Magnus effect is commonly explained using Bernoulli principle. However, taking the lift on a rotating cylinder as an example, the velocity difference is caused by the extra work done by the rotating cylinder but not by the pressure difference, the Bernoulli principle is basically energy conservation along a streamline. However here the energy is not conserved due to external work done. So is Bernoulli principle abused in explaining Magnus effect?
[Physics] Is Magnus effect a corollary of Bernoulli principle
aerodynamicsbernoulli-equationfluid dynamicsliftprojectile
Related Solutions
There's no problem with the Bernoulli effect, only with the way it's understood and explained. It's usually explained with mistakes, like the need for asymmetrical airfoil and equal flow time above and below, and without mentioning the need to deflect the direction of airflow.
Here's the best light-math explanation I've seen. Also study this section that directly answers your question.
EDIT: It is easy to find wrong pictures like this:
as opposed to a correct one like this (from the link above):
So the answer to your question is: All of the lift depends on the Bernoulli principle, because speed and pressure are in trade-off, but the physics need to be correctly understood.
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:
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.
The airfoil deflects air downward and by Newton's 3rd law an equal and opposite force (lift) is applied to the wing.
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).
Best Answer
This is a most excellent and astute question. Ultimately it comes down to experiment: the model below works pretty well for many fluids. What this must mean therefore is that the loss is small enough that each particle of fluid, in flowing past the region of disturbance, loses a fraction of its energy that is small enough that it doesn't upset the energy balance underlying the Bernoulli equation greatly. At the same time the viscosity is big enough that the cylinder can sustain the circulation in the flow.
Once one has established a circulation in a flow, the circulation will linger - sometimes almost indefinitely, with very little further input of energy. The Vorticity Equation shows this. You can therefore think of a situation where the cylinder just "happens" to be sitting in a flow with circulation without worrying about how that circulation arose. One can then come up with a lossless mathematical model that does indeed show the Magnus effect, which, judging from the depth and astuteness in your question, you may already ken. In this model, the flow has a certain assumed circulation and one does not think about how this circulation came about. I am talking about the an inviscid, irrotational flow whose complex velocity potential is:
$$\Omega : \{z\in\mathbb{C}|\;|z| > a\}\to \mathbb{C};\;\Omega(z) = a\, v^*\left(\frac{z}{a} + \frac{a}{z}\right) + \frac{\Gamma}{2\,\pi\,i}\,\log z\tag{1}$$
where $v\in\mathbb{C}$ is the fluid's velocity a long way from the cylinder (i.e. the cylinder is steeped in an initially uniform flow) and $\Gamma\in\mathbb{R}$ is the circulation. The cylinder's cross section is the region $\{z\in\mathbb{C}|\;|z| \leq a\}$.
This is a perfectly steady state flow, and, once the circulation is set up, it sustains itself. There is no loss anywhere in the model, so you apply the Theorem of Blasius to calculate the lift, which is simply the quantitative calculation around the cylinder of the wonted Bernoulli's theorem argument.
So you could imagine the following thought experiment (not practical). You have a magic fluid whose viscosity you can switch on and off at will. You have a uniform flow in this fluid and you steep your cylinder in it, and, with internal motors or whatever, you spin the cylinder and the lossy viscosity will set up a circulation in the flow. Then you switch the viscosity off suddenly. The circulation will linger in the flow and now, in the absence of loss, you can do the calculation above and see that there is indeed a lift. Note that one always needs a circulation to generate nonzero lift.
Question by OP
It is still the Bernoulli argument that describes the origin of the pressure gradients and thus, ultimately, the force. The circulation simply introduces an asymmetry to the flow that then makes the sum of pressures nonzero.
The Blasius Theorem is equivalent to the Bernoulli principle as shown as follows: on a section $\mathrm{d} z$ of a contour around the body in the complex plane, the pressure force by the Bernoulli principle is:
$$-\frac{i\,\rho}{2}\,(|v|^2-|\mathrm{d}_z\Omega|^2) \,\mathrm{d}\,z^*\tag{2}$$
where $v$ is as defined in (1) and $\rho$ the fluid density. Here, as in (1), the vector direction is shown by the phase of the complex number. The contour around the body's edge is a streamline, so that the stream function (imaginary part of the complex potential) is constant along it. Therefore, around the edge of the body, $|\mathrm{d}_z\Omega|^2 = (\mathrm{Re}(\mathrm{d}_z\Omega))^2 = (\mathrm{d}_z\Omega))^2$ so that, on summing (2) around the closed contour to find the nett force we wind up with something near to an ordinary contour integral (on witnessing that $\oint v^2\,\mathrm{d}z=0$):
$$F^* = -\frac{i\,\rho}{2} \oint (\mathrm{d}_z\Omega))^2\,\mathrm{d}\,z = \pi\,\rho\,\sum \text{residues of }(\mathrm{d}_z\Omega))^2\text{ at poles within cylinder}\tag{3}$$
which is readily worked out to be $F^*=-i\,v^*\,\rho\,\Gamma$ so that $F=i\,v\,\rho\,\Gamma$, i.e. at right angles to the flow. So you can see that the result as worked out from the Bernoulli principle tells you that the lift is proportional to the circulation.