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Exhaust Speed At Back Of Turbofan Engine  
User currently offlinesmartt1982 From United Kingdom, joined Nov 2007, 225 posts, RR: 0
Posted (4 years 4 months 1 week 5 days 22 hours ago) and read 8865 times:

Hiya

I am a bit confused about something. It was my understanding that turbofan engines are used on airliners due to their exhaust speed being more in tune with the speed that the actual aircraft flies. I was also of the understanding that the cone at the back of some jet engines is to allow the exhaust to slow down and increase to ambient pressure so that it is as close to atmosphere pressure as possible.

Is it that the speed of sound due to the exhaust temp is so high it does not reach supersonic at all?, when it reaches the ambient outside temp will it then be supersonic due to the temp drop?. Silly question but of the hot exhaust that is being ejected where is the thrust being produced from, is it the velocity or the pressure that is causing the equal and opp reaction.

Like on concorde at supersonic, I don't believe the afterburners were engaged all the time, was it simply a convergent/divergent duct that kept them supersonic?, presumambly it was the exhaust velocity that was causing the reaction

Silly probably very obvious answer to the question but thanks

steve

15 replies: All unread, jump to last
 
User currently offlinerwessel From United States of America, joined Jan 2007, 2368 posts, RR: 2
Reply 1, posted (4 years 4 months 1 week 5 days 14 hours ago) and read 8746 times:
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Basically you don't want a supersonic jet flow from the engine for either efficiency or noise reasons. Basically the overall exhaust velocity is high subsonic on essentially all subsonic jets. The core flow is often hot enough to be mildly supersonic relative to ambient, but is locally subsonic. The mixing of that with ambient air is a significant source of noise.

The majority of the exhaust stream (the bypass flow) is not heated any meaningful amount. Also, any excess heat in the exhaust stream (essentially all from the core), is just wasted energy, but is usually too difficult to capture*.

The idea is to *decrease* the pressure of the stream to atmospheric - any excess** is just wasted. That's basically an issue only for the core flow on a high bypass engine, the fan simply doesn't increaser the pressure of the airflow that much (and the short duct behind it is more than enough to drop the pressure back to ambient). The idea is to convert the excess pressure to velocity. It's velocity times mass that generates thrust. Pressure does nothing for you. Think of it this way: any pressurized (above ambient) gas flow from the back of the engine will immediately expand *sideways*. Of course in real engines, there will always be a bit of a mismatch between the pressure of the exhaust stream and ambient, but the idea is to minimize that.



*In situations where weight is not an issue, for example, in ground based gas-turbine powered generating plants, the exhaust stream is often used to boil water in a "combined cycle" design, and that steam then used to drive another (steam) turbine and generator. This can increase the efficiency of the plant by as much as 50%. Additional turbine stages can accomplish some of the same, but as the gas pressure and temperature falls (at each turbine stage), this becomes more and more difficult.

**You don't want to go the other way either, and expand the gas stream to below atmospheric, but that's not usually an issue for subsonic outlets.


User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 2, posted (4 years 4 months 1 week 5 days 6 hours ago) and read 8540 times:

Quoting smartt1982 (Thread starter):
It was my understanding that turbofan engines are used on airliners due to their exhaust speed being more in tune with the speed that the actual aircraft flies.

Correct. The closer the exhaust velocity is to the aircraft velocity, the higher the propulsive efficiency (but the lower the thrust). You need to strike a balance. Turbojets are *way* too high above aircraft velocity for good efficiency at normal commercial jet speeds. Accelerating a lot of mass a little bit is much more efficient than accelerating a little bit of mass by a lot.

Quoting smartt1982 (Thread starter):
I was also of the understanding that the cone at the back of some jet engines is to allow the exhaust to slow down and increase to ambient pressure

It's actually there to decrease the pressure, not increase.

Quoting smartt1982 (Thread starter):
Is it that the speed of sound due to the exhaust temp is so high it does not reach supersonic at all?

Yes, for commercial jets. Also, there's no convergent/divergent nozzle, which is what you'd need to get supersonic exhaust.

Quoting smartt1982 (Thread starter):
when it reaches the ambient outside temp will it then be supersonic due to the temp drop?

By the time it reaches ambient temperature, is has slowed down way too much to be supersonic.

Quoting smartt1982 (Thread starter):
is it the velocity or the pressure that is causing the equal and opp reaction.

This is a very complicated question, but the short answer is that it's pressure. The velocity is the result of the pressure. The pressure is what's actually pushing the airplane.

Quoting smartt1982 (Thread starter):
Like on concorde at supersonic, I don't believe the afterburners were engaged all the time, was it simply a convergent/divergent duct that kept them supersonic?

Yes.

Quoting smartt1982 (Thread starter):
presumambly it was the exhaust velocity that was causing the reaction

No. The velocity is a byproduct of the reaction, not the cause.

Quoting smartt1982 (Thread starter):
Silly probably very obvious answer to the question but thanks

Not nearly as silly as you think...momentum flows and control volumes around jet engines can get really complex.

Quoting rwessel (Reply 1):
It's velocity times mass that generates thrust. Pressure does nothing for you.

Pressure is what gives you velocity. The way the math falls out, velocity shows up and pressure doesn't, but because they're directly coupled that doesn't actually tell you which one is causing the thrust. The physical mechanism of thrust transfer from the airflow to the engine (thence to the airplane) is pressure. The velocity comes about because of equal and opposite reactions...if the air is pushing on the engine, the engine is pushing on the air. Air is getting pushed, so it speeds up.

Tom.


User currently offlinerwessel From United States of America, joined Jan 2007, 2368 posts, RR: 2
Reply 3, posted (4 years 4 months 1 week 5 days 5 hours ago) and read 8531 times:
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Quoting tdscanuck (Reply 2):
Pressure is what gives you velocity. The way the math falls out, velocity shows up and pressure doesn't, but because they're directly coupled that doesn't actually tell you which one is causing the thrust. The physical mechanism of thrust transfer from the airflow to the engine (thence to the airplane) is pressure. The velocity comes about because of equal and opposite reactions...if the air is pushing on the engine, the engine is pushing on the air. Air is getting pushed, so it speeds up.

Well, it's the conversion of pressure into velocity that generates the thrust, and certainly the opposite reaction to that is carried by the (back) pressure from the gas to some part of the engine. To an extent this is a semantic quibble, but would you seriously call the thrust produced by throwing a stone to be caused by the (back) pressure of the rock on my hand? Yes, that's the transfer mechanism, but the thrust is created by the acceleration of the mass (rock). In either case the transfer mechanism is specific to the implementation (I can kick the rock with my foot, or use iron rocks and a magnetic accelerator, or ionized rocks and a magnetic field, or put the rocks in a blowgun, and there won't be a hand, or pressure on it, to act as the transfer mechanism), the thrust is created by the acceleration of the mass. Of course for all practical jet engines, it's pretty much the same mechanism, although alternatives are imaginable (hypothetically an ion drive “jet” is perfectly plausible, for example).

In any event, I interpreted the OP's question as applying to the exhaust stream after it exited the engine proper, in the sense that the pressurized air would "push" against the back of the aircraft in some fashion, sort of like an inflatable bag lifting a mass. A similar misconception to the one many people had about rockets not working in space due to a lack of a medium to "push against."


User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 4, posted (4 years 4 months 1 week 5 days 4 hours ago) and read 8526 times:

Quoting rwessel (Reply 3):
Well, it's the conversion of pressure into velocity that generates the thrust

That's not strictly true...in something like a pop-bottle rocket before you pull the launch pin you've got thrust with no velocity, because the exhaust gas is trapped. This is exactly the same propulsion physics at work in jet engines.

Quoting rwessel (Reply 3):
To an extent this is a semantic quibble, but would you seriously call the thrust produced by throwing a stone to be caused by the (back) pressure of the rock on my hand?

I would, so yeah, it's a semantic quibble.

Quoting rwessel (Reply 3):
Yes, that's the transfer mechanism, but the thrust is created by the acceleration of the mass (rock). In either case the transfer mechanism is specific to the implementation (I can kick the rock with my foot, or use iron rocks and a magnetic accelerator, or ionized rocks and a magnetic field, or put the rocks in a blowgun, and there won't be a hand, or pressure on it, to act as the transfer mechanism), the thrust is created by the acceleration of the mass.

But, again, you can get thrust without velocity (or acceleration). Use your magnetic accelerator against a really big rock and you still get thrust, even if the rock doesn't move. It's the interacting force (magnetic, pressure, electric, whatever) that's creating the thrust...the fact that the expelled mass is accelerated is a byproduct of the particular engine architecture, not a physical requirement for all thrust-generating systems. The thing that's actually pushing on you, in the case of a jet engine, is pressure.

Tom.


User currently offlinerwessel From United States of America, joined Jan 2007, 2368 posts, RR: 2
Reply 5, posted (4 years 4 months 1 week 5 days 4 hours ago) and read 8523 times:
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Quoting tdscanuck (Reply 4):
But, again, you can get thrust without velocity (or acceleration). Use your magnetic accelerator against a really big rock and you still get thrust, even if the rock doesn't move. It's the interacting force (magnetic, pressure, electric, whatever) that's creating the thrust...the fact that the expelled mass is accelerated is a byproduct of the particular engine architecture, not a physical requirement for all thrust-generating systems. The thing that's actually pushing on you, in the case of a jet engine, is pressure.

Unfortunately "reactionless" drives are purely science fiction at the moment.

You can't get thrust without accelerating some (other) mass. If the other mass is large enough, its net acceleration can be very small relative to yours, but it's still there. Let's say you had a lump of iron the size of a planet (on the order of 10**20 tonnes), you're in a 10t spacecraft, and flip on a magnet, and you start accelerating towards the lump. The lump is accelerated (towards you) with the same force as your spacecraft, but its motion is 19 orders of magnitude smaller than yours because of its much larger mass.

It’s that acceleration of the (other) mass that creates the thrust, although there obviously needs to be a transfer mechanism involved somewhere. In my book the acceleration is primary (as it must *always* be there), the transfer mechanism is secondary (since any one of a variety of mechanisms might be used – although rather obviously a practical system must have one).


User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 6, posted (4 years 4 months 1 week 5 days 3 hours ago) and read 8509 times:

Quoting rwessel (Reply 5):
You can't get thrust without accelerating some (other) mass.

Sure you can. If I tie my big magnet to my big rock and turn it on, I've got tons of thrust and nothing is accelerating at all.

I agree with you completely that you can't *accelerate* your vehicle without accelerating something else, but we're just talking thrust here...there's no physics problem with having static thrust without accelerating anything.

Quoting rwessel (Reply 5):
In my book the acceleration is primary (as it must *always* be there), the transfer mechanism is secondary

Except it doesn't always have to be there (the static case). Also, I think the transfer mechanism is primary in this particular discussion because of the originating question:

Quoting smartt1982 (Thread starter):
Silly question but of the hot exhaust that is being ejected where is the thrust being produced from, is it the velocity or the pressure that is causing the equal and opp reaction.

Tom.


User currently onlinevikkyvik From United States of America, joined Jul 2003, 10108 posts, RR: 26
Reply 7, posted (4 years 4 months 1 week 5 days 2 hours ago) and read 8488 times:
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Quoting rwessel (Reply 3):
To an extent this is a semantic quibble, but would you seriously call the thrust produced by throwing a stone to be caused by the (back) pressure of the rock on my hand?

I laughed out loud at this one, because about a year ago, I went over the same exact question in my head for awhile, trying to sort it out in my head.

Yes, I would definitely consider the pressure to be the "conduit" or whatever for the thrust. The velocity of the gas might certainly cause the appropriate pressures (or does the pressure cause the velocity   ), but the actual force generated by the molecules of gas is transferred to the engine through its pressure.

So for the stone-throwing example, yes, your hand is being pushed by the stone (equal-opposite to your hand pushing the stone), and there you have the force on yourself. To achieve that pressure (and hence that force), you're required to accelerate the stone (which you do through the pressure of your hand on it).

For magnetic thrust, wouldn't the conduit or mechanism simply be the magnetic field? In the turbofan, the mechanism would be the pressure field.

Very similar to the whole Newton vs. Bernoulli discussion of lift. They're both valid, but ultimately, without the pressure differential, the wing wouldn't feel lift.



"Two and a Half Men" was filmed in front of a live ostrich.
User currently offlinerwessel From United States of America, joined Jan 2007, 2368 posts, RR: 2
Reply 8, posted (4 years 4 months 1 week 5 days 2 hours ago) and read 8478 times:
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Quoting tdscanuck (Reply 6):
Sure you can. If I tie my big magnet to my big rock and turn it on, I've got tons of thrust and nothing is accelerating at all.

And now I've tripped over the whole force vs. work thing... *argh*

But to clear up the mess we've made for the OP...

- A jet engine produces thrust my accelerating mass (mostly air) backwards.

- This involves generating higher than ambient pressures inside the engine.

- The engine converts that pressure into velocity.

- The equal and opposite reaction is transferred to the engine by the pressure of that gas, not unlike the air escaping from the open end of a balloon.

- A (theoretical) design goal would be to convert all the excess pressure (relative to ambient) to velocity before letting the stream exit the engine, any gas exiting the engine at higher than ambient pressure represents lost energy.

Quoting tdscanuck (Reply 6):
Also, I think the transfer mechanism is primary in this particular discussion because of the originating question:

Quoting smartt1982 (Thread starter):
Silly question but of the hot exhaust that is being ejected where is the thrust being produced from, is it the velocity or the pressure that is causing the equal and opp reaction.

I was more influenced by the title of the thread.


User currently offlineflipdewaf From United Kingdom, joined Jul 2006, 1577 posts, RR: 0
Reply 9, posted (4 years 4 months 1 week 4 days 23 hours ago) and read 8457 times:
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It may be better to think of the acceleration of the gasses as "work done" rather than acceleration. The aim is to get all of the work done heating the air into work done accelleratin it.

With the situation of the large rock that won't move you miss some very important points, the large rock will move no matter how small the resulting force is on it but I think you are neglecting the fact that the rock and thrusting device are In fact attached through the ground, if your thrusting device is a ram then thee will be a compression thrust in the ram but also through friction a tension force alongthe ground thus making the net thrust on the rock zero. In a frictionless environment the rock would move.

Fred


User currently offlinejetmech From Australia, joined Mar 2006, 2699 posts, RR: 53
Reply 10, posted (4 years 4 months 1 week 4 days 19 hours ago) and read 8360 times:

Quoting rwessel (Reply 3):
Well, it's the conversion of pressure into velocity that generates the thrust
Quoting rwessel (Reply 3):
but would you seriously call the thrust produced by throwing a stone to be caused by the (back) pressure of the rock on my hand?

I'd also be leaning towards pressure as the main mechanism producing "thrust" in a jet engine. Thrust is a force, which is often thought of as a property existing at a point. Pressure is merely a force distributed over a finite area. The fan raises the static pressure of the incoming air slightly above ambient. When this pressure difference (Force / Area) across the fan is divided by the area of the fan, the areas cancel out leaving the force part only.

Quoting rwessel (Reply 3):
but the thrust is created by the acceleration of the mass (rock).
Quoting rwessel (Reply 5):
It’s that acceleration of the (other) mass that creates the thrust
Quoting rwessel (Reply 8):
- A jet engine produces thrust my accelerating mass (mostly air) backwards.
- This involves generating higher than ambient pressures inside the engine.
- The engine converts that pressure into velocity.

I think the two mechanisms are related. Two common formulas to work out thrust in Newtons are,

(1) Force (N) = Mass flow rate (kg/s) *[exhaust velocity - inlet velocity] (m/s)
(2) Force (N) = Mass (kg) * acceleration (m/s^2)

For fluids in a closed conduit, there must be a pressure difference along the conduit if you want the fluid to flow. If you apply a pressure differential to a "parcel" of fluid in the conduit, you will get the required force to cause this parcel of fluid to move - remembering that a pressure differential divided by the cross sectional area of the fluid parcel results in a net force.

Thus, the fan of a jet engine increases the static pressure of the incoming air above ambient, with the pressure differential across the fan producing the desired quantity of thrust. We now have a pressure differential across the "parcel" of air in the fan duct (closed conduit) between the back face of the fan and the fan duct exit.

This pressure differential, being higher than the one between the inlet and front face of the fan ensures that the air passing through the fan leaves at a higher velocity then it enters. Thus, in addition to the pressure difference mechanism, we also get "thrust" from formula (1).

In addition, acceleration is defined as a change in velocity. We definitely have a velocity change (increase) across the fan, which is achieved via a process of acceleration. Thus, in addition to the pressure difference mechanism, we can also get "thrust" from formula (2).

Nonetheless, I think it is the pressure difference across the fan that is the key. The other mechanisms are by-products of generating the pressure difference, which luckily enough, can be used to give the same calculated thrust force.

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently onlinevikkyvik From United States of America, joined Jul 2003, 10108 posts, RR: 26
Reply 11, posted (4 years 4 months 1 week 4 days 16 hours ago) and read 8318 times:
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Quoting jetmech (Reply 10):
When this pressure difference (Force / Area) across the fan is divided by the area of the fan, the areas cancel out leaving the force part only.
Quoting jetmech (Reply 10):
remembering that a pressure differential divided by the cross sectional area of the fluid parcel results in a net force.

Not to be nitpicky, but I think you mean pressure difference multiplied by area.

Pressure divided by area would give you force/area^4.



"Two and a Half Men" was filmed in front of a live ostrich.
User currently offlinerwessel From United States of America, joined Jan 2007, 2368 posts, RR: 2
Reply 12, posted (4 years 4 months 1 week 4 days 11 hours ago) and read 8212 times:
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Quoting jetmech (Reply 10):
I think the two mechanisms are related. Two common formulas to work out thrust in Newtons are,

(1) Force (N) = Mass flow rate (kg/s) *[exhaust velocity - inlet velocity] (m/s)
(2) Force (N) = Mass (kg) * acceleration (m/s^2)

These are, of course, the same formula, but looked at from two slightly different perspectives. And you'll notice that you've defined the thrust as coming from the velocity change of the mass passing through the engine. What do I win?   

The thrust (in the work sense) comes from the acceleration of the working mass. A jet engine uses pressure to accomplish that - and while that's fundamental to a conventional jet engine, it's *not* fundamental to the production of thrust.

As I said from the start, this is something of a semantic quibble - a jet engine's primary* purpose is the accelerate the air stream - from the point of view of understanding why an airplane can move down the runway, we need to understand the acceleration of the working mass. If we want to discuss *how* a jet engine produces that acceleration, we need to understand (among other things) the pressures inside the engine. Considering the acceleration of the working mass to be a byproduct of the pressure changes inside the engine, is, IMO, correct only if we’re discussing how a jet engine works (or how to build one), but confuses purpose and mechanism when we’re talking about what a jet engine does when hanging on the wing of an airliner.


*we also expect it to perform some secondary tasks, like drive accessories (hydraulic pumps, generators, oil pumps), provide bleed air, etc.


User currently offlinejetmech From Australia, joined Mar 2006, 2699 posts, RR: 53
Reply 13, posted (4 years 4 months 1 week 4 days 9 hours ago) and read 8173 times:

Quoting vikkyvik (Reply 11):
Not to be nitpicky, but I think you mean pressure difference multiplied by area.

You are entirely correct! I have to be much more careful writing in Tech / Ops after a long day   .

Quoting rwessel (Reply 12):
These are, of course, the same formula

Yep. The right hand side gives exactly the same unit.

Quoting rwessel (Reply 12):
As I said from the start, this is something of a semantic quibble - a jet engine's primary* purpose is the accelerate the air stream

Fair enough. I was trying for my benefit as well as anyone else's to try and put down in words the interrelation between the various mechanisms at play inside a jet engine. I agree that it does degenerate very quickly into semantics, but I feel it is worth the pain to gain a deeper understanding of how things work. Speaking of semantics, and tying oneself into knots, I had another think about this formula last night, and what I stated are the implications for the fan of a jet engine;

Quoting jetmech (Reply 10):
(1) Force (N) = Mass flow rate (kg/s) *[exhaust velocity - inlet velocity] (m/s)
Quoting jetmech (Reply 10):
This pressure differential, being higher than the one between the inlet and front face of the fan ensures that the air passing through the fan leaves at a higher velocity then it enters

Assuming fan blades are airfoils, and are thus designed to generate a pressure difference like the wing of an aircraft, it would seem that there is far greater potential to increase the velocity of the air approaching the fan, as opposed to the velocity of the air leaving the fan.

At ranges of angles of attack where one is developing an appreciable lift coefficient, and before the onset of appreciable flow separation, the majority of the lift force comes from the decreased static pressure (below ambient) on the suction side of the airfoil. The suction side of an airfoil is analogous to the "front" face of the fan.

The pressure side of the airfoil produces a region of increased (above ambient) static pressure, with the pressure side of an airfoil being analogous to the "back" face of the fan. What is fascinating is the fact that the suction side of the fan has far greater potential to develop a pressure differential ( with respect to ambient ) compared with the pressure side of the fan.

The maximum pressure differential (with respect to ambient) that the pressure side of the airfoil (fan) can develop is limited to a pressure coefficient of positive one (+1), which only occurs at the stagnation point. The rest of the pressure face of the fan will have a pressure differential less than one (+1).

However, on the suction side of the airfoil (fan), the peak and average pressure coefficient can easily be much less than negative one (-1). If we now take this information into consideration along with the physics of fluid flow in a closed conduit, it appears that the majority of the velocity change across the fan may occur as the air approaches the fan.


JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlinejetmech From Australia, joined Mar 2006, 2699 posts, RR: 53
Reply 14, posted (4 years 4 months 1 week 4 days 9 hours ago) and read 8158 times:

Quoting jetmech (Reply 13):

For some reason, A'net has decided to chop off the last paragraph of my previous post after I added the picture! The last paragraph was along the lines as follows;

Thus, it appears that the velocity increase induced upon the air approaching the fan may be far greater than that applied to the air as it leaves the fan. If this is so, then according to the formula, we are generating negative thrust as the "inlet" velocity has exceeded the "exhaust" velocity. We know this is obviously not true, so it leaves the tricky question as to the exact definition of inlet and exhaust velocity.”

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineflipdewaf From United Kingdom, joined Jul 2006, 1577 posts, RR: 0
Reply 15, posted (4 years 4 months 1 week 3 days 23 hours ago) and read 8090 times:
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Quoting jetmech (Reply 14):

Nono, I think what you are actually saying is that the pressure difference between ambient and the front face of the fan is greater than the pressure difference between the front face and the rear face meanin that the overall acceleration of the air (and therefore thrust) is predominantly before the fan but there is still some after?

Maybe try it with disc theory and blade analysis as well to see what they say.

Fred


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