For a source flow, show that it is a physically possible incompressible flow except at origin. Also, show that it is irrotational.

For a source flow, show that it is a physically possible incompressible flow except at origin. Also, show that it is irrotational.

techAir Asked on 31st December 2018 in Aerodynamics.
Add Comment
  • 1 Answer(s)

    For a source flow, velocity \(\mathop V\limits^ \to   = {V_r}\mathop {{e_r}}\limits^ \to   = \frac{\Lambda }{{2\pi r}}\mathop {{e_r}}\limits^ \to  \)In polar co-ordinates :, the curl of velocity field will be \[\nabla  \cdot \mathop V\limits^ \to   = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {r{V_r}} \right) + \frac{1}{r}\frac{{\partial {V_\theta }}}{{\partial \theta }}\]\[\nabla  \cdot \mathop V\limits^ \to   = \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {r\frac{\Lambda }{{2\pi r}}} \right] + \frac{1}{r}\frac{{\partial \left( 0 \right)}}{{\partial \theta }}\]\[\nabla  \cdot \mathop V\limits^ \to   = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {\frac{\Lambda }{{2\pi }}} \right) + 0 = 0\]Therefore, the flow is a physically possible incompressible flow, except at origin where r =0.

    Also,\[\nabla  \times \mathop V\limits^ \to   = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to  }&{r\mathop {{e_\theta }}\limits^ \to  }&{\mathop {{e_z}}\limits^ \to  }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{{V_r}}&{r{V_\theta }}&{{V_z}}\end{array}} \right| = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to  }&{r\mathop {{e_\theta }}\limits^ \to  }&{\mathop {{e_z}}\limits^ \to  }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{\frac{\Lambda }{{2\pi r}}}&0&0\end{array}} \right|\]\[\nabla  \times \mathop V\limits^ \to   =  – r\mathop {{e_\theta }}\limits^ \to  \left( {\frac{{\partial 0}}{{\partial r}} – \frac{{\partial \Lambda /2\pi r}}{{\partial z}}} \right) + \mathop {{e_z}}\limits^ \to  \left( {\frac{{\partial 0}}{{\partial r}} – \frac{{\partial \Lambda /2\pi r}}{{\partial \theta }}} \right) = 0\]Therefore, \(\nabla  \times \mathop V\limits^ \to   = 0\) (everywhere in the flow).

    Worldtech Answered on 31st December 2018.
    Add Comment
  • Your Answer

    By posting your answer, you agree to the privacy policy and terms of service.