PYQs

Consider the motion of fluid in cylindrical coordinates. The equation of continuity is given by:

\[\nabla \cdot(\rho \vec{q})+\frac{\partial \rho}{\partial t}=0\]

For incompressible fluid and steady flow, \(\quad \nabla \cdot(\vec{q})=0\)
\(\implies \frac{1}{r} \frac{\partial}{\partial r}\left(rq_{r} \right)+\frac{1}{r} \frac{\partial}{\partial \theta}(q_\theta)+\frac{\partial}{\partial z}(q_z)=0\)
For axi-symmetric flow, \(\frac{\partial}{\partial \theta}=0\)
\(\begin{aligned} \Rightarrow & \frac{1}{r} \frac{\partial}{\partial r}\left(r q_{r}\right)+\frac{\partial}{\partial z}(q_z)=0 \\ \Rightarrow & \frac{\partial}{\partial r}\left(r q_{r}\right)+r \frac{\partial}{\partial z}\left(q_{z}\right)=0 \end{aligned}\)
Now the condition that \(r q_{r} d z-r q_{z} dr\) may be an exact differential. Let it be equal to \(d \psi\).
\(\implies \quad r q_{r} d z-r q_z d r=d \psi=\frac{\partial \psi}{\partial r} dr+\frac{\partial \psi}{\partial z} d z\)
\(\implies rq_{r}=\frac{\partial \psi}{\partial z}\) and \(-r q_{z}=\frac{\partial \psi}{\partial r}\)
\(\implies q_{r}=\frac{1}{r} \frac{\partial \psi}{\partial z}\) and \(q_{z}=\frac{-1}{r} \frac{\partial \psi}{\partial r}\)
which satisfy continuity equation. Also, the streamlines are given by:
\(\frac{d r}{q_{r}}=\frac{d z}{q_z}\)
\(\implies rq_rdz-rq_zdr=0\)
\(\implies d \psi=0\)
\(\implies \psi = constant\)
\(\implies \psi\) exists due to equation of continuity.

If flow is irrotational, then \(\phi\) (potential) exists, such that
\(\vec{q} = - \nabla{\phi}\)
\(q_{r}=-\frac{\partial \phi}{\partial r}\), \(q_{z}=-\frac{\partial \phi}{\partial z}\)
Also,
\(q_{r}=\frac{1}{r} \frac{\partial \psi}{\partial z}\), \(q_{z}=-\frac{1}{r}\frac{\partial \psi}{\partial r}\)
Also,
\(\frac{\partial}{\partial z}\left(\frac{\partial \phi}{\partial r}\right)=\frac{\partial}{\partial r}\left(\frac{\partial \phi} {\partial z}\right)\)
\(\frac{\partial}{\partial z}\left(\frac{-1}{r} \frac{\partial \psi}{\partial z}\right)=\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial \psi}{\partial r}\right)\)
\(\frac{-1}{r} \frac{\partial^{2} \psi}{\partial z^{2}}=\frac{1}{r} \frac{\partial^{2} \psi}{\partial r^{2}}-\frac{1 \partial \psi}{r^{2} \partial r}\)
\(\frac{\partial^{2} \psi}{\partial z^{2}}-\frac{1 \partial \psi}{r \partial r}+\frac{\partial^{2} \psi}{\partial r^{2}}=\theta\)

\[\begin{aligned} \text { Complex potential, }w&=U_{0}\left(z+\frac{a^{2}}{2}\right) \\ \text { at } z=r e^{i \theta}, w &=U_{0}\left(r e^{i \theta}+\frac{a^{2}}{r} e^{-i \theta}\right) \\ w&=u_0\left[r \cos \theta+i r \sin \theta+\frac{a^{2}}{8} \cos \theta-i \frac{a^{2}}{r} \sin \theta\right] \\ w&=u_0\left[r \cos \theta+\frac{a^{2}}{r} \cos \theta\right]+i u_{0}\left[r \sin \theta-\frac{a^{2}}{r} \sin \theta\right] \end{aligned}\]

Since \(w=\phi+i \psi\),

\(\implies \phi=U_{0}\left[r \cos \theta+\frac{a^{2}}{r} \cos \theta\right]\) and \(\psi=u_{0}\left[r \sin \theta-\frac{a^{2}}{r} \sin \theta\right]\)

\[\begin{aligned} q &= \left|\frac{d w}{d z}\right| \\ &= u_{0} \vert 1-a^{2}/{z^{2}} \vert \\ &=u_{0}\vert-a^{2}e^{-i 2 \theta}/r^{2} \vert \\ \implies q &=u_{0}\vert1-e^{-i 2 \theta}\vert \end{aligned}\]

At stagnation point, \(q=0\),

\(\implies 1-e^{-i 2 \theta} =0\) \(\implies \cos 2 \theta-i \sin 2 \theta=1\) \(\implies \cos 2 \theta=1\) and \(\sin 2 \theta=0\) \(\implies 2 \theta= 2n \pi\) and \(2 \theta = n \pi\), \(n \in Z\) \(\implies \theta = m \pi\), \(m \in Z\)