Can Vmg exceed the V of a jet exhaust?
 

As a lowly PPL who occasionally sits behind a piston engine, a question about jet turbines has been bothering me: Is the velocity of the output ie exhaust of a jet engine equal to the aircraft's Vmg in still air once it is cruising? If it's not, then how does an aircraft achieve, say, Mach 2 with a jet exhaust slower than Mach 2? If it does, is it something to do with acceleration? Please excuse the ignorance!

A little bit rusty, and hopefully there will be some better answers to come, but here goes.

The aircraft can't fly faster than the exhaust velocity of the jet engine.

The thrust of the jet engine is mass flowrate x (exhaust velocity - inlet velocity)

If the aircraft moves faster than the exhaust you would get negative thrust, and the aircraft would go into reverse.

The exhaust gases must have some rearward component, relative to the ambient air the aircraft is flying through. In that case, an aircraft at mach 2, would indeed have a speed differential compared to it's exhaust gases, of greater than the aircraft's true airspeed. But be careful about speeds here. There's an aircraft at 600 ktas (mach 1?) at -57 deg C, and there's exhaust gases at 600 ktas (mach .8?) at +900 deg C.

Same as a piston/propellor combination. With respect to the air mass the aircraft is flying in, if some air isn't being sent backwards, there's no forward thrust.

So as well as exiting the nacelle at some speed, the exhaust air exits both compressed and hot, with those factors adding to thrust?

RJM,

renard already gave the basic answer and the basic formula.

A given mass of air per second comes in through the intake (at what is basically the airspeed) and gets ejected through the exhaust. So to get thrust (forward force) you have to accelerate that air backwards, i.e., it the jet velocity has to be higher than the airspeed.

(Note that "mass flow in" = "mass flow out" ; the mass of the fuel added is negligeable. It's the energy of the burning fuel that accelerates the air, plus the shape of the exhaust nozzle.)

"...the exhaust air exits compressed..."
Ideally, no. You let it expand in the nozzle to the outside atmospheric pressure, thereby accelerating it.
If the pressure at the end of the nozzle is still higher than the atmospheric pressure, the jet will expand further behind the nozzle, without contributing more thrust, so some of the energy in the jet is wasted.
In practice, unless you have a variable nozzle, the nozzle diameter is optimised for cruise, and you accept there will be some loss in the off-design case.

CJ

F=MA.......... Basic Physics.:ok:

Force (Thrust) = Mass (of the Air) X Acceleration (of the Air).

Thrust (Force) is what we need for sustained flight. As mentioned earlier, the Mass of air "in" is the same as the Mass of air "out", and is thus a constant (ignoring the negligible addition of mass due to fuel added). Thus, if we want Thrust, we must Accelerate the air to a higher speed than that when entering the engine.

Regards,

Old Smokey

Thanks all. You've answered all aspects of my question. :ok:

Theoretically, ignoring for example drag on the airframe and internal engine parts, you can fly a small bit faster than the exhaust velocity, due to the added mass of the burnt fuel. How much depends on the ratio of air and fuel masses. At some point the deceleration of the air mass and the acceleration of the fuel reach an equilibrium, and you don't accelerate any further.

At the extreme of this air/fuel ratio, when reducing the air intake down to zero, you get a rocket engine. Rocket engines can accelerate to virtually any speed (if the fuel wouldn't run out at some point), irrespective of the exhaust velocity.

The underlying physical principle is the law of conservation of impulse. Newton's second law, as outlined above, only follows from that.

In reality however, with all the various drag forces acting on the aircraft, and with the small amount of fuel compared with the air mass, you will always fly a fair bit slower than the exhaust velocity.

Just as well I read your post to the end before commenting....
You're right, of course.

Easier to understand, almost... you can use the "balloon" analogy. Nothing in, only "mass out".

What's not always stressed enough in the tale, is that for a jet engine it's very nearly all "mass in = mass out" (the added fuel mass being pretty well negliable), so the energy added by the fuel HAS to be converted into velocity, so the nozzle is a vital part of the system.

CJ

Hmm, I think not...
 

I think many of the posts above are incorrect (now thats asking for it!)

Consider the basic physics for a rocket - not quite the same but still a reaction motor. It is well understood that a rocket can only attain its own exhaust velocity if its full-to-empty mass ratio is equal to or better than "e" (2.718...), efficiencies aside. But if better than "e", it can and will exceed its own exhaust velocity and in fact needs to, to escape earth. This basic relationship does not depend on whether there is air present or not, other than the fact that rocket motor efficiency is degraded by the presence of external air pressure acting against the nozzle - the basic physics still applies.

Perhaps people are forgetting that what happens to the exhaust gases after leaving the nozzle is irrelevant.

This misunderstanding is what caused eminent scientists (including the Royal Observatory Astronomer Royal, no less) in the 1930s/40s to declare "Space travel is impossible". So you are in good company.

James,
I would say we simply shouldn't have dragged rocket engines into the discussion....
Rocket engines carry their own 'reaction mass' with them, which is then accelerated rearwards "from standstill" (relative to the vehicle).
Jet engines differ in that they get their 'reaction mass' from the outside (neglecting the tiny mass contribution of the fuel), coming in a the speed of the vehicle. So it has to be accelerated to a greater speed at the exhaust to provide a reaction (thrust). The thrust formula is fundamentally different.

CJ

Momentum, chaps
 

Force (thrust) is rate of change of momentum. That's the time derivative of the product of mass and velocity coming out versus going in.

Thrust = [d(mv)/dt at exhaust] - [d(mv)/dt at intake]

In the case of constant exhaust and intake velocity, this is renard's formula from post 2.

Thrust = (v_exhaust - v_intake) dm/dt

Fuel burnt adds a little to mass, bleed air takes a bit away.

-

For a rocket, the product of m and v at the intake is zero, and the change of momentum comes entirely from the exhaust speed and the burn rate of the fuel and oxidiser.

Thrust = (v_exhaust m/s) x (burn rate - Kg/s)

-

How fast you end up going (horizontally) is controlled just by drag.

If drag is very small, then a little bit of thrust is enough to reach a speed that can be much higher than v_exhaust: in fact the exhaust velocity hasn't got much to do with it, other than for efficiency.

In the limit that drag is very high you get a test rig, where there's a substantial static force on the pylon, but no motion (OK - with the exception of a tiny change in earth rotation).

CJ - I think most of us are well aware of the differences between rocket motors and jet turbines; specifcally in the fact that the rocket carries its own reaction mass with it while the jet turbine gathers it up as it goes. No problem.

But the point of referring to the rocket motor is that it is a reaction motor, like a jet. A jet creates thrust from the reaction of expelled mass, as does a rocket. The exhaust nozzle does not know if the reaction mass has come from a tank or from the air in front of it. The point of referring to rockets was to point out that there is no fundamental reason why a reaction motor cannot travel faster than its own exhaust velocity, which is what the original Poster asked.

If I recall correctly, even the the most powerful chemical rocket fuel pairs (hyrdrogen/flourine) cannot attain an exhaust velocity as high as that required for orbit (sorry, do not have the figures) and yet a single stage to orbit rocket is feasible and was being developed. That project was dropped not because the basic physics was wrong.

Furthermore, the notion that the reaction force is related to the airspeed implies that engine thrust diminishes as air speed increases - even during take off roll. This is clearly not the case.

Thank you to the original Poster - a good discussion!

For a jet engine, there is a fundamental reason, and that is that the reaction mass flow enters at the same speed as that of the 'reaction motor' (the aircraft) and has to be expelled at a higher velocity to create thrust.

I think what you've missed is that the mass flow into the engine rapidly increases with the airspeed at takeoff. More mass to accelerate, more thrust.

CJ

a bit confusing
 

if you regard the a/c at rest and the air moving past it (at the air speed of the a/c natch), to gain thrust (ie accelerate or just overcome drag to maintain same speed), each 'parcel' of air ingested must be accelerated, so it leaves the exhaust at a speed higher than the airspeed. So in the moving a/c reference, the exhaust has a rearwards velocity relative to the undisturbed air.

Sorry CJ but I don't buy your maths. Methinks thou does't confuse a moving and static frame of reference (or something).

I remain convinced that a turbine aircraft can fly faster than its exhaust gas velocity and can see no reason why not, once you abandon the intuitive feeling that the exhaust gas has some effect after it has left the engine. I am happy to be proved wrong and I was rather hoping a mechanical engineer or physicist might come to my rescue (where are they when you need one...??) But the arguments presented on this thread so far do not convince me I am wrong. As Poirot would say: "Ze truth will reveal itself"

As a matter of interest, I had a quick squiz at the jet turbine data sheets and amongst the heaps of technical data, I could find no mention of exhaust gas velocity. Surely if it were a limiting factor in the engines use, it would be specifed?

The other point to note is that high bypass turbofans have low speed (subsonic) gas flows (albeit at a high mass transfer) and yet a high bypass turbofan can fly at M0.85 (easily) and perhaps more - as far as I can see, the only limit to higher speeds is if there is enough thrust to overcome the steeply rising drag.

I am away from the net for a day or two, so I will not be participating in this interesting discussion for a while.

Sorry James, it's not my maths. The basic formula is already in post #2 (by renard). Otherwise Google is your friend, there's quite a lot about the subject on the the web, and clearer than I could write it.

The trouble starts when you begin describing it in words... everybody has his own way for "getting one's mind around it". After that, some find a moving reference frame easier, others a static one. The end result is the same.

Big fans (much like props) accelerate a lot of air a little, but they still accelerte it.

CJ

CJ - it sems to be just you and I left in this thread but I would really like the original poster to get the right information, since he/she posed the original question.

I have no problem with the maths in post #2 regarding mass flow and velocities. If you care to, just add a value of Mach 1 to all velocity variables on both sides of the equation (or any other velocity). Obviously, the equation still holds and Sir Isaac is happy. You may well end up with a negative exhaust velocity but all that means is that the exhaust gases are momentarily "chasing" the aircraft as they slow down - remember, they cannot affect the thrust measured at the nozzle after leaving the nozzle. By adding a velocity to the equation, all you are doing is changing from a static reference frame to a moving reference frame.

The key word is velocity, a vector quantity. There is nothing wrong with a negative velocity - it just signifies movement in the opposite direction, whereas a negative speed is undefined.

Consider for a moment the practical implications of your position: As soon as airspeed exceeds exhaust velocity, thrust goes negative (according to post #2) So in a steep dive, all our turbine driver needs to do is pull off the power to get thrust reversal! But his panel would have placards warning him of the extreme danger of throttling back in level flight, in case the dreaded negative thrust kicks in.. etc.etc.

I think awblains post says it all - it warrants a careful re-read. He/she clearly has a better grasp of the theoretical mechanics than I.

"If drag is very small, then a little bit of thrust is enough to reach a speed that can be much higher than v_exhaust: in fact the exhaust velocity hasn't got much to do with it, other than for efficiency."

To really prove this thing once and for all, does anyone out there know what the maximum exhaust velocity is of the exhaust gases in the engine powering the SR71 at M3? I would venture to say below M3 but of course we need the figures in m/sec, not Mach numbers.

http://www.me.com/ro/yaarpanjabi/Gal...12461771980001

See if that helps.

Cheers

A real life example
 

ypj, thanks but I could not open the link but am curious..

CJ - As you suggested, I looked a bit further on the net. Lets look at some approximate numbers to illustrate the point.

Take a JT9D - I found some Static sea level figures (not verified but believable):

Fan airflow: 1248 lb/sec @ 885 ft/sec (=270m/s)
Turbine exhaust: 247lb/sec@1,190 ft/sec (=363m/s)
Thrust 43,500 lbs.

Fan contribution to thrust: approx 3.5 x turbine exhaust contribution (by applying mass flows & velocities) i.e fan thrust is dominant.

Lets bolt this JT9D onto a heavy airliner and take off. We will apply the incorrect notion that forward airspeed must be subtracted from the engine exhaust velocity.

The captain figures out that he needs to rotate the plane at 130kt (67m/s). He also knows that his airspeed in the controlled airspace is 250kt (129m/s). He also has a barber pole on his ASI at 350kt (180m/s).

Does our captain know that when he reaches his rotate speed his trusty JT9s have already lost almost a quarter of the thrust they had when he first applied power at the start of his take off run (270m/s-67m/s)? And he has not even left the ground!

And does he know if he now loads his plane up so that Vr is increased to 175kt (90m/s) he will have lost a full third of his static thrust at a time when he really needs that thrust? Aha, easy answer - as your take off weight increases, REDUCE your rotate speed to make more thrust available!!

But wait, it gets worse. He accelerates to 250kts, maintaining (near) sea-level due to traffic congestion. Shucks, now the JT9D is only producing one half of its static thrust (270m/s-129m/s). And he has not even begun his climb yet. He asks, "Gee, with the air-resistance at 250kt, is there ANY surplus thrust left to climb with?

And could he get to barber-pole speed (180m/s), where his thrust will have now diminished to only one third of the static thrust?

The more I research this thing, the more I see it has been discussed - I think it is an old chestnut. The great Wiki has an entry which supports your view CJ but it is debated at length on the discussion tab. I think the vast majority are comfortable with the incorrect but intuitive view that forward airspeed directly reduces thrust. And, yes, I am ignoring the host of highly complex issues of air behaviour in the inlet ducts and other jet engine performance aspects about which I know nothing.