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Assume a two-stream problem having one portion of the boundary the fuel stream with fuel mass fraction =, and another portion of the boundary the oxidizer stream with oxidizer mass fraction =,. For example, if the oxidizer stream is air and the fuel stream contains only the fuel, then Y O , O = 0.232 {\displaystyle Y_{O,O}=0.232} and Y F , F ...
Thus, to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction.
The stoichiometric proportions of the above equation is given by 1 kg of CH 4 + (64/16) kg of O 2 → (1+ 64/16) kg of products The transport equations for the fuel and oxygen mass fractions are + = (.) + [2]
For example, oxygen makes up about 8 / 9 of the mass of any sample of pure water, while hydrogen makes up the remaining 1 / 9 of the mass: the mass of two elements in a compound are always in the same ratio. Along with the law of multiple proportions, the law of definite proportions forms the basis of stoichiometry. [1]
The combustion of a stoichiometric mixture of fuel and oxidizer (e.g. two moles of hydrogen and one mole of oxygen) in a steel container at 25 °C (77 °F) is initiated by an ignition device and the reactions allowed to complete. When hydrogen and oxygen react during combustion, water vapor is produced.
where w C, w H, w S, w O refer to the mass fraction of each element in the fuel oil, sulfur burning to SO 2, and AFR mass refers to the air-fuel ratio in mass units. For 1 kg of fuel oil containing 86.1% C, 13.6% H, 0.2% O, and 0.1% S the stoichiometric mass of air is 14.56 kg, so AFR = 14.56. The combustion product mass is then 15.56 kg.
The stoichiometric concentration of methane in oxygen is therefore 1/(1+2), which is 33 percent. Any stoichiometric mixture of methane and oxygen will lie on the straight line between pure nitrogen (and zero percent methane) and 33 percent methane (and 67 percent oxygen) – this is shown as the red stoichiometric line.
Mass fraction can also be expressed, with a denominator of 100, as percentage by mass (in commercial contexts often called percentage by weight, abbreviated wt.% or % w/w; see mass versus weight). It is one way of expressing the composition of a mixture in a dimensionless size ; mole fraction (percentage by moles , mol%) and volume fraction ...