0
$\begingroup$

Background

The flow field around a half-infinite body is defined by a potential flow consisting of a parallel flow with velocity $u = U_{\infty}$ and a source of magnitude Q at position $x = a$, $y = b$

Parallel flow situation: we have $u = U_{\infty}$ and $v = 0$,

$$\phi_{paralell}(x,y) = u_{\infty}x + v_{\infty}y \Rightarrow \phi(x,y) = ux$$

Source flow situation: we have $u = \frac{Q}{2\pi}\frac{x}{x^2+y^2}$ and $v = \frac{Q}{2\pi}\frac{y}{x^2+y^2}$,

$$\phi_{source}(x,y) =\frac{Q}{2\pi} \ln (r) =\frac{Q}{2\pi} \ln (\sqrt{x^2+y^2})$$

$$ \implies \phi_{paralell} + \phi_{source} = \phi_{total}$$

With a source with magnitude ”Q” located at $x = a$, $y = b$. The superimposed velocity potential reads: $$ u_{\infty}(x-a) + \frac{Q}{2\pi} \ln \Big(\sqrt{(x-a)^2+(y-b)^2}\Big) $$ We can then calculate the velocity field $\bar{u} = (u,v)^T$ from the velocity potential $\phi$

$$u = \frac{\partial \phi}{\partial x}=u_{\infty} +\frac{Q}{2\pi} \frac{(x-a)}{(x-a)^2 + (y-b)^2} \quad \textbf{(1)}$$

and $$v = \frac{\partial \phi}{\partial y} = \frac{Q}{2\pi} \frac{(y-b)}{(x-a)^2 + (y-b)^2}$$

The problem

Calculate the position of all stagnation points $(u = v = 0)$. For the following questions assume that the source is placed at the origin, $a = b = 0$. Use cylindrical coordinates, $x = r \cos(\theta)$ and y = $r \sin(\theta)$ for your calculations.

The answer At the stagnation point ($SP$) the velocity is zero, $u = v = 0$. Evaluating this condition for the velocity field found above results for

  • the x-direction in: \begin{align} u &= 0 \\ \iff 0 &= u_{\infty} + \frac{Q}{2\pi } \frac{1}{x_{SP} - a} \quad \textbf{(2)}\\ \iff x_{SP} &= -\frac{Q}{2\pi u_{\infty}} + a \quad \textbf{(3)} \end{align}

  • the y-direction in: $$v = 0 \iff y_{SP} = b$$

Here is what I don't understand

how can we go from $\textbf{(1)}$ to $\textbf{(2)}$ and $\textbf{(3)}$? What are the mathematical and physical explanations behind

$\endgroup$
4
  • $\begingroup$ It recognizes that the sp is going to be at y=0. $\endgroup$ Jun 21, 2019 at 17:15
  • $\begingroup$ Thank you for your comment @ChetMiller. For me that explain why $y_{SP} = b$ (which I uncorrectly wrote originally and corrected) but not the x-direction component, or there is something I'm missing $\endgroup$
    – ecjb
    Jun 21, 2019 at 17:20
  • $\begingroup$ My mistake. It is at y = b. Why do you feel that the algebra does not lead to the desired result? $\endgroup$ Jun 21, 2019 at 17:54
  • $\begingroup$ the y = b is clear to me. How you get the result of $x_{sp}$ is not clear $\endgroup$
    – ecjb
    Jun 21, 2019 at 17:56

1 Answer 1

1
$\begingroup$

$$u = u_{\infty} +\frac{Q}{2\pi} \frac{(x-a)}{(x-a)^2 + (y-b)^2} \quad \textbf{(1)}$$

$$0 = u_{\infty} +\frac{Q}{2\pi} \frac{(x_{sp}-a)}{(x_{sp}-a)^2 + (y_{sp}-b)^2} \quad \textbf{}$$

$$0 = u_{\infty} +\frac{Q}{2\pi} \frac{(x_{sp}-a)}{(x_{sp}-a)^2 + 0} \quad \textbf{}=u_{\infty} +\frac{Q}{2\pi}\frac{1}{(x_{sp}-a)}$$

$\endgroup$

Your Answer

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

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