For a vortex flow, u=4x/(x^2+y^2) and v=–4y/(x^2+y^2), calculate (a) The time rate of change of the volume of a fluid element per unit volume. (b) The vorticity.

For a vortex flow, $$u = \frac{{4x}}{{{x^2} + {y^2}}}$$ and $$v = \frac{{ – 4y}}{{{x^2} + {y^2}}}$$, calculate (a) The time rate of change of the volume of a fluid element per unit volume. (b) The vorticity.

Asked on 3rd January 2019 in
On changing the equation to polar co-ordinates $\begin{array}{l}x = r\cos \theta \\y = r\sin \theta \\{V_r} = u\cos \theta + v\sin \theta \\{V_\theta } = – u\sin \theta + v\cos \theta \end{array}$$\begin{array}{l}u = \frac{{4y}}{{({x^2} + {y^2})}} = \frac{{4r\sin \theta }}{{{r^2}}} = \frac{{4\sin \theta }}{r}\\v = \frac{{ – 4x}}{{({x^2} + {y^2})}} = \frac{{4r\cos \theta }}{{{r^2}}} = \frac{{ – 4\cos \theta }}{r}\\{V_r} = \frac{4}{r}\cos \theta \sin \theta – \frac{4}{r}\cos \theta \sin \theta = 0\\{V_\theta } = \frac{{ – 4}}{r}{\sin ^2}\theta – \frac{4}{r}{\cos ^2}\theta = \frac{{ – 4}}{r}\end{array}$
Time rate of change of volume of a fluid element per unit volume is given as$\begin{array}{l}\nabla .\mathop V\limits^ \to = \frac{1}{r}\frac{{\partial r{V_r}}}{{\partial r}} + \frac{1}{r}\frac{{\partial {V_\theta }}}{{\partial \theta }}\\\nabla .\mathop V\limits^ \to = \frac{1}{r}\frac{{\partial (0)}}{{\partial r}} + \frac{1}{r}\frac{{\partial \left( { – c/r} \right)}}{{\partial \theta }} = 0 + 0 = 0\end{array}$ (b) Vorticity = $\begin{array}{l}\nabla \times \mathop V\limits^ \to = {e_z}\left[ {\frac{{\partial \left( { – c/r} \right)}}{{\partial r}} – \frac{c}{{{r^2}}} – \frac{1}{r}\frac{{\partial \left( 0 \right)}}{{\partial \theta }}} \right]\\ = {e_z}\left[ {\frac{c}{{{r^2}}} – \frac{c}{{{r^2}}} – 0} \right]\\ = 0\end{array}$$$\nabla \times \mathop V\limits^ \to = 0$$ except at origin, since it is singular at origin.