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A previous question by David Zaslavsky was a request for a broad, "how things work" type of explanation of the lift of an airfoil. The answers given there are enlightening, but don't address a more specific question I have.

First a summary of my current understanding. There is no logical reason to assume that parcels of air take equal time to travel above and below the wing. If we did make such an assumption, Bernoulli's principle would make all kinds of incorrect predictions. "How things work" explanations (e.g., this one) often speak as though Bernoulli's principle is one effect, and then there are some other effects, such as conservation of momentum; it's implied that these effects are additive, and it's often stated that the Bernoulli effect is negligible. But in fact Bernoulli's principle is, under certain hypotheses, equivalent to conservation of energy. Conservation of energy and conservation of momentum are not effects to be added. They're two physical laws, both of which need to be obeyed.

It's only under certain hypotheses that Bernoulli's principle is equivalent to conservation of energy. It requires nonviscous flow, and the version given in freshman physics textbooks also assumes incompressible flow, although there is also a version for compressible flow. The WP article on Bernoulli's principle states that it's usually valid for Mach numbers lower than about 0.3, which I guess might be valid for birds and some small planes (cruise speed for a Cessna 172 is Mach 0.25), but not passenger jets.

Based on this, I can envision two logical possibilities:

(1) The hypotheses of Bernoulli's principle hold for birds and some small planes. The equation gives at least an approximately correct result for the net force on the wing, but there is no obvious, simple way to know what velocities to assume.

(2) The hypotheses of Bernoulli's principle fail in these cases.

Which of these is correct? If #2, which hypothesis is it that fails? Is the incompressible-flow version inapplicable but the compressible-flow version OK?

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related: physics.stackexchange.com/q/51503/4552 –  Ben Crowell Jun 21 '13 at 13:27
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2 Answers 2

First to Address the Equivalence of Bernoulli's Equation to Conservation of Momentum:

There are (at least) three popular explanations for lift on an airfoil:

  1. Faster air on the top has lower static pressure than slow moving air on the bottom. The resulting pressure difference multiplied by the area is equal to the lift.

  2. The airfoil deflects air downward and by Newton's 3rd law an equal and opposite force (lift) is applied to the wing.

  3. Bound circulation on the wing generates lift due to the Kutta--Joukowski theorem.

All three are equivalent.

Bernoulli's equation is derived from conservation of momentum (Navier-Stokes equations) with the assumption that the velocity has a potential function. Bernoulli's principle is merely the mechanism for the equal and opposite force to be applied to the wing in explanation #2.

Circulation is required for there to be any downward deflection of air. Without circulation, the flow would not exit smoothly at the trailing edge. Bound circulation is a result of boundary layers forming on the top and bottom surfaces. (Also a result of conservation of momentum)

To get a decent estimate of the lift curve of an airfoil at small angles-of-attack, panel methods are used to solve for the tangential velocity at the surface of an airfoil. This tangential velocity is then fed into Bernoulli's equation to get the pressures. Integrating the pressure over the surface of the airfoil we can find the lift.

These panel method's are inviscid but add just the right amount of circulation to get a realistic solution. (satisfying the Kutta condition)

While all three of the explanations above are valid, they are just rewording the unfulfilling explanation of the reason for lift: Conservation of Momentum.

Range of Applicability of Bernoulli's Equation

The incompressible version of Bernoulli's equation, $1/2 \mathbf{U}\cdot\mathbf{U} + p/\rho = const$, is valid:

  • along streamlines
  • along vortex lines (lines parallel to $\omega = \nabla \times \mathbf{U}$)
  • and everywhere in irrotational flow ($\omega = 0$)

(for details see Ch. 2 of Viscous Fluid Flows by Frank M. White)

For airplanes, you can trace a streamline far in front of the aircraft into a region where $\mathbf{U} = const$ and hence $\omega = 0$. This means that the $const$ in Bernoulli's equation is the same everywhere and the equation can be used to find pressures anywhere on the surface of the wing.

In practice, if the Bernoulli equation is used, the velocities are found by panel methods. (If you are doing a viscous simulation you probably are calculating the pressures directly without using Bernoulli's equation.) The velocities from the panel method are analogous to the edge velocity at the top of the boundary layer near the surface. Any good viscous flow book will show that for boundary layers $dp/dy \approx 0$, where $y$ is oriented normal to the surface. This is also true for compressible flows up to about $M \approx 5$.

This method of using Bernoulli's equation to find the pressures (and by extension, forces) over the surface is a good approximation in many cases. When the results are not accurate, it is typically a result of inaccurate velocity information. For example, due to their inviscid nature, panel methods are not very good at describing separated flow such as an airfoil at high angle-of-attack.

Compressible vs. Incompressible

$M < 0.3$ as the limit of incompressible flow is thrown around a lot but typically without any justification. Consider gas at rest with density $\rho_0$ that is then accelerated isentropically to Mach number $M$. The density of the gas will change in this new state and is given by:

$$\frac{\rho_0}{\rho} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{1/(\gamma-1)}$$

As it turns out, in air $M \approx 0.3 \rightarrow \rho_0/\rho = 0.95$ and it is then assumed for practical purposes that if the density changes by no more than 5% the flow can be assumed to be incompressible.

You are correct that there are compressible versions of Bernoulli's equation. The results of calculations using the compressible equation should be a good approximation as long as the inputted velocities are accurate and the streamlines can be traced into the freestream (i.e. not separated flow).

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This doesn't really address the question, which is about the range of applicability of Bernoulli's principle as a good approximation. –  Ben Crowell May 2 '13 at 23:03
    
@BenCrowell Sorry for the late response but I have updated my answer to better address your question. –  OSE Jan 24 at 19:04
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To think of Bernoulli's principle as opposed to something else is not right. Bernoulli's principle can be thought of as the reason for the something else.

Take a look at John Denker's explanation. It's the best I've seen.

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+1: Yes. It's depressing how often smart people still get this wrong. It's like a lot of these bad ideas came from people overreacting to other bad ideas and the real physics got lost somewhere. –  Michael Brown May 2 '13 at 14:42
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Denker's discussion is great. What I'm getting from it is that the hypotheses of Bernoulli's principle are a pretty good approximation for bird wings and for low-speed flight in some prop planes, but they are a poorer approximation for higher-speed flight unless higher-order corrections are added. –  Ben Crowell May 2 '13 at 15:47
    
@Ben: Right, as you get to Mach numbers above maybe 0.5. –  Mike Dunlavey May 2 '13 at 16:08
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