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 $v = \frac{Q}{2\pi}\frac{y}{x^2+y^2}$,

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

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

Calculate the stream line $\psi_{stag} = const$, which goes through the stagnation point and defines the surface $r(\theta)$, of the semi-infinite body


Assuming the source being in the origin at position $a = 0$, $b = 0$, the stream function reads:

\begin{align*} \psi_{total} &= \psi_{parallel} + \psi_{source}\\ &= u_{\infty}(y-b) + \frac{Q}{2\pi} \arctan \frac{y-b}{x-a}\\ &= u_{\infty}y + \frac{Q}{2\pi} \arctan \frac{y}{x} \end{align*}

Substituting cartesian coordinates by polar coordinates $x = r \cos \theta$, $y = r \sin \theta$ the stream function can be rewritten as: \begin{align*} \psi &= u_{\infty}r \sin \theta + \frac{Q}{2\pi} \arctan \frac{\cancel{r} \sin \theta}{\cancel{r} \cos \theta}\\ &= u_{\infty}r \sin \theta + \frac{Q}{2\pi} \underbrace{\arctan \frac{ \sin \theta}{ \cos \theta}}_{=\theta, \forall \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]}\\ &= u_{\infty}r \sin \theta + \frac{Q}{2\pi}\theta \end{align*}

The stagnation point in polar coordinates is $$r = \frac{Q}{2\pi u_{\infty}}, \theta = \pi \quad \textbf{(1)}$$ The value of the stream function at the stagnation point is found to be:

\begin{align*} \psi = u_{\infty}\frac{Q}{2\pi u_{\infty}} \underbrace{\sin \theta}_{=0} + \frac{Q}{2\pi}\underbrace{\theta}_{=\pi}\\ = \psi_{SP} = \frac{Q}{2} \end{align*}

The streamline through the stagnation point is also defining the surface of the semi-infinite body:

\begin{align*} \psi(r, \theta) &= \psi_{SP}\\ \implies u_{\infty}r\sin(\theta) + \frac{Q}{2\pi}\theta &= \frac{Q}{2}\\ u_{\infty}r\sin(\theta) &= \frac{Q}{2} - \frac{Q}{2\pi}\theta\\ u_{\infty}r\sin(\theta) &= \frac{Q}{2\pi}(\pi - \theta)\\ r(\theta) &= \frac{1}{u_{\infty} \sin \theta}\frac{Q}{2\pi}(\pi - \theta) \end{align*}

my question

What is the mathematical / physical justification to be able to that that $\theta = \pi$ in (1)


1 Answer 1


The angle $\theta$ is measured about the point source, with the + x direction taken as $\theta=0$. The stagnation point is situated to the left of the point source, which corresponds to $\theta=\pi$. Just draw a diagram.


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