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Also, what is the maximum fuel:air ratio possible in a combustion chamber of a jet engine?

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    $\begingroup$ Would you please provide some sources for your information? $\endgroup$
    – Jpe61
    Jan 25 at 22:58
  • $\begingroup$ I think this question should be rolled back, the latest edit of ditching the f/a ratio of 1/4-1/3 totally changes the question. As it stands now, this is a very good question, but it is not what Call Me Dosya asked originally. $\endgroup$
    – Jpe61
    Jan 28 at 18:00
  • $\begingroup$ If one would take a look at the %fuel/air combustible range for Jet A1 kerosene (Engineering Toolbox), it is around 0.7 to 5%. So 1/4 to 1/3 would be way too rich. $\endgroup$ Jan 29 at 2:24
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Edit: to honor my original answer, I shall keep it as it is. The question was edited so much I think it would have constituted a whole new question. In it's original form, the Q presented jet engine f/a ratio as 1/4 - 1/3 and compared it to the stoich ratio of 1/15.

Your numbers seem way off, except for the stoichiometric fuel/air ratio (for kerosene).

Please see the following NASA article: Beginner's Guide to Propulsion and EngineSim Fuel and Air Relationships Activity

A quote from there:

For jet engine combustion, the important parameter is the fuel/air ratio (f/a) which is the ratio of the mass of fuel to the mass of air being burned in the engine.

The stoichiometric ratio of 1/15 can also be written as approximately 0.067

The article further elaborates on the subject:

For example, at sea level an airliner taking off at 375 mph with the throttle at 100% has an f/a of .017

This 0.017 translates into about 1/59

No matter how you look at the ratio, mass or volume, in no case will a jet engine run at f/a of 1/4~1/3. The mixture will simply be too rich to ignite and burn.

Here is a link to a Wikipedia graph about the range of typical jet engine air:fuel ratios:

Combustion stability limits of aircraft gas turbine.svg

Please note the graph displays the ratio in reverse order from what was used before in this answer, ranging from ~25/1 to ~180/1.

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We have to remember air is only 20% oxygen. When you write the stoichimetric formula for the chemical reaction (using oxygen) on a molar basis, you get the combustion ratio by converting the fuel and oxidizer from moles to mass.

To get the combustion ratio using air, you divide the oxygen mass by its fractional proportion in air.

Interestingly, the stoichimetric fuel/oxygen ratio (by mass) is around 1:3. 3 divided by 0.21 gives us around 15 for kerosene and air.

Turbojets burn fuel in combustion "cans" within the flammable fuel/air percentage limits and use the heat from that combustion to warm more air passing through the the core.

Bypass fans can make these ratios even higher. For example, the Ge 90 has a bypass ratio of 9 to 1. Of the 10% of the total air mass flow going through the core, its air to fuel ratio is around 33 to 1, much higher than stoichiometric.

The total air to fuel ratio (by mass) is 330 to 1!

One can see why fuel combustion "cans" were invented, and why fan jets are becoming more popular.

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  • $\begingroup$ When calculating a/f ratio, you can not take into account the bypass air. It is the same as if one would calculate the a/f ratio of a propeller acft and count into it the air the propeller is moving 😉 $\endgroup$
    – Jpe61
    Jan 28 at 18:06
  • $\begingroup$ @Jpe61 that is true. The total air moved is interesting. Prop thrust may be more related to its lift that "air moved". More to write about. $\endgroup$ Jan 28 at 20:33
  • $\begingroup$ Lift is pretty much the air moved. $\endgroup$
    – Jpe61
    Jan 29 at 10:32

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