I Pressure and Lift around a Wing

1. Apr 6, 2018

fog37

Hello everyone,

I am pondering on the airflow around a wing:

When air flows over a wing, part of the air goes below the wing and part over the wing. Even when the air is moving, the pressure at a point in space is always isotropic. The pressure on top of the wing is slightly lower than than the free stream pressure. This pressure distribution produces a net force $L _{top}$ on top of the wing directed downward. The pressure at the bottom of the wing is slightly larger than the free stream pressure farther away from the wing and produces a net lift force $L _{bottom}$. The difference between these two forces, one pointing down and one pointing up, produces the overall lift force $L_{total}= L _{top} - L_{bottom}$ directed upward.

Does the static air located far above the wing rush toward the top of wing since it is at a higher pressure than the air closer to the wing? Also, the higher pressure air below the wing should push both on the bottom surface of the wing and also on the air farther below the wing. Is that what happens?

Thanks!

2. Apr 6, 2018

FactChecker

The airflow over the wind does get drawn downward. The net result is that the airflow gets diverted downward. This can be looked at as essential for the lift force on the wing. They are equal and opposite forces and reactions -- the airflow is diverted down and the wing is pushed up.
Yes.

3. Apr 6, 2018

fog37

Thanks a lot! This is how I envision it:

It is clear, from Bernoulli's equation, that fast moving air has a lower isotropic pressure than slower moving air. That said, I am not clear why from a molecular standpoint? Does anyone have a clear interpretation?

4. Apr 6, 2018

5. Apr 6, 2018

fog37

Thanks. I read that. I will read again.

6. Apr 6, 2018

FactChecker

Good. You should be aware that any simple answer to your question is over-simplified and will have to be discarded if you get deeper into the subject. The only real way to calculate lift is to apply Computational Fluid Dynamics (CFD), which traces tiny bits of air around the wing while calculating all the pressures and forces on it. Even that is very limited. Wind tunnel tests are done to verify and modify the CFD results to get an aerodynamic model. The flight tests are done in a careful sequence to make sure that flight results and aerodynamic model agree before more dangerous flights are attempted.

7. Apr 6, 2018

fog37

You are right. I don't want to look to much into it. I am just trying to grasp some qualitative understanding of the situation.

Thanks.

8. Apr 6, 2018

rootone

The shape of the wing forces air to go 'down', so therefore the wing obtains a force going 'up' (lift).

9. Apr 7, 2018

Staff: Mentor

[Mod Note:]
A severely wrong post and several responses were deleted.

Also, let's please keep this focused on the OP's questions and perhaps most importantly respect the OP's request to keep this as basic as possible. Not every discussion about how a car is propelled (for example) needs to get into the thermodynamics of the Otto cycle.

Last edited: Apr 7, 2018
10. Apr 8, 2018

sophiecentaur

It amazes me that this inescapable fact seems to be accepted in discussions about a helicopter rotor but not accepted for a wing travelling in a straight line. The details of how the downward motion of air is achieved is very complicated and, as has been pointed out, there are many different levels for the analysis. You don't need an aerofoil to produce lift but it's just a good design which involves relatively little drag.

11. Apr 8, 2018

FactChecker

That's an interesting point. I suspect that the helicopter version is accepted in an over-simplified form as though its blades operate like a screw digging into the air rather than as wings with a profile like an airplane wing. (Many early attempts at flight made that mistake.)

I remember seeing an experiment where a toy helicoptor inside a box begins to hover within the box. The weight of the box remains identical to the weight when the helicoptor just sat on the floor with no blade rotation. So obviously the weight of the hovering helicoptor was transfered by the air downdraft to the floor. I wonder if an equivalent experiment can be done with a model airplane. I think that it would be much more difficult to set up in such a convincing way.

12. Apr 8, 2018

sophiecentaur

I would be surprised (very) at an experiment that could show Newton's Laws of motion don't apply.

13. Apr 8, 2018

FactChecker

Sorry for the confusion. (I believe in Newton's laws as much as you do.) The example of a hovering helecopter, only supported by air is relatively simple. But it would be hard to keep a wing in place in a wind tunnel without some supports that complicate the logic of the conclusion.

14. Apr 8, 2018

sophiecentaur

I was not disagreeing with you. What you describe is supporting Newton.
A wind tunnel would need to be designed from scratch with facilities for measuring its weight. A lot of trouble to prove what can be proved in other, easier ways.

15. Apr 8, 2018

olivermsun

For the purposes of your demonstration, what would be wrong with using a 2-d wing section supported by the sides of the wind tunnel?

16. Apr 8, 2018

FactChecker

The point is to show that the force of the air on the bottom of the box is exactly the same as the weight of the plane that is floating above. Any support and forces from it can be all accounted for, but it is not as clear a demonstration. The hovering helicopter which requires no attachments is ideal.

17. Apr 8, 2018

olivermsun

I am confused as to the point of this demonstration.
The plane is floating, so it follows that the downward force exerted by the plane is equal to its weight.
Why is it also necessary to demonstrate that the box enclosing the plane weighs the same whether the plane is floating or parked?

18. Apr 8, 2018

FactChecker

I'm afraid that I have diverted this thread too much. I just thought that the helicopter demonstration was such a good demonstration of "equal and opposite" in the aerodynamics context that it impressed me at the time. It wouldn't bother me if a monitor cut this out.

19. Apr 9, 2018

sophiecentaur

A narrow wind tunnel would be cheaper but there would be an issue with the interaction of the air flow and the sides / corners. The way N3 applies to these things is easily observed when you see the result of turning a heavy boat with a rudder. There is a visible disturbance of the water, way into the inside of the curve, with eddies moving inwards by a significant distance. That 'Momentum' is very visible, unlike the air that trails behind an aircraft.

20. Apr 9, 2018

olivermsun

I didn't say a "narrow" wind tunnel, I said a 2-d wing section — you have a relatively wide section and tunnel where the wing does not vary across the flow, and then you try to observe a 2-d flow away from the edges of the tunnel.

The streamlines are typical made more visible by injecting smoke or other tracers.

21. Apr 9, 2018

sophiecentaur

Oh yes. A misunderstanding. I expect most basic wing design and optimisation can be done with 2D sections. What happens at the ends can be very relevant though - as demonstrated by all these fancy new wings on passenger jets with the clever bits on the tips. (You can tell I am speaking not as an aerodynamicist)
But, of course, the streamlines would need a massive space to develop fully and to reveal a downward motion that is actually providing the N3 lift.

22. Apr 9, 2018

rcgldr

Correct. Using the air as a frame of reference (instead of the wing), the air accelerates downwards and it's pressure decreases as it approaches the plane swept out by a wing as a wing passes through the air. As the air passes through the plane swept out by a wing, there's an increase in pressure with little or no increase in speed. This pressure jump corresponds to the energy added to the air by a wing passing through a volume of air. With this increase in pressure, the air continues to accelerate downwards as it's pressure returns to ambient. The speed of the air when it's pressure returns to ambient is called the "exit velocity". Nasa has an article on propellers that explains this, although the diagrams ignore the viscous interaction with the surrounding air. Note that from the air's frame of reference, Bernoulli is violated as the air flows through the plane swept out by the wing, because work is performed by the wing by increasing it's pressure. The Nasa article mentions this.

https://www.grc.nasa.gov/www/k-12/airplane/propanl.html

23. Apr 13, 2018

fog37

So, inside a fluid is moving in the direction $x$ at uniform speed v, a fluid particle (small blob of fluid composed of many molecules) will feel an isotropic pressure $p_0$ from all the other fluid particles around it. Neglecting depth effect on pressure, if the fluid particle is brought to rest (at a stagnation point), the new pressure the particle would experience is increased to $p_0+\rho \frac {v^2} {2}$. But without bringing the fluid parcel to rest, the pressure will only be and remain $p_0$ and be isotropic and the fact that the fluid is moving at speed $v$ in the x direction will not show it effect.

Does that mean that the pressure $p_0$ would be the same as when the fluid is at rest even if the fluid is actually moving unless we bring it to a stop?

24. Apr 14, 2018

Your nomenclature is somewhat nonstandard. Let me try rewriting this. Let's assume for a moment that a given flow is steady, inviscid, and adiabatic. A given fluid element moving with velocity $V$ in an incompressible flow will have total (or stagnation) pressure
$$p_0 = p + \dfrac{1}{2}\rho V^2.$$
We typically call $p_0$ (sometimes written $p_t$) the total or stagnation pressure, which is the pressure that would be felt if the flow was isentropically (i.e. no heat added, not energy dissipated) brought to rest. $p$ is the static pressure and is what is felt by a surface immersed in said fluid whether it is moving or not. The $V^2$ term is dynamic pressure and isn't truly a pressure but a kinetic energy term. The whole statement really amounts to a conservation of energy statement. In an isentropic flow, $p_0$ is a constant, so if you bring the velocity to zero, $p = p_0$ and the surface at the stagnation point feels that pressure. Anywhere else in the flow, and the pressure felt by the surface is $p$, which is some fraction of $p_0$ based on the value of the dynamic pressure.

The static pressure (sometimes called thermodynamic pressure) is always isotropic. It acts equally in all directions. This is universally true of static/thermodynamic pressure. Now, if you are talking about the actual stress tensor in the fluid, that is not necessarily isotropic in all cases, but the pressure itself is.

25. Apr 14, 2018

fog37

I "think" we are on the same page. There is one and only one pressure which is always isotropic, from the standpoint of a fluid parcel. In a fluid at rest (hydrostatics), the pressure would be $p$ and only vary with depth. If the fluid was in motion, the fluid parcel would still feel the same pressure $p$ unless we brought the flow to a stop. At the stopping point (stagnation point) the pressure, still isotropic, would become $p_0$ and be larger since the momentum of the flow is now converted in extra pressure. What has always confused me was the fact that, either at rest or in motion, the fluid parcel would apparently feel the same pressure, the thermodynamic pressure, unless it was brought to rest.
Bernoulli's equation is often read as the local isotropic pressure as being different for different flow speeds. that does not seem to agree with what I say above. If the fluid is moving fast or slow, the pressure a parcel exerts on other parcels and the pressure the parcels exert on is just $p_0$ since the pressure increase only derives from converting the momentum into force.
That said, I see how in a tube narrowing, based on Bernoulli's equation, there must be a gradient of pressure to accelerate the fluid (faster in the narrowing). So, in that case, the pressure $p$ seems to be different for different speeds...