68 replies to this topic

[WPT]

Some years ago I posted a comment about the thrust distribution for the SR-71. Namely that at M3 cruise the inlet developed much more thrust than the engine itself. I had read about this in a book, "Modern Combat Aircraft Design" by Klaus Huenecke. The Concorde was also covered. Well, I didn't just have guns to the left of me and guns to the right of me, but guns all around me. Since I don't know much about fluid dynamics, all I could do in response was to post what Huenecke had written, which failed to convince. I folded. I wish now to unfolded. Below is a URL for a forum that covers the subject.

• http://www.pprune.or...ake-thrust.html
• Read post #107 (page 6) thru post #149 (page8)
• Post #144 has a link to a Rolls Royce book. Be sure to read Chapter 20
• Post #12 (page 1) has a diagram of the Concorde propulsion system that appeared in Huenecke's book

Hope this does the trick. WPT

[gruntguru]

A very enjoyable read. Learned heaps! Thanks WPT.

[Kelpiecross]

Very simple and elementary stuff - basic jet engine theory.

(Or maybe I should write WHAT!)

I prefer the point of view (expressed by some people) that it is the burning kerosene that does all the work - and leave it at that for the present.

I would have to add that I have never seen (at least at a first reading) such a unexpected and totally anti-intuitive result. Most of the thrust is produced before any fuel is burnt at all? As ever I am not saying I don't believe it - but is there a simpler, words-of one-syllable method of explaining this in a more commonsense, logical way?

Edited by Kelpiecross, 05 June 2013 - 13:29.

[WPT]

The engine be the prime mover, keeps the flow thru the system what it needs to be so the other parts can do their thing, too. Work the diffuser example. Huenecke in his book states that the 8% thrust thrust figure for the engine in the Concorde at M2.2 is the force felt by the engine mounts. WPT

[Magoo]

It's all in how you say it.

[Wuzak]

Huenecke in his book states that the 8% thrust thrust figure for the engine in the Concorde at M2.2 is the force felt by the engine mounts. WPT

Surely all the engine thrust is transferred through the engine mounts?

[Kelpiecross]

Surely all the engine thrust is transferred through the engine mounts?

I presume the statement means that the intake area that produces most of the thrust is not part of the engine - although I would imagine that if the intake area were attached to the engine all the thrust would be through the engine mounts.
This whole business of the intakes "sucking" the plane along is still a total puzzle to me.

[Paolo]

This whole business of the intakes "sucking" the plane along is still a total puzzle to me.

The intake "sucks" the plane along thanks to pressure distribution.
That peculiar pressure distribution happens because there is kerosene being burnt in the engine and accelerating air out of the nozzle, as Kelpiecross noted.
You shut down the engine -> pressure distribution changes -> no "sucking" anymore, drag instead.

[WPT]

In the Concorde neither the inlet or ejector nozzle are attached to the engine (same is true for the SR-71). Hunecke, IMHO, made the motor mount statement to make clear his meaning that the engine itself is only producing 8% of the total thrust at M2.2.

Kelpiecross: Think you would have no trouble in agreeing that if air flows over a wing surface, then the wing would develope lift. How is that any different in saying if air is flowing through a diffuser, then the diffuser developes thrust? Of course, without the engine to cause the flows, nothing happens. I found information on the B-70 (M3 at 70,000feet) and was able to do some rough calculations for its inlets (two inlets, each serving three engines). The subsonic diffuser produces mighty thrust. WPT

[bigleagueslider]

The J58-P4 engine used on the SR-71 had a variable geometry inlet nozzle that bypassed the intake airflow around the compressors at high mach speeds. It is very difficult to get a turbine engine to function with supersonic airflows. The airflow ahead of the compressor section must normally be slowed to subsonic by diffusion in the inlet duct. The exhaust flow must also be accelerated to a velocity greater than the local airflow at the nozzle discharge location, otherwise no net thrust force would result.

[Kelpiecross]

The exhaust flow must also be accelerated to a velocity greater than the local airflow at the nozzle discharge location, otherwise no net thrust force would result.
[/quote]

I am not sure that this statement is correct - but I would not wager any vital parts of my anatomy on that.
Considering a plane (or whatever) being propelled by a rocket engine or by firing projectiles out the back - the most efficient speed of the rocket gases/cannonballs is that of the plane - that is; the gases etc. are stationary after being ejected from the vehicle - all the kinetic energy being applied to the ejected gases or cannonballs etc. is given to the vehicle. If the gases are faster than the vehicle - some of the applied kinetic energy is wasted on the gases not the vehicle. Likewise ejected gases slower than the vehicle follow the vehicle along - again wasting kinetic energy on the exhaust.

[Greg Locock]

your anatomy is safe, good catch.

[gruntguru]

The exhaust flow must also be accelerated to a velocity greater than the local airflow at the nozzle discharge location, otherwise no net thrust force would result.


I am not sure that this statement is correct - but I would not wager any vital parts of my anatomy on that.
Considering a plane (or whatever) being propelled by a rocket engine or by firing projectiles out the back - the most efficient speed of the rocket gases/cannonballs is that of the plane - that is; the gases etc. are stationary after being ejected from the vehicle - all the kinetic energy being applied to the ejected gases or cannonballs etc. is given to the vehicle. If the gases are faster than the vehicle - some of the applied kinetic energy is wasted on the gases not the vehicle. Likewise ejected gases slower than the vehicle follow the vehicle along - again wasting kinetic energy on the exhaust.

You are right and wrong.
1. There must be some difference (however small) between the velocity of the gases exiting the engine, and the vehicle speed otherwise there will be no thrust.

2. For the case of a jet engine it is possible (in theory only) for the exhaust velocity to be matched to the vehicle velocity provided the intake velocity is higher than the vehicle velocity (the vehicle is being sucked along). Obviously this won't work for rockets.

[pugfan]

You are right and wrong.
1. There must be some difference (however small) between the velocity of the gases exiting the engine, and the vehicle speed otherwise there will be no thrust.

2. For the case of a jet engine it is possible (in theory only) for the exhaust velocity to be matched to the vehicle velocity provided the intake velocity is higher than the vehicle velocity (the vehicle is being sucked along). Obviously this won't work for rockets.

For part 1. Isn't there an increased mass however because fuel has been added?

[Kelpiecross]

You are right and wrong.
1. There must be some difference (however small) between the velocity of the gases exiting the engine, and the vehicle speed otherwise there will be no thrust.

2. For the case of a jet engine it is possible (in theory only) for the exhaust velocity to be matched to the vehicle velocity provided the intake velocity is higher than the vehicle velocity (the vehicle is being sucked along). Obviously this won't work for rockets.

On point 1. - I think there is some confusion here (possibly on my part) about velocities etc. relative to the engine and the vehicle etc. With a rocket engine, if the exhaust velocity relative to the engine is the same as the vehicle speed relative to its surroundings (thus giving the exhaust gases a zero velocity relative to the same reference frame as the vehicle) surely this gives the maximum propulsive efficiency and, I presume the maximum thrust. Your statement appears not to say this. In another topic on this forum (I forget which one) you recently pointed this out to me.

On point 2. - This idea of jet engines "sucking" themselves along (especially at supersonic speeds) is a concept I am still grappling with. Generally speaking things are not "sucked' along -the main example of this is the reverse situation to a deflating air balloon "rocket". If a container holding a vacuum has an orifice opened the inrushing air does not suck it along - similarly a hose underwater attached to the inlet of a pump does not suck itself along. So the idea that a jet engine can actually suck itself along is an odd concept to me at the moment.

[Zoe]

Rocket engines don't really work that way. It is basically the recoil of the masses the engine ejects; actually its energy, that generates the thrust.

Otherwise they couldn't be working in vacuum-

Zoe

[gruntguru]

Pug and KC.
You guys have made me think.

On the fuel mass - yes although it is usually neglected. On say a turbofan it may be two orders of magnitude less than the air mass.

Neglecting fuel - if the air enters and leaves the engine at the freestream velocity, there will be no thrust - you could replace the engine with a frictionless pipe and see the same result. There must be a momentum change ie the air leaving the engine must have a higher velocity relative to the engine than the air entering it.

For the rocket. Yes the exhaust only needs to be travelling backwards relative to the rocket - at any speed. This is because the exhaust mass was originally stationary relative to the rocklet before the engine "threw" it backwards thereby changing its momentum. (Unlike the jet engine exhaust which was originally part of the freestream.) EDIT. This would mean I was wrong about a jet engine being able to "suck" itself along. The intake airspeed (prior to any interaction with the engine) can only be at the freestream.

Edited by gruntguru, 05 July 2013 - 06:02.

[Canuck]

According to the 2nd link, in the notes about the ejector design "once again, the exit velocity must match the flight speed".

Edited to add:

In the post above, the difference in rocket vs jet operation is posited as being a result of the relative intake air speed differences to thrust (0 and equal). If you follow through the links, it's made clear that the intake air velocity is held relatively constant (.3 - .5 Mach) irrespective of flight speed. This implies that the intake air at the engine (so post-inlet) is slower than exhaust speed, thereby giving you your relative difference between in and out. Beats me how the thing taxis and takes off though if this is the theory of operation.

Edited by Canuck, 06 July 2013 - 01:13.

[bigleagueslider]

For part 1. Isn't there an increased mass however because fuel has been added?

Technically, it is correct that injecting fuel into an airflow passing through a duct will increase the total mass flow at the duct exit. But it likely won't result in any real thrust increase without combustion of the fuel. The thrust produced from the exhaust flow of a turbojet engine is the result of thermal energy being added to the airflow mass from the heat released by combustion of the fuel.

Here's an extract from wikipedia: "The rate of flow of fuel entering the engine is very small compared with the rate of flow of air. If the contribution of fuel to the nozzle gross thrust is ignored, the net thrust is: F_N = \dot{m}_{air} (v_e - v)

The velocity of the jet (ve) must exceed the true airspeed of the aircraft (v) if there is to be a net forward thrust on the aircraft. The velocity (ve) can be calculated thermodynamically based on adiabatic expansion."

Turbofan engines are different than turbojets. A turbofan is basically a ducted fan driven by a turboshaft. Liquid rocket engines are also different. They produce thrust by imparting momentum to a mass flow of fuel and oxidizer combustion by-products, and then expanding and ejecting that gas from a nozzle.

[gruntguru]

If you follow through the links, it's made clear that the intake air velocity is held relatively constant (.3 - .5 Mach) irrespective of flight speed. This implies that the intake air at the engine (so post-inlet) is slower than exhaust speed, thereby giving you your relative difference between in and out. Beats me how the thing taxis and takes off though if this is the theory of operation.

For thrust calculations the "engine" must be considered as including all components from the air entry point through to the final exit from the nozzle. By this definition the inlet air will always be travelling at the freestream velocity.

During supersonic flight the intake ducting slows the intake air before it enters the turbojet power unit. During taxiing and low speed flight the intake air is accelerated through the intake ducting - up to the speed required at the powerplant intake.

[Canuck]

So the note about exhaust velocity matching flight speed is incorrect then?

[gruntguru]

Yes - matched speed = zero thrust.

Efficiency improves as you approach matched speed but mass flow must be increased to maintain thrust levels. At a matched speed the mass flow required is infinity so clearly not feasible.

[Kelpiecross]

Yes - matched speed = zero thrust.

Efficiency improves as you approach matched speed but mass flow must be increased to maintain thrust levels. At a matched speed the mass flow required is infinity so clearly not feasible.

So as the plane speed approaches the exhaust speed the thrust approaches zero but the propulsive efficiency approaches 100%? And zero plane speed gives maximum thrust and zero efficiency?

[gruntguru]

maybe saying = zero acceleration is a better way to put it, otherwise you may lead people to think, that a plane will "fall out of the sky" if this condition occurs.

The situation is worse than zero acceleration. Zero thrust means nothing to counteract drag so the plane slows down (at which point the thrust will increase) however at zero thrust it is the same as having no engine at all.

This is why actual jet engines have exhaust velocity significantly higher than airspeed.

Edited by gruntguru, 08 July 2013 - 23:12.

[pugfan]

No doubt I'm missing something fundamental as I haven't really thought this through and existing designs don't have the sort of geometry this implies but can this not be thought of in terms of pressure?

A high pressure/low velocity exhaust at the nominal outlet balances the mass flow of a low pressure/high velocity inlet and results in a net forward force?

Edited by pugfan, 09 July 2013 - 04:45.

[gruntguru]

by virtue of burning fuel the exit max flow will always be higher then the free stream mass flow, so even if both are multiplied by the same velocity, there will be a resultant force (thrust).

Agreed but very small. The fuel mass flow is much less than the air mass flow and can reasonably be neglected in thrust calculations.

[bigleagueslider]

Agreed but very small. The fuel mass flow is much less than the air mass flow and can reasonably be neglected in thrust calculations.

I also agree about the minor contribution to thrust of simply adding a relatively small fuel mass to the total flow. Here's what Wikipedia has to say about it (if you accept it as a legitimate technical reference):

"The rate of flow of fuel entering the engine is very small compared with the rate of flow of air. If the contribution of fuel to the nozzle gross thrust is ignored, the net thrust is:
FN = m(dot)air x (Vj - V)
The speed of the jet Vj must exceed the true airspeed of the aircraft V if there is to be a net forward thrust on the airframe".

[saudoso]

If I center the reference system on the plane and make a quick momentum accounting I guess it is right. It seems to me that if Vj is smaller than V the aircraft is sucking air from the tailpipes and being pushed backwards.

[Greg Locock]

If I center the reference system on the plane and make a quick momentum accounting I guess it is right. It seems to me that if Vj is smaller than V the aircraft is sucking air from the tailpipes and being pushed backwards.

An astronaut steps out of an orbiting shuttle. His speed is 30000 mph, and rotates to face the direction of travel. He throws a spanner away from him, forward, at a relative speed of 1 mph. does his speed decrease or increase?

[saudoso]

An astronaut steps out of an orbiting shuttle. His speed is 30000 mph, and rotates to face the direction of travel. He throws a spanner away from him, forward, at a relative speed of 1 mph. does his speed decrease or increase?

It's not valid analogy since the spanner was travelling at 30000mph with him in the first place. He didn't collect it at 30000mph - that would have a completelly different result.

Your example is valid for a reaction engine. The jet engine collects air from one side and throws it to the other.

[Greg Locock]

yeah i'm doing this step by step.

So, he catches a spanner thrown to him from in front of him, at a relative speed of 0.5 mph, and then throws it behind hm at 1 mph.

That's all a jet does (ignoring the fuel mass), throws the air out the back faster than it comes in the front. Everything else is details.

[saudoso]

Ok, the 30000mph speed is worth nothig. Ditch it.

He picks the spanner at 0.5mph and throws it opposite at 1mph. There was a speed change in the oposti direction of the spanner travel direction.

So I guess we are on the same page?

[Greg Locock]

Yes we are, I didn't like the wording of the wiki quote, "The speed of the jet Vj must exceed the true airspeed of the aircraft V if there is to be a net forward thrust on the airframe"

in what frame of reference is Vj? As demonstrated by the spaceman example, V is actually irrelevant.

[saudoso]

On that example it's assumed that the inbound velocity of the air flow into the intake is equal to the forward velocity of the aricraft V.

[gruntguru]

Yes we are, I didn't like the wording of the wiki quote, "The speed of the jet Vj must exceed the true airspeed of the aircraft V if there is to be a net forward thrust on the airframe"

in what frame of reference is Vj? As demonstrated by the spaceman example, V is actually irrelevant.

The only absolute here is that the air exiting the jet must be travelling backwards relative to the air entering the jet (ie increase momentum in the direction opposite to travel. The air entering the jet is always travelling at V relative to the aircraft so Vj must be >V if there is to be thrust generated. To that extent V is relevent.

[Greg Locock]

...indeed, which is where the confusion about speeds vs velocities and frames of reference comes in. If they want to talk about speed, or magnitude of velocity, great, but if the aircraft has a +ve velocity then the exiting jet will always have a -ve velocity relative to the aircraft if the x axis points forward, or a smaller, and possible -ve, velocity in ground reference. In aircraft frame of reference the incoming air also has a -ve velocity, so the wiki is actually wrong if the frame of reference is the natural one of aircraft centric, forward motion +ve.

OTOH I routinely use -ve velocity for forward motion (because the x axis points towards the rear of the car) and in that case the jet exhaust speed is always > vehicle velocity.

Or I may be completely wrong, I'm off to the pub.

[saudoso]

Switching of the engine in fact will greatly increase drag and result in an even greater loss of speed.

Having no engine at all means having a hollow on it's place not blocking the air flow in any way.

Edited by saudoso, 12 July 2013 - 13:00.

[bigleagueslider]

An astronaut steps out of an orbiting shuttle. His speed is 30000 mph, and rotates to face the direction of travel. He throws a spanner away from him, forward, at a relative speed of 1 mph. does his speed decrease or increase?

First of all, the orbital velocity of the Space Shuttle in a typical low earth orbit (LEO) of around 125 miles up was approximately 17,500mph. The space shuttle did not have sufficient thrust capability to achieve an orbital velocity anywhere close to 30,000mph. As for a crewman conducting an extra-vehicular activity (EVA) deciding to throw (or more correctly "gently lob") a wrench forward into the direction of travel, releasing it at a relative velocity of 1mph, then due to the lack of any significant gravitational or aero effects, the force applied to the wrench by the crewman throwing it would need to be opposed by an equivalent opposing force acting on the crewman. One would have to take into account the large relative mass inertia difference between the wrench and crewman, as well as the resulting acceleration/deceleration that inertia force produced on the wrench or crewman.

The crewman would definitely slow down a tiny bit, while the wrench would also speed up a tiny bit. Of course, there is also the technicality of what constitutes a "direction of travel" in reference to the crewman. Since the orbiting crewman is travelling a roughly circular path, rather than a linear path.

Interesting topic.

[gruntguru]

I just didn't agree with the statement that equal velocity equals ZERO thrust and that this is the same as switching off the engine, that's all.

I agree that the fuel mass is small compared to the air mass, but small and zero is not the same - at least in my book.
As seen in the other thread, this at times can lead to some wrong conclusions and reasoning - infinitesimal small is not zero either.

I agree that fuel massflow is significant and must be considered for the hypothetical case where the jet exhaust is at the same velocity as the inlet.

For normal jet engine operation, neglecting fuel massflow will affect thrust calculations by perhaps a couple of percent and can therefore be neglected for any "back of the envelope" calculations.

Small is never zero but often negligible in the engineering "real world".

[gruntguru]

One would have to take into account the large relative mass inertia difference between the wrench and crewman, as well as the resulting acceleration/deceleration that inertia force produced on the wrench or crewman.

Put simply - the COG of the crewman-wrench system will maintain its previous velocity. So if the crewman weighs 100kg and the wrench 1kg, when the wrench is thrown at 1 mph (apologies for mixed units), the crewman will be moving at 1/100 mph in the opposite direction. (both velocities measured wrt the crewman-wrench COG).

The fact the crewman-wrench COG is travelling a curved path can be disregarded - it will continue to travel that path. Interestingly the absolute velocity of the crewman has reduced so he will move to a lower orbit and the wrench will move to a higher orbit. Eventually the crewman and the wrench will be on opposite sides of the earth!

Will their shared COG still be where it would have been, had the wrench not been thrown????

[Kelpiecross]

Put simply - the COG of the crewman-wrench system will maintain its previous velocity. So if the crewman weighs 100kg and the wrench 1kg, when the wrench is thrown at 1 mph (apologies for mixed units), the crewman will be moving at 1/100 mph in the opposite direction. (both velocities measured wrt the crewman-wrench COG).

The fact the crewman-wrench COG is travelling a curved path can be disregarded - it will continue to travel that path. Interestingly the absolute velocity of the crewman has reduced so he will move to a lower orbit and the wrench will move to a higher orbit. Eventually the crewman and the wrench will be on opposite sides of the earth!

Will their shared COG still be where it would have been, had the wrench not been thrown????

I suspect that they would still share the same COG. Presumably a rocket could be treated in the same way by considering the unchanged initial COG of rocket plus fuel. To extend this idea - a rocket which reached a nearby star would still have its COG at its starting point on the launch pad. The rocket has gone to a star but its COG hasn't gone anywhere. This seems to be an odd concept but I suspect it is true in theory.


On the "curved path" question - Uncle Albert would say that the spaceman and his bloody wrench think they are following straight paths. Space is curved? - what is curved? There is bloody nothing there. I have always thought that there must be an, at present, unknown mechanical, nuts-and-bolts alternative to concepts like "curved space". I hope the spaceman is not going to discover that he needs that wrench.

Edited by Kelpiecross, 15 July 2013 - 04:39.

[bigleagueslider]

Put simply - the COG of the crewman-wrench system will maintain its previous velocity. So if the crewman weighs 100kg and the wrench 1kg, when the wrench is thrown at 1 mph (apologies for mixed units), the crewman will be moving at 1/100 mph in the opposite direction. (both velocities measured wrt the crewman-wrench COG).

The fact the crewman-wrench COG is travelling a curved path can be disregarded - it will continue to travel that path. Interestingly the absolute velocity of the crewman has reduced so he will move to a lower orbit and the wrench will move to a higher orbit. Eventually the crewman and the wrench will be on opposite sides of the earth!

Will their shared COG still be where it would have been, had the wrench not been thrown????

In order for the wrench (which has a fixed mass and velocity) to move away from the crewman (who also has a fixed mass and velocity) requires that there is a transfer of momentum between the two bodies, and it must be in equilibrium if there is no added energy input. Technically, you would need to consider the fact that there is some small amount of energy contributed by the muscle contractions of the crewman's throwing arm.

As for the orbital degradation of the wrench or crewman in LEO, this would be entirely due to drag from the presence of a micro-atmosphere at this altitude. Consider that a satellite placed in GEO has no relative velocity difference with respect to the earth's surface, yet its orbit does not decay over time.

[gruntguru]

In order for the wrench (which has a fixed mass and velocity) to move away from the crewman (who also has a fixed mass and velocity) requires that there is a transfer of momentum between the two bodies, and it must be in equilibrium if there is no added energy input. Technically, you would need to consider the fact that there is some small amount of energy contributed by the muscle contractions of the crewman's throwing arm.

Only if you wish to analyse energy transfers and conservation of energy. Conservation of momentum applies regardless of energy transfers "within" the system (ie between the man and the wrench)

As for the orbital degradation of the wrench or crewman in LEO, this would be entirely due to drag from the presence of a micro-atmosphere at this altitude. Consider that a satellite placed in GEO has no relative velocity difference with respect to the earth's surface, yet its orbit does not decay over time.

I wasn't talking about orbital degradation. Consider the system I described without any friction.

My gut feeling (as stated by KC) is that the COG of the man-wrench system will continue to orbit the earth on the same path it did prior to throwing the wrench . . . . . or could it be that along with the change in altitude of the two bodies there is a small net energy transfer from the earth and its gravitational field?

[gruntguru]

I suspect that they would still share the same COG. Presumably a rocket could be treated in the same way by considering the unchanged initial COG of rocket plus fuel. To extend this idea - a rocket which reached a nearby star would still have its COG at its starting point on the launch pad. The rocket has gone to a star but its COG hasn't gone anywhere. This seems to be an odd concept but I suspect it is true in theory.

I suspect that your suspicion is correct! With the exception that during the initial phase of the flight, most of the momentum of the rocket exhaust is actually transferred to the Earth - so the COG of the Earth moves away from the rocket. The Earth must be considered part of the system when applying conservation of momentum during the launch and somewhat beyond (perhaps for as long as the exhaust gases continue to fall back to earth).

[saudoso]

Not exactly.

If the rocket was shot like a cannonball or all the burning happened inside a barrell the resulting COG of Rocket/Earth would be static.

But as the rocket will keep burning for a long time, releasing particles with diferent speeds and directions during a long time, the resulting CG of all this expelled mass (mess?) will be a really fast moving target.

[Kelpiecross]

If you specify a frame of reference like the rocket ship and its fuel only or the rocket, its fuel and the earth the COG for each FOR I suspect the COG is fixed no matter what complications or diversions you might consider.

[saudoso]

As long nothing else adds momentum to the system componets, what's got a very slight chance of happening in such a big system.

We can be pretty sure that the COG of the Universe still seats where the big bang singularity once was though

[Greg Locock]

As long nothing else adds momentum to the system componets, what's got a very slight chance of happening in such a big system.

We can be pretty sure that the COG of the Universe still seats where the big bang singularity once was though ;)

But is the universe spinning, and if so relative to what?

[saudoso]

But is the universe spinning, and if so relative to what?

Don't get me thinking about what's encompassing the Universe, and then what's around that other thing and so on. Pretty please?

[gruntguru]

But is the universe spinning, and if so relative to what?

If it is, we should be able to measure a resulting centrifugal force.

Relative to all the other universes of course.