Calculate time rate of change of volume and the vorticity for a source flow.

For a source flow ,calculate

(a).The time rate of change of the volume of a fluid element per unit volume.

(b) The vorticity.

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    (a)\[\nabla .\mathop V\limits^ \to   = \frac{1}{{\partial V}}\frac{{D\left( {\partial V} \right)}}{{Dt}}\]In polar-coordinates we have \[\nabla .\mathop V\limits^ \to   = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {r{V_r}} \right) + \frac{1}{r}\frac{{\partial {V_\theta }}}{{\partial \theta }}\]Let\[\begin{array}{l}x = r\cos \theta \\y = r\sin \theta \\{V_r} = u\cos \theta  + v\sin \theta \\{V_\theta } =  – u\sin \theta  + v\cos \theta \end{array}\]Let velocity field is given by\[\begin{array}{l}u = \frac{{cx}}{{({x^2} + {y^2})}} = \frac{{cr\cos \theta }}{{{r^2}}} = \frac{{c\cos \theta }}{r}\\v = \frac{{cy}}{{({x^2} + {y^2})}} = \frac{{cr\sin \theta }}{{{r^2}}} = \frac{{c\sin \theta }}{r}\end{array}\]therefore\[\begin{array}{l}{V_r} = \frac{c}{r}{\cos ^2}\theta  + \frac{c}{r}{\sin ^2}\theta  = \frac{c}{r}\\{V_\theta } =  – \frac{c}{r}\cos \theta \sin \theta  + \frac{c}{r}\cos \theta \sin \theta  = 0\\\nabla .\mathop V\limits^ \to   = \frac{1}{r}\frac{{\partial \left( c \right)}}{{\partial r}} + \frac{1}{r}\frac{{\partial \left( 0 \right)}}{{\partial \theta }} = 0\end{array}\](b) Vorticity is given by\[\begin{array}{l}\nabla  \times \mathop V\limits^ \to   = {e_z}\left[ {\frac{{\partial {V_\theta }}}{{\partial r}} + \frac{{{V_\theta }}}{r} – \frac{1}{r}\frac{{\partial {V_r}}}{{\partial \theta }}} \right]\\\nabla  \times \mathop V\limits^ \to   = {e_z}\left[ {0 + 0 – 0} \right] = 0\end{array}\]Therefore, the flow field is ir-rotational. 

    techAir Answered on 3rd January 2019.
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