Source (sink) flow

Potential function for 2D source of strength m at r = 0:

$$\phi = \frac{m}{2\pi }\ln r \notag$$

It satisfies $\nabla ^2\phi = 0$ (check Laplace's equation in polar coordinate in the keyword search utility), except at $r = \sqrt{x^2 + y^2} = 0$ (so must exclude r = 0 from flow)

1.

Question: Derive the equations for the velocity field for the 2D source.

(a)

Hint: expression for the gradient in polar coordinates (use the keyword utility: coordinate system - velocity vector)

2.

Question: Evaluate the outward volume flux.

(a)

Hint: Consider a contour which contains the 2D source.

(b)

Hint: Use Gauss theorem to deform the contour into a small circle of radius $\varepsilon$ around the source.

(c)

Hint: Evaluate the flux in the direction normal to the circle (radial velocity) and integrate along the circle.

If $m < 0 \Rightarrow$ sink. Source with strength $m$ located at $(x_{0},y_{0})$:

$$\phi = \frac{m}{2\pi }\ln\sqrt {\left( {x - x_0 } \right)^2 + \left( {y - y_0 } \right)^2} \notag$$

$$\phi = \frac{m}{2\pi }\ln r \mbox{\ (Potential function) \ } ... ...w}} \psi = \frac{m}{2\pi }\theta \mbox{ \ (Stream function)\ }$$

3D source - Spherical coordinates

A spherically symmetric solution: $\phi = \frac{1}{r}$ (verify $\nabla ^2\phi = 0$ except at $r = 0$)

Define 3D source of strength $m$ located at $r = 0$:

$$\phi = - \frac{m}{4\pi r}, \mbox{\ then \ } V_r = \frac{\part... ...rtial r} = \frac{m}{4\pi r^2}, V_\theta , V_\varphi = 0 \notag$$

1.

Question: Evaluate the net outward volume flux.

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