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For an incompressible potential flow around a smooth rigid body, is it true that the pressure on the surface of the body is proportional to $a\cos^2\theta+b$ where $\theta$ is the angle the inward unit surface normal vector makes with the velocity of the flow at infinity for some constants $a$ and $b$?

The reason for my conjecture is the following two examples.

  1. The incompressible potential flow around a sphere and a cylinder both assume the above relation for the pressure on the surface of the rigid body.

  2. Suppose a column of particles with an infinitesimal cross section area $dA$ collide with a facet with its normal vector forming an angle $\theta\in\big[0,\frac\pi2\big]$ with the particle flow direction vector. The particles bounces off the facet completely elastically. The momentum change is in the normal direction of the facet, and the speed of change is then $2\rho v^2\cos\theta dA$, where $\rho$ is the density of the air flow and $v$ the speed of it. The area upon which this momentum change occurs is $\frac{dA}{\cos\theta}$. Divide the first quantity by the second, we get the pressure $p(\theta):=2\rho v^2\cos^2\theta$. Now the early arriving particles bounce off of the surface normally and collide completely elastically with the late arriving particles and bounce back towards the surface again. By symmetry, the average particle velocity near the surface vanishes in the surface normal direction but its component tangent to the surface remains. Macroscopically, the fluid on average as a whole moves along the tangent of the surface. Alternatively we can assume the complete inelastic collision of the air molecule with the surface, so that the momentum normal to the surface completely dissipates only the tangential component is unmolested so the air molecules after the collision move parallel along the surface. In this case, it is clear $p(\theta):=\rho v^2\cos^2\theta$ which is half of the previous value as the surface normal momentum transferred is half of that in the elastic case. In the case of fractional elastic collision, the $p(\theta):=(1+\alpha)\rho v^2\cos^2\theta$ where $\alpha\in[0,1]$ is the coefficient of collision elasticity.

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  • $\begingroup$ @D.Halsey: It is just my wild guess. See my edited question for my inspiration. $\endgroup$ – Hans Feb 6 '19 at 22:32
  • $\begingroup$ Incompressible potential flow satisfies Laplace's equation. Anything you just make up is highly unlikely to. $\endgroup$ – D. Halsey Feb 6 '19 at 22:39
  • $\begingroup$ @D.Halsey: I understand the Laplace equation quite well. The three examples in my question support my guess. Do you know an explicit example to the contrary? I wonder what the solution is for an ellipse or ellipsoid. I do not have it at hand, though I may be able to solve it either with some special functions or with the elliptical coordinate. There must be an explicit solution already published somewhere. $\endgroup$ – Hans Feb 6 '19 at 23:26
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    $\begingroup$ You should be able to find solutions for ellipses or flat plates, since they result from a simple conformal mapping (Joukowski). $\endgroup$ – D. Halsey Feb 7 '19 at 0:11
  • $\begingroup$ Since a flat plate presents a constant angle to the flow, your equation would predict a constant pressure, which is certainly not correct. $\endgroup$ – D. Halsey Feb 7 '19 at 0:20
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Here is a counterexample.

Consider the 2 dimension problem. Let $e^{i\alpha}\zeta$ where $\alpha$ is a real number and $\zeta$ a complex number. It is a complex velocity potential function generating a constant velocity field pointing towards $e^{i\alpha}$. Let $\omega:=e^{i\alpha}\zeta+\frac{e^{-i\alpha}}{\zeta}$. $\omega(\zeta=e^{i\theta})=0, \,\forall\theta$. $\omega(\theta)$ is the solution for the potential flow since it satisfies the boundary on the unit circle $\zeta=e^{-i\theta},\,\forall\theta$.

(to be continued)

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The question asks about incompressible potential flow, and a conjecture that the OP made that the pressure on a body is always proportional to $a\cos^2\theta+b$, where $\theta$ is related to the local surface angle at the point where the pressure is to be calculated.

In the comments, I suggested that it would be a simple task to find cases that would disprove the conjecture, and I didn't consider that any answer beyond that would be warranted. However, I now think it is worthwhile to explain why no simple equation would be applicable.

Incompressible potential flow is governed by Laplace's equation, which is an elliptic partial-differential equation. All such equations have the property that solutions at any point depend on conditions everywhere, not just at the point being calculated. Physically, this means that pressures at any point on a body depend on the entire shape of the body, not just the local surface angle, and is related to the fact that incompressible flows can be considered as having infinite speed of sound.

To get more specific, equations like the OP's would predict that changing the shape of the aft portion of an airfoil would change the pressures only in that aft portion. This is shown to be wrong in the following figure taken from this report https://www.nrel.gov/docs/fy05osti/36335.pdf. The figure gives potential flow results for an airfoil for three values of flap deflection. In spite of only changing the aft shapes (the flaps), the pressure distributions are clearly changed over the entire airfoil surface. enter image description here

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  • $\begingroup$ So far an explicit counterexample has not been presented yet, the simplicity of the task notwithstanding. The dependence argument above is insufficient to disprove the conjecture. It is not true that the change in the Laplace function value at one point has to affect that at another point, especially at the boundary. In fact, the boundary and the value upon it can assume almost arbitrary value, with some regularity condition. Moreover, $a$ and $b$ depend on the whole problem, they are only invariant along the boundary. Also is the plot from an Laplace equation or considering more factors? $\endgroup$ – Hans Feb 8 '19 at 0:13
  • $\begingroup$ @Hans: According to the reference, the airfoil pressure distributions were obtained using a panel method, which is a highly accurate (but not exact) numerical procedure involving integral equations on the airfoil boundary. These are totally equivalent to solving the Laplace equation with Neumann boundary condition (zero normal velocity). I believe this is a suitable counterexample, and that there is no way that your equation could produce the results from even one of the flap angles, much less all 3. $\endgroup$ – D. Halsey Feb 8 '19 at 0:30
  • $\begingroup$ @Hans: Your description of air molecules bouncing off the surfaces is more appropriate for hypersonic flows. At smaller Mach numbers, the fluid is modeled as a continuum; it flows rather than bouncing. And the pressure on the surface does indeed depend on the entire shape, not just the local angle. $\endgroup$ – D. Halsey Feb 8 '19 at 0:35
  • $\begingroup$ The panel method looks like a straightforward numerical implementation of the double layer potential Green's function method for the Neumann boundary value problem of the Laplace equation. Some simple manipulation could easily generate 3 or arbitrarily more similar looking curves. Copy and paste the following script into www.wolframalpha.com and see the plot: plot 1.2cos^2(pi/2*(1-e^(-3x)))-0.2, cos^2(pi/2*(1-e^(-3x))), 0.8cos^2(pi/2*(1-e^(-3x)))+0.2, x = 0 to 1.1 $\endgroup$ – Hans Feb 8 '19 at 4:46
  • $\begingroup$ Regarding my colliding molecule model, the bouncing is not a problem. The bouncing only manifest when there is only one incoming thin column of particles. When you shoot an infinitely many columns in parallel, one column of particle bouncing off the surface will hit the late comers bounce back towards the surface and back again but move steadily along the surface tangent. The sum of the normal components close to the surface cancel and the tangential component is undissipated. So the fluid moves macroscopically along the surface tangent. I have added this explanation to Point 2 of my question. $\endgroup$ – Hans Feb 8 '19 at 6:33

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