I’m sure that I must be missing something simple here, but this has always confused me. It seems to me that the term “thrust reverser” is a misnomer. Perhaps my definition of thrust is too narrow or perhaps my understanding of thrust reversers is lacking. I couldn’t find anything helpful in a rather extensive Google session.
I understand how a turbofan engine works (I think). When a plane lands the engines are brought to idle, the thrust reversers are deployed, and the engines are spooled up in power. (I am assuming clam shell type reversers here; I realize that there are other types.) With the reversers deployed let’s say the engine is taken up to 10,000 lbs of thrust. Let’s look at just one engine here.
It seems to me that there is now 10,000 lbs of forward thrust from the ejected mass of air from the engine. This mass of air then hits the thrust reverser, deflecting the mass of air forward. Now it seems that there is 10,000 lbs of reverse thrust applied at that point. So now we have a net thrust of zero (10,000 lbs forward balanced by the 10,000 lbs reverse thrust). And it seems like there is 20,000 lbs of “force” simply straining the connection between the engine and the thrust reverser. Finally that high speed mass of forward moving air hits the relatively low speed slipstream air. This seems to me that this simply causes a whole lot of aerodynamic drag which helps to slow the plane.
So my questions are:
Is my understanding of how thrust reversers work accurate?
Am I wrong about the braking being caused by “drag” rather than “thrust”?
That exhaust stream has to go somewhere. The clamshell reverser sends it in a slightly forward angle so there is a net reverse thrust that is a fraction of the original thrust of the engine.
rsa your mechanics are correct but there is one slight flaw in your logic. The “ejected mass of air” from the exhaust never acts on anything except the few inches of air between the exhaust and the clamshell. It is effectively contained and still a part of the engine until directed forward and away from the engine by the thrust reverser.
As proof of this, I can (and do) back my airplane away from the gate using reverse thrust. The available thrust is much less than forward thrust, but it is definitely reverse thrust and not just drag.
Thanks pilot141. If I had thought about it more, I would have realized that I have been on airplanes many times when the thrust reversers were used to taxi “in reverse”. So obviously calling it drag is not correct. I was making the distinction on the basis that air is compressible so that once the exhaust is uncontained it should not be called thrust. I guess I’ll go with calling it thrust with some significant efficiency losses.
Maybe a fluid dynamics expert will be along to explain in detail.
Well how about this: I have a high-pressure water hose.Well I don’t actually, but it’d be cool if I did. If I spray it someplace, I get blown backwards. If the hose is coiled up in a circle, this still happens to me.
Basically, until it leaves the engine, I say you can consider your high-velocity air to be part of a closed system, and thus ignore any odd directions it happens to take before it leaves the aircraft. If it leaves going back, you go forwards. If it leaves forwards, you go back.
Oh yeah, and it really does provide a lot of rear thrust. I remember reading here some advice - staying calm in a crisis is imperative, but there’s two times you’re in a hurry in a jet: explosive decompression, and uncommanded thrust reverser deployment.
I took a basic course in fluid mechanics a couple of semesters ago. When you’re looking at streams of moving fluids, there are two things to keep in mind:
Force is equal to the mass flow rate times the change in velocity (F = m[sub]dot[/sub][symbol]D[/symbol]v)
All internal forces are balanced, and can be neglected.
A jet engine produces thrust equal to m(v[sub]out[/sub]-v[sub]in[/sub]). In normal flight, both v[sub]out[/sub] and v[sub]in[/sub] are in the same direction, so only the magnitude of the velocities has to be considered. The force reaction is exerted within the engine and is transmitted to the plane, pushing it forward.
After landing, the sum of the v[sub]out[/sub] vectors from the top and bottom of the deployed clamshells (as you can see in Johnny L.A.'s photo) points forward, and is greater than v[sub]in[/sub]. Therefore, the resulting reaction acts opposite to the direction of motion, slowing the plane down.
