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airfoil rising


Zet

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Zet, please note that you have to restrain the object both against the drag and the lift in the restrained case.

JC has the right of it +1

 

However please also note we are talking about the lift and drag of aircraft in the atmosphere.

 

The aircraft has kinetic energy, which is replaced by the engine when under power.

When gliding the aircraft looses both gravitational and kinetic energy, eventually coming to rest on the ground.

No glider flies forever.

 

You are describing the action in a wind tunnel, where the energy comes from the fan that drives the airflow.

In the free atmosphere the energy comes from the natural forces that created the wind, which (I think) are usually temperature differentials.

 

As regards you question comparing the energy in lifting a heavy object v a lighter one.

 

The lift force is determined by geometry so will not be able to lift a heavy object so far or so fast ( as you have shown in your diagrams) so energy conservation is preserved.

That's right.

 

In the frame of still air my case 1 would be equivalent to a heavier aircraft flying level at constant speed and case 2 would be equivalent to a lighter aircraft gaining altitude but otherwise at the same horizontal velocity.

 

The heavier aircraft would require more power, and of course be adding more kinetic energy to the still air with respect to that frame than in case 2.

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... so far or so fast ( as you have shown in your diagrams) ...

 

 

 

How far do I say (or show) the airfoils rise?

 

And how fast do I say (or show) the airfoils rise?

 

Huh?

 

 

 

 

 

... under power ...

 

 

... more power ...

 

 

 

In the restatement of the issue (please see post #46 in this thread) the air moves horizontally and not the airfoil. There is no “thrust” or “power” moving the airfoil as part of the conservation of energy analysis of the issue in this form.

 

 

 

1. Generally speaking, if you restrain an airfoil both horizontally and vertically you will have higher drag than the same set up where the airfoil rises.

 

2. When the airfoil rises there will be slightly less angle of attack from the reference frame of the wing, so induced drag should be reduced. Less energy will be removed from the free stream, so the overall velocity will remain higher than in the first case

 

Energy is of course conserved in each case. The gravitational/potential energy gain in the second case is accounted for as a loss of the free stream energy, along with other losses associated with drag, which together will be less than the drag associated losses in the first case.

 

 

If I understand you correctly, you are saying that when the airfoil is prevented from rising the velocity of the moving air decreases more and when the airfoil is allowed to rise the velocity of the moving air decreases less.

 

Is that right?

 

503_ar09.gif

 

If, in the end, there is more kinetic energy in the case where there is also more gravitational potential energy and if, in the end, there is less kinetic energy in the case where there is also less gravitational potential energy (... if I understand you correctly ...) then how do you reach the conclusion that the conservation of energy analysis has been resolved and the energy in this scenario has been shown to be conserved?

 

?

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If I understand you correctly, you are saying that when the airfoil is prevented from rising the velocity of the moving air decreases more and when the airfoil is allowed to rise the velocity of the moving air decreases less.

 

Is that right?

 

503_ar09.gif

 

If, in the end, there is more kinetic energy in the case where there is also more gravitational potential energy and if, in the end, there is less kinetic energy in the case where there is also less gravitational potential energy (... if I understand you correctly ...) then how do you reach the conclusion that the conservation of energy analysis has been resolved and the energy in this scenario has been shown to be conserved?

 

?

First bolded...that's correct.

 

Second bolded...I am not showing energy is conserved, I am assuming the law of conservation of energy holds, and know there are generally frictional type losses, inefficiencies, in any real system. No thought experiment will prove it wrong.

Edited by J.C.MacSwell
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First bolded...that's correct.

 

Second bolded...I am not showing energy is conserved, I am assuming the law of conservation of energy holds, and know there are generally frictional type losses, inefficiencies, in any real system. No thought experiment will prove it wrong.

Can't the problem be simplified by:

1) making the fluid incompressible (as in a hydrofoil instead of an airfoil)

2) making the hdrofoil density the same as fluid density

3) making the fluid viscosity = 0 (no energy loss due to friction)

 

IIUC, 1) would avoid any change in fluid energy due to

compression (in air, change in internal energy is P dV).

Also, 3) would avoid change in fluid energy due to frictional heating.

2) would make the gravitational potential energy constant.

1. Generally speaking, if you restrain an airfoil both horizontally and vertically you will have higher drag than the same set up where the airfoil rises.

 

2. When the airfoil rises there will be slightly less angle of attack from the reference frame of the wing, so induced drag should be reduced. Less energy will be removed from the free stream, so the overall velocity will remain higher than in the first case

 

Energy is of course conserved in each case. The gravitational/potential energy gain in the second case is accounted for as a loss of the free stream energy, along with other losses associated with drag, which together will be less than the drag associated losses in the first case.

 

How is the loss of the free stream energy, mentioned here:

 

gravitational/potential energy gain in the second case is accounted for as a loss of the free stream energy,

 

 

 

accomplished. Does the velocity of the free stream decrease with respect to its

value before encountering the foil. If so, then this is, essentially, what Zet said in post#46 which

says:

 

when the airfoil rises there must be a decrease in the horizontal velocity of the wind

 

-regards,

Larry

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Zet

How far do I say (or show) the airfoils rise?

 

And how fast do I say (or show) the airfoils rise?

 

Huh?

 

You didn't (put numbers to it), I did.

 

But variations in speed, acceleration, and distance must all be taken into account in an energy analysis (balance).

 

You have not done this.

 

This is part of what makes energy analyis not as simple as we might like.

 

Here is another reason.

 

Your comparison of bouyancy forces and airflow lift forces is flawed.

 

 

This new conversation between cxxLjevans, studiot, and MigL leads right back to my original issue with the conservation of energy analysis of an airfoil. I’ll try to present my problem again in light of this conversation and in a different way.

 

234_ar02.gif

 

If a solid is submerged in a more dense fluid then the force from the pressure pushing down on the top of the object plus the weight of the object will be less than the force from the pressure pushing up on the bottom of the object.

 

This is energy. Energy is the ability to do work. This is gravitational potential energy.

 

No the pressure difference is not gravitational potential energy.

 

If the (bouyancy) force leads to movement of the object then it will accelerate the object, imparting the kinetic energy of movement to that object.

In addition, once the object has been moved upwards some distance (any distance however small) then it also imparts gravitational potential energy.

 

All this is fine and dandy and leads to the following energy balance.

 

As soon as a net bouyancy force exists (however small) it accelerates the object upwards (however slowly), thus imparting kinetic and potential energy to the object as noted above.

 

However this does not happen with an airfoil lift.

 

It is well known that an aircraft has to achieve a certain minimum speed in order to be able to take off at all.

 

Translated into air motion past a stationary airfoil this means that if you free stand an airfoil on a support and blow air past it, there will be a minimum airstream speed, below which nothing will happen.

 

 

An aside here to Larry jevens

This is why we need friction. The airfoil would simply blow backwards off the support in the absence of friction as we brought the airspeed up above zero.

 

So if we consider any airstream speed in the sub takeoff range, a lift force will be generated, but the airfoil will not rise.

This is unlike the bouyancy force situation described above.

 

No one doubts that the airstream imparts energy to an object in its path.

But this energy is always initially purely kinetic.

Any potential energy transferred comes later.

 

The second question is "What is the form of the energy loss from the airstream?"

 

Again this is complicated since the airstream may possess linear kinetic energy, rotational energy, static head (potential) energy, gravitational potential energy and thermal energy.

 

This is why an energy balance is less than easy for flight calculations.

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Does the velocity of the free stream decrease with respect to its

value before encountering the foil. If so, then this is, essentially, what Zet said in post#46 which

says:

 

when the airfoil rises there must be a decrease in the horizontal velocity of the wind

 

-regards,

Larry

Unless there is some source of energy to maintain it, yes it decreases. The energy does have to come from somewhere.

 

The point is that it does not have to decrease more than when the airfoil is fixed. It is a more efficient case, since work is actually being done as the gravitational potential energy is gained.

 

An analogy would be sitting in a car pointed up the hill with some pressure on the gas but also on the brake, restrained from going anywhere. If you take your foot off the brake, you head up the hill. There is no need to wonder why you don't need any more fuel than what you were using. You still need fuel but possibly less than before, even though you are now gaining potential energy as you get higher on the hill.

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... I am not showing energy is conserved, I am assuming the law of conservation of energy holds ...

 

 

 

Okay.

 

Your post #49 appears, to me, to be an analysis of the situation. And so you seemed, to me, to reach a conclusion and not merely asserting conservation of energy. This is a Science forum where the Law of Physics are, of course, assumed to be true. And so “energy is conserved” can be simply be asserted.

 

However, if one desires to understand how energy is conserved in a given situation (as I do) then analysis is required.

 

---

 

 

 

Can't the problem be simplified by:

1) making the fluid incompressible (as in a hydrofoil instead of an airfoil)

2) making the hdrofoil density the same as fluid density

3) making the fluid viscosity = 0 (no energy loss due to friction)

 

IIUC, 1) would avoid any change in fluid energy due to

compression (in air, change in internal energy is P dV).

Also, 3) would avoid change in fluid energy due to frictional heating.

2) would make the gravitational potential energy constant.

 

 

Yes.

 

In this thought experiment (post #46) the “airfoil” can be in any fluid. It can be in a gas or a liquid. The logic, and the issue, remain the same.

 

It should be assumed, for the ease of analysis, that the fluid is incompressible.

 

If the “airfoil” is neutrally buoyant (the same density as the fluid it displaces) then, in the end, in the rising case there in an increase in kinetic energy (in the form of the rising “airfoil” and the fluid displaced downwards) that does not occur in the non-rising case, and so there must be a decrease in another form of energy in the rising case that does not occur in the non-rising case. The issue essentially remains the same. (And this simplifies things in that the vertical kinetic energy is no longer then later transformed into gravitational potential energy.)

 

And, yes, it should be assumed this all occurs in a hypothetical frictionless world.

 

Thank you for the improvements (simplifications) to this thought experiment.

 

 

 

How is the loss of the free stream energy, mentioned here:

 

gravitational/potential energy gain in the second case is accounted for as a loss of the free stream energy,

 

accomplished.

 

 

 

If I read J.C.MacSwell’s post correctly, he says that the velocity of the wind will decrease more in the non-rising case than in the rising case (which means the velocity of the wind does not offset the difference in the kinetic energies of the rising versus non-rising “airfoils” but only adds further to the imbalance in energy conservation between the two cases) ... but ... then he suggests (somehow) that this does balance things out.

 

---

 

 

 

But variations in speed, acceleration, and distance must all be taken into account in an energy analysis (balance).

 

 

 

When I throw a ball is thrown up into the air how fast the ball is traveling when it leaves my hand and how quickly the ball decelerates and how far the ball travels do not need to be included to analyze how energy is conserved in this situation.

 

To include these values is fine, but it is not necessary, and they can potentially (if one was so inclined) be used as a distraction from the actual issue.

 

Energy is conserved because the decrease in kinetic energy of the rising ball is matched by an equal increase in gravitational potential energy (plus the increase in the kinetic energy of the displaced air plus any increase in thermal energy). Energy is conserved and this is clear. And it is clear how energy is conserved. No numbers are needed for the how of it.

 

 

 

No the pressure difference is not gravitational potential energy.

 

 

 

When I drop a ball off of my roof the more dense ball falls and the less dense air is displaced upwards. There is an increase in kinetic energy and a decrease in gravitational potential energy.

 

When a submerged solid in a container of fluid rises the less dense solid rises and the more dense fluid is displaced downwards. There is an increase in kinetic energy and a decrease in gravitational potential energy.

 

It does not matter if the falling more dense material is a solid or a fluid and it does not matter if the rising less dense material is a solid or a fluid. In both cases there is an increase in kinetic energy and a decrease in gravitational potential energy.

 

If you have a solid submerged in fluid and it is in free space far from any gravitational fields then there is no buoyant force on that solid.

 

To get a buoyant force on a submerged object you need to be in a gravitational field.

 

And to get “buoyant potential energy” where the overall upward force is greater than the overall downward force then you need to have a less dense object in a more dense fluid.

 

This is a form of gravitational potential energy.

 

 

 

In addition, once the object has been moved upwards some distance (any distance however small) then it also imparts gravitational potential energy.

 

 

 

 

No.

 

It does not “impart” gravitational potential energy. It loses gravitational potential energy.

 

 

 

 

This is why we need friction.

 

 

 

No.

 

If the one airfoil (as described in post #46) is able to move vertically but not horizontally then it is not friction that keeps it in place. It must be in some sort of track (a frictionless track it is now here so stipulated) that allows it to move vertically but keeps it in place horizontally.

 

 

 

 

Again this is complicated since the airstream may possess linear kinetic energy, rotational energy, static head (potential) energy, gravitational potential energy and thermal energy.

 

 

 

 

In the thought experiment in post #46, in the end, in the one case there is more gravitational potential energy than in the other.

 

And so, for energy to be conserved, there needs to be a decrease in another form of energy in the rising case that does not occur in the non-rising case.

 

What is it?

 

Is it unknowable?

 

The only energy change option I can see in the system is a decrease in the motion of the wind (and so a decrease in kinetic energy). And the only mechanical reason I can see for this happening is an increase in drag when the airfoil rises as opposed to when it is held in place. But there is a problem with this logic when it comes to differently shaped airfoils of equal masses and volumes. And (!) if J.C.MacSwell is right and the opposite is factually physically true (there is more drag when the airfoil is held in place than when it is allowed to rise) this only compounds the conservation of energy analysis problem.

 

---

 

 

 

 

The point is that it does not have to decrease more than when the airfoil is fixed.

 

 

 

 

In a closed system, if there is an increase in one form of energy (such as gravitational potential energy), then there needs to be a decrease in another form of energy (such as kinetic energy) for energy to be conserved.

 

If not the velocity of the wind, then what?

 

?

Edited by Zet
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When I throw a ball is thrown up into the air how fast the ball is traveling when it leaves my hand and how quickly the ball decelerates and how far the ball travels do not need to be included to analyze how energy is conserved in this situation.

 

The physics of projectiles have nothing to to with the physics of airfoils.

 

So please do not claim to be following the laws of physics as here

 

 

This is a Science forum where the Law of Physics are, of course, assumed to be true.

 

Of course they must be correctly applied, which you have singularly failed to do.

 

This must be your best (or worst?) pronouncement.

 

 

studiot, on 02 Feb 2015 - 12:31 PM, said:snapback.png

 

This is why we need friction.

 

 

 

No.

 

If the one airfoil (as described in post #46) is able to move vertically but not horizontally then it is not friction that keeps it in place. It must be in some sort of track (a frictionless track it is now here so stipulated) that allows it to move vertically but keeps it in place horizontally.

 

So are you really saying that is not physically restrained against the real physical drag force or are you just wasting everyone's time?

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---

 

 

 

 

 

 

In a closed system, if there is an increase in one form of energy (such as gravitational potential energy), then there needs to be a decrease in another form of energy (such as kinetic energy) for energy to be conserved.

 

If not the velocity of the wind, then what?

 

?

There is more energy lost (eventually degrading to thermal energy) from the creation of vortices in the vertically fixed (restrained) case. The potential energy gain still comes from the wind.

 

I think this may help you:

 

Lets add a third case to the two I outlined in my post 49.

3. Same airfoil, same set up, except we will change the angle of attack, not by allowing it to rise as in case 2 (which changes the apparent angle) but by actually angling the nose down just to the point it will not rise.

 

This will give us less drag than both 1. (highest drag) and 2. (now intermediate)

 

So if you want to compare just case 2 and 3, it should work the way you feel it should.

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If we take one point at a time we might get somewhere useful.

 

Let us re-examine this in the light of some physics.

 

Zet

This new conversation between cxxLjevans, studiot, and MigL leads right back to my original issue with the conservation of energy analysis of an airfoil. I’ll try to present my problem again in light of this conversation and in a different way.

 

234_ar02.gif

 

If a solid is submerged in a more dense fluid then the force from the pressure pushing down on the top of the object plus the weight of the object will be less than the force from the pressure pushing up on the bottom of the object.

 

This is energy. Energy is the ability to do work. This is gravitational potential energy.

 

And if the solid is allowed to move then the less dense solid will rise and the more dense fluid will fall. There will be an increase in kinetic energy and an equal decrease in gravitational potential energy.

 

---

 

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.

 

So both mass and energy are conserved within the system.

 

Now let us look at you contention that

 

"There will be an increase in kinetic energy and an equal decrease in gravitational potential energy."

 

So the block floats to the surface.

 

Then what?

 

Then it stops!

 

So what now has more kinetic energy?

 

Similarly some liquid falls.

 

Again then what?

 

Again it stops.

 

So now we have a lower potential energy and zero kinetic energy in the system.

 

So where did the energy go, if it was conserved?

 

Keeping to the topic of potential energy you seem to be arguing at cross purposes with yourself here.

 

 

Zetsnapback.png

 

Studiot

In addition, once the object has been moved upwards some distance (any distance however small) then it also imparts gravitational potential energy.

 

 

 

 

No.

 

It does not “impart” gravitational potential energy. It loses gravitational potential energy.

 

 

Studiot The full quote

If the (bouyancy) force leads to movement of the object then it will accelerate the object, imparting the kinetic energy of movement to that object.

In addition, once the object has been moved upwards some distance (any distance however small) then it also imparts gravitational potential energy.

 

Had your read and quoted the full of my text (was two lines too many) it would have made sense, in that it would have been clear that the word 'it' refers to the bouyancy force imparting (gravitational) potential energy to the floating object.

 

You have emphatically denied this as shown above.

 

You are happy to assert that falling objects loose gravitational potential energy,

yet seem unable to accept the rising ones gain it.

Edited by studiot
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So where did the energy go, if it was conserved?

 

 

 

A body will continue in motion (even if that motion is redirected in an elastic collision) unless something stops that motion.

 

When the submerged solid rises (and eventually reaches the top of the fluid) and when the fluid is displaced downwards, they will both will continue moving unless something else stops them.

 

In a frictionless world (as it is so been stipulated here for the purposes of analysis) when the submerged solid reaches the top of the container of fluid it will (due to the fact it is upward vertical motion) move up beyond its preferred position (lowest energy state) at the top of the container of fluid and then it will fall back down, below its preferred energy state (since it now has downward vertical motion), and it will continue to bob up and down like this forever unless something else (such as friction) stops it. It retains all of the kinetic energy (in oscillation with potential energy as it bobs above and then below its preferred position and comes to periodic and temporary stops) it gained from the equal loss in gravitational potential energy.

 

In your example, where the rising solid and displaced fluid come to a stop, it is friction that must bring them to a stop. And when friction decreases kinetic energy there is an equal increase in thermal energy. Energy is conserved.

 

To recap (in your friction filled example which is not the thought experiment presented here):

 

The less dense object rises and the move dense fluid is displaced downwards.

 

This is a decrease in gravitational potential energy.

 

The submerged solid and the displaced fluid are set into motion.

 

This is an equal increase in kinetic energy.

 

The solid and the fluid come to a stop.

 

This is a decrease in kinetic energy.

 

It is friction that brings them to a stop.

 

There is an equal increase in thermal energy.

 

Energy is conserved.

 

It is all very simple and all very clear.

 

 

 

You are happy to assert that falling objects loose gravitational potential energy, yet seem unable to accept the rising ones gain it.

 

 

 

If I’m standing in my backyard and I throw a ball up in the air, the ball will rise and it will slow down as it rises. The more dense ball displaces the less dense fluid downwards.

 

There is an increase in gravitational potential energy and a decrease in kinetic energy.

 

In this case the rising object gains gravitational potential energy.

 

There is no lack of “acceptance” of this on my part.

 

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.

 

This almost seems axiomatic.

 

 

 

 

Had your read and quoted the full of my text (was two lines too many) it would have made sense ...

 

 

 

I did read both lines. And I only quoted the second because the first sentence is right (there will be an increase in kinetic energy) and the second one is wrong (there will not be an increase in gravitational potential energy).

 

If what you are saying (in the two lines) is true (which is it not) then you are saying when a submerged solid rises in a container of fluid due to “buoyant potential energy” then energy is not conserved. I’m not saying this, you essentially are. You are saying there is both an increase in kinetic energy and an increase in gravitational potential energy. And so the one does not offset the other. If you are right (which you are not) then energy is not conserved.

 

No?

 

If we are still in disagreement about this I would be willing to try to clarify it further. But, again, this is such basic physics and such basic logic that everything I have said in this reply almost seems axiomatic.

 

---

 

Does anyone have any ideas on how energy is conserved when the airfoil rises in the thought experiment in post #46 (or as modified and improved in post #54)?

Edited by Zet
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---

 

Does anyone have any ideas on how energy is conserved when the airfoil rises in the thought experiment in post #46 ?

It all comes from increased drag, just as you suspected.

 

In one case you are paying for lift but not using it to get work done, as you are restraining the airfoil, not allowing any motion. Efficiency is 0. 100% waste.

 

In the other you are paying for lift and getting something from it. Efficiency > 0. < 100% waste.

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Zet

It is friction that brings them to a stop.

 

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

 

 

Zet

This almost seems axiomatic.

 

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.

 

Zet

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.

 

 

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.

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

 

 

 

Huh?

 

I said: Friction was not what keeps the airfoil(s) in the thought experiment in post #46 in place. If the one airfoil is kept in place horizontally but is allowed to move vertically, then it is not friction that keeps it in place. Friction would work in both directions. (Unless you are proposing that there is horizontal friction and not vertical friction. Which seems weird. And is unnecessary. The airfoil, it can be stipulated, in in a frictionless track that allows it to move vertically and not horizontally.)

 

And, I said: In your thought experiment in post #62 where the rising submerged object eventually comes to a stop, for this to happen there must be friction.

 

I don’t understand your point.

 

---

 

 

 

There is a difference in the energy flows for the system, the rising body and the fluid ...

 

 

 

Okay.

 

Perhaps I missed your point.

 

In post #55 you write:

 

If the (bouyancy) force leads to movement of the object then it will accelerate the object, imparting the kinetic energy of movement to that object.

 

In addition, once the object has been moved upwards some distance (any distance however small) then it also imparts gravitational potential energy.

 

All this is fine and dandy and leads to the following energy balance.

 

As soon as a net bouyancy force exists (however small) it accelerates the object upwards (however slowly), thus imparting kinetic and potential energy to the object as noted above.” (bold added)

 

Okay.

 

 

If

 

you are saying when object rises in a fluid, that for that object in

and of itself there is an increase in gravitational potential energy,

 

while,

 

depending on whether the fluid the object displaces is of a greater

or lesser density, then, overall there will be an increase or a decrease

in gravitational potential energy

 

then I agree with you.

 

 

Where I then have a problem is where you say:

 

All this is fine and dandy and leads to the following energy balance.

 

As soon as a net bouyancy force exists (however small) it accelerates the object upwards (however slowly), thus imparting kinetic and potential energy to the object as noted above.

 

If I understand you now correctly (as spelled out in the previous piece of logic), then, yes, your second sentence above makes perfect sense.

 

But the prior sentence then does not.

 

If I now understand you correctly, you are not “balancing” out energies in the “following” sentence. You are saying, in this second sentence, when a less dense object rises in a more dense fluid that for that object in and of itself (and without any reference to the rest of the system) there is both an increase in kinetic energy and an increase in gravitational potential energy. There is no reference to the greater amount of mass of fluid that “falls” or “descends” or is “displaced downwards.” There is no “balancing” of the increase in kinetic and gravitational potential energies of the solid in and of itself with the increase in kinetic energy of the descending fluid and the decrease in gravitational potential energy of the descending fluid which must be for an “energy balance.” Saying there are two increases in energy with the rising submerged solid is not an “energy balance.”

 

When I read this I assumed since this was supposed to be an “energy balance” that you were attempting to balance the energies, and the only “gravitational potential energy” change in this system that leads to a “balance” is the decrease in gravitational potential energy of the descending fluid (which apparently, if I now understand you correctly, you did not include but rather where just listing the two increases with the object). And so, if I now understand you correctly, this was a mistaken reading of what you wrote on my part. (If so, apologies.)

 

So, ...

 

Either

 

I misunderstood you and thought you were trying to balance the

increases with the decreases in energy, when in fact you were just

listing the energy increases for the submerged solid and that alone,

in which case your statement “... and leads to the following energy

balance” is flat out incorrect

 

Or

 

you were claiming to balance out the energies, and so your statement

“... and leads to the following energy balance” is genuine but in the

“following” sentence you fail to balance anything out and just list the

two energy increases for one part of the system.

 

 

And, so, yes, perhaps I missed your point. And if I did, I apologize.

 

And, if so, now to you: where is the “balance” in the “following”?

 

And if there is none, then what is your point?

 

 

---

 

 

 

 

No fluid falls if you blow it horizontally past an airfoil.

 

 

 

In the rising case it does.

 

 

---

 

To J.C.MacSwell:

 

I’m still thinking about the efficiency angle. I’ve never thought about it from this perspective before. My initial gut feeling is that this isn’t going to work. But I need to think it through and will.

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In my first year in University, my roommate used to go to bed every night muttering the Chemical Engineer's mantra.

He attributed his high success in the end of year exams to this being the only thing he could remember and he applied it remorselessly.

 

Input = Output plus Accumulation

 

Since then I have been conducting successful engineering balances for over forty years, force balances, momentum balances, energy balances, mass balances.....

So please don't lecture me on how to conduct a balance.

 

The trick with balances of any sort is to know what to apply it to.

If you try to apply it to the whole system as you are, you will often miss out important contributors, as you have done.

 

At least you have now understood what the balance applies to, although I explicitly stated what energy changes I was referring to.

 

 

Zet

In a frictionless world (as it is so been stipulated here for the purposes of analysis) when the submerged solid reaches the top of the container of fluid it will (due to the fact it is upward vertical motion) move up beyond its preferred position (lowest energy state) at the top of the container of fluid and then it will fall back down, below its preferred energy state (since it now has downward vertical motion), and it will continue to bob up and down like this forever unless something else (such as friction) stops it. It retains all of the kinetic energy (in oscillation with potential energy as it bobs above and then below its preferred position and comes to periodic and temporary stops) it gained from the equal loss in gravitational potential energy.

 

In your example, where the rising solid and displaced fluid come to a stop, it is friction that must bring them to a stop. And when friction decreases kinetic energy there is an equal increase in thermal energy. Energy is conserved.

 

Congratulations on realising that the body must bob up and down for a period.

However this will not be forever since you have neglected the atmosphere above the container.

Both the surface of the liquid and the body itself will oscillate up and down, generating pressure pulses in the atmosphere above.

This is a dissipative process that will eventually leak away all the kinetic energy in radiated wave motion in the atmosphere.

 

 

Studiot

No fluid falls if you blow it horizontally past an airfoil.

 

 

 

 

 

Zet

In the rising case it does.

 

Absolutely not.

Horizontal motion is just that by definition.

 

Before we discuss friction, have you heard of D'Alembert's Paradox?

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... forty years ... So please don't ...

 

 

 

I have not spent forty years balancing things but I do speak English. I know the meaning of the word “balance.” And I know what the concept of “balance” means in the context of conservation of energy.

 

I thought (mistakenly) that you were saying that when a less dense object rises in a more dense fluid that there is an increase in gravitational potential energy. There is not. There is an increase in the gravitational potential energy of the rising object, but there is a greater decrease in the gravitational potential energy of the falling fluid. And so, there is a decrease in gravitational potential energy. And the “balance” is that there is an equal increase in kinetic energy in the form of the rising solid and falling fluid (assuming no loses to friction).

 

In your “balance” there are two energy increases and no energy decreases.

 

 

 

If you try to apply it to the whole system as you are, you will often miss out important contributors, as you have done.

 

 

 

Huh?

 

You left out the decrease in the gravitational potential energy of the fluid in your “energy balance.”

 

How can you now say I’m the one who is missing the important contributors?

 

 

 

However this will not be forever since you have neglected the atmosphere above the container.

 

 

 

If the fluid is a liquid and if the container of fluid is both in a gravitational field and in a vacuum, then the risen solid will bob forever (in a frictionless world).

 

 

 

... discuss friction ...

 

 

 

No.

 

There is no friction in this thought experiment. And I’m not going to go down another tangent with you (where, in all likelihood, I’ll make another mistake. I make a lot of them. And I don’t mind admitting when I do.)

 

 

I suspect you can’t answer the simple question posed in post #46.

 

And I suspect what you are doing is bringing up every possible tangential concept in order to muddy the waters and make it seems as if this question is so complicated that it is beyond any of us ordinary folk’s ability to address and so we should just leave it alone and walk away.

 

One way to “resolve” an issue is to answer the question, and another way to “resolve” an issue is to make it go away.

 

If I’m wrong about your motives, I apologize.

 

And you can easily prove I’m wrong about your motives by simply answering the question.

 

-----

 

503_ar05.gif

 

 

In the one case there is an increase in gravitational potential energy, and in the other case there is not.

 

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

 

What is it?

 

(The only energy change option, I can see, is a decrease in the velocity of the wind (and so a decrease in kinetic energy) when the airfoil rises. And the only mechanical reason I can see for this is an increase in drag when the airfoil rises. But, there is a problem with this logic when it comes to differently shaped airfoils of equal masses and volumes. And, so, this is where I am stuck. This is where, for me, the conservation of energy analysis remains unresolved.)

 

?

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Zet

If the fluid is a liquid and if the container of fluid is both in a gravitational field and in a vacuum, then the risen solid will bob forever (in a frictionless world).

 

What will happen to any liquid exposed to an extensice vacuum at its free surface as you have drawn it?

 

It will evaporate very rapidly.

 

 

Studiot

Before we discuss friction, have you heard of D'Alembert's Paradox?

 

Zet

No.

 

There is no friction in this thought experiment. And I’m not going to go down another tangent with you (where, in all likelihood, I’ll make another mistake. I make a lot of them. And I don’t mind admitting when I do.)

 

 

I suspect you can’t answer the simple question posed in post #46.

 

 

If you weren't so determined to proove yourself absolutely right and everyone else absolutely wrong you might actually learn something.

 

D'Alemberts Paradox, (some call it D'Alembert's Theorem) refers to ideal fluids. That is incompressible fluids without friction.

 

Briefly put, it states that ideal flow past any object creates zero drag in the direction of flow.

 

So you see this is highly relevent to your simple airfoil model.

 

The question in post#46 was a restatement of your original question and has been answered many times by several posters in this thread.

 

Your question is basically

 

"Where does the energy to lift the airfoil come from ?"

 

The energy comes from whatever source drives either the airfoil through the fluid or the fluid past the airfoil.

However it is really difficult to match a particular parcel of energy in the fluid with a particular amount of lift energy.

 

 

I will try one more time by offering this diagram and very simple explanation.

 

post-74263-0-52947400-1423087980_thumb.jpg

 

Edit to be completed.

 

OK so how much energy to raise the airfoil unit distance ?

 

This one is easy it's airfoil mass times 1metre times g

 

But how do you relate that to the fluid?

 

That's not so easy since each parcel of fluid is blowing past the airfoil and adding in a small part of this energy before it is replaced by another parcel. And how far into the fluid do you consider your airfoil drawing energy?

 

Well one way is to calculate the lift force.

 

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.

 

The fluid is considered ideal which means it is incompressible and inviscid.

 

The fluid is approaching section ABB'A'A horizontally so its velocity, U equals Ux its horizontal velocity and its vertical velocity is zero at this section in front of the airfoil.

 

Physical observation shows that the streamlines turn downwards as they pass the airfoil as shown.

 

So when the fluid emerges from section DCC'D'D they have a horizontal and a vertical velocity.

 

Now the volume rate of flow through each section = area cross section times the velocity and since both sections are vertical (sorry about my sketch) they have they admit zero volume flow vertically.

 

Since the fluid is incompressible the volumetric flowrate through each section is the same.

 

U1A1 = U2A2

 

But 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.

 

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

 

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

 

This is the same formula obtained in Jukowski's theorem by complex integration, but obtained much more easily, if clumsily.

Edited by studiot
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What will happen to any liquid exposed to an extensice vacuum at its free surface as you have drawn it?

 

It will evaporate very rapidly.

 

 

 

If you weren't so determined to proove yourself absolutely right and everyone else absolutely wrong you might actually learn something.

 

D'Alemberts Paradox, (some call it D'Alembert's Theorem) refers to ideal fluids. That is incompressible fluids without friction.

 

Briefly put, it states that ideal flow past any object creates zero drag in the direction of flow.

 

So you see this is highly relevent to your simple airfoil model.

 

The question in post#46 was a restatement of your original question and has been answered many times by several posters in this thread.

 

Your question is basically

 

"Where does the energy to lift the airfoil come from ?"

 

The energy comes from whatever source drives either the airfoil through the fluid or the fluid past the airfoil.

However it is really difficult to match a particular parcel of energy in the fluid with a particular amount of lift energy.

 

 

I will try one more time by offering this diagram and very simple explanation.

 

attachicon.giflift1.jpg

 

Edit to be completed.

 

OK so how much energy to raise the airfoil unit distance ?

 

This one is easy it's airfoil mass times 1metre times g

 

But how do you relate that to the fluid?

 

That's not so easy since each parcel of fluid is blowing past the airfoil and adding in a small part of this energy before it is replaced by another parcel. And how far into the fluid do you consider your airfoil drawing energy?

 

Well one way is to calculate the lift force.

 

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.

 

The fluid is considered ideal which means it is incompressible and inviscid.

 

The fluid is approaching section ABB'A'A horizontally so its velocity, U equals Ux its horizontal velocity and its vertical velocity is zero at this section in front of the airfoil.

 

Physical observation shows that the streamlines turn downwards as they pass the airfoil as shown.

 

So when the fluid emerges from section DCC'D'D they have a horizontal and a vertical velocity.

 

Now the volume rate of flow through each section = area cross section times the velocity and since both sections are vertical (sorry about my sketch) they have they admit zero volume flow vertically.

 

Since the fluid is incompressible the volumetric flowrate through each section is the same.

 

U1A1 = U2A2

 

But 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.

 

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

 

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

 

This is the same formula obtained in Jukowski's theorem by complex integration, but obtained much more easily, if clumsily.

 

 

Blah, blah, blah ...

 

 

-----

 

503_ar05.gif

 

In the one case there is an increase in gravitational potential energy, and in the other case there is not.

 

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

 

What is it?

 

?

 

I don’t think you can answer this question.

 

?

 

 

.

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

 

What is it?

 

?

 

I don’t think you can answer this question.

 

?.

 

Wouldn't that (lifting) energy be subtracted from the 'thrust' ...or whatever horizontal force is adding energy?

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Wouldn't that (lifting) energy be subtracted from the 'thrust' ...or whatever horizontal force is adding energy?

That's what I would guess; however, if the energy comes from the horizontal thrust, then

that begs the question of how a horizontal force causes a change in vertical potential energy.

IOW, how can something pushing horizontally cause movement of something vertically?

 

OTOH, if the energy comes from the fluid, then how is that energy substracted from the

fluid. Does the fluid drop, decreasing is potential energy? Does its speed decrease?

Does its static pressure decrease? If so, then how?

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That's what I would guess; however, if the energy comes from the horizontal thrust, then

that begs the question of how a horizontal force causes a change in vertical potential energy.

IOW, how can something pushing horizontally cause movement of something vertically?

 

OTOH, if the energy comes from the fluid, then how is that energy subtracted from the

fluid. ....

I don't see why that begs the question; something, such as an airfoil, redirects the horizontal force.

 

On the other hand (or on the other side of the interface between the airfoil and the air), probably there are some effects upon air pressure and air speed. But I wouldn't say the energy is coming "from" the air/fluid.

 

~

Edited by Essay
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Ljevans

IOW, how can something pushing horizontally cause movement of something vertically?

 

Essay

something, such as an airfoil, redirects the horizontal force.

 

If you place an onject at the base of a wedge and push it horizontally onto the wedge what happens?

 

Zet

Blah, blah, blah ...

 

 

So you confirm that it is a waste of time talking to you.

 

So this is the last time I will tell you:

 

If some external agent does not put enough energy into the system for flight you cannot take it out.

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(In the thought experiment in post #46 the airfoil is more dense than the fluid. Now it has been changed so that the solid and the surrounding fluid are of the same density. The logic, and the issue, are essentially the same.)

 

 

446_ar10.gif

 

 

 

 

Wouldn't that (lifting) energy be subtracted from the 'thrust' ...or whatever horizontal force is adding energy?

 

 

 

 

Yeah.

 

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

 

And when the moving fluid encounters the “airfoil” it will move more quickly over the top than it does under the bottom (as has been experimentally demonstrated).

 

And so there will be more of a decrease in pressure pushing down on the top and less of a decrease in pressure pushing up on the bottom (as per Bernoulli’s principle).

 

This will cause the otherwise neutrally buoyant “airfoil” to rise. And so there will be an increase in kinetic energy in the form of the vertically rising “airfoil” and in the form of the displaced fluid downwards.

 

And so, for energy to be conserved, there must be a decrease in another form of energy.

 

503_ar11.gif

 

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.

 

---

 

 

 

IOW, how can something pushing horizontally cause movement of something vertically?

 

 

 

Different decreases in pressures on the top and bottom as per Bernoulli’s principle, and so a “dynamic buoyant force” (or whatever it should be called).

 

 

 

 

OTOH, if the energy comes from the fluid, then how is that energy substracted from the

fluid. Does the fluid drop, decreasing is potential energy? Does its speed decrease?

Does its static pressure decrease? If so, then how?

 

 

 

If the fluid is incompressible, then whether the “wind” moves up or down or in circles or whatever after encountering the “airfoil” then there is no change in the gravitational potential energy of the fluid in and of itself.

 

503_ar11.gif

 

Again, the obvious answer is that there must be a decrease in the velocity of the “wind” when the “airfoil” rises that does not occur when the “airfoil” kept from rising. (And so the increase in vertical kinetic energy of the rising “airfoil” and falling displaced fluid is offset by an equal decrease in the kinetic energy of the “wind”).

 

 

 

 

But I wouldn't say the energy is coming "from" the air/fluid.

 

 

 

I think it must.

 

There is an increase in vertical kinetic energy and so there must be a decrease in another form of energy (... for energy to be conserved). And in this simple scenario there is really only one energy change option: a slowing of the “wind” and so a decrease in kinetic energy in this form.

 

?

 

 

.

 

 

 

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