Aerofoil Question

1. Jul 16, 2015

Consider this aerofoil. Let the leftmost point be S and rightmost point be E. Let the topmost point be T and a point on the bottom of the foil be B.
Let the wind blow as shown in the figure above.
Bernoulli's principle says that the air blowing along S-T-E must travel a path longer than than that travelled by air blowing along S-B-E, and for the time taken to reach E must be the same, the wind blowing through the top of the foil must have a higher velocity and hence the pressure on the top of the foil is lesser than that on the bottom part of the foil and hence the plane takes off.
But, why should the air blowing along S-T-E and S-T-B take the same time to reach E?
Let me frame this question in a different way. Consider two particles in air coming towards the foil, one which will take the path S-T-E (for convenience sake let this be particle 1) and the other which will take the path S-B-E (let this be 2). Until they reach the foil, they have the same velocities. But after they reach the foil, either particle 1 speeds up, or particle 2 slows down, or both happen. Thus, at least one of them accelerates. According to Newton's first law, which says that no object accelerates unless an external force is applied, there must be a force acting on this particle. Where does this force arise from and for what reason does it arise?

2. Jul 16, 2015

No. Bernoulli's principe does not say this. Nothing in physics says this. It flat out is not correct. In fact, the air moving over the top move much faster than below and leaves the trailing edge quite a bit before a particle reaching the leading edge at the same time but traveling underneath reaches the trailing edge. This is commonly called the equal transit time fallacy.

Also, I am not sure why any of that was relevant to this necrothread.

Last edited: Jul 16, 2015
3. Jul 16, 2015

A.T.

Check out the video below at 0:38, to see if the smoke stripes split by the leading edge reach the trailing edge at the same time.

Last edited: Jul 16, 2015
4. Jul 16, 2015

Staff: Mentor

The others took care of the rest, but:
Bernoulli's principle does describe why a packet of air might speed up or slow down (that's basically its entire point): it's the pressure that provides the force to accelerate the packet of air.

5. Jul 16, 2015

Eh, but it doesn't really, though. It does establish the relationship between pressure and velocity but doesn't provide any means for calculating a flow field in such a situation without knowing one of them a priori or without another equation to describe either the pressure or the velocity field.

6. Jul 16, 2015

Staff: Mentor

I'm not sure what your point is, but it doesn't seem related to the very basic question I was answering.

7. Jul 16, 2015

My point is that Bernoulli's equation doesn't really deal with forces at all. It ultimately deals with energy. It's all related, of course, but for someone who clearly doesn't understand what Bernoulli's equation is (Rishi), using it to explain forces is likely misleading.

8. Jul 18, 2015

Bernoulli's principle says that the air blowing along S-T-E must travel a path longer than than that travelled by air blowing along S-B-E, and for the time taken to reach E must be the same
Sorry, I did not notice that I typed the words "Bernoulli's principle" in the beginning
my question is all about why time taken should be the same for air blowing along S-T-E and S-B-E.

9. Jul 18, 2015

You say that the pressure provides the force to accelerate the packet of air, and at the same time, the acceleration of the particle causes the change in pressure (because of acceleration, there is difference in velocities of the packets of air along S-T-E and S-B-E, and hence pressure decreases according to bernoulli's principle)
This says nothing about why the time taken should be the same

10. Jul 18, 2015

A.T.

The time isn't the same. See video in post #3.

11. Jul 18, 2015

rcgldr

More like coexistent. Air accelerates from higher pressure zones to lower pressure zones, and absent external forces (so that the total energy remains constant), Bernoulli's equation relates the reduction in pressure and increase in speed (speed^2) as air accelerates from a higher pressure zone towards a lower pressure zone. Bernoulli doesn't explain why pressure differentials exist around a wing, only how the air will react once those pressure differentials exist.

I'm not sure how Bernoulli's equation applies to curved flows, since the centripetal (perpendicular to the flow) component of acceleration is related to a pressure differential, but not to a change in speed, only a change in direction, while at the same time the existence of a pressure differential related to centripetal acceleration is going to have an impact on the speed of the air in the direction of flow.

12. Jul 18, 2015

RandomGuy88

Bernoulli's equation applies along a streamline. When the flow is curved, the pressure gradient is perpendicular to the streamlines and is balanced by the centrifugal force resulting from the fact that the fluid has mass.

13. Jul 18, 2015

Bernoulli's equation does not apply to this kind of flow because the flow is turbulent.
Bernoulli's equation applies only to those situations where the fluid is incompressible, which is clearly not the case from what the video shows.

Last edited: Jul 18, 2015
14. Jul 18, 2015

RandomGuy88

Are you talking about the video of the smoke flow over the airfoil?
What makes you think that flow is not incompressible, because it definitely is.

There is a compressible version of the Bernoulli's equation.

15. Jul 18, 2015

There has been nothing in this discussion about compressible or turbulent flows, and the curvature of a streamline does not imply turbulence or compressibility, and therefore does not preclude the use of Bernoulli's equation.

16. Jul 18, 2015

The flow is of course turbulent, otherwise the velocities of air packets would not change. Think about it.
I did not say that it is turbulent because of the curvature of a streamline.

17. Jul 18, 2015

RandomGuy88

The flow in the video above is incompressible.

No one is saying anything about the flow not being turbulent. There is clearly turbulence in the wake of the airfoil due to trailing edge separation.

18. Jul 18, 2015

HEY, you can see from the video that the lines are closer to each other on the top part of the foil than the ones on the bottom. Thus the pressure should be higher at the top if you do not consider the velocity part. Thus you can not say anything about the net effect (considering both the velocity and this compression effect), because the engines are the actual ones that help the plane to fly.

19. Jul 18, 2015

I just wanted to mention that point about turbulence because I wanted to say that Bernoulli's equation could not be applied

20. Jul 18, 2015

RandomGuy88

I don't think anything in this post makes any sense.

Pressure is lower on the top of the airfoil not higher. The streamlines get closer together as a result of mass conservation. And the streamlines being closer together is not an indication of compressibility.

What net effect?

What do you mean "the engines are the actual ones that help the plane to fly" ?

21. Jul 18, 2015

What is it exactly that you think turbulence means? I'm not sure you are clear on the concept. Further, a turbulent boundary layer doesn't mean you can't apply Bernoulli's equation to an airfoil as long as you apply it along the inviscid streamlines outside the boundary layer. That's done frequently and you can calculate lift very effectively that way from a given flow field.

22. Jul 19, 2015

This is turbulent because there is a change in pressure with space (the pressure does change as the air moves towards the foil).
Turbulence does occur when you have variation of pressure with space and time.
This is What actually happens in the case of an aerofoil (look at the second diagram).

Last edited: Jul 19, 2015
23. Jul 19, 2015

Lines getting closer does indicate compressibility. Can you tell me what else can possibly indicate compressibility.

24. Jul 19, 2015

Without the engines, the plane can't fly.
This is what I meant by that statement.

25. Jul 19, 2015

Compressibility, on the other hand, is generally indicated by the Mach number. Any flow in which $M \geq 0.3$ is typically considered compressible. Any slower than that and compressibility effects are negligible and the flow may be safely treated as incompressible. Streamlines can get closer together or farther apart in either case. The other indicator (or rather, definition) of compressibility is that $\nabla \cdot \vec{v} \neq 0$ in a flow field.