Derive the velocity potential for a doublet.

Derive the velocity potential for a doublet.

Worldtech Asked on 4th November 2019 in Aerodynamics.
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    For a doublet,the  stream function is \[\psi=\frac{-k}{2\pi}\frac{\sin\theta}{r}\]

    \[V_{r}=\frac{\partial \phi}{\partial r}=\frac{1}{r}\frac{\partial \psi}{\partial \theta}\]

    Here \[\psi=\frac{-k}{2\pi}\frac{\sin\theta}{r}
    \\\Rightarrow \frac{\partial \psi}{\partial \theta}=\frac{-k}{2\pi}\frac{\cos\theta}{r}
    \\\frac{\partial \phi}{\partial r}=\frac{1}{r}\left ( \frac{-k}{2\pi}\frac{\cos\theta}{r} \right )=\frac{-k}{2\pi}\frac{cos\theta}{r^{2}}\]

    On integrating with respect to ‘r’

    \[\phi=\left ( \frac{-k}{2\pi}cos\theta \right )\left ( -\frac{1}{r} \right )
    \\\Rightarrow \phi=\frac{k}{2\pi}\frac{\cos\theta}{r}\]

    techAir Answered on 4th November 2019.
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