How can the lift equation be derived using Bernoulli's law?

[CaptainPickle]

Hi there.

I am trying to derive the lift equation for aircrafts through Bernoullis law. I am having some trouble though, since I wind up with the differences between the force that acts on the top of the wing and the force that acts under the wing instead of the net force, so to speak. Could someone give me a hint?

Here is my derivation:
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A gas flowing through a tube is observed. The difference in the gas' mechanical energi is equal to the work of the gas:

∆E_mek=A⇔∆E_pot+∆E_kin=A

We look at each part of the equation one at a time:

∆E_pot=m∙g∙h_2-m∙g∙h_1=m∙g∙(h_2-h_1)

We know that mass is equal to density times volume:

∆E_pot=ρ∙V∙g∙(h_2-h_1)

Now we look at the kinectic energy:

∆E_kin=1/2∙m∙v_2^2-1/2∙m∙v_1^2=1/2∙m∙(v_2^2-v_1^2 )

We substitute with density and volume again:

1/2∙ρ∙V∙(v_2^2-v_1^2 )

Now we look at the work:

A=F_1∙∆x-F_2∙∆x

We know that force is equal to preassure times area:

A=p_1∙a∙∆x-p_2∙a∙∆x

An areal times a length is the same as a volume:

A=(p_1-p_2 )∙V

Now we asseble the different equations into one:

∆E_pot+∆E_kin=A

ρ∙V∙g∙(h_2-h_1 )+1/2∙m∙(v_2^2-v_1^2 )=(p_1-p_2 )∙V

Which can be reduced to:

p_1+ρ∙g∙h_1+1/2∙ρ∙v_1^2=p_2+ρ∙g∙h_2+1/2∙ρ∙v_2^2

Or:

p_1+ρ∙g∙h_1+1/2∙ρ∙v_1^2=Konstant

Now we look at a wing. The difference in height between the top of the wing and the bottom of the wing is minimal. We choose to say that Δh = 0.

Then we have:

∆p=1/2∙ρ∙(v_o^2-v_u^2)

Preassure is equal to Force per area:

p=F/A

We choose to say that the area underneath the wing is the same as on the top of the wing:

∆F/A=1/2∙ρ∙(v_o^2-v_u^2)

∆F/A∙A=1/2∙ρ∙(v_o^2-v_u^2 )∙A⇔∆F=1/2∙ρ∙(v_o^2-v_u^2 )∙A

We now multiply the equation with a dimensionless constant to indicate the how aerodynamic the wing is:

∆F=1/2∙ρ∙(v_o^2-v_u^2 )∙A*Cl

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This is what I end up with. And I would like to end up with the speed the air is flowing over the wing, instead of the difference.

Do you guys understand my problem? English is not my first language, so I have used the physical and mathematical notation I know. I don't know if it differs from English notation.

Thanks in advance.

 

[K^2]

You cannot derive lift equation from Bernoulli's Law. You are interested in the pressure at the interface, and flow velocity at interface is zero, so you can't use Bernoulli's.

Lift is usually derived using Kutta-Joukowski theorem. Look it up.

 

[lalbatros]

I would like to note that the Kutta-Joukowski theorem is related to the Bernouilli's law.
This is not surprising since the Bernouilli's law related pressure and velocity.
See the "heuristic argument" section on http://en.wikipedia.org/wiki/Kutta–Joukowski_theorem#Heuristic_argument".

However, the Bernouilli's law does not explain the lift.
Instead, it simply relates the lift to the "circulation" of the velocity around the airfoil.
How this "circulation" comes about is another story.

In addition, D'Alembert's paradox is that for irrotational, inviscid and incompressible flow there should be no lift and therefore, this "circulation" should be zero!

Therefore, there can only be a lift if at least one of these three assumptions are removed, at least in somewhere in the fluid flow. This why the boundary layer plays the role described by Prandtl.


See also:
http://www3.kis.uni-freiburg.de/~peter/teach/hydro/hydro05.pdf"
http://books.google.com/books?id=Gg... fluid dynamics&pg=PA145#v=onepage&q&f=false"
"[URL D'Alembert's paradox[/URL]

 

[lalbatros]

CaptainPickle,

I did not read your derivation.
However, have you tried to factorize your last result?
Maybe you will get what you wanted?

Michel

 

[rcgldr]

Kutta-Joukowski theorem

This theorem is used to calculate lift for "idealized flow", usually in two dimensions. Calculating lift in real world situations usually involves some simplified implementation of Navier Stokes equations or somthing similar, which are more complicated and go beyond Bernoulli in that they take into account the work performed on the air (Bernoulli equation makes the assumption that no external work is performed on a streamline), and these models also deal with turbulent flow (Bernoulli doesn't handle these cases either).

In addition, D'Alembert's paradox is that for irrotational, inviscid and incompressible flow there should be no lift and therefore, this "circulation" should be zero!

The flow for an inviscid fluid is indeterminate, since in inviscid (zero viscosity) fluid, there's are no interactions between streamlines. D'Alembert's paradox arises from one possible flow. It's also equally likely that a cylinder of fluid with a diameter equal to the size of an object flowing through the fluid, flows at the same speed as the object, with the surrounding fluid stationary.

Lift = 1/2∙ρ∙v²∙A∙Cl

Note this is a simplification that assumes Cl is a constant. It ignores the fact that Cl can be a function of velocity, if there's sufficient range of velocity (Reynolds number, turbulence, flow separation, ...). As mentioned in the first part of this post, Cl isn't calculated from Bernoulli, but normally from actual measurements or Navier Stokes or something similar.

 

[lalbatros]

rcgldr,

I agree that the Navier-Stokes equation must be able to predict the lift of a plane.
However, I was interrested by the hidden part CaptainPickle's question: Why the lift?
And my interrest was first to understand it for myself.
I understood that Prandtl made the link between the Navier-Stokes equation and the lift, by precisely solving d'Alembert's paradox, specially in the boudary layer.

As a student, I was really shocked to learn about the airfoil lift during the course on complex variable dealing with the conform transformations and their applications.
Just as if the lift of an airfoil had to be explained by complex numbers!
This was like an intellectual agression to me!

Despite its incomplete story, the "heuristic derivation" of the Kutta–Joukowski theorem brings me back to my intuition acquired when I was playing with toy planes as a kid. It had always been obvious for me that a toy plane was "lifted from below", or in other terms that it was an action-reaction story. It seems that the Bernoulli's principle is precisely telling that relation between the velocity (action) and the pressure (reaction).

However, it is also clear that the Bernoulli's principle cannot predict the full motion, and only the Navier-stokes equation can go in that direction. Therefore, I would be quite interrested to learn about one more "pedagogic step" to explain in simple terms "why there is a lift". Clearly the Kutta–Joukowski theorem is only half an answer, but it is pedagogical.

I think the missing part might be as simple as specifying realistic boundary conditions along the airfoil.
Would you have some suggestion or some reading to get a short answer to my kid question?

Thanks,

Michel

 

[K^2]

This theorem is used to calculate lift for "idealized flow", usually in two dimensions. Calculating lift in real world situations usually involves some simplified implementation of Navier Stokes equations or somthing similar, which are more complicated and go beyond Bernoulli in that they take into account the work performed on the air (Bernoulli equation makes the assumption that no external work is performed on a streamline), and these models also deal with turbulent flow (Bernoulli doesn't handle these cases either).

Actually, in a real simulation, you are going to use both. You will use Navier-Stokes equations to solve for actual flow, but you will then apply Kutta-Joukowski Theorem to the flow to find total lift.

The main reason for that is the difficulty in precise numerical computation of pressure at the exact interface between airfoil and the fluid/air using finite element analysis, and that's what you need to compute lift forces directly. In contrast, as long as you can assume 2-dimensional flow with no turbulence (AoA < critical) it's very easy to get a good estimate for the circulation.

Though, there are some applications where you need to find exact distribution of forces on the foil. Then you have to byte the bullet and find pressure at interface.