How does the weight of airplanes affect atmospheric pressure?

[DrewD]

http://www.grc.nasa.gov/WWW/k-12/airplane/bernnew.html

This site was posted on PF and I read it a while back and felt that it was informative. It has a simple explanation on the first page and some deeper stuff on the second. My recollection is that it says that either a freebody force explanation or bernoulli fluid laws can work. But it helps to clarify the common mistakes that are made. I haven't the time to reread it now, so I hope that it is as I remember (if its not let's pretend that they changed it since I read it)

 

[boneh3ad]

So again, same limitation, isn't it? You can find the lift by using the velocity profile to find pressure or momentum change, but neither tell you what that velocity profile will look like.

Yes. For that you can apply the Kutta condition and get your answer for the case of no separation. Otherwise you need the full Navier-Stokes equations.

 

[rcgldr]

I'm not sure how accurate this would be. Part of the lift is due to a non-Bernoulli interaction between air and a wing that increases the mechanical energy of the air (wrt ambient air).

What interaction is that?

Explain that

After a wing passes through a volume of air, the affected air ends up with a non-zero exit velocity (the velocity of the affected air when it's pressure returns to ambient). Using the unaffected air as a frame of reference, the affected air's mechanical energy is increased by a wing, which means work is done, which violates the rule for Bernoulli.

A similar process occurs with a propeller or rotor, except that the pressure differential is greater than a typical wing, and there is a greater amount of induced flow (the inwash ahead of the propeller or rotor). From a NASA article:

... at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli's equation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed ...

propanl.htm

 

[boneh3ad]

After a wing passes through a volume of air, the affected air ends up with a non-zero exit velocity (the velocity of the affected air when it's pressure returns to ambient). Using the unaffected air as a frame of reference, the affected air's mechanical energy is increased by a wing, which means work is done, which violates the rule for Bernoulli.

A similar process occurs with a propeller or rotor, except that the pressure differential is greater than a typical wing, and there is a greater amount of induced flow (the inwash ahead of the propeller or rotor). From a NASA article:

... at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli's equation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed ...

propanl.htm

Yes, you refer to the wake. However, this really relates to drag, not so much to lift.

 

[rcgldr]

Yes, you refer to the wake. However, this really relates to drag, not so much to lift.

The direction of the "wake" is mostly downwards (lift), and somewhat forwards (drag). For a propeller, the "wake" is mostly aftwards (thrust), and somewhat circular (drag).

The force exerted by gravity onto the aircraft is in turn exerted by the aircraft onto the air, and eventually, that force is exerted by the air onto the Earth's suface (forces do not vanish), via a continuous impulse (downward shifting of air) that originated at the aircraft.

The other law of physics involved with an aircraft (in level flight) is that the impulse (force x time) exterted by the aircraft onto a volume of air over some period of time results in a corresponding change in momentum of that air, and in the real world, during the interaction between aircraft and air, the downwards component of velocity is not offset by a reduction of pressure, so there's a net "exit velocity", representing the increase in mechanical energy of the air affected by an aircraft.

This NASA article includes an image of the downwash effect on a cloud:

downwash.htm

 

[K^2]

Bernoulli's principle is one way to calculate the lift on a wing in certain situations (the wing cannot be separated, for example). Given a velocity distribution over a wing, you can use Bernoulli's equation to deduce the pressures on the wing and hence the lift. It says nothing about how you find sai velocities or the best shape of a wing. Bernoulli's equation is merely a tool; it cannot explain lift completely.

Bernoulli Equation doesn't explain lift at all. It can be used as a good estimate, under certain conditions, but in order to get exact lift, you need pressure at the actual boundary. And at the actual boundary the flow velocity is precisely zero and Bernoulli Effect is also zero.

The only reason Bernoulli Equation works to estimate lift is because viscosity of the air is low, and if you are working in regime where air can be treated as incompressible, the pressure gradient near the boundary is low. So you can take air flow near the boundary, and use Bernouli Equation there to say that pressure at the boundary will be the same. At mach number << 1, you will be correct.

There is one more problem with Bernoulli in that it is difficult to implement near the critical angle of attack. As the separation layer creeps up the wing, relevance of Bernoulli Effect to pressure at the boundary decreases further. So you really want a different method of estimating lift, and that's where Kutta Condition and Kutta-Joukowski Theorem come up, as you point out.

But ultimately, it's the pressure differential on the wing surface that's the direct cause of the lift, and you cannot apply Bernoulli Principle at the boundary. You can use it as a motivation to look for an effect, but if you want to actually explain it, you can't do much better than, "Boundary conditions of the problem." Because that's the real reason for the pressure differential.

What does the thrust have to do with the lift?

Everything.

Please, explain flight of a glider, which generates no thrust.

 

[CWatters]

K^2 wrote:

Please, explain flight of a glider, which generates no thrust.

Back to flying school for you!

If a glider has no source of thrust what on Earth do you think balances the force of drag? Everyone knows that gravity provides the thrust needed by a glider to overcome drag.

It's very easy to demonstrate that lift is proportional to speed. Just turn your engines off, slow down and try and stay up there!

Nice video of someone converting excess height to excess speed to excess lift and back to excess height again...



I'm an ex glider pilot. What you fly?

 

[K^2]

I've flown C-172 and a T-6 Texan, though, the later only briefly. So yeah, I know how to trade altitude for speed and vice versa.

While a glider does use longitudinal component of weight to counter drag in steady flight, it doesn't fall under definition of thrust. Under standard aeronautical definitions, thrust and weight are two distinct forces, even when they are not orthogonal. I know what you mean, but it just doesn't seem to be related to what Lunar-Scooter is talking about.

Lift is proportional to square of the air speed, yes. But your air speed isn't proportional to the thrust. Sure, as soon as you kill the engine, if you don't start trading altitude for it, you'll start losing speed. But you won't lose lift the instant you lost thrust. You'll lose it gradually, as you lose the air speed, which, of course, will force a descent, unless you increase angle of attack.

The sum of longitudinal forces, including drag, thrust, and conditionally weight, determines longitudinal acceleration of the plane. While the instantaneous air speed is what determines the lift. Another good demonstration of this is when you are flying under constant thrust and hit a gust of wind. Your thrust doesn't change, but your lift certainly does.

Point is, lift is independent of thrust. Yes, thrust is kind of important for flying, but it doesn't contribute to lift in any way. These are two completely separate forces generated in two completely separate ways.

 

[boneh3ad]

Bernoulli Equation doesn't explain lift at all. It can be used as a good estimate, under certain conditions, but in order to get exact lift, you need pressure at the actual boundary. And at the actual boundary the flow velocity is precisely zero and Bernoulli Effect is also zero.

The only reason Bernoulli Equation works to estimate lift is because viscosity of the air is low, and if you are working in regime where air can be treated as incompressible, the pressure gradient near the boundary is low. So you can take air flow near the boundary, and use Bernouli Equation there to say that pressure at the boundary will be the same. At mach number << 1, you will be correct.

There is one more problem with Bernoulli in that it is difficult to implement near the critical angle of attack. As the separation layer creeps up the wing, relevance of Bernoulli Effect to pressure at the boundary decreases further. So you really want a different method of estimating lift, and that's where Kutta Condition and Kutta-Joukowski Theorem come up, as you point out.

But ultimately, it's the pressure differential on the wing surface that's the direct cause of the lift, and you cannot apply Bernoulli Principle at the boundary. You can use it as a motivation to look for an effect, but if you want to actually explain it, you can't do much better than, "Boundary conditions of the problem." Because that's the real reason for the pressure differential.

Have you read anything I've said? I agree (mostly) with you. I do need to point out that, as borne out by both theory and experiments, the wall-normal pressure gradient across a boundary layer is effectively zero, so if you use the displacement thickness to add to the shape of the airfoil and then run an inviscid simulation, you will get effectively the exact pressures at the wall. There have been many years of theory and experiments confirming this in both air and water. This is why Bernoulli's equation works for incompressible, unseparated flows. There are also compressible forms of the Bernoulli equation that solve this, but even they suffer from the same limitations.

Like I said, Bernoulli's principle is a tool that can be used I calculate lift, not the law that explains its origin. On that we agree.

 

[boneh3ad]

The direction of the "wake" is mostly downwards (lift), and somewhat forwards (drag). For a propeller, the "wake" is mostly aftwards (thrust), and somewhat circular (drag).

This isn't true. The wake trails behind a foil passing through a fluid. On a plane. It is a long disturbance in the fluid behind and slightly below the plane. True it does have a vertical component, which is that downwash that corresponds to the opposite of the lift, however the wake does not travel downward for long. It is mostly directly rearward.

The force exerted by gravity onto the aircraft is in turn exerted by the aircraft onto the air, and eventually, that force is exerted by the air onto the Earth's suface (forces do not vanish), via a continuous impulse (downward shifting of air) that originated at the aircraft.

Yes we know, forces balance. That's a fundamental law. The downward momentum flux of the wake of an airfoil is directly proportional to lift, and that lift is often calculable using Bernoulli (but no always, of course).

Some additional lift is lost due to the effect of wingtip vortices on a real wing, but even that, through the laws of physics, would still be shown in the balance of momentum in the downwash with the lift. That also over complicated the answer to the simple (on its surface) "how does an airplane really fly?"

 

[sophiecentaur]

K^2 wrote:


. . . . .

Nice video of someone converting excess height to excess speed to excess lift and back to excess height again...



I'm an ex glider pilot. What you fly?

I couldn't get much from that movie because the camera was (almost certainly) mounted on the wing so how could you tell its speed or height from the picture. Whilst I believe the theory, how does showing a change of angle wrt the horizon prove anything?

 

[sophiecentaur]

This isn't true. The wake trails behind a foil passing through a fluid. On a plane. It is a long disturbance in the fluid behind and slightly below the plane. True it does have a vertical component, which is that downwash that corresponds to the opposite of the lift, however the wake does not travel downward for long. It is mostly directly rearward.



Yes we know, forces balance. That's a fundamental law. The downward momentum flux of the wake of an airfoil is directly proportional to lift, and that lift is often calculable using Bernoulli (but no always, of course).

Some additional lift is lost due to the effect of wingtip vortices on a real wing, but even that, through the laws of physics, would still be shown in the balance of momentum in the downwash with the lift. That also over complicated the answer to the simple (on its surface) "how does an airplane really fly?"

Presumably the faster the plane flies, the slower the downwash would be. This most be along the same lines as when boat is planing; the faster it goes, the less of the hull is below the surface. Impulse is force times time. The force must be equal to the weight of the plane / boat but at high speed, that force acts on a section of air for a shorter time so the air moves downward slower (all other things being equal).

The effect of downwash can be pretty catastrophic for light aircraft flying through the flight path of large aircraft as they land (at 'low' speed).

 

[K^2]

It's mostly the vortex state that causes problems for other aircraft, not the downwash.

 

[DaveC426913]

It's mostly the vortex state that causes problems for other aircraft, not the downwash.

Same thing. Downwash is inner half of each vortex.

http://eis.bris.ac.uk/~glhmm/gfd/MHutchinson_airplane.jpg

 

[sophiecentaur]

Cracking picture. I can see the rotational motion is much greater than the bulk downward motion. Wouldn't like to fly into that in a Cessna!

 

[DaveC426913]

Cracking picture. I can see the rotational motion is much greater than the bulk downward motion. Wouldn't like to fly into that in a Cessna!

Well, the rotational motion is simply the downward moving air combined with the corresponding upward moving air to fill the void. So downward and rotational should be essentially the same.

 

[sophiecentaur]

That figures. I now realize there can be no net motion. It's just a matter of how big this vortex region is. ?

 

[K^2]

There IS a net motion. If there was no net motion, there would be no net lift. You see where the could would be if it wasn't for the plane on that picture? All that cleared space is the net downward air displacement.

Same thing. Downwash is inner half of each vortex.

Well, first of all, it's not the same thing, because you can reduce vortices, with winglets, for example, but you can't reduce downwash.

But more to the point, yes, you aren't going to come across one without coming across the other, but the downwash is not what's dangerous. If it was just downwash, you'd correct for it by increasing angle of attack, throttling up, and then you continue as normal. You can get hit by a downwards jet of air at high altitude. Happens all the time, and quite a few of these are significantly worse than any downwash.

Vortex, on the other hand, will at best start flipping you over without any chance of correction until you are clear. At worst, it will overstress the aircraft. Vortex state is what gets you. Not the downwash.

 

[DaveC426913]

There IS a net motion. If there was no net motion, there would be no net lift. You see where the could would be if it wasn't for the plane on that picture? All that cleared space is the net downward air displacement.

Well, true. You can look at it as downwash slightly greater than upwash, or you can look at it as one big vortex with net downward motion.

Either way, true, the net motion is downward. Though I suspect the speeds in the vortex greatly exceed the speed of the downwash.

Well, first of all, it's not the same thing, because you can reduce vortices, with winglets, for example, but you can't reduce downwash.But more to the point, yes, you aren't going to come across one without coming across the other,...

Yeah, they're both components of the same thing. But yeah, you can look a them independently. Which is what you were saying originally, though it was not apparent at first.

but the downwash is not what's dangerous. If it was just downwash, you'd correct for it by increasing angle of attack, throttling up, and then you continue as normal. You can get hit by a downwards jet of air at high altitude. Happens all the time, and quite a few of these are significantly worse than any downwash.

Vortex, on the other hand, will at best start flipping you over without any chance of correction until you are clear. At worst, it will overstress the aircraft. Vortex state is what gets you. Not the downwash.

Agreed. Downwash leaves a stable craft. Vortex demolishes the craft's stability.

 

[HallsofIvy]

Pixie dust, it's all pixie dust!

 

[K^2]

Though I suspect the speeds in the vortex greatly exceed the speed of the downwash.

There would have to be regions like that to pull the condensate upwards as evident on the photo, yes.

 

[boneh3ad]

Well, true. You can look at it as downwash slightly greater than upwash, or you can look at it as one big vortex with net downward motion.

Either way, true, the net motion is downward. Though I suspect the speeds in the vortex greatly exceed the speed of the downwash.

For visualization purposes, sure. Really that isn't what goes on though. The wingtip vortices and the downwash are entirely different phenomena that just happen to be generated in the same flow field.

If you could somehow hypothetically remove the wingtip vortices from the picture, you would still get downwash or else your plane will fall from the sky. This is evident in the fact that you still get downwash (stronger actually) when winglets or raked wingtips are used to minimize the wingtip vortices.

A more accurage description is that the vortices are superposed onto the same flow field as the downwash. The vortices will not necessarily be moving down at the same rate as the downwash either, as they leave the wingtips largely level (relative to the wing) and end up bending downward under the influence of the downwash.

Yeah, they're both components of the same thing. But yeah, you can look a them independently. Which is what you were saying originally, though it was not apparent at first.

No they aren't. Wingtip vortices are a result of having a finite wing. Downwash is a necessary byproduct of lift. They are two fundamentally different phenomena generated for different reasons that just happen to coexist as a result of the operation of a finite airplane wing.

Another interesting distinction: downwash will die out after a while after the plane passes. The wingtip vortices persist for quote some time until eventually they are dissipated by the viscosity of air. Still, the can persist for miles behind a plane before this happens.

 

[A.T.]

Wingtip vortices are a result of having a finite wing. Downwash is a necessary byproduct of lift.

Aren't vortices a necessary byproduct of downwash (of a finite extend)? Can you accelerate air downwards locally without generating vortices? If a wing produces no downwash (symmetric airfoil with alpha=0), does it still create significant vortices?

 

[boneh3ad]

Aren't vortices a necessary byproduct of downwash (of a finite extend)? Can you accelerate air downwards locally without generating vortices? If a wing produces no downwash (symmetric airfoil with alpha=0), does it still create significant vortices?

Yes and no. In a practical sense, one will always come with the other. However, let's say you had an infinite, 2-D wing. You can certainly still generate downwash, but there will be no vortices generated. Generally, you can have the downwash without the wingtip vortices (in theory), but you can't have wingtip vortices without downwash since the wing must be generating lift to create the vortices.

They are certainly related phenomena, but not the same phenomena.

 

[sophiecentaur]

I'm not quite clear on this but some air must be displaced downwards, or at least there has to be a downward force on it to balance the lift force. This will produce a large vortex as air from above flows into follow it. By this time, the plane will have passed, of course, so it escapes the consequence. I imagine this could be why helicopters are less efficient when hovering because they get their own downwash (?).

In a global sense, I imagine the overall atmospheric pressure is increased to some extent by the weight of all the planes that are aloft at any time. Though, the degree of this effect would depend on their speeds (as in the case of planing boats).

 

[PYR]

For a NON Bernoulli explanation of lift you can also look at a 10 min video on you tube at:

how do planes fly - a non Bernoulli explanation




It explains how lift is due to the wing moving air downwards (action) and lift is the reaction of air on the wing (3rd law of Newton).

 

[rcgldr]

I imagine this could be why helicopters are less efficient when hovering because they get their own downwash.

When in a hover, the air has enough time to established a strong induced downwash above the rotor, requring the rotor impart even more speed to the air, which consumes more energy. Worse yet for some helicopters is to vertically decend into their own downwash. Aerobatic model helicopters don't have this issue becuase the thrust to weight ratio is 5 or greater.

In a global sense, I imagine the overall atmospheric pressure is increased to some extent by the weight of all the planes that are aloft at any time.

Consider the earth, atmosphere and every object in the atmosphere as a closed system. Assuming that the objects are not accelerating vertically, then the total average force applied to the surface of the Earth equals the weight of the atmosphere and the weight of all the objects supported by the atmosphere.

You can reduce this to a single model flying or hovering within a large box. As long as the center of mass of the system is not accelerating vertically, then the net downwards force on the box equals the weight of the air and the model. The downwards force is exerted via a pressure differential, less at the top, more at the bottom.