hr) 10. 18 10. 35 10. 36 9. 94 8. 29 Surface losses from system (kJ/hr) 0. 63 0. 64 0. 66 0. 66 0. 61 Other losses (kJ/hr) 64. 15 60. 93 58. 05 56.

29 68. 58 However, a different trend is observed comparing between the percentages of the oxygen and carbon dioxide constituents of the flue gases as obtained experimentally. These are as represented in table 3 below. The results show that the air factor increases as the burner factor (air setting) increases as expected. Similarly the amounts of oxygen increase in the same direction. This should be the obvious result since increasing air factor means increasing the amount of air (including oxygen available for combustion). Consequently, the amount of oxygen coming at the outlet would increase.

A reverse case is however observed for carbon dioxide constituents (figure 1). This is due to the fact that, with increased air setting, much air is available than may be required in accordance with the air/fuel ratio of 17. 2 (that is, as required by stoichiometry). The resulting effect produces a situation that it seem as if, the amount of gas to produce the carbon dioxide has been reduced. Consequently, the carbon dioxide obtained becomes increasingly limited in quantity. This also adds to the reason for increased oxygen percentage in the flue gas.

The values presented in table 3 can be further compared with the result obtained by running the program FLAME to determine the percentages of the flue gas composition theoretically. The theoretical result is as presented in table 4 below and drawn in the graph in figure 2. The comparison reveals that the trend does not follow the same direction. Figures 1 and 2 reflects this better. Looking at the figures, the percentage of oxygen increases as the air factor increases both for the experimental values (figure 1) and the theoretical values (figure 2).

However, the was a difference in the trend for carbon dioxide. The reason for this is not far fetched. The theoretical values could not simulate the real or actual situation as revealed by the experimental result. While the experimental results is a reflection of the fact that increasing the air factor means increasing the amount of air more than that needed by stoichiometry. The implications of this is revealed in the direction of the graph for carbon dioxide. This is what the theoretical program results is incapable of depicting.


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