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turbine, the air is mixed with fuel and burned at constant pressure. The chemical bonds of the fuel atoms are destroyed and released energy heats up the exhaust gases. The default fuel in the microturbine is natural gas, even though the microturbine can be modified to run on swamp gas, kerosene, propane and other fuels. Natural gas does not consist of one substance. It is a mixture, where the main components are four different hydrocarbon fuels and nitrogen and carbon dioxide. The exact mass percentage varies for each mixture, but a common type of natural gas can have the following composition: Methane CH4 Ethane C2H6 Propane C3H8 N-Butane C4H10 Nitrogen N2 Carbon dioxide CO2 81% 7.9 % 4.2 % 4.7 % 1 % 1.2 % Table 1: Mass percentage of a common natural gas mixture To get the right chemical mixture after combustion, stoichiometric reaction formulas are used. We assume that the existing nitrogen and the carbon dioxide in the air and in the fuel as well as the water in the air are all inert gases, i.e. they do not react with anything (no NOx) or contribute to the chemical reactions. We also assume complete combustion with an abundance of air, i.e. there are no remains of the fuel in the products. The stoichiometric relationship gives: molar mass M, i.e. the mass for one mole. Then we can rewrite the equation to: CH4 +2O2 →CO2 +2H2O (5.8.1) The formula above shows the relationship in molar quantities. To get a mass flow balance we want to change the units. To convert the equation above to mass flow units we use the substances’ (5.8.2) The combustion occurs with an abundance of air, i.e. more air than needed according to the stoichiometric equations. This results in the existence of air in the products as well as in the reactants. Now we can put up equations for the mass flows of the different components in the flue gas leaving the combustion chamber. MMM CH4 +2⋅ O2 O2 → CO2 CO2 +2⋅ H2O H2O MCH MCH MCH 4 4 4 M m&CO ,out =m&CO ,in + CO2 ⋅m&fuel,CH +m&fuel,CO 22M42 CH 4 M m&H O,out =m&H O,in +2⋅ H2O ⋅m&fuel,CH 22M4 m&O ,out =m&O ,in −2⋅ O2 22M4 CH 4 M CH 4 ⋅m&fuel,CH m&N2,out =m&N2,in +m&N2,fuel 39 (5.8.3) (5.8.4) (5.8.5) (5.8.6) The total mass balance equation is then: m&out =m&in +m&fuel If the other hydrocarbon fuels are used, their individual contributions are added to the mass flow equations of the products. (5.8.7)PDF Image | Modelling of Microturbine Systems
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