PYQs

First Part: The complex potential \(w\) at any point \(P(z)\) is given by \(w=-m \log (z-a)-m \log (z+a)+2 m \log z \ldots (1)\)

\[\implies w=m\left[\log z^{2}-\log \left(z^{2}-a^{2}\right)\right]\]

\(\implies \phi+i \psi=m\left[\log \left(x^{2}-y^{2}+2 b y\right)-\log \left(x^{2}-y^{2}-a^{2}+2 b y\right)\right],\) as \(z=x+i y\)

Equating the imaginary parts, we have \(\begin{array}{c} \psi=m\left[\tan ^{-1}\left\{2 x y /\left(x^{2}-y^{2}\right)\right\}-\tan ^{-1}\left\{2 x y /\left(x^{2}-y^{2}-a^{2}\right)\right\}\right] \\ \psi=m \tan ^{-1}\left[\frac{-2 a^{2} x y}{\left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right)}\right], \text { on simplification } \end{array}\)

The desired streamlines are given by \(\psi=\) constant \(=m \tan ^{-1}(-2 / \lambda) .\) Then we obtain

\[(-2 / \lambda)=\left(-2 a^{2} x y\right) /\left[\left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right)\right] \quad \text { or } \quad\left(x^{2}+y^{2}\right)^{2}=a^{2}\left(x^{2}-y^{2}+\lambda x y\right)\]

Second Part: We have,

\[\begin{aligned} &\begin{aligned} \frac{d w}{d z} &=-\frac{m}{z-a}-\frac{m}{z+a}+\frac{2 m}{z}=-\frac{2 a^{2} m}{z(z-a)(z+a)} \\ \therefore~ &q=\left\vert\frac{d w}{d z}\right\vert=\frac{2 a^{2} m}{\vert z \\vert z-a\vert\vert z+a\vert}=\frac{2 a^{2} m}{r_{1} r_{2} r_{3}} \end{aligned}\\ &\begin{array}{llll} \text { where } & r_{1}=\vert z-a\vert, & r_{2}=\vert z+a\vert & \text { and } & r_{3}=\vert z\vert \end{array} \end{aligned}\]

The object system consists of source \(+m\) at \(A\) \((o,a)\), i-e. at \(z=ia\) and sink \(-m\) at \(z=i b\). The image system consists of source \(+m\) at \(A^{\prime}\left(z ={-i a}\right)\) and sink \(-m\) at \(B^{\prime}(z=-i b)\) w.r.t the positive line \(OX\) which is rigid boundary.

The complex potential due to object system with rigid boundary is equivalent to the object system and its image system with no rigid boundary.

\[\begin{aligned} \therefore w&=-m \log \left(z-i a\right)+m \log (z-i b)\\ &- m \log (z+i a)+m \log (z+i b) \\ \implies &=-m \log \left(z^{2}+a^{2}\right)+m \log \left(z^{2}+b^{2}\right) \end{aligned}\]

Now, \(\begin{aligned} \frac{dw}{d z} &=-2 m \\ &=\left[\frac{1}{z^{2}+a^{2}}-\frac{1}{z^{2}+b^{2}}\right] \\ &=\frac{2 m z\left(a^{2}-b^{2}\right)}{\left(z^{2}+a^{2}\right)\left(2^{2}+b^{2}\right)} \\ \implies q &=\vert\frac{d \omega}{d z}\ \vert =\frac{2 m\left(a^{2}-b^{2}\right) \vert z \vert }{\vertz^{2}+a^{2}\ \vert \vertz^{2}+b\ \vert } \end{aligned}\)

For any point on \(x\) -axis, we have \(z=x\) so that \(q=\frac{2 m x\left(a^{2}-b^{2}\right)}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}\)

This is the expression for velocity at any point on \(x\)-axis. Let \(P_{0}\) be the pressure at \(x=\infty\). By Bernoulli equation for steady motion.

\[\therefore \quad \frac{P}{\rho}+\frac{1}{2} q^{2}=c\]

In view of \(p=\rho_{0}, q=0\) when \(x=\infty\) we get \(c=p_{0} / \rho\). \(\frac{P_{0}-P}{\rho}=\frac{1}{2} q^{2}\)

Required pressure \(P\) on boundary is given by:

\[\begin{aligned} P &=\int_{-\infty}^{\infty}\left(P_{0}-P\right) d x \\ &=\int_{-\infty}^{\infty} \frac{1}{2} \rho q^{2} d x \\ &=\frac{1}{2} \rho \int_{-\infty}^{\infty} \frac{4 m^{2} x^{2}\left(a^{2}-b^{2}\right)^{2}}{\left(x^{2}+a^{2}\right)^{2}\left(x^{2}+b^{2}\right)^{2}} dx \\ &=4 \rho m^{2}\left(a^{2}-b^{2}\right)^{2} \int_{0}^{\infty} \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)^{2}\left(x^{2}+b^{2}\right)^{2}} \\ &=4 m^{2} \rho \int_{0}^{\infty}\left[\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\left\{\frac{1}{x^{2}+b^{2}}-\frac{1}{x^{2}+a^{2}}\right\}-\frac{a^{2}}{\left(x^{2}+a^{2}\right)^{2}}\right. \left.-\frac{b^{2}}{\left(x^{2}+b^{2}\right)^{2}}\right] d x. \\ &= 4 m^{2} f\left[\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\left\{\frac{\pi}{2 b}-\frac{\pi}{2 a}\right\}-\frac{\pi}{4 a}-\frac{\pi}{4 b}\right] \\ &=\frac{\pi \rho m^{2}(a-b)^{2}}{a b(a+b)} \\ &= 4 m^{2} f\left[\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\left\{\frac{\pi}{2 b}-\frac{\pi}{2 a}\right\}-\frac{\pi}{4 a}-\frac{\pi}{4 b}\right] \\ &=\frac{\pi \rho m^{2}(a-b)^{2}}{a b(a+b)} \end{aligned}\]

(We have used below results) \(\begin{aligned} \int_{0}^{\infty} \frac{d x}{x^{2}+a^{2}} &= \left[\frac{1}{a} \tan -\frac{x}{a}\right]_{0}^{\infty} \\ &=\frac{\pi}{2 a} \\ \text{Also }, \int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)^{2}}&=\frac{1}{a^{3}} \int_{0}^{\infty} \cos ^{2} \theta d \theta ; x=a \tan \theta \\ &=\frac{x^{4}}{2 a^{3}} \int_{0}^{\pi / 2}(1+\cos 2 \theta) d \theta=\frac{\pi}{2} \cdot \frac{1}{2 a^{3}} \\&=\frac{\pi}{4 a^{3}} \end{aligned}\)

First Part:

The complex potential \(w\) at any point \(P(z)\) is given by:

\(w=-m \log (z-a)-m \log (z+a)+2 m \log z \quad \cdots \quad (1)\) or, \(w=m\left[\log z^{2}-\log \left(z^{2}-a^{2}\right)\right]\) or \(\phi+i \psi=m\left[\log \left(x^{2}-y^{2}+2 i x y\right)-\log \left(x^{2}-y^{2}-a^{2}+2 i x y\right)\right],\) as \(z=x+i y\)

Equating the imaginary parts, we have \(\begin{array}{c} \psi=m\left[\tan ^{-1}\left\{2 x y /\left(x^{2}-y^{2}\right)\right\}-\tan ^{-1}\left\{2 x y /\left(x^{2}-y^{2}-a^{2}\right)\right\}\right] \\ \psi=m \tan ^{-1}\left[\frac{-2 a^{2} x y}{\left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right)}\right], \text { on simplification. } \end{array}\)

The desired streamlines are given by \(\psi=\) constant \(=m \tan ^{-1}(-2 / \lambda)\). Then we obtain \((-2 / \lambda)=\left(-2 a^{2} x y\right) /\left[\left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right)\right] \\ \text { or } \\\ \left(x^{2}+y^{2}\right)^{2}=a^{2}\left(x^{2}-y^{2}+\lambda x y\right)\)

Second Part: From (1), we have

\[\frac{d w}{d z}=-\frac{m}{z-a}-\frac{m}{z+a}+\frac{2 m}{z}=-\frac{2 a^{2} m}{z(z-a)(z+a)}\]

\(q=\vert\frac{d w}{d z}\ \vert =\frac{2 a^{2} m}{ \vert z \vert \vert z-a \vert \vert z+a \vert }=\frac{2 a^{2} m}{r_{1} r_{2} r_{3}}\), where

\(r_{1}= \vert z-a \vert\), \(r_{2}= \vert z+a \vert\) and \(r_{3}= \vert z \vert\)

First Part:

The complex potential \(w\) at any point \(P(z)\) is given by:

\(w=-m \log (z-a)-m \log (z+a)+2 m \log z \quad \cdots \quad (1)\) or, \(w=m\left[\log z^{2}-\log \left(z^{2}-a^{2}\right)\right]\) or \(\phi+i \psi=m\left[\log \left(x^{2}-y^{2}+2 i x y\right)-\log \left(x^{2}-y^{2}-a^{2}+2 i x y\right)\right],\) as \(z=x+i y\)

Equating the imaginary parts, we have \(\begin{array}{c} \psi=m\left[\tan ^{-1}\left\{2 x y /\left(x^{2}-y^{2}\right)\right\}-\tan ^{-1}\left\{2 x y /\left(x^{2}-y^{2}-a^{2}\right)\right\}\right] \\ \psi=m \tan ^{-1}\left[\frac{-2 a^{2} x y}{\left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right)}\right], \text { on simplification. } \end{array}\)

The desired streamlines are given by \(\psi=\) constant \(=m \tan ^{-1}(-2 / \lambda)\). Then we obtain \((-2 / \lambda)=\left(-2 a^{2} x y\right) /\left[\left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right)\right] \\ \text { or } \\\ \left(x^{2}+y^{2}\right)^{2}=a^{2}\left(x^{2}-y^{2}+\lambda x y\right)\)

Second Part: From (1), we have

\[\frac{d w}{d z}=-\frac{m}{z-a}-\frac{m}{z+a}+\frac{2 m}{z}=-\frac{2 a^{2} m}{z(z-a)(z+a)}\]

\(q=\vert\frac{d w}{d z}\ \vert =\frac{2 a^{2} m}{ \vert z \vert \vert z-a \vert \vert z+a \vert }=\frac{2 a^{2} m}{r_{1} r_{2} r_{3}}\), where

\(r_{1}= \vert z-a \vert\), \(r_{2}= \vert z+a \vert\) and \(r_{3}= \vert z \vert\)