0
$\begingroup$

In my Fluid Mechanics class, we arrived at an equation for 2D compressible flow (using perturbation theory):

$$\left(1-M^{2}\right)\frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}=0$$

where $\phi$ is the perturbed fluid potential ($\vec v=\vec U+\vec\nabla(\phi)$), $\vec U=U\vec e_{x}$ is a constant flow and $M=U/c_s$ is the Mach number.

I wanted to resolve this equation for an arbitrary contour defined by $$ f_\pm=\begin{cases}f(x) & \quad\text{if }x\in[0,l] \\ 0 & \quad \text{if } x\in]l,+\infty[\end{cases} $$

where $f_{+}$ is for $y>0$ and $f_{-}$ for $y<0$. and you also asume $\frac{df}{dx}\ll(\text{the other non pertubed quantities})$

enter image description here

My teacher only solved the supersonic case using the fact that for $M>1$ you get the wave equation. He found his boundary conditions like this:

$$\frac{df}{dx}=\tan(\alpha)=\frac{\nabla\phi|_{y}}{U+\nabla\phi|_{y}}$$ enter image description here

wich gives neglecting the quadratic perturbed quantities:

$$\frac{\partial\phi}{\partial y}|_{y=f'(x)}\simeq\frac{\partial\phi}{\partial y}_{y=0}\simeq Uf'(x).$$

My problem is that I have tried solving this equation (using Fourier transform) for a subsonic flow using the same boundary condition (and a bounded solution). And while I found a solution, this solution doesn't seem very physical or accurate when I plot it on python.

So I was wondering if my boundary conditions are incorrect and if they are what are the correct one?

$\endgroup$

1 Answer 1

1
$\begingroup$

For the subsonic case you get Laplace's equation rather than the wave equation. The solution method is quite different and too complicated to describe here. It comes down to solving a singular integral equation, for which some slick techniques were devised, explained in any undergrad aerodynamics text

$\endgroup$
2
  • $\begingroup$ are the boundary conditions the same ( $\frac{\partial\phi}{\partial y}_{y=0}\simeq Uf'(x)$ )? and can you solve for those boundaries using the fourrier transform? $\endgroup$
    – Jhon Jack
    Commented Jun 21, 2017 at 23:19
  • $\begingroup$ The boundary conditions are the same, but there is no transformation from a subsonic flow to a supersonic flow. If you look for solutions by separation of variables you will see they are quite different. $\endgroup$
    – Philip Roe
    Commented Jun 22, 2017 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.