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.

Convergence?
 

I think the discussion so far is getting close. Any incorrectness in the discussion so far appears to be based on terminology and modulated to account for intuition.

The gas entering the engine must leave the engine faster: unless there is acceleration there isn't any forward thrust. The exhaust jet must move backwards compared with the swallowed air. All viewers agree: people on the ground, any people floating in the air, people on the plane.

That does means that the exhaust jet has to travel in the opposite direction to the aircraft as far as the undisturbed local air is concerned, although not necessarily at any particular speed. (Thanks to the following post for prompting the edit). Viewed from the undisturbed air, the exhaust gas must move backwards to produce a forward thrust. That's not so for a rocket.

-

The extreme case of exhaust speed (efficiency) is probably the space shuttle main engine: mass flow rate at liftoff is 1000 lb/s (from a ~500s burn and ~500,000lb of fuel/oxidiser per engine); quoted thrust is 480,000 lbs/2.1 Meganewtons. 500 Kg/s mass ejected and 2.1 Meganewtons thrust implies an exhaust speed of 2,100,000/500 = 4200 m/s, which isn't going to change much with height. However, orbital speed is about 7500 m/s. So, above some point in the climb, the exhaust still moves in the same direction as the shuttle compared with the earth, by about 3 km/s when the fuel is cutoff.

-

For low-speed thought experiments, you need to be careful when quoting numbers, because the speed that air hits the fan is faster than the pitot airspeed from the nose: the airflow has already been sucked faster when it enters the nacelle.

High school physics....
 

To keep the jet aircraft moving at a constant velocity, you need a force which balances the drag force.

That force is obtained from the engine exhaust.

F = Ma

That is, the force (thrust) developed by the engine depends on the mass of air it accelerates, and the size of that acceleration.

The force generated by the engine is in the forwards direction, so the acceleration of the air must be in the reverse direction.

So the air coming out of the back of the engine is moving faster than the air coming in the front.

Relative to the engine, the speed of the air coming in the front is the aircraft's airspeed; so the air going out the back has been accelerated relative to that velocity; so the air coming out of the engine is moving faster than the aircraft's forward speed.

For a rocket, the situation is completely different - you're just throwing mass out of the back to generate the force, and you don't care about its velocity (because it's starting velocity in the rocket's frame was zero, so any non-zero exit velocity means that the gases have been accelerated, meaning a forward thrust has been generated).

Not hard to understand, surely?

Out of the mouth of babes....
In this case I'm strictly referring to the post title... "high school physics".
Well done, Bryce!

Go and stand on a skateboard with a bag of sand.
Throw it backwards, and you'll move forwards. Rocket engine.

Now stand on your skateboard and have somebody throw a bag of sand towards you. Catch it and you'll move backwards. Put your own energy into it and throw it backwards faster than it arrived. You'll move forwards. Jet engine.

CJ

Out of interest I ran the figures for the CF6 through NASAs educational software and it came up with the following
It should be noted that the air meeting the compressor face is not at free stream velocity (speed of the aircraft). The inlet is designed to slow the airflow so that it is presented to the compressor at a speed of .5M or less, to avoid choking the compressor. The ram drag above, is the drag associated with slowing down the free stream air as the air is brought inside the inlet. Even the SR-71 at 3.2M has the air delivered to the compressor at approx .5M.

Of interest also is the J-58 installed in the SR-71. At 3.2M, 54% of the thrust is provided by the differential pressure between the internal and external surfaces of the inlet spike. Of the remainder, 17% is provided by the engine and 29% by the ejector (the afterburner, which is in ram jet mode at this speed - air being tapped off the 4th stage compressor, diverted around the turbine, and injected into the afterburner)

I've tried reformatting with little luck - hope you can make sense of it.

OK, I am convinced now! The post from Brian Abraham above shows quite clearly how a forward speed increase at sea level directly and signifcantly reduces net thrust, which concurs with the principle that exhaust speed cannot be less than forward speed. So I am happy to admit I have learned something, which in retrospect I should have been able to figure out.

Thank you to those posters who so patiently took the trouble to explain it and to those who were able to resist the temptation to be condescending.

Most importantly, I thank the original poster for initiating the discussion and I hope he/she enjoyed the postings.

Ok, so now that we have got that sorted out, let’s look a bit more closely at the propelling nozzle.

Most texts tell us that the nozzle produces forward thrust by accelerating the air rearwards.

But if we examine it closely we will find that the aerodynamic force on the nozzle is actually trying to tear it off the jet pipe and push it rearwards.

Anyone who does not believe this can try the following simple experiment.
Buy an ice cream cone and throw away the ice cream (OK you can eat it if you really must)

Now bite off the pointed end to leave a convergent nozzle.

Put the narrow end between your lips and blow through it.

You will find that a slow warm airstream comes out of it. (because it is a divergent duct)

Now put the wide end between your lips and blow through it.

This time you will get a faster colder airstream. (because it is now a convergent duct)

The above results are exactly what Bernouli would lead us to expect.


Now release your grip on the cone a little bit and blow really hard until you no longer hold the nozzle.

If the nozzle is producing thrust it will force its way into your mouth.

But it will actually fly out and away from you.

This is because the aerodynamic force on the nozzle is drag acting downstream.

So if the force on the nozzle is drag, how does it increase the thrust?

This question is not unrelated to the subject of this thread.

No ! (or is this a wind up ??)
 

Pressure forces air out of the cone. It is that pressure acting on the cone which propels it away. In a jet, what ever force that accelerates the air must have a reaction, ie thrust. For a rocket you integrate the pressure over the internal space, same would work for anything else, as it must unless you want an argument with Mr Newton.

Ummm... Keith - without trying to be offensive: your question, though stated reasonably enough, is not actually very reasonable.

You say 'most texts state ...' then go on to discount the information in those texts.

Maybe you meant to say 'I don't really understand this, please explain', but it reads as 'the texts are wrong - explain that!'.

You're not (I expect) denying that there is thrust (I mean, we've all seen that jet aeroplanes fly, right?) - just questioning where in the engine it appears. Which surfaces, that is, are acted upon by the exhaust to propel the aircraft?

I don't know. I would send you back to the texts you mention - if they're decent books, they'll tell you, or tell you which other book to read. From my reading, Cumpsty and Kerrbrock's books are decent starting points.

If you want a guess, based on having read a bit about how gas turbines work (but by no means being an expert of any kind) - any compressor fan (large effect in high bypass engines), plus the nozzle (small effect in high bypass engines).

Mr Optimistic

I have no argument with Mr Bernouli.

And as far as I can see I have no argumnet with anything that you have said.

Agreed

Agreed.
But it is an aerodyanmic force acting rearwards so we (usually) call it drag.


Agreed, provided the acceleration is rearward.

Agreed.

Now let's get back to what I actually said.

Many people believe that the force acting on the nozzle is thrust. They are wrong it is actually drag.

It is made up of friction forces, pressure forces and inertial forces acting on the forward facing (internal) surfaces of the nozzle. But the overall force acting on the nozzle is in a rearward direction.

You are correct in saying that the thrust is the Newton3 reaction to the pressure that is accelerating the air through the nozzle. But this thrust force is not acting on the nozzle.

Now the question becomes, "So if the thrust isn't acting on the nozzle, what is it acting on"?

BRYCEM
I'm not saying that the texts are wrong. Some cover this aspect of the subject and some do not.

I'm saying that many people draw the wrong conclusions from the texts. The majority of people walking away after completing a JAR ATPL Theory course (and a good many other courses) probably think that the force on the propelling nozzle is thrust.

I could of course go and read more books, but my purpose in asking the question is not get an answer. It is to get the readers to think a bit more about the subject.

I could of course go and read more books, but my purpose in asking the question is not get an answer. It is to get the readers to think a bit more about the subject. Today 13:12
I could of course go and read more books, but my purpose in asking the question is not get an answer. It is to get the readers to think a bit more about the subject. Today 13:12
Keith - some people (not me, obviously) might construe that as you 'wasting their time', and general clever-dickness.

If you already know the answer to the question, please supply it; otherwise, what exactly are you trying to achieve, except education by irritation?

On modern (high bypass) engines, the bulk of the thrust is obtained at the fan.

I'm done with this, BTW.

Some people first define what they're talking about. A con-di nozzle above Mach 1 usually produces at least some of the thrust.

On Concorde (like the SR-71 quoted earlier) most of he thrust came from... yes, the intakes.

On the SR-71, above Mach 3 under certain circumstances, the engine itself actually produced drag, evidenced by it moving backwards on the suspension points.

CJ

It's a long time ago but if my memory hasn't faded too much are we thinking about a choked nozzle and an increase of pressure behind the nozzle and therefore the "thrust" acting on the surface area of the nozzle itself?

It's past midnight here so I'll sleep on it and get back to this in the morning.

FE Hoppy,

Most of the maths just deals with a jet engine as a black pipe (rather than a black box :) ) with a hole at the front and a hole at the back. The slightly more subtle maths adds front and back surface and local pressure at the "interface" into the equation.

We now are getting into the internal dynamics.... I'll leave that to the experts. About the only thing I'm familiar with in that field is the classic rocket engine with a con-di nozzle with Mach 1 in the throat. Took me enough time to get my mind round it at the time, when actually "playing" with real ones.

CJ

Thrust mechanism
 

Origin of most of the thrust in subsonic engines: the fan bearings.
The fan blades accelerate the air backwards, and Sir Isaac pulls the fan blades forward, dragging the aircraft along with them.

In a rocket/turbojet/ramjet, it's the combustion chamber pressure, pushing on the closed front end of the chamber, reduced by viscous forces as the exhaust streams along the nozzle, and in some cases a turbine.

Thrust from shuttle engine again: about 400,000lb. The chamber pressure quoted is about 2800 psi, the throat of the exit of the chamber is roughly 24 inches across (about 430 square inches), which would give 1,200,000lb: so the price of stabilizing (by slowing and expanding) the flow out of the nozzle seems to be 800,000lb of thrust. The nozzle is thus pulled off the engine with 800,000lb of force, while the combustion chamber is pushed up into the engine by 1,200,000lb, giving a net 400,000lb. This suggests that the link between the chamber and the nozzle need to be tougher than that between the engine and its mount. I think that makes sense. Rocket scientists?

Taking the simple approach - the expanding burning gases will take the path of least resistance, which in this case is out the rear of the engine, since the front is 'blocked' by the highly compressed air entering the engine. The difference in pressure at the nozzle exit (and fan-bypass exhaust) is the thrust produced.

The amount of fuel burnt is proportional to the additional energy imparted to the airflow (less internal friction loss, etc).

-GY

So if the force on the nozzle is drag, how does it increase the thrust?

I think where Keith is going is to look at

you integrate the pressure over the internal space

A common presentation is to look at the pressure forces throughout the engine. Especially for the FJ folk, a lot of the net thrust comes from careful and clever intake design (the airflow has to be slowed down to subsonic flow at the engine intake face) and then there are bits and pieces throughout the engine where local net forces are forward or aft.

The bit coming out the back is only part of the equation ...

BRYCEM
If you manage to control your emotions for a moment you will find that over-dosing on grump is clouding your vision.

Your statement that

is ture but not relevant to the question

We are talking about the nozzle at the back end of the jet pipe.

By the time the air reaches this nozzle it has already passed through the low pressure turbine so its effect on the fan is almost zero (back pressure effects only).


Cristiaan J
We are not talking about a condi nozzle, just a plain old convergent one.

FE Hoppy
If we were taliking about a condi-nozzle then some of the thrust would be acting on that part of the nozzle that is downstream of the throat.

Awblain.
Your statement that

is along the right lines.

JT
You are of course correct (as usual).

But I think that your statement about

probably leaves many readers none the wiser.



The best answers (those from which we learn the most) are often not the ones that we get from books, but the ones that we get from within our own heads. Spending a little bit of time pondering this type of question can often be very useful (even if the questions are sometimes rather irritating).

But I think that your statement about


Quote:
integrating the pressures......
probably leaves many readers none the wiser.


Keith's wisdom supersedes my writing with the brain only partly engaged ..

Folk should recall that fluid flows are about pressures. Pressure gradients and deltas generate fluid flow. Static and flowing fluids generate forces on things like engine bits (and wings, tails, fuselages, etc.) by exerting pressure on surfaces.

Pressure acting over a surface (area) results in a force. Depending on the net orientation of the bit of surface in question (in, say, an engine) the force associated with the fluid pressure will have a net forward or aft direction (and, mostly, a lot laterally .. with which we are not terribly interested).

When one adds up all the bits of forces so calculated, one gets a net thrust (forward or aft).

"Integration" is a mathematics buzzword which really just means "adding up all the little bits and pieces" in the calculation. It comes from the integral calculus which is a really neat way of doing a lot of this stuff but it's not necessary for the pilot folk to have any competence in the mathematics per se.

Thanks, John.... you've now probably confused even more of the readers even more....

Integrating from intake to exhaust is a nice idea but, more often than not, you don't have enough pressure data to do more than an approximation.

To me, T = m* x deltaV does it every time, with the small refinements where needed.

Otherwise, I just like looking at shock diamonds, or at the plume during a shuttle launch, when the air pressure becomes less than the pressure at the nozzle end.

CJ

hmm #2
 

The high pressure in an engine or jet would tend to push the back end off: nothing to do with drag force on the nozzle surface, just pressure acting on the rear surface. Nozzle turns pressure into flow speed. Blow up a balloon, turn it around so it faces away from you and let go of the untied end. Would it fly away from you owing to drag over the aperture or into your face because of the thrust ? Put a weak joint between the nozzle of a rocket motor and the case and see what happens (owing to the pressure inside the case) - from half a mile away that is. And the adiabatic expansion or compression on flow through a cone would cause no discernable heating or cooling (convergent at subsonic speeds would in any case heat). Don't need de laval nozzle theory here. NB look at most rocket plumes, they bulge out ie exit pressure is still above ambient. Small drag losses in a nozzle I suspect are just accounted for by an 'engineering factor' in practice.