1
$\begingroup$

I'm trying to solve two-dimensional potential flow over airfoils with the finite element method, using the stream function formulation ($\Delta\psi = 0$, $u = -\partial\psi/\partial y$, $v = \partial\psi/\partial x$) and superposition to account for and enforce the Kutta condition (zero velocity) at the trailing edge. Following the approach of de Vries we have [1]

\begin{equation} \psi = \psi_1 + b*\psi_2 + c*\psi_3 \end{equation}

where

\begin{equation} \psi_1: \Delta\psi = 0,\quad \psi_{\Gamma_2} = -U_\infty y,\quad \psi_{\Gamma_1} = 0\end{equation} \begin{equation} \psi_2: \Delta\psi = 0,\quad \psi_{\Gamma_2} = 1,\quad \psi_{\Gamma_1} = 0\end{equation} \begin{equation} \psi_3: \Delta\psi = 0,\quad \psi_{\Gamma_2} = 0,\quad \psi_{\Gamma_1} = 1\end{equation}

and $\Gamma_2$ is the external boundary of the domain with horizontal velocity $U_\infty$, and $\Gamma_1$ the boundary of the airfoil. Using superposition, and the fact that $u_T=v_T=0$ at the point of the trailing edge $T$, we should be able to determine the constants $b$ and $c$ with (equations 16a and 16b in [1])

\begin{equation} u_T = u_{1T} + bu_{2T} + cu_{3T} = 0\end{equation} \begin{equation} v_T = v_{1T} + bv_{2T} + cv_{3T} = 0\end{equation}

All this seems resonable until one considers the fact that $\psi_2$ and $\psi_3$ will be linearly dependent $(\psi_3 = 1-\psi_2)$ so the system $\{u_T,v_T\}$ is not determined. Is there something I'm missing here or have misunderstood? Clearly the author has used this approach successfully, but I just can't see how it can work in this case.

[1] G. de Vries and D. H. Norrie, "The Application of the Finite-Element Technique to Potential Flow Problems", J. Appl. Mech 38(4), 798-802, 1971, DOI: 10.1115/1.3408957

$\endgroup$
  • $\begingroup$ It isn't obvious to me why $\psi_3 = 1 - \psi_2$. Can you explain this? Unfortunately, I don't have access to the paper. Is there a publicly-accessible copy somewhere? $\endgroup$ – Bill Greene Feb 9 at 19:17
  • $\begingroup$ Consider Laplace equation on the unit line with Dirichlet BCs 0 and 1 on the end points, the one solution will be x and the other 1-x. This will extend problems in higher dimensions with only two boundaries. The paper is available here. $\endgroup$ – WillG Feb 10 at 0:48
  • $\begingroup$ The problem that you are proposing only has as solution $u=x$, assuming that the left end is located at $x=0$. $\endgroup$ – nicoguaro Feb 10 at 7:17
  • $\begingroup$ To clarify what I meant, applying BCs 0/1 for $\psi_2$ gives you the solution $\psi_2=x$, and similarly BCs 1/0 gives $\psi_3=1-x$, thus $\psi_3 = 1-\psi_2$. $\endgroup$ – WillG Feb 10 at 7:29
  • $\begingroup$ Thanks for the paper link. I understand about the two solutions for the 1D case. But why do you say those solutions are "linearly dependent?" You can't get the second solution by multiplying the first by a constant. Have you actually run into a specific problem applying the authors' approach? $\endgroup$ – Bill Greene Feb 10 at 12:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.