PYQs

Given,

(i) Calculation of vorticity:

velocity \(\vec{q}=\left(z-\frac{2 x}{r}\right) \hat{i}+\left(2 y-3 z-\frac{2 y}{r}\right) \hat{j}+\left(x-3 y-\frac{2 z}{r}\right) \hat{k}\) \(=u \hat{i}+v \hat{j}+\omega \hat{k}\)

Now,

\[\nabla \times \vec{q}=\begin{vmatrix}\hat{i} & \hat{j} & \dot{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z-\frac{2 x}{r} & 2 y-3 z-\frac{2 y}{r} & x-3 y-\frac{2 z}{r}\end{vmatrix}\]

So,

\[\begin{aligned} \nabla \times \vec{q}=& \hat{i}\left(-3+\frac{2 z}{r^{2}} \times \frac{y}{r}+3-\frac{2 y}{r^{2}} \times \frac{z}{r}\right)-\hat{j}\left(1+\frac{2 z}{r^{2}} \times \frac{x}{\sigma}-1-\frac{2 x}{r^{2}} \times \frac{z}{r}\right) \\ &+\hat{k}\left(+\frac{2 y}{r^{2}} \times \frac{x}{r}-\frac{2 x}{r^{2}} \times \frac{y}{\sigma}\right) \\ =& 0 \end{aligned}\]

\(\therefore\) fluid does not possess vorticity.

(ii) Calculation of circulation:

Given,

\(u=z-\frac{2x}{r}\), \(v=2y-3z-\frac{2y}{r}\), \(w=x-3y-\frac{2z}{r} \ldots (1)\)
\(C: x^2+y^2=9, z=0\)

Let \(x= 3 \cos \theta\), \(y= 3 \sin \theta\), \(z=0 \ldots (2)\)

\(\implies\) \(dx= -3 \sin \theta d \theta\), \(dy= -3 \cos \theta d \theta\), \(dz=0 \ldots (3)\)

Now, the circulation around the circle \(C\) is given by:

\(\Gamma=\int_{C} \vec{q} \cdot d \vec{r}=\int_{0}^{2 \pi} \vec{q} \cdot \overrightarrow{d r}\)
\(\implies \Gamma=\int_{0}^{2 \pi} u d x+v d y+w d z\)
\(\implies \Gamma = \int_{0}^{2 \pi} (0-2 \cos \theta)(-3 \sin \theta d \theta) + (6 \sin \theta - 2 \sin \theta - 2 )(3 \cos \theta d \theta) + 0\) \(\implies \Gamma= \int_{0}^{2 \pi} 6 \sin \theta \cos \theta d \theta + 12 \sin \theta \cos \theta d \theta\) \(\implies \Gamma= \int_{0}^{2 \pi} (2 \sin \theta \cos \theta \d \theta)\)
\(\implies \Gamma = \int_{0}^{2 \pi} \sin 2 \theta d \theta\) \(\implies \Gamma=0\)

Thus, the circulation is 0.

The differential equations of streamlines and vortex lines are respectively: \(\begin{array}{l}\frac{d x}{u}=\frac{d y}{v}=\frac{d z}{w}\end{array} \quad \ldots (1)\) \(\begin{array}{2} \frac{d x}{\xi} =\frac{d y}{\eta}=\frac{d z}{\zeta}\end{array} \quad \ldots (2)\) (1) and (2) will intersect orthogonally iff \(\begin{array} u \xi+v \eta+w \zeta=0 \\ \implies \mu\left(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial 2}\right)+v\left(\frac{\partial u}{\partial z}-\frac{\partial \omega}{\partial x}\right)+\omega\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)=0\end{array}\) This implies that \(u d x+v d y+w d z \text { is }\) a perfect differential.

\[\begin{array}{rl} \therefore u d x+v d y+w d z&=\mu d \psi \\ & =\mu\left(\frac{\partial \psi}{\partial x} d x+\frac{\partial \psi}{\partial y} d y+\frac{\partial \psi}{\partial z} d x\right). \\ \implies u, u ,w &=\mu\left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right)\end{array}\]

Let row of vertices be taken along \(x-axis\).

So, we have vortices of strength \(k\) each points \((0,0)\), \((\pm 2a,0)\), \((\pm 4a,0), \ldots\) and those of strength \(-k\) each at points \((\pm a,0)\), \((\pm 2a,0), \ldots\)

Thus, the complex potential of entire system is given by:

\[\begin{aligned} w & =\frac{i k}{2 n}\left[\operatorname{log} z+\operatorname{log}(z-2 a)+\operatorname{log}(z+2a)+\ldots\right] \\ &- \frac{i k}{2 n}\left[\operatorname{log}(z-a)+\log (z+a)+ log(z-3a)+\cdots \right] \\ &=\frac{ik}{2 \pi} \log \left( \frac{z\left[z^{2}-(2 a)^{2}\right]\left[z^{2}-(4 a)^{2}\right)] \ldots}{\left.\left[z^{2}-a^{2}\right]\left[{z}^{2}- (3a^{2}\right)\right] \ldots} \right) \\ &=\frac{i k}{2\pi} \log \left( \frac{\left.\frac{z}{2 a}\left[1-(\frac{z}{2 a}\right)^{2}\right]\left[1-\left(\frac{z}{4 a}\right)^{2}\right]c\ldots }{\left[1-\left(\frac{z}{a}\right)^{2}\right]\left[1-\left(\frac{z}{3 a}\right)^{2}\right) \ldots}\right) \\ &=\frac{i k}{2 \pi} \log \left( \frac{\sin \pi z / 2 a}{\cos (\pi 2 / 2 a)} \right) \\ \implies w &= \frac{i k}{2 \pi} \log \left(\tan \left(\frac{\pi z}{2 a}\right) \right) \end{aligned}\]

This will determine velocity potential and stream function.

Vorticity \((\Omega)=\) curl \(q\) \(\begin{aligned}\Omega &= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ u & v & w \end{vmatrix} \\ &= \left(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial z}\right) \hat{\imath}+\left(\frac{\partial u}{\partial z}-\frac{\partial w}{\partial x}\right) \hat{j} +\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right) \hat{k} \end{aligned}\)

Now, since \(\Omega\) is constant, therefore

\[\begin{aligned} \frac{\partial w}{\partial y}-\frac{\partial v}{\partial z}&=c_{1} \ldots (1) \\ \frac{\partial u}{\partial z}-\frac{\partial w}{\partial x}&=c_{2} \ldots (2) \\ \frac{\partial v}{d x}-\frac{\partial u}{\partial y}&=c_{3} \ldots (3) \end{aligned}\]

Differentiating (2) by \(z\) and (3) by\(y\) and subtracting, we get:

\[\begin{aligned} \frac{d^{2} u}{\partial z^{2}}-\frac{\partial^{2} w}{\partial z \partial x}-\frac{\partial^{2} v}{\partial y d x}+\frac{\partial^{2} u}{d y^{2}}&=0 \\ \frac{\partial^{2} u}{\partial y^{2}}+\frac{d^{2} u}{\partial z^{2}}-\frac{\partial^{2} v}{\partial x \partial y}-\frac{\partial^{2} w}{d x \partial z}&=0 \\ \frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}-\frac{\partial}{\partial x}\left[\frac{\partial) v}{\partial y}+\frac{\partial w}{\partial z}\right] &= 0 \ldots (4) \end{aligned}\]

Acccording to the equation of continuity,

\(\begin{aligned} \frac{\partial u}{d x}+\frac{d v}{d y}+\frac{d w}{d z}=0 \\ \therefore-\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}+\frac{d w}{d z} \end{aligned}\) Putting this in (4), we get

\(\begin{aligned}\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}-\frac{\partial}{\partial x}\left(-\frac{\partial u}{\partial x}\right)=0 \end{aligned}\) \(=\quad \frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^2 u}{\partial z^{2}}+\frac{\partial^{2} u}{\partial x^{2}}=0\) \(\therefore \nabla^{2} u=0\) Similarly we can show \(\nabla^{2} v=0\) and \(\nabla^{2} \omega=0\).

Hence proved.

Let the axis of the cylinder be \(x-axis\) and vortices are situated at \(P(r, \theta)\) and \((r,-\theta)\) respectively.

Also, let strength at \(P=k\) and strength at \(Q=-k\). Clearly, line \(PQ\) is perpendicular to the \(x-axis\).
The image system due to \(P\) consists: \(\rightarrow\) Vortex of strength \(-k\) at \(\frac{a^{2}}{r}\left(P^{\prime}\right),\) where \(a\) is radius of cylinder, \(\rightarrow\) Vortex at centre of cylinder of strength \(k\).

Image systen due to \(Q\) consists: \(\rightarrow\) Vorex of strength \(k\) at \(\frac{a^{2}}{r}\text{say Q}\) \(\rightarrow\) Vortex of stremth \(-k\) at center of cylinder.

Vortices at the centre of cylinder cancel out.

At any point \(S(2)\), the complex potential \(W\) is given by:

\[\begin{aligned} W=\frac{i k}{2 \pi}[& \log \left(z-r e^{i \theta}\right)-\log \left(z-\frac{a^{2}}{r} e^{i \theta}\right)-\log \left(z-r e^{-i \theta}\right) \\ &\left.+\log \left(z-\frac{a^{2}}{r} e^{-i \theta}\right)\right] \end{aligned}\]

Complex potential at \(P(r, \theta)\) will be: \(W^{\prime}=W\) - potential due to \(P\) \(W^{\prime}=\frac{i k}{2 \pi}\left[-\log \left(z-\frac{a^{2}}{r} e^{i \theta}\right)-\log \left(z-i e^{-i \theta}\right)+\log \left(z-\frac{a^{2}}{r} e^{-i \theta}\right)\right]\) \(W^{\prime}=\frac{i k}{2 \pi}\left[\log \frac{\left(z-\frac{a^{2}}{r} e^{-i \theta}\right)}{\left(z-r e^{-i \theta}\right)\left(z-\frac{a^{2}}{r} e^{i \theta}\right)}\right]\) At \(P\), \(z=r e^{i \theta}\) \(W^{\prime}=\frac{i k}{2 \pi}\left[\log \frac{\left(r e^{i \theta}-\frac{a^{2}}{r} e^{-i \theta}\right)}{\left(r e^{i \theta}-r e^{-i \theta}\right)\left( e^{i \theta}-\frac{a^{2}}{r} e^{i \theta}\right)}\right]\) As \(W^{\prime}=\phi+i \psi\) \(\therefore \psi=\frac{k}{2 \pi} \log \frac{\mid r e^{i \theta}-\frac{a^{2} e^{-i \theta}}{r}|}{\left|r e^{i \theta}-r e^{-i \theta}\right| |r e^{i \theta}-\frac{a^{2}}{r}^{i \theta} \mid}\)

Streamlines are given by \(\psi=\) constant \(\frac{k}{2 \pi}\left[\log \frac{\vert r^{2} e^{i \theta}-a^{2} e^{-i \theta}\vert /r}{ \vert { 2 i r \sin \theta \mid \vert \frac{r^{2} e^{i \theta}-a^{2}e^{i \theta}} {r} } \vert}\right]=C_{1}\)

\(\log \frac{\mid\left(r^{2}+a^{2}\right) \sin \theta i+\left(r^{2}-a^{2}\right) \cos \theta|}{|2 i r \sin \theta|\left|\left(r^{2}-a^{2}\right) \cos \theta+i\left(r^{2}-a^{2}\right) \sin \theta\right|}=C_{2}\) \(\Rightarrow \frac{\left.\left(\left(r^{2}+a^{2}\right)^{2}-\left(r^{2}-a^{2}\right)^{2}\right) \sin ^{2} \theta+r^{2}-a^{2}\right)^{2}}{4 r^{2} \sin ^{2} \theta\left(r^{2}-a^{2}\right)^{2}}=C_{3}\) \(\begin{aligned} \frac{4 r^{2} a^{2} \sin ^{2} \theta+\left(r^{2}-a^{2}\right)^{2}}{4 r^{2} \sin ^{2} \theta\left(r^{2}-a^{2}\right)^{2}} &=C_{3} \\ 4 r^{2} a^{2} \sin ^{2} \theta+\left(c^{2}-a^{2}\right)^{2}=& C_{3} 4 r^{2} \sin ^{2} \theta \left(r^{2}-a^{2}\right)^{2} \end{aligned}\)

If we take \(C_{3}=\frac{1}{4 b^{2}}, b\) is constant \(4 r^{2} a^{2} \sin ^{2} \theta=\left(r^{2}-a^{2}\right)^{2}\left[4 r^{2} \sin ^{2} \theta \frac{1}{4 b^{2}}-1\right]\) \(4 r^{2} a^{2} \sin ^{2} \theta=\frac{\left(x^{2}-a^{2}\right)^{2}}{b^{2}}\left(r^{2} \sin ^{2} \theta-b^{2}\right)\) \({4 r^{2} a^{2} b^{2} \sin ^{2} \theta=\left(r^{2}-a^{2}\right)^{2}\left(r^{2} \sin ^{2} \theta-b^{2}\right)}\)

Hence, proved.

The differential equations of streamlines and vortex lines are respectively: \(\begin{array}{l}\frac{d x}{u}=\frac{d y}{v}=\frac{d z}{w}\end{array} \quad \ldots (1)\) \(\begin{array}{2} \frac{d x}{\xi} =\frac{d y}{\eta}=\frac{d z}{\zeta}\end{array} \quad \ldots (2)\) (1) and (2) will intersect orthogonally iff \(\begin{array} u \xi+v \eta+w \zeta=0 \\ \implies \mu\left(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial 2}\right)+v\left(\frac{\partial u}{\partial z}-\frac{\partial \omega}{\partial x}\right)+\omega\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)=0\end{array}\) This implies that \(u d x+v d y+w d z \text { is }\) a perfect differential.

\[\begin{array}{rl} \therefore u d x+v d y+w d z&=\mu d \psi \\ & =\mu\left(\frac{\partial \psi}{\partial x} d x+\frac{\partial \psi}{\partial y} d y+\frac{\partial \psi}{\partial z} d x\right). \\ \implies u, u ,w &=\mu\left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right)\end{array}\]