1
$\begingroup$

How does it relates to stoichiometry ? What are the possible extinction accidents related to air/fuel mixture ratio ?

$\endgroup$
  • $\begingroup$ Do you have a specific engine in mind? In general it can be pretty broad. There are both lean burn (excess oxygen) and rich burn (excess fuel) combustors out there. Even for a specific engine the air/fuel ratio can vary depending on operating condition. $\endgroup$ – Daniel K Oct 11 '19 at 2:02
  • $\begingroup$ @DanielKiracofe The engine I have in mind is the SNECMA Turbomeca Larzac 04-C6 en.wikipedia.org/wiki/SNECMA_Turbomeca_Larzac $\endgroup$ – Arnaud BUBBLE Oct 16 '19 at 7:58
0
$\begingroup$

This cannot be answered specifically, the stoichiometric fuel/air ratio (FAR) is based on burning all oxygen, while for cooling purposes the airflow is generally much higher; therefore the FAR will be lower than the stoichiometric FAR. Note that you do not want to burn fuel at stoichiometric conditions as this produces the most nitrogen oxides. Fuel is being burnt rich and then quickly burnt lean.

Note that the FAR is depending on the speed, altitude and power setting (and the type of engine). To quantify the FAR, you would need to model the engine and simulate the performance. As an example (very simple to show the effect) a turbine exit temperature sweep for different altitudes and Mach numbers is shown below for a simple turbojet engine (the engine linked in the comments of the question is practically a turbojet engine considering the very low bypass ratio). You see that the FAR is variable (top graph is the burner exit FAR, bottom is the stoichiometric FAR):

Turbojet FAR calculations using Gas turbine Simulation Program GSP


I cannot address the second part of the question regarding the accident rate, that does not seem to be relevant.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.