# Find the stagnation point around semi-infinite body (superposition of parallel flow and source)

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

• It recognizes that the sp is going to be at y=0. Jun 21, 2019 at 17:15
• 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
– ecjb
Jun 21, 2019 at 17:20
• My mistake. It is at y = b. Why do you feel that the algebra does not lead to the desired result? Jun 21, 2019 at 17:54
• the y = b is clear to me. How you get the result of $x_{sp}$ is not clear
– ecjb
Jun 21, 2019 at 17:56

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