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In the event of an emergency shutdown of an engine (while in the air),

  • are there systems which prevent movement of turbines (brakes) in case of internal damage?

  • Or will that cause aerodynamic drag and therefore not practical?


Clarification:

I am thinking of a case where debris within the turbine will cause more damage (e.g., severing of hydraulic lines, electrical lines, etc.) if the turbine continues to spin. I know that if debris was to enter, the damage caused would already be catastrophic, but I’m curious if such emergency braking systems are in place and whether or not they are applied automatically.

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  • $\begingroup$ The idea I had in mind is if a fire broke out; presumably in the air; stopping everything ( immediately ) could (excluding the obvious, it needs to be running since it's in the air) the equipment could I guess get damaged? (I am not sure though) It would make sense if they would slowly shutdown instead of a total immediate stop; at least to me that is; I will follow this post; good question! $\endgroup$ Jan 18, 2023 at 6:59
  • $\begingroup$ If a fire breaks out, the fuel line to the engine will be shut down and the engine will stop to rotate "per inertia". To stop it with a brake would request a massime brake due to the huge inertia of the spinning parts and it would simply be a waste of weight and a source of an (extremely dangerous) failure case. Brakes are applied for example to helicopter's rotors but only to avoid their spinning on ground due to the wind and they can be applied only when the rotor has come to an almost complete stop. If you want, I could elaborate this comment in an answer putting some numbers in it. $\endgroup$
    – sophit
    Jan 18, 2023 at 7:16

2 Answers 2

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No. Assuming the rotating bits are undamaged, both the engine core and the fan section will windmill in the airflow. The benefit of stopping the engine spools is not worth the complexity, cost, weight and failure modes introduced.

In any case, you want the engine windmilling because that's one method to get it re-lit if you're above the maximum altitude for using the APU for starting, and for some reason you can't start by cross bleed (like say, double engine flameout at high altitude). There will some minimum speed to need to achieve to get the core spinning fast enough for a windmill relight.

If the engine flames out while at high power, "core lock" can happen, where the engine case shrinks faster than the compressor/turbine causing the blades to brush the walls of the case. If you get below some speed, it can allow the core to stop completely until the spool shrinks enough to release the blades. You may not be able to restart the engine even off the starter, air turbine starters making very little torque, until that happens.

Windmilling tendency depends on the strength of the ram air flow through the engine core, which depends on the compressor inlet configuration and the service loads of things like pumps driven from the core spool. With most engines, the engine will windmill at just about any speed, but on others, the core will stop at some low speed, and you will need to maintain some margin above that to keep N2 from going to zero to guarantee relight capability.

So, overall, a windmilling engine is desirable.

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As promised, here a couple of numbers (very simple equations, don't be scared).


are there systems which prevent movement of turbines (brakes).

A brake basically converts the kinetic energy possessed by whatever it brakes into heat.

In this specific case we want to brake the spinning part of a turbofan. The kinetic energy of something rotating is given by a formula very similar to the well known $½mV²$:

$E_k=½I\omega²$

where $\omega$ is the rotating speed and $I$ is the moment of inertia of body; $I$ depends not only on the mass of the body but also on its distribution around the rotating axis.

Unfortunately I couldn't find any value of $I$ for a real turbofan/jet (I suppose it's a confidential information), so we make a ballpark estimation of the rotating masses: going from the inlet to the outlet, we normally find the fan, several compressor stages and several turbine stages. Even if they have a more or less complicated shape, from a moment of inertia point of view each of those stages is basically a rotating disk; and when they are considered all together, they become a cylinder (or sort of). For a cylinder of radius $r$ and mass $m$, the moment of inertia is simply $I=½mr²$ and its kinetic energy is therefore:

$E_k=¼mr²\omega²$

Now, all of the stages previously introduced have obviously a different radius and at least two different rotating speeds so we just use an average value for them: we are not going to get a precise value for the energy but at least its order of magnitude should be fine.

Let's use then a modern PW1100, as found for example on the A320, making the following assumptions:

  • We consider the mass $m$ of the rotating parts as ⅓ of the total mass; ⅔ of non-rotating mass should be fine considering that for each rotating stage there's a stator stage plus the combustion chamber plus all the pipes, the outer case and other structural components; $m=2850/3=950kg$.
  • For $r$ we should take into account that the compressor stages get smaller, the turbine stages get bigger and that there's a big fan of 1m radius in front of everything; we just use an average value of some 0.3m.
  • the low-pressure stages rotate at 10'000rpm and the high pressure stages rotate at 20'000rpm; we just use the averaged value (of the squared, since in the equation for the $E_k$, $\omega$ is "²") i.e. $\omega=16000rpm=1650rad/s$.

Finally:

$E_k=¼ \times 950 \times 0.3² \times 1650²=60MJ$

Again, this is a very rough ballpark estimate, so the $6$ in front of "$0MJ$" is wrong but for sure we can say that the rotating parts of a modern turbofan posses a kinetic energy of some tens of millions of Joule.

Is it big? Well, at landing, the whole A320 has a kinetic energy of $½mV²=½\times 64000 \times 80²=205MJ$. Some of this energy is dissipated by the friction of the wheels with ground and some by the aerodynamic drag but most of it (some 80%) transforms in heat in the brakes. Comparing this value with our rough estimation, we've then obtained that the kinetic energy of a turbofan is more or less ¼ (let's round it a bit) of the kinetic energy of the whole landing airplane it is mounted on.

Since the A320 has four braked wheels, a brake on the shaft of a turbofan, would look something like this (source)

 A320 main landing gear

Being a part which would be used maybe once every some billions of flight hours, it would be pretty big, heavy and actually dangerous in case of failure.

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