airfoil rising

Essay · February 5, 2015

In this thought experiment there is no “thrust.” The “wind” is simply stipulated to be in in motion before encountering the “airfoil.”

You may not want to call it thrust, but there is some very large (I assume) external force acting upon the airfoil.

Istm, the energy is coming from your stipulation (or from thrust), which is shunted in various horizontal and vertical components of the "wind" you speak about.

I think Studiot, in post 18, first explained this.

Some small fraction of the "very large" thrust (or whatever your stipulation is for the force imbalance) becomes the source of energy for 'lifting' or slightly counteracting gravity or 'offsetting' some gravitational potential energy.

~

Zet · February 5, 2015

You may not want to call it thrust

Is it “thrust”?

Did I use that word wrong?

If so, I apologize.

If a ball is rolling across the floor, and so is in motion, and there is nothing else acting on it, is it simply “in motion” or is there “thrust”?

And if there is motion within a fluid and there is nothing else acting on it is it simply “in motion” or is there “thrust”?

Yes, the original formulation of this question was about a moving airfoil, and so I ended up in an extended conversation about “thrust” that I didn’t feel helped answer my question. So, in post #46 I changed it so that it was the fluid that is in motion and not the airfoil. I had hoped that this would eliminate this conversation about “thrust” and better highlight my issue.

For the wind to be in motion there must have been something that set it in motion. And so, if I understand the word correctly, something must have “thrust” it into motion.

But once it is in motion, I believe, if there is nothing else acting on it, then we are beyond “thrust” and, at this point, it is simply “in motion.”

No?

Again, if I used the term incorrectly I apologize.

Perhaps I may have seemed to suggest that the wind will last indefinitely, in which case, you may be suggesting that there needs to be “thrust” to maintain this.

If I gave you that impression, I did not mean to.

The wind is in motion. It passes over and under the airfoil. And then it moves on. What happens to it after that (when it eventually comes to a stop) is beyond the scope of this thought experiment.

(And how it was originally set or “thrust” into motion is beyond the scope of this thought experiment.)

?

Edited February 5, 2015 by Zet

studiot · February 6, 2015

Essay

Some small fraction of the "very large" thrust (or whatever your stipulation is for the force imbalance) becomes the source of energy for 'lifting' or slightly counteracting gravity or 'offsetting' some gravitational potential energy.

That is close enough.

+1

J.C.MacSwell · February 6, 2015

The obvious answer is a slowing of the “wind” (and so a decrease in kinetic energy in this form) when the “airfoil” rises which does not also occur when the “airfoil” is held in place and does not (cannot) rise.

---

Why does it not occur anyway? (it occurs more so...but baby steps)

The airfoil is slowing the wind. Drag is being created. What makes you think it slows the wind more by allowing it to rise?

If I do a brake stand in my car, revving the engine at full power, and finally take my foot off the brake, are you going to ask "I wonder where the energy came from to accelerate the car?"

Essay · February 6, 2015

Is it “thrust”?

Did I use that word wrong?

If so, I apologize.

If a ball is rolling across the floor, and so is in motion, and there is nothing else acting on it, is it simply “in motion” or is there “thrust”?

...............but, but, but, ...there was thrust! ...and the energy of a ball simply rolling (with no extra continuous input of energy) couldn't be translated into lifting power. ~imho

...snip....

The wind is in motion. It passes over and under the airfoil. And then it moves on. What happens to it after that (when it eventually comes to a stop) is beyond the scope of this thought experiment.

(And how it was originally set or “thrust” into motion is beyond the scope of this thought experiment.)

?

You explained that very well, and I think I see why you're expecting a different ( or supplemental) explanation for the lifting effect.

Mainly, you're basing this on a "thought experiment" ...and specifically, you've left of the "source" for the force imbalance.

If the "wind" just passes by and "moves on," as you say, then the airfoil won't lift. The wind must be continuous, and of sufficient velocity to be effective.

I think you'll find that to create a constant velocity wind over the path (or the stationary space) of your airfoil, you'll need a source of acceleration somewhere in your system. I'd expect that as soon as the source of acceleration stopped, that the sufficient velocity would become insufficient, and slow enough that the airfoil would no longer be lifted.

But I studied much more biology and chemistry, than physics, so I'm open to more ideas about this. But in "PChem" (or Physical Chemistry) classes (3 semesters), the importance of defining the system, or knowing the parameters of the system, was repeatedly emphasized.

I think that overlooking some parameter or boundary or definition of the system under consideration is a flaw with many thought experiments.

~

MigL · February 6, 2015

Although studiot has attempted to explain things very clearly, I get the impression we need to simplify the thought experiment further.

Get in your car and drive on a highway at 100km/hr. Now roll down the window ( wait till summer ) and stick your arm out such that your hand makes a 30 deg. angle of incidence to the direction of travel. We can look at the situation as either your hand moving through the air or the air moving past your stationary hand, it doesn't matter which.

You will note that you feel two components of force on your hand, a large component pushing your hand backwards ( drag ) and a smaller component pushing your hand upwards ( lift ), and to keep your hand 'stationary' in the frame of the window, you have to exert considerable force down and forward or your hand will go flying back and up. This is the thrust or motive energy you must provide to keep your crude 'airfoil' hand 'stationary' ( at 100 km/hr ) in the frame of the window.

However, if you let your hand rise in the vertical direction, you do not have to supply the downward thrust or push to keep your hand vertically stationary, just the forward component of thrust.

Now since moving hand/stationary air is perfectly equivalent to stationary hand/moving air, we can look at the forces the air applies to the stationary hand and the hand that is allowed to rise. Similar ly to the previous case the air applies less force to the hand if the hand is allowed to rise.

In effect, the air doesn't have to push upwards and loses less energy ( it is deflected less ) than the vertically constrained case..

arc · February 6, 2015

Zet, have you ever walked head on into a strong wind and had to lean into it to keep from being pushed backwards past your center of balance. And then when you turned around you struggled to where you needed to hurry to keep from being push forward onto your face.

That was the kinetic energy in the moving air being converted to pressure.

In your airfoil/fluid dynamic model some of that energy is increased pressure (aka heat) in the fluid near the airfoils surface. It is lost when it passes off the end of the surface and rejoins a lower state, but it is continually replenished at the airfoils leading edge. The difference between the two airfoils is some of that pressure (heat) is exchanged for kinetic when the airfoil rises. All is in balance.

Edited February 6, 2015 by arc

Zet · February 7, 2015

Please excuse me for not initially directly addressing the recent comments, but, rather, let me, if it is okay, restate the issue, in a much more literal way and in light of the recent comments, which will hopefully address them in course and eliminate as much extraneous elements as I am able in order to highlight the specific problem I’m having.

---

There is an L shaped container of fluid.

This is in a closed system and in a frictionless world.

The fluid is incompressible and has zero viscosity.

In the horizontal part of the container there is a submerged solid.

The solid is up above the floor of the container and fixed in place on top of a pole.

The solid is neutrally buoyant (it has the same density as the fluid it displaces).

And, at the end of the horizontal part of the container there is a door and it is closed.

The door is opened and the fluid flows out of the container.

All of the fluid drains out of the container.

In the end, there is a decrease in gravitational potential energy and there is an equal increase in kinetic energy. (Again, it is frictionless world for the ease of analysis.)

Energy is conserved.

The same thing is done again.

But this time, the submerged solid is curved on the top and flat on the bottom.

As the fluid drains out of the container and so moves over and under the submerged solid, due to the shape of the solid there is a greater decrease in pressure on the top and a lesser decrease in pressure on the bottom (as per Bernoulli).

So there is now an overall upward force on the otherwise neutrally buoyant solid.

The solid is fixed to the end of the pole and does not (cannot) move.

All of the fluid drains out of the container.

In the end, there is a decrease in gravitational potential energy and there is an equal increase in kinetic energy.

Energy is conserved.

The same thing is done again for a third time.

But this time the submerged solid is not fixed in place on top of the pole but, rather, is in a frictionless track that allows it to move vertically but keeps it in place horizontally.

As the fluid drains out of the container and so moves over and under the submerged solid, due to the shape of the solid there is a greater decrease in pressure on the top and a lesser decrease in pressure on the bottom (as per Bernoulli).

So there is now an overall upward force on the otherwise neutrally buoyant solid.

It is free to rise and so it will. And after it rises, and before all of the fluid drains out of the container, it is fixed in place at its new height.

All of the fluid drains out of the container.

---

In the second case, where the solid is fixed in place and does not (cannot) rise, in the end, there is a decrease in gravitational potential energy in the form of the falling fluid and there is an equal increase in kinetic energy in the form of the falling fluid. Energy is conserved.

In the third case, there is the same decrease in gravitational potential energy in the form of the falling fluid and there is also an increase in kinetic energy in the form of the falling fluid.

However, in the third case, where the solid rises, in the end, there is also an increase in gravitational potential energy in the form of the risen solid (and there was also an increase the vertical kinetic energy in the form of the rising solid and displaced fluid downwards), and so there must be a decrease in another form of energy does not occur in the second case.

What is it?

Less kinetic energy in the falling fluid? (Is it: the moving fluid slows down when the submerged solid rises as opposed to when it is held in place?).

Less thermal energy (even though there is no friction)? (Is it: there must be a decrease in friction (and so it cannot be stipulated that there is no friction in this thought experiment) when the submerged solid rises as opposed to when it is held in place?)

Less of some other form of energy? (What could it be?)

?

---

If I do a brake stand in my car, revving the engine at full power, and finally take my foot off the brake, are you going to ask "I wonder where the energy came from to accelerate the car?"

If my car is running (and so there is a decrease in chemical potential energy) then there must be an equal increase in another form of energy (or energies) for energy to be conserved.

Whether this increase is in the form of kinetic and thermal energies or if this increase in all just in the form of thermal energy, either way, energy is conserved.

What makes you think it slows the wind more by allowing it to rise?

When the airfoil rises there is an increase in gravitational potential energy (and vertical kinetic energy).

When the airfoil does not rise there is not an increase in gravitational potential energy (and there is not the same increase in vertical kinetic energy).

And so, when the airfoil rises there must be a decrease in another form of energy that does not occur when the airfoil does not (cannot) rise … for energy to be conserved.

What is it?

---

... you've left of the "source" ...

You are right. I did. I thought being abstract and just stipulating the fluid is “in motion” would be enough. Perhaps I was wrong about that. Hopefully this more literal restatement of the case (with the cause of the motion of the fluid, gravitational attraction, included) works better.

---

... angle of incidence to the direction of travel.

If an airfoil has an angle of attack, then the bottom of the airfoil collides with the fluid and the fluid is set in motion downwards and the airfoil is set in motion upwards. There is an increase in vertical motion and an equal decrease in horizontal motion (assuming no friction). Energy is conserved.

... simplify ...

In this thought experiment there is no angle of attack. And this was done on purpose to highlight better the specific issue I’m having with the conservation of energy analysis of a rising airfoil.

... the air doesn't have to push upwards and loses less energy ( it is deflected less ) than the vertically constrained case..

I believe you and I are in agreement that there must be a decrease in the velocity of the wind (and so a decrease in kinetic energy) when the airfoil rises as opposed to when it does not (since there are no other energy change options, ... at least that I can see). And I believe you and are in agreement that the only mechanical reason for this to happen (that I can see) is that there must be an additional amount of drag when the airfoil rises that does not occur when it is held in place (post #47).

Correct?

---

The difference between the two airfoils is some of that pressure (heat) is exchanged for kinetic when the airfoil rises. All is in balance.

I’m not sure I understand what you are saying.

Are you saying that when an airfoil rises (and so there is an increase in vertical kinetic energy and gravitational potential energy) that there is an equal decrease in thermal energy?

If so, I’m not so sure how this would work (mechanically-wise).

---

Thank you all for trying to help me out with this.

---

?

.

Edited February 7, 2015 by Zet

MigL · February 7, 2015

No, I believe I said there is less thrust needed to keep the airfoil 'in place' ( only horizontally ) when it is allowed to rise, than to keep it 'in place' ( horizontally and vertically ) when it is not.

In effect, the air loses less kinetic energy if the airfoil is allowed to rise than if it is not.,

Consider a fan ( vertical airfoils ).

If it is stationary it imparts all its rotational energy to the airstream and accelerates it.

If it is allowed to move, like a propeller, it imparts a lot less energy to the airstream, but rather 'screws' through the air.

Edited February 7, 2015 by MigL

Essay · February 7, 2015

Please excuse me for not initially directly addressing the recent comments, but, rather, let me, if it is okay, restate the issue, in a much more literal way and in light of the recent comments, which will hopefully address them in course and eliminate as much extraneous elements as I am able in order to highlight the specific problem I’m having.

You are right. I did (leave off the 'source'). I thought being abstract and just stipulating the fluid is “in motion” would be enough. Perhaps I was wrong about that. Hopefully this more literal restatement of the case (with the cause of the motion of the fluid, gravitational attraction, included) works better.

===

Thank you all for trying to help me out with this.

Thanks, though I can’t believe I misspelled “off” in that point from post 80.

This new example works as well, or better, since it uses actual fluid. Regardless of the numbers in the examples that you posted, I think you’d find small differences in the fluid exiting the tube.

Especially in terms of overall velocity, along with differences in laminar flow and vorticity, and very slight differences in heat content, the exiting fluid should show that the energy balances. And, as you suggested (that some kinetic and/or thermal differences might occur), they've been listed earlier in the thread in terms of the different turbulence that is generated for the two cases of a stationary or ‘sliding’ airfoil. Turbulence, or the lack of it, is where your 'missing' energy resides, I suspect.

And speaking of turbulence, you should be able to keep your system “frictionless” for the most part (laminar flow and sliding airfoils), but only include friction as it relates to generating turbulence and the effects of turbulence.

~

Edited February 7, 2015 by Essay

arc · February 7, 2015

I’m not sure I understand what you are saying.

Are you saying that when an airfoil rises (and so there is an increase in vertical kinetic energy and gravitational energy) that there is an equal decrease in thermal energy?

If so, I’m not so sure how this would work (mechanically-wise).

In the example I gave you, when you walk towards or just stand against the wind you are in opposition of the winds kinetic energy, and you can physically experience this energy as pressure on your body.

But this fluid pressure is also a function of heat and heat is, as you know, caused by friction. So when the fluid is interacting with the airfoil the wing's particular design will determine how much of this kinetic energy will be converted to pressure and heat. Some of its kinetic energy is converted to physical pressure and by that heat from friction by the fluid's molecules that are very near or in contact with the airfoils surface.

When you turned around and walked in the direction that the wind was blowing you had reduced the pressure that you experienced from the wind. If you were in a sailboat you would be saying that you converted some of the energy of pressure that was on the sail from being tied to a dock into kinetic by allowing the sailboat to move with the wind or the path of least resistance, the pressure on the sail would go down as the sailboat's speed increased just as you had done by walking in the same direction the wind was moving.

The airfoil does a similar thing. The airfoil is a machine that creates its own lift. It does this by manipulating the fluid environment around itself. When it interacts with a high velocity fluid it produces a path of least resistance for itself in a vertical direction. As it moves upwards its direction's kinetic energy is derived through a reduction of pressure (aka heat from friction) from the moving fluid. The wings surface experiences less pressure and by that less heat as it moves up. But here's the problem you have. The gravitational potential energy you want to attain in your airfoil must be provided by energy input. Some type of energy must replace the energy that was used for that lift and the forward trust that allows the airfoil to operate its velocity dependent mechanism.

You are trying to define a partial system or a subset, and you have removed certain aspects of physical reality to simplify your model to your liking. When you do this it becomes fictitious and unreliable, more for your entertainment and less for everyone else's benefit or reward.

Everything Studiot has told you was first rate science and answered your questions, and more important was for your benefit. Please reread his posts.

There is an L shaped container of fluid.

This is in a closed system and in a frictionless world.

The fluid is incompressible and has zero viscosity.

In the horizontal part of the container there is a submerged solid.

The solid is up above the floor of the container and fixed in place on top of a pole.

The solid is neutrally buoyant (it has the same density as the fluid it displaces).

And, at the end of the horizontal part of the container there is a door and it is closed.

The door is opened and the fluid flows out of the container.

All of the fluid drains out of the container.

In the end, there is a decrease in gravitational potential energy and there is an equal increase in kinetic energy. (Again, it is frictionless world for the ease of analysis.)

Energy is conserved.

The same thing is done again.

But this time, the submerged solid is curved on the top and flat on the bottom.

As the fluid drains out of the container and so moves over and under the submerged solid, due to the shape of the solid there is a greater decrease in pressure on the top and a lesser decrease in pressure on the bottom (as per Bernoulli).

So there is now an overall upward force on the otherwise neutrally buoyant solid.

The solid is fixed to the end of the pole and does not (cannot) move.

All of the fluid drains out of the container.

In the end, there is a decrease in gravitational potential energy and there is an equal increase in kinetic energy.

Energy is conserved.

The same thing is done again for a third time.

But this time the submerged solid is not fixed in place on top of the pole but, rather, is in a frictionless track that allows it to move vertically but keeps it in place horizontally.

As the fluid drains out of the container and so moves over and under the submerged solid, due to the shape of the solid there is a greater decrease in pressure on the top and a lesser decrease in pressure on the bottom (as per Bernoulli).

So there is now an overall upward force on the otherwise neutrally buoyant solid.

It is free to rise and so it will. And after it rises, and before all of the fluid drains out of the container, it is fixed in place at its new height.

All of the fluid drains out of the container.

---

In the second case, where the solid is fixed in place and does not (cannot) rise, in the end, there is a decrease in gravitational potential energy in the form of the falling fluid and there is an equal increase in kinetic energy in the form of the falling fluid. Energy is conserved.

In the third case, there is the same decrease in gravitational potential energy in the form of the falling fluid and there is also an increase in kinetic energy in the form of the falling fluid.

However, in the third case, where the solid rises, in the end, there is also an increase in gravitational potential energy in the form of the risen solid (and there was also an increase the vertical kinetic energy in the form of the rising solid and displaced fluid downwards), and so there must be a decrease in another form of energy does not occur in the second case.

What is it?

Less kinetic energy in the falling fluid? (Is it: the moving fluid slows down when the submerged solid rises as opposed to when it is held in place?).

Less thermal energy (even though there is no friction)? (Is it: there must be a decrease in friction (and so it cannot be stipulated that there is no friction in this thought experiment) when the submerged solid rises as opposed to when it is held in place?)

Less of some other form of energy? (What could it be?)

In your model above, the second one cannot move up so there is an increase in drag by the moving fluid as the fluid attempts to utilize the airfoils design. The airfoil pulls against its mounting like a kite would pull on its string. There is a increase of friction on the airfoil and this means heat was generated. The drag was slowing down the nearby fluid. So in the second example; a very small portion of the containers kinetic or gravitational energy was converted to heat.

In the third example you added a gravitational potential energy storage device which would be the same as just removing the airfoil from the container once it rose to the top. It should have been allowed to return to the bottom and thus balance the system in the end. The way it is, it was raised and it still has gravity acting on it. And we all know, eventually, gravity always wins. At some point in the future that system will decay and the airfoil or just its atoms will fall. So all you did was delay the experiment until this happens.

studiot · February 7, 2015

arc

In the third example you added a gravitational potential energy storage device which would be the same as just removing the airfoil from the container once it rose to the top. It should have been allowed to return to the bottom and thus balance the system in the end. The way it is, it was raised and it still has gravity acting on it. And we all know, eventually, gravity always wins. At some point in the future that system will decay and the airfoil or just its atoms will fall. So all you did was delay the experiment until this happens.

arc, your comments would be fine except for one thing.

Yes I said earlier that a glider will eventually return to Earth (thereby loosing its potential energy) and that is true here as well.

However Zet's (correct) point was that this energy must have come from the fluid and therefore the fluid must have lost some energy.

The point is that the story does not stop there, since some agent must have input the energy into the fluid in the first place or it would not be moving.

Further continued lift requires a continued supply of moving fluid and therefore continued external energy input.

Edited February 7, 2015 by studiot

cxxLjevans · February 7, 2015

arc, your comments would be fine except for two things.

1)Yes I said earlier that a glider will eventually return to Earth (thereby loosing its potential energy) and that is true here as well.

However Zet's (correct) point was that this energy must have come from the fluid and therefore the fluid must have lost some energy.

But for the second thing. Zet is also adamant that he has specified zero fluid friction and here his understanding of fluid mechanics falls down.

I thought friction was caused by viscosity and 0 viscosity (and 0 friction, IIUC) is used to derive Euler's equation:

http://en.wikipedia.org/wiki/Inviscid_flow#Reynolds_number

Since Euler assumed frictionless flow was OK, I can see why Zet assumed it was.

Edited February 7, 2015 by cxxLjevans

J.C.MacSwell · February 7, 2015

I think it is not uncommon to assume lift and associated (vortex inducing) drag for airfoils with inviscid flow, by making assumptions about the separation points and wake (which is outside of the potential flow and cannot be entrained with it, as there are no shear forces and no friction), for the purpose of analyzing the flow outside the boundary layer of a real airfoil in steady flow.

The assumptions are necessary (for the reasons stated by Studiot, without them no drag and no lift), and with them you can get a pretty good approximation of the lift to induced drag, above a baseline of skin friction and form drag.

When an airfoil is allowed to rise this reduces it's effective and apparent angle of attack (I prefer to take the no lift angle as zero, which is different from Zet's, but either can be used) Within normal range, until the stall point is approached, induced drag is close to linear correlation with lift and angle of attack.

The problem is assuming no induced drag (no lift associated drag). That's free lunch. You are assuming energy conservation does not hold from the start, so of course energies won't balance...you can use turbines and propellors based on this and design perpetual motion machines...free lunch.

Zet, for your L shaped thought experiment...

Assuming you get lift and associated drag, but otherwise neglect skin friction and form drag:

When the airfoil is constrained that is the highest lift case...and highest drag. The result is lower exit velocity of the flow, but higher kinetic energies remaining overall, due to vortices from the induced drag associated with the higher lift.

When the airfoil is allowed to rise that is a lesser lift case...and lesser drag. The result is higher exit velocity of the flow, though less K.E. overall, due to less vortices from the reduced induced drag associated with lift... and the potential gained from raising the airfoil being the difference.

Edited February 7, 2015 by J.C.MacSwell

arc · February 7, 2015

And yet to shout at me for saying that friction is necessary.

What is (almost) axiomatic is that any body that moves away from the commen centre of mass will gain gravitational potential energy, whatever the cause of of that movement.

The gravitational field is a conservative field, which means that the difference of potential energy of any body between two positions depends only upon the positions and not on the path taken to reach them.

That is axiomatic in potnetial theory.

You seem so desperate to prove me ( and the rest of physics wrong) that you are not listening to what I said.

There is a difference in the energy flows for the system, the rising body and the fluid (which incidentally I have never accepted neceessarily 'falls'. No fluid falls if you blow it horizontally past an airfoil.).

You are confusing all these energy flows.

Which is why I keep repeating that an energy analysis is the difficult way.

In the rising case it does.

This is what I mean about describing a partial model and removing certain aspects of physical reality to simplify it. The fluid moving down does so due to gravity. Without gravity you would not have a buoyant system to study in the first place but since we can disconnect certain aspects lets pretend we can have a buoyant like movement vertically through a fluid without gravitational effects on the fluid. Well I see a lot of fluid being ejected out the top of the model or maybe to the sides. So this doesn't work, gravity must be taken into account through the pressure and the friction it imposes on the model whether it is vertical or horizontal.

studiot · February 7, 2015

This is what I mean about describing a partial model and removing certain aspects of physical reality to simplify it. The fluid moving down does so due to gravity. Without gravity you would not have a buoyant system to study in the first place but since we can disconnect certain aspects lets pretend we can have a buoyant like movement vertically through a fluid without gravitational effects on the fluid. Well I see a lot of fluid being ejected out the top of the model or maybe to the sides. So this doesn't work, gravity must be taken into account through the pressure and the friction it imposes on the model whether it is vertical or horizontal.

Airfoil action is not to do with gravity.

Discussion of bouyancy only clouds the issue airfopil lift and bouancy are very different and not mutually exclusive.

Air moves up as well as down, hence the term circulation.

It is the circulation that cause the lift and the interaction of the obstruction (airfoil) and the moving fluid that causes the circulation.

Circulation is theoretically possible in a frictionless fluid, but uncontrollable.

It is friction that allows us to control the circulation, thereby achieving consistent and sustained lift.

I have been trying to find a simple explanation wihtout higher mathematics.

Here are some nice animations.

http://www.diam.unige.it/~irro/profilo4_e.html

arc · February 7, 2015

Airfoil action is not to do with gravity.

Discussion of bouyancy only clouds the issue airfopil lift and bouancy are very different and not mutually exclusive.

Air moves up as well as down, hence the term circulation.

It is the circulation that cause the lift and the interaction of the obstruction (airfoil) and the moving fluid that causes the circulation.

Circulation is theoretically possible in a frictionless fluid, but uncontrollable.

It is friction that allows us to control the circulation, thereby achieving consistent and sustained lift.

I have been trying to find a simple explanation wihtout higher mathematics.

Here are some nice animations.

http://www.diam.unige.it/~irro/profilo4_e.html

I agree completely with that. I should have reference the particular post, #62, your "block in a bucket"

"Your block-in-a-bucket here is an example of an isolated system.

This means that neither any mass nor any energy is allowed into or out of the system."

And zet's#63;

However, if the rising object is less dense and if the fluid displaced downwards is more dense then there is rather an acceleration and a gain in kinetic energy and a loss in gravitational potential energy.

So, no, I am not willing to “accept” that in the case of a solid rising in a fluid due to the upward buoyant force on it that there is a gain in gravitational potential energy.

My apologies Studiot I was thinking that exchange was still in the "block-in-a-bucket" model. I get a headache trying to discern the differences with each new example. I would rather just work the one (airfoil) over and over than go from an airfoil to a primarily buoyant example where something may slip by the analysis (pun). I think I better just sit back and be a spectator on this one.

studiot · February 7, 2015

One very important question to answer, if you are considering an energy analysis.

It is easy to state that the 'energy comes from the fluid',

but which bit of fluid?

The fluid above, below or to the side of the airfoil?

The fluid that has yet to arrive at the leading edge of the airfoil?

The fluid that has already passed the airfoil?

The fluid that is just passing the airfoil?

All the fluid (in the atmosphere)?

Some combination of these?

It is easy to state that the energy loss causes the fluid to slow down.

But wait, if the fluid that has passed the airfoil is now moving more slowly than the new fluid arriving, then fluid is building up somewhere in the system.

But the fluid has been declared incompressible.

Continuity (Laplaces equation) is violated.

And if the energy is being transferred from the moving fluid to the airfoil, what proof is offered that the agent driving the fluid doesn't simple drive harder (ie add more energy to the fluid) to replace the transferred energy?

Any who really want to understand the energy interactions of flight need to be able to consider and answer these questions and others.

Controlled flight is a very complicated system and attempts to simplify usually remove some essential link in the chain, thereby breaking the whole chain.

I think this is the problem with Zet's attempt at simplification.

Edited February 7, 2015 by studiot

Essay · February 8, 2015

As I mentioned earlier, you can balance the energy by accounting for slight differences in the exit fluid, between each different example, mostly in the form of velocity and vorticity (turbulence); and that friction can’t be totally ignored (especially regarding turbulence and its effect on heating …and lift), except where ignoring friction will acceptably remove various, design-related, or materials-related, confounding factors. We all seem to agree on this generally; however….

While using the words ‘turbulence’ and ‘vorticity’ synonymously, on my part, might have been confusing too;

I’m wondering about the terminology describing how the “energy comes from the fluid.”

Wouldn't it be better to speak about how the distribution of the fluid imparts force, rather than speaking about the fluid as if it was transferring some electrical energy or some other, unknown, ‘anit-grav’ energy, via direct contact with the airfoil?

Energy moves the fluid (or airfoil), and forces are transferred, and then we can look at what energy was required to redistribute the fluid that created those forces (and created effects such as lift) to see how it all balances; but to say the energy comes from the fluid …seems misleading to me.

But now that I've struggled to express that nuance, I can see why it probably doesn't change much for explaining where the “missing” or ‘hidden’ or lifting energy comes from

...or goes to.

Never mind the semantics….

~

Zet · February 8, 2015

the air loses less kinetic energy if the airfoil is allowed to rise than if it is not.,

I have no reason to doubt you. But, if you are right, this, I believe, only compounds my problem.

This would mean that, in the end, in the case where there is more gravitational potential energy there is also more kinetic energy and, in the end, in the case where there is less gravitational potential energy there is also less kinetic energy.

No?

---

Turbulence, or the lack of it, is where your 'missing' energy resides, I suspect.

I have a bad habit of streamlining the description of the example in order to get to, highlight, the part of it I’m having trouble with (e.g. “simply stipulating the fluid is in motion and not including how is was set in motion”). I think this more literal approach seems to work better.

I know that there is not just horizontally moving fluid after it encounters the airfoil but also swirling fluid (and downwardly moving fluid). (By just illustrating horizontally moving air I was just trying to (simplify and) get to the point that the fluid must be moving faster in the one case and slower in the other case after encountering the airfoils for energy to be conserved … as it is the only energy change option I can see). The point (that I was trying to make) is the motion (in whatever direction and of whatever type) of the fluid must be less in the one case and more in the other case to offset the different amounts of gravitational potential energies in the two cases in the end (unless there is another energy change option available and I just can’t see it).

I hope this more literal drawing works better.

---

.

But this fluid pressure is also a function of heat and heat is, as you know, caused by friction

In a friction filled world, when a fluid is in motion over the surface of a solid, the faster the relative velocity means the greater the skin friction and so the more thermal energy is generated.

If a body remains at rest while a fluid flows around it then the relative velocity between the fluid and the solid is greater than if the solid moves along somewhat with and in the direction of the moving fluid. And so if it remains at rest (again, in a friction filled world) there is more skin friction and so more thermal energy generated and if it moves along with the fluid somewhat there is less skin friction and so less thermal energy generated.

Friction both generates thermal energy and decreases kinetic energy. And so, in the resting case there is a greater increase in thermal energy and an equal greater decrease in the kinetic energy of the fluid and in the moving along somewhat case there is a lesser increase in thermal energy and an equal lesser decrease in the kinetic energy of the fluid.

Energy is conserved.

And if a body remains at rest while a fluid flows over it there is no increase in kinetic energy in the form of the moving body while if moving fluid sets the body into motion along with it (if the force from the pressure from the fluid on the body is the cause of that motion) then there is an increase in kinetic energy in the form of the moving solid and there is an equal decrease in kinetic energy in the form of the moving fluid.

Energy is conserved.

Are you saying that the more or less skin friction there is between a fluid and a solid the more or less pressure there is from that fluid on that solid?

(If so, I’ve never heard of this. Do you have a link?)

I read (and reread) your post several times, but I didn’t understand the point of “fluid pressure is also a function of heat.”

I have never heard of this before. And I would be grateful for a link (!).

(The temperature of a fluid affects the density of the fluid and so, in turn, affects the pressure of that fluid at a given depth. But I don’t think you’re saying this?)

(And if I have misunderstood you, I apologize. I’m trying.)

As it moves upwards its direction's kinetic energy is derived through a reduction of pressure (aka heat from friction) from the moving fluid.

Are you saying that when a (initially) horizontally moving fluid encounters a horizontally fixed airfoil there is a decrease in the amount of skin friction between them if that airfoil moves vertically that does not occur when the airfoil remains also vertically fixed in place and so less thermal energy is generated when the airfoil rises?

If so (but I don’t believe this is or could be the case, unless the velocity of the fluid slows down in the rising case in a way that it doesn’t in the non-rising case (which is what I’ve been saying I think must occur for energy to be conserved all along)) then, since friction both increases thermal energy and decreases kinetic energy, in the non-rising case there would be a decrease in kinetic energy in the form of the moving fluid and an equal increase in thermal energy, while in the rising case there would be a lesser decrease in kinetic energy in the form of the moving fluid and an equal lesser increase in thermal energy. (Energy is, would be, conserved.)

If this is what you’re saying (that there less skin friction between the moving fluid and the airfoil when the airfoil vertically rises) then this would balance out in and of itself.

---

In this thought experiment, the horizontally fixed and otherwise neutrally buoyant airfoils have different decreases in pressures on the top and bottom of them (as per Bernoulli) when the fluid is in motion, which, in turn, creates an overall upward (dynamic buoyant) force. If allow to rise, the airfoil will.

And when the airfoil rises (due to this “dynamic buoyant force”) there is an increase in vertical kinetic energy and in gravitational potential energy that does not occur when the airfoil remains in place.

For energy to be conserved there must be a decrease in another form of energy in the rising case that does not occur in the non-rising case.

What is it?

So in the second example; a very small portion of the containers kinetic or gravitational energy was converted to heat.

Yep (in a friction filled world).

And this balances out in and of itself.

In the third example you added a gravitational potential energy storage device which would be the same as just removing the airfoil from the container once it rose to the top. It should have been allowed to return to the bottom and thus balance the system in the end. The way it is, it was raised and it still has gravity acting on it. And we all know, eventually, gravity always wins. At some point in the future that system will decay and the airfoil or just its atoms will fall. So all you did was delay the experiment until this happens.

No.

In this thought experiment, in the end, there is an increase in gravitational potential energy in the one case that does not occur in the other. And so, for energy to be conserved, there must be a decrease in another form of energy in the one case that does not occur in other.

If, after the end of this thought experiment, the one higher up airfoil then falls back down to its original position (where the other airfoil remained) then this greater amount of gravitational potential energy in the one system than in the other becomes a greater amount of kinetic energy in the one system than in the other. (And, you could take it a step further, and you could point out when the falling airfoil comes to a stop there is a decrease in kinetic energy. But there will be an equal increase in thermal energy. And, so, in the one system is more thermal energy than in the other ... unless there was a decrease in another form of energy that does not occur in the other when the imbalance between the two cases in gravitational potential energy was first created.)

This simply begs the question.

When the airfoil rises there must be a decrease in another form of energy that does not occur when it does not … for energy to be conserved.

What is it?

---

Since Euler assumed frictionless flow was OK, I can see why Zet assumed it was.

Whether it is acceptable to stipulate zero viscosity in a thought experiment or not, I don’t know. I trust you that it is.

But, if it’s not, it really doesn’t change anything. The logic, and the issue, remain the same.

It just makes this thought experiment slightly more complicated if we have to stipulate that “the fluid has the lowest amount of viscosity theoretically possible, which is something greater than zero” rather than “the viscosity of the fluid is zero.”

If we stipulate the former, then since the viscosity is so minuscule its impact on this system will then be so minor and so it can be justifiably ignored for the ease of analysis, while if we stipulate the latter, then since there is no viscosity to factor in then it is also not a factor.

The logic, and the issue, remain the same.

---

When the airfoil is allowed to rise that is a lesser lift case...and lesser drag. The result is higher exit velocity of the flow, though less K.E. overall, due to less vortices from the reduced induced drag associated with lift... and the potential gained from raising the airfoil being the difference.

This could be it!

And so my illustrated simplification of having the fluid after it encounters the airfoils continue on only horizontally was not just trivial, it is a problem.

So, if I understand you correctly, in the end there is overall more motion the non-rising case and there is overall less motion in rising case.

And it’s just that the difference in kinetic energy between the two cases necessary for energy to be conserved is found in the swirl of the fluid itself (less by the difference in horizontal motion).

And, again if I understand you correctly, while, in the end, there is a greater decrease in the horizontal motion of fluid in the non-rising than in the rising case, this is offset and surpassed by an greater increase in the swirling of the fluid in itself in the non-rising case and a lesser increase in the swirling of the fluid in the rising case.

And so, in the end, there is less kinetic energy in the rising case (in the form of the moving fluid in all its various motions) and more kinetic energy in the non-rising case (in the form of the moving fluid in all its various motions) and this offsets the different amounts of gravitational potential energy between the two cases (and by a precisely equal amount).

Ta-da!

If this is right, then this resolves this portion of the conservation of energy analysis!

Thank you very much.

Do you have a link(s) for this?

---

Now, the big question is: does anyone else think J.C.MacSwell (post #89) is right?

And, if so, does anyone have a link for that?

?

(Essay: I wrote this before reading your post #94. I will read it now and respond to it next time.)

Cheers!

arc · February 8, 2015

I'm sorry Zet but I feel I am causing you and your quest for an answer more harm than help. I am going to just step back and learn from the others here. This has become complicated and I feel has gone well beyond my very humble abilities. You should understand that even though this complex and dynamic process is difficult to put into conversational language it is well represented mathematically and proved in every successful flight. I believe Studiot and the others will succeed in furnishing you with an adequately derived response. Good luck.

Edited February 8, 2015 by arc

Essay · February 8, 2015

I have a bad habit of streamlining the description of the example in order to get to, highlight, the part of it I’m having trouble with (e.g. “simply stipulating the fluid is in motion and not including how is was set in motion”). I think this more literal approach seems to work better.

I know that there is not just horizontally moving fluid after it encounters the airfoil but also swirling fluid (and downwardly moving fluid). (By just illustrating horizontally moving air I was just trying to (simplify and) get to the point that the fluid must be moving faster in the one case and slower in the other case after encountering the airfoils for energy to be conserved … as it is the only energy change option I can see). The point (that I was trying to make) is the motion (in whatever direction and of whatever type) of the fluid must be less in the one case and more in the other case to offset the different amounts of gravitational potential energies in the two cases in the end (unless there is another energy change option available and I just can’t see it).

I hope this more literal drawing works better.

... I agree... [snip]...

This could be it!

And so my illustrated simplification of having the fluid after it encounters the airfoils continue on only horizontally was not just trivial, it is a problem.

So, if I understand you correctly, in the end there is overall more motion the non-rising case and there is overall less motion in rising case.

And it’s just that the difference in kinetic energy between the two cases necessary for energy to be conserved is found in the swirl of the fluid itself (less by the difference in horizontal motion).

And, again if I understand you correctly, while, in the end, there is a greater decrease in the horizontal motion of fluid in the non-rising than in the rising case, this is offset and surpassed by an greater increase in the swirling of the fluid in itself in the non-rising case and a lesser increase in the swirling of the fluid in the rising case.

And so, in the end, there is less kinetic energy in the rising case (in the form of the moving fluid in all its various motions) and more kinetic energy in the non-rising case (in the form of the moving fluid in all its various motions) and this offsets the different amounts of gravitational potential energy between the two cases (and by a precisely equal amount).

Ta-da!

If this is right, then this resolves this portion of the conservation of energy analysis!

Thank you very much.

Do you have a link(s) for this?

---

Now, the big question is: does anyone else think J.C.MacSwell (post #89) is right?

And, if so, does anyone have a link for that?

....

Cheers!

I agree that seems to be the correct answer to your question. It's helped me to understand this better also, so thanks for pursuing the details on the KE and its partition into the velocity and turbulence (swirl) of the exit fluid. I was sure that you'd find the answer (or balance of energy) there in the vortices of the exit fluid, but biology is more my focus, so I don't know about good citations for this type of physics.

You have great drawings too. For the two cases above, however, I suspect the turbulence that you show would be different, depending on whether the airfoil is fixed or can lift freely ...as several have discussed to some degree already.

If you find any more information about this or have more thoughts, keep us posted ...it's always good to clarify old ideas or learn something new.

~

Edited February 8, 2015 by Essay

studiot · February 8, 2015

As regards to the 'experiment' in post#83

Thermal transfer due to friction and turbulence are simply red herrings.

In this experiment the OP has chosen to soapbox and simply demonstrated his misunderstanding of energy conservation in general and Bernouilli's theorem in particular.

This experiment can be addressed by a correct application of the simple form of Bernoulli's theorem, containing three terms, such as would be taught in the first semester of an engineering fluids course.

I very much doubt that turbulence is involved. Does anyone know if Bernoulli's theorem can be applied to turbulent motion?

Bernoulli's theorem is an energy balance for any point in the fluid and it states that:

The sum of all the energies of the fluid = a constant minus any work done by the fluid

In this case there are three energy terms

The static head, the dynamic head and the gravitational head.

At the outset the first two are zero and no work is being done.

This establishes the value of the constant.

When the fluid is initially allowed to flow no work is done and all the gravitational energy head energy is exchanged for potential and dynamic head.

When the obstruction is released if it rises at all, and it may not, some work is done and so the constant in the Bernoulli equation is reduced by this amount.

Now, oh wizards of fluids 101,

1) Is it possible to solve this system and distribute the changes in the three fluid energy terms without further information other than fluid physical properties?

2) I noted that the obstruction may not rise at all, even when it is free to do so. What controls this?

As regards to thermal transfer.

Consider an aircraft that takes off from the ground at 20^oC and climbs to 30,000 feet.

Why does the pilot worry about wing icing if the object is heated by friction?

I look forward to Zet's correct Bernoulli analysis.

cxxLjevans · February 8, 2015

[snip]

> To do this look at my diagram. I have drawn two

> streamlines BC and AD and two more (B'C' and A'D') exactly

> 1 metre alongside so they form a square box section stream

> tube. The streamlines are above and below a 1 metre

> section of airfoil. So everything in the third dimension

> is measured per metre.

[snip]

> Both sections have the same area so

>

> U1x = U2x

>

> This is the obstruction does not slow the fluid

> horizontally, in accordance with D'Alembert. What does

> happen is a vertical downward velocity Uy is imparted to

> the fluid.

How? Maybe this:

http://en.wikipedia.org/wiki/Lift_%28force%29#Flow_deflection_and_Newton.27s_laws

provides an explanation for the downward velocity; however,

I haven't read it closely. I'd guess a downward flow is

caused by a positive angle of attack, like that experienced

by your flattened hand angled up and held outside of a

speeding car window.

In addition, this page:

http://www.grc.nasa.gov/WWW/K-12/airplane/shed.html

shows a "downwash" which looks like it *may* correspond to

your Uy at the exit to your stream tube.

However, my question (and I suspect Zet's ) is what happens

when there's 0 angle of attack and the only lift is provided

by the difference in the static presssure between the upper

and lower surfaces of the airfoil. Bernoulli said there'd

be a change in pressure caused by the fluid moving faster

over the top and that produces the lift. And (like Zet),

I'm wondering where the energy comes from causing that rise?

Does the fluid flowing under the wing slow down or decrease

in pressure or both in order to compensate for the energy

required to raise the foil?

>

> The mass flowrate equals the fluid density times the

> volumetric flowrate,

>

> and the momentum flowrate equals the mass flowrate times

> the horizontal or vertical velocity.

>

> At section1 there is zero vertical momentum in the fluid

> but at section 2 there is a downwards momentum flow of

>

> (pUxA2)Uy

>

Now, if we assume the flow conditions used to derive

Bernoulli's equation:

* Velocity head

* Elevation head

* Pressure head

remain constant (where these terms are as defined here:

http://en.wikipedia.org/wiki/Hydraulic_head

)

then, since the Velocity head has increased (since Ux

remains constant but Uy increases from 0) the pressure head

must decrease to compensate. Hence, the airfoil is sucked

backward (I guess you could say that is part of the drag).

> But a momentum flow is the rate of change of momentum and

> this is the definition of a force.

>

> The force exerted on the airfoil is the lift force and

> this is equal to in magnitude but opposite in direction to

> this force on the fluid.

>

> So the lift force = Fy = -(pUxA2)Uy

>

What about the difference in pressure between the upper

surface of the air foil and the lower surface of the

airfoil? I thought this was the reason given by the

Bernoulli equation to explain lift. Indeed, the page:

http://en.wikipedia.org/wiki/Lift_%28force%29#Increased_flow_speed_and_Bernoulli.27s_principle

claims:

For any airfoil generating lift, there must be a pressure

imbalance, i.e. lower average air pressure on the top than

on the bottom. Bernoulli's principle states that this

pressure difference must be accompanied by a speed

difference.

-regards,

Larry

studiot · February 8, 2015

Thank you , Larry for your response.

Larry Evans

How? Maybe this

and

provides an explanation for the downward velocity; however,
I haven't read it closely. I'd guess a downward flow is
caused by a positive angle of attack, like that experienced
by your flattened hand angled up and held outside of a
speeding car window.

I wonder why you did not read it closely?

Here is an extract from the generally lucid Wiki article:

so that the airfoil exerts a downward force on the air as it flows past

Perhaps you would like to stop and ponder how or what it means for a passive solid body to 'exert a force on the air'

The look again at my post#93 (it was a short one).

Perhaps this is why momentum is considered rather than force in fluid texts?

I have not said that there is no downwash.

Quite the opposite, although I did not call it that.

The downwash is the 'down' part of the circulation (surprise surprise)!

Pressure differences above and below the obstruction, in front of it and bhind it as well?

Of course there are.

I have never said any different.

But what pressure are you discussing,

Dynamic pressure or static pressure?

And are you aware of the huge variation of these along the section of the obstruction?

There are some much better diagrams in my link in post#91 than the ones in the Wiki article.

As regards the Wiki article application of Bernoulli's theorem, they are discussing horizontal flow in an infinite atmosphere.

This is not the correct statement of the theorem or application of it to post#83.

As a matter of interest the correct application of Bernoulli's theorem in the Wiki case should include the circulation, a point often missed in explanations including Wiki itself. This Wiki case is actually much more difficult that that of post#83.

You need to fully understand Bernoulli's theorem to apply it correctly.

What did you make of the questions I posed in post#98, in particular this one?

Studiot

This experiment can be addressed by a correct application of the simple form of Bernoulli's theorem, containing three terms, such as would be taught in the first semester of an engineering fluids course.

I very much doubt that turbulence is involved. Does anyone know if Bernoulli's theorem can be applied to turbulent motion?

A while ago we had a very long thread by someone who insisted that a particular 'explanation' of airfoil action was due to certain physical laws only and the others were wrong or didn't apply.

Of course the truth is that they all always apply so if one apparantly contradicts another there is something amiss with the analysis.

Sensible persons use the easiest set of laws to obtain the answer, not the most loved or the most difficult.

This is the situation in many walks of mechanics.

Edited February 8, 2015 by studiot

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