Question

I understand that a stoichiometric mixture contains a balanced mixture of air and fuel and both fuel and air are completed consumed. For example, the ideal stoichiometric mixture for propane is approximately 4.01% of the fuel which is equivalent of 23.91:1 ARF by mass.

I also understand each fuel has its own explosive limit. As an example of Propane, the lower explosive limit is 2.1% (lean mixture) and upper explosive limit is 9.5% (rich mixture).

The explosive pressure (or energy) will be larger with 4.01% of fuel mixture compared with 2.1% of fuel mixture assuming the mixture is in a defined container with no fuel saturation. Following this pattern, if the gas mixture reaches to its rich state, fuel level at 9.5%, would the explosive pressure (or energy) be greater than the stoichiometric mixture? Can this be calculated?

Answer 1

Yes, the explosive pressure would be greater than the stoichiometric mixture if the gas mixture reaches its rich state at 9.5% fuel level. This can be calculated using the ideal gas law, which states that for a given amount of gas at a constant temperature and pressure, the volume is directly proportional to the number of moles of gas.

Step-by-step explanation:

Yes, the explosive pressure (or energy) would be greater than the stoichiometric mixture if the gas mixture reaches its rich state at 9.5% fuel level. This can be calculated using the ideal gas law, which states that for a given amount of gas at a constant temperature and pressure, the volume is directly proportional to the number of moles of gas.

In this case, the moles of fuel (propane) in the mixture will be higher in the rich state compared to the stoichiometric mixture, resulting in a larger explosive pressure. However, it's worth noting that explosive pressure also depends on factors like the container and ignition source.