1
$\begingroup$

In fluid Mechanics,

The superimposed stream function of point source and sink is: $\psi=-\frac{Qcos\theta_1}{4\pi}+\frac{Qcos\theta_2}{4\pi}$

Graphical image of the function

and for a sink - source doublet, we need to show that as: $l\rightarrow 0$

we get:

$\psi=\frac{m}{r}sin^2\theta$
where: $m=\lim_{l\rightarrow0}\frac{Ql}{4\pi}$

I know it's a simple geometric limit problem , but somehow I can't derive the desired result.

Thanks!

$\endgroup$
2
$\begingroup$

The positions of the sink and source are $(-l/2,0)$ and $(l/2,0),$ respectively.

Hence,

$$\psi = \frac{Q}{4 \pi}(\cos \theta_2 - \cos \theta_1) \\= \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{(x-l/2)^2 + y^2}}- \frac{x+l/2}{\sqrt{(x+l/2)^2 + y^2}}\right) \\ = \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{x^2 + y^2 -lx +l^2/4}}- \frac{x+l/2}{\sqrt{x^2 + y^2 +lx + l^2/4}}\right) $$

In terms of polar coordinates, $r = \sqrt{x^2 +y^2}$ and

$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 - \frac{lx -l^2/4}{r^2}\right)^{-1/2}- \frac{x+l/2}{r}\left(1 + \frac{lx +l^2/4}{r^2}\right)^{-1/2}\right).$$

Using the Taylor expansion of the square root terms for small $l$ we obtain

$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 + \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)- \frac{x+l/2}{r}\left(1 - \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)\right) \\ =\frac{Q}{4 \pi}\left(-\frac{l}{r}+ \frac{lx^2}{r^3} +O(l^2) \right) \\= \frac{Ql}{4 \pi r}\left(-1+ \frac{x^2}{r^2} \right)+O(l^2),$$

and with $x = r\cos \theta,$

$$\psi = \frac{Ql}{4 \pi r}\left(-1+ \cos^2 \theta \right)+O(l^2) \\ = -\frac{Ql}{4 \pi r}\sin^2 \theta+O(l^2).$$

Taking the limit as $l \to 0$ we get

$$\lim_{l \to 0} \psi = -\frac{m}{r}\sin^2 \theta$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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