3
$\begingroup$

Why isn't fuel injected before the compressor or intake chamber?

The combustion chamber gets a lot of design and clever thinking to be able to mix fuel and compressed air in a short distance/space. This adds more weight and length to the overall engine.

I understand that changing that way of injecting fuel would need a new design and engineering to make it happen, e.g. to prevent blade corrosion, but nevertheless, could it be done?

If someone or a company did it already, and it didn't work, what were the reasons?

$\endgroup$
  • 1
    $\begingroup$ Suck-Squeeze-Bang-Blow, if you inject before the squeeze, you may have problems with proper fuel atomization. $\endgroup$ – Ron Beyer Mar 29 '17 at 21:14
  • $\begingroup$ @RonBeyer I don't think the question is about four-stroke piston engines. $\endgroup$ – David Richerby Mar 30 '17 at 9:35
  • $\begingroup$ In the original Brayton engine, a piston compressed a volume of pre-mixed air and gas... $\endgroup$ – xxavier Mar 30 '17 at 9:53
  • 4
    $\begingroup$ @DavidRicherby I'm not talking about one either, the process is the same for jet engines and two-stoke engines too. in a jet engine the fan sucks, the compressor squeezes, the igniter bangs, and the exhaust blows. $\endgroup$ – Ron Beyer Mar 30 '17 at 11:53
  • 1
    $\begingroup$ “if you inject before the squeeze, you may have problems with proper fuel atomization”, except that's precisely what spark-ignited (gasoline) engines do! $\endgroup$ – Jan Hudec Mar 30 '17 at 18:24
2
$\begingroup$

tl;dr The flame would progress upstream into the compressor until it and damage it, fatally disrupt the carefully balanced flow and stall the engine, or both.

long answer: In a turbine engine, there is a continuous flame where injected fuel is ignited by the already burning fuel downstream. Now you can't match the flow speed and the flame propagation speed exactly, because a small disruption could then blow out the engine and because both depend on the power setting in non-trivial way.

So in normal operation, the flame can propagate upstream until it gets to the point where the fuel is not yet sufficiently mixed with air to burn and there it stays. If you injected the fuel before the compressor, the flame would progress into it and as it would engulf the last stage, it would damage it and it would also increase the pressure beyond what the preceding stages could maintain, which would stall the compressor and cause the engine to stop, either immediately due to lack of oxygen, or by causing more damage with each stall cycle.

Note that spark-ignition (gasoline and LPG and CNG) engines usually do inject fuel before compression, but it limits their compression ratio to around 12:1 with high-octane gasoline, 14:1 with clever tricks. If the compression ratio was higher, the fuel-air mixture would ignite from the compression heat before the piston reaches the top-dead-centre and act against the rotation. In contrast compression-ignition (Diesel) engines have compression ratio at least 14:1 and inject the fuel only when it should start burning.

While early turbines had rather low compression ratios, the newest generation is getting into the range comparable with Diesel engines where the fuel might ignite due to the compression heat alone—they still use sparks for ignition, because the starter won't build up as high pressures in them, but it clearly shows they can't inject the fuel before it should start burning.

$\endgroup$
  • 1
    $\begingroup$ Interesting... In the original Brayton gas engine, that worked at constant pressure too, but with pistons, a stack of wire screens was necessary to prevent the propagation of the flame from the burning charge back to the 'receiver', that was full of a mixture of air and gas... $\endgroup$ – xxavier Mar 30 '17 at 20:10
  • 1
    $\begingroup$ @xxavier, I didn't know that, but it confirms the explanation. A fine wire mesh does stop flame propagation, but it would also cause significant resistance to the air flow. Not injecting the fuel before the point it should start burning is way simpler. $\endgroup$ – Jan Hudec Mar 31 '17 at 12:52

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.