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Combustion
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Combustion is defined as rapid oxidation accompanied by heat and usually light.
Deflagration is defined as a subsonic combustion that propagates through a gas or along the surface of an explosive at a rapid rate driven by the transfer of heat^{[1]}
Detonation is defined as a supersonic combustion through a medium before it has been reached by the flame front^{[2]}
In combustion, a fuel + an oxidizer come together to produce some products + light + heat.
Contents
Combustion reaction[edit]
In a combustion reaction, some "fuel" will react with air and release chemical energy and produce CO_{2} and H_{2}O as by-products. During the process of combustion, air typically oxidizes with fuel, but in essence pure oxygen or any oxidizing agent may also contribute. When examining combustion reactions it is essential to know what is oxidizing with the fuel. It is important to recognize the constituents that air is composed of. For the most part air is composed of 21% oxygen, 78% nitrogen, and 1% of other constituents by volume. For calculations we can say 21% oxygen and 79% nitrogen. Since we have a ratio of 79 moles nitrogen to 21 moles oxygen we use (O_{2}+3.76N_{2}) for air. The internal combustion engine became popular during the industrial revolution (1860's) and has been the main contributor of atmospheric CO_{2}. Anthropogenic carbon dioxide emissions has brought atmospheric concentrations of CO_{2} to approximately 406 ppm (December 2017). Because carbon dioxide is one of the leading greenhouse gases, there has been an emergent need of more sustainable energy sources. It may be necessary to weigh our different options of fuel use while we transition into clean, sustainable energy because it is not currently economically viable to retrofit our current energy infrastructure while we are so dependent on fossil fuels. That being said, we can measure amounts carbon dioxide released into the atmosphere for the combustion of different fuels and the amount of energy it would create to help us reduce our emissions of CO_{2} during this transition.
Theoretical air is the oxygen required for complete combustion to take place. However, in practice the air used is 2-3x that of theoretical air. As a final note, dew point is the temperature where water vapor begins to condense. This value is found by calculating the partial pressure of the water vapor. In practice, it is important to maintain a temperature above the dew point.
The standard [math]PV=NRT[/math] equation can be rephrased as [math]P_{wv}=N_{wv}RT/V[/math] to determine the partial pressure of water vapor. From here, the equation can be written as [math]P_{wv}=P_{tot}(NP_{wv}/NP_{tot})[/math], where P_{wv} represents the partial pressure of the water vapor, N_{wv} represents the moles present in the water vapor, R is a constant, T is the temperature, and V is the volume.
Heat from combustion can be found from the equation [math]Q=H_{p}-H_{R}[/math], where H_{P} is the enthalpy of the products and H_{R} IS the enthalpy of the reactants.
Commonly used combustible fuels[edit]
Fuel | Chemical formula |
---|---|
Gasoline(Octane) | C_{8}H_{18} |
Diesel(average formula) | C_{12}H_{23} |
Ethanol | C_{2}H_{6}O |
Propane | C_{3}H_{8} |
Butane | C_{4}H_{10} |
Wood | C_{6}H_{12}O_{6} |
Coal(simple) | C_{5}H_{4}O |
Natural Gas(average formula) | C_{1.2}H_{4} |
Balanced combustion equations for common fuels[edit]
(All balanced equations assume no leftover oxygen and combustion did not result in the formation of NO_{x})
Gasoline(Octane): C_{8}H_{18} + 12.5(O_{2} + 3.76N_{2}) [math]\to[/math] 8CO_{2} + 9H_{2}O + 47N_{2}
Diesel(average formula): C_{12}H_{23} + 17.75(O_{2} + 3.76N_{2}) [math]\to[/math] 12CO_{2} + 11.5H_{2}O + 66.74N_{2}
Ethanol: C_{2}H_{6}O + 3(O_{2} + 3.76N_{2}) [math]\to[/math] 2CO_{2} + 3H_{2}O + 11.28N_{2}
Propane: C_{3}H_{8} + 5(O_{2} + 3.76N_{2}) [math]\to[/math] 3CO_{2} + 4H_{2}O + 18.8N_{2}
Butane: C_{4}H_{10} + 6.5(O_{2} + 3.76N_{2}) [math]\to[/math] 4CO_{2} + 5H_{2}O + 24.44N_{2}
Wood: C_{6}H_{12}O_{6} + 6(O_{2} + 3.76N_{2}) [math]\to[/math] 6CO_{2} + 6H_{2}O + 22.56N_{2}
Coal: C_{5}H_{4}O + 5.5(O_{2} + 3.76N_{2}) [math]\to[/math] 5CO_{2} + 2H_{2}O + 20.68N_{2}
Natural Gas: C_{1.2}H_{4} + 2.2(O_{2} + 3.76N_{2}) [math]\to[/math] 1.2CO_{2} + 2H_{2}O + 8.272N_{2}
Air/Fuel Ratio (AF)[edit]
The Air/Fuel Ratio or AF is the ratio of mass of air to mass of fuel:
AF= m_{air}/m_{fuel}
This is found by finding the product of the molar weight of components and their coefficient in the balanced chemical formula for combustion, and summing these components to find the molar mass of air. The components in air are usually O_{2} and N_{2}.
The same process is done to find the molar mass of the fuel.
Example:
C_{2}H_{4} + 5(O_{2} + 3.76N_{2}) [math]\to[/math] 2CO_{2} + 2H_{2}O + 18.8 N_{2}
AF = m_{air}/m_{fuel} = (5*32g/mol + 5*3.76*28g/mol)/ (2*12g/mol+4*1g/mol) = 300 g /28 g = 10.7
Carbon Dioxide Example[edit]
To determine how much carbon dioxide is released from a combustion reaction, it is necessary to know the molecular mass of what is being combusted and to have a balanced chemical equation. In finding this information, you are determining the CO_{2} intensity of a certain material (i.e., how much CO_{2} is released from combustion per unit material.
Find the amount of CO_{2} produced in grams from the combustion of 1 gram of wood.
Chemical formula for wood: C_{6}H_{12}O_{6}, with a molar weight of 12(6) + 12(1) + 6(16) = 180 grams per mol of wood.
CO_{2} has a molar weight of (12 + (16 X 2) = 44 grams/mol.
Combustion reaction assuming pure oxygen environment
C_{6}H_{12}O_{6} + O_{2} [math]\to[/math] CO_{2} + H_{2}O
After balancing
C_{6}H_{12}O_{6} + 6O_{2} [math]\to[/math] 6CO_{2} + 6H_{2}O
From the balanced equation: 6_{moles CO2} X 44 grams/mol = 264 gCO_{2}
How much CO_{2} is produced by the combustion 1 tonne of wood?
From here, all that is needed is a ratio between the masses of the reactants and the masses of what you are calculating for.
[math]{{180g\over264g}={10^6g\over{X}}}[/math]
Solving for x provides how much carbon dioxide is produced in the combustion.