0
$\begingroup$

$\require{cancel}$

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

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

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

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

answer:

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)

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

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.