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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

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  • $\begingroup$ It recognizes that the sp is going to be at y=0. $\endgroup$ Commented 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
    Commented 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? $\endgroup$ Commented 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
    Commented Jun 21, 2019 at 17:56

1 Answer 1

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

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

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

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