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Potential Flow

PYQs

Potential Flow

1) For a two-dimensional potential flow, the velocity potential is given by $ϕ = x^{2} y = x y^{2} + \frac{1}{3} (x^{3} - y^{3})$ . Determine the velocity components along the direction $x$ and $y$ . Also, determine the stream function $ψ$ and check whether $ϕ$ represents a possible case of flow or not.

[2018, 15M]

Let $\vec{q} = u \hat{i} + v \hat{ȷ}$
Then,
$u = - \frac{\partial ϕ}{\partial x}$
$⟹ u = - (2 x y - y^{2} + x^{2})$
$⟹ u = y^{2} - x^{2} - 2 x y$

Also,
$v = - \frac{\partial ϕ}{\partial y}$
$\Rightarrow v = - (x^{2} - 2 x y - y^{2})$
$⟹ v = y^{2} - x^{2} - 2 x y$
We know that $ϕ + i ψ$ is an analytic function and salisfies Cauchy Riemann equations.

So, $\frac{\partial ψ}{\partial y} = u$ and $\frac{\partial ψ}{\partial x} = v$
$\Rightarrow \frac{\partial ψ}{\partial x} = y^{2} - x^{2} + 2 x y$
Integrating wrt $x$ :
$ψ = x y^{2} - \frac{x^{3}}{3} + x^{2} y + f (y)$

Now, $\frac{\partial ψ}{\partial y} = 2 x y + x^{2} + f^{'} (y)$ , and
$\frac{\partial ψ}{\partial y} = - u$

$\Rightarrow x^{2} + 2 x y + f^{'} (y) = x^{2} + 2 x y - y^{2}$
$\Rightarrow f^{'} (y) = - y^{2}$
$\frac{d f}{d y} = - y^{2}$
$\Rightarrow f = \frac{- y^{3}}{3} + c$

So,
$ψ = x y^{2} - \frac{(x^{3} + y^{3})}{3} + x^{2} y + c$
$ψ = 0$ at origin $⟹ c = 0$

So, $ψ = x y^{2} + x^{2} y - \frac{(x^{3} + y^{3})}{3}$

Checking possible flow:
Since $\frac{\partial u}{\partial x} + \frac{\partial y}{\partial y} = - 2 x - 2 y + 2 y + 2 x = 0$
$⟹$ Equation of continuity is satisfied,
$⟹$ It is a possible flow.

2) If the velocity of an incompressible fluid at the point $(x, y, z)$ is given by $(\frac{3 x z}{r^{5}}, \frac{3 y z}{r^{5}} \cdot \frac{3 z^{2} - r^{2}}{r^{5}})$ , $r^{2} = x^{2} + y^{2} + z^{2}$ then prove that the liquid motion is possible and that the velocity potential is $\frac{z}{r^{3}}$ . Further, determine the streamlines.

[2017, 15M]

Here $u = \frac{3 x z}{r^{5}}$ , $v = \frac{3 y z}{r^{5}}$ , $w = \frac{3 z^{2} - r^{2}}{r^{5}} = \frac{3 z^{2}}{r^{5}} - \frac{1}{r^{3}} \dots$ (1) where
$r^{2} = x^{2} + y^{2} + z^{2} \dots$ (2)

From $(2), \partial r / \partial x = x / r, \partial r / \partial y = y / r, \partial r / \partial z = z / r$ From (1), ( 2 ) and (3), we have

\frac{\partial u}{\partial x} = 3 z [\frac{1}{r^{5}} + (- 5 x) r^{- 6} \frac{\partial r}{\partial x}] = \frac{3 z}{r^{5}} - \frac{15 x^{2} z}{r^{7}}

\frac{\partial v}{\partial y} = 3 z [\frac{1}{r^{5}} + (- 5 y) r^{- 6} \frac{\partial r}{\partial y}] = \frac{3 z}{r^{5}} - \frac{15 y^{2} z}{r^{7}}

\frac{\partial w}{\partial z} = \frac{6 z}{r^{5}} - 15 z^{2} r^{- 6} \frac{\partial r}{\partial z} + 3 r^{- 4} \frac{\partial r}{\partial z} = \frac{6 z}{r^{5}} - \frac{15 z^{2}}{r^{6}} \cdot \frac{z}{r} + \frac{3}{r^{4}} \cdot \frac{z}{r} = \frac{9 z}{r^{5}} - \frac{15 z^{3}}{r^{7}}

∴ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = \frac{15 z}{r^{5}} - \frac{15 z}{r^{7}} (x^{2} + y^{2} + z^{2}) = \frac{15 z}{r^{5}} - \frac{15 z}{r^{7}} \times r^{2} = 0

Since the equation of continuity is satisfied by the given values of $u, v$ and $w,$ the motion is possible. Let $ϕ$ be the required velocity potential. Then

\begin{array}{l} d ϕ = \frac{\partial ϕ}{\partial x} d x + \frac{\partial ϕ}{\partial y} d y + \frac{\partial ϕ}{\partial z} d z = - (u d x + v d y + w d z), by definition of ϕ \\ = - [\frac{3 x}{r^{5}} d x + \frac{3 y z}{r^{5}} d y + \frac{3 z^{2} - r^{2}}{r^{5}} d z] = \frac{r^{2} d z - 3 z (x d x + y d y + z d z)}{r^{5}} \\ Thus, d ϕ = \frac{r^{3} d z - 3 r^{2} z d r}{{(r^{3})}^{2}} = d (\frac{z}{r^{3}}), using (2) \end{array}

Integrating,

ϕ = z / r^{3}

[Omitting constant of integration, for it has no significance in $ϕ$ ]

In spherical polar coordinates $(r, θ, ϕ),$ we know that $z = r \cos θ .$ Hence the required potential is given by $ϕ = (r \cos θ) / r^{3} = (\cos θ) / r^{2}$ . We now obtain the streamlines. The equations of streamlines are given by

\frac{d x}{u} = \frac{d y}{v} = \frac{d z}{w} i.e., \frac{d x}{3 x z / r^{5}} = \frac{d y}{3 y z / r^{5}} = \frac{d z}{(3 z^{2} - r^{2}) / r^{5}}

So,

\frac{d x}{3 x z} = \frac{d y}{3 y z} = \frac{d z}{3 z^{2} - r^{2}} \dots (4)

Taking the first two members of (4) and simplifying, we get

d x / x = d y / y

⟹ d x / x - d y / y = 0

Integrating, $\log x - \log y = \log c_{1}$ l.e. $x / y = c_{1}, c_{1}$ being a constant $\dots (5)$

Now, each member in $(4) = \frac{x d x + y d y + z d z}{3 x^{2} z + 3 y^{2} z + 3 z^{3} - r^{2} z} = \frac{x d x + y d y + z d z}{3 z (x^{2} + y^{2} + z^{2}) - r^{2} z}$

= \frac{x d x + y d y + z d z}{3 z (x^{2} + y^{2} + z^{2}) - z (x^{2} + y^{2} + z^{2})} = \frac{x d x + y d y + z d z}{2 z (x^{2} + y^{2} + z^{2})}, by (2) \dots (6)

Taking the first member in (4) and (6), we get

\frac{d x}{3 x z} = \frac{x d x + y d y + z d z}{2 z (x^{2} + y^{2} + z^{2})} or \frac{2}{3} \frac{d x}{x} = \frac{1}{2} \frac{2 x d x + 2 y d y + 2 z d z}{x^{2} + y^{2} + z^{2}}

Integrating, $(2 / 3) \times \log x = (1 / 2) \times \log (x^{2} + y^{2} + z^{2}) + \log c_{2}$ or $x^{2 / 3} = c_{2} {(x^{2} + y^{2} + z^{2})}^{1 / 2}, c_{2}$ being an arbitrary constant. $\dots (7)$ .

The required streamlines are the curves of intersction of (5) and (7).

3) A simple source of strength $m$ is fixed at the origin $O$ in a uniform stream of incompressible fluid moving with velocity $U \vec{i}$ , show that the velocity potential $ϕ$ at any point $P$ of the stream is $\frac{m}{r} - U r \cos θ$ , where $O P = r$ and $θ$ is the angle which $O P$ makes with the direction $i$ . Find the deferential equation of the streamlines and show that the lie on the surfaces $U r^{2} \sin^{2} θ - 2 m \cos θ =$ constant.

[2016, 15M]

Given, strength of source = $m$ .

$⟹$ The complex potential at any point $z$ is given by, $w = - m \log z$

$\begin{aligned} ∴ \frac{d w}{d z} & = \frac{- m}{z} = \frac{- m}{r e^{i θ}} = \frac{- m}{r} e^{- i θ} \\ ⟹ \frac{d w}{d z} & = \frac{- m}{r} (\cos θ - i \sin θ) = \frac{- m \cos θ}{r} + \frac{i m \sin θ}{r} \end{aligned}$ $⟹ \frac{d w}{d z} = \frac{- m x}{r^{2}} + i \frac{m y}{r^{2}} = - u_{m} + i v_{m}$

where $u_{m}$ and $v_{m}$ are velocity components due to source at $m$ . $∴ u_{m} = \frac{m x}{r^{2}}, v_{m} = \frac{m y}{r^{2}}$

Also, fluid is moving with velocity $U \hat{i}$ . $∴$ Total velocity. $\vec{q} = (U + \frac{m x}{r^{2}}) \hat{i} + (\frac{m y}{r^{2}}) \hat{j} = u \hat{i} + \hat{j} v$ Let $ϕ$ be the velocity potential. $∴ u = \frac{- \partial ϕ}{\partial x}$

\Rightarrow \frac{- \partial ϕ}{\partial x} = U + \frac{m x}{r^{2}}

\Rightarrow d ϕ = (- U - \frac{m x}{r^{2}}) d x

Integrating both sides, we get $f = - U x - m \log r + f (y)$ where, $f (y)$ is arbitrary function of $y$ Also, $\frac{- \partial ϕ}{\partial y} = v = \frac{m y}{r^{2}} \dots (i)$

\Rightarrow d ϕ = \frac{- m y}{r^{2}} d y

⟹ ϕ = - m \log r + g (x) \dots (i i)

where $g (x)$ is an arbitrary function of $x .$ From (i) and (ii) $ϕ = - U x - m \log r$

\Rightarrow ϕ = - U r \cos θ - m \log r

Now, equation of streamlines is given by:

\frac{d x}{u} = \frac{d y}{v}

\Rightarrow \frac{d x}{u + \frac{m x}{r^{2}}} = \frac{d y}{\frac{m y}{r^{2}}}

$\Rightarrow \frac{d y}{d x} = \frac{\frac{m y}{r^{2}}}{(U + \frac{m x}{r^{2}})}$ , which is the differential equation of streamlines.

4) The space between two concentric spherical shells of radii $a$ , $b (a < b)$ is filled with a liquid of density $ρ$ . If the shells are set in motion, the inner one with velocity $U$ in the $x - d i r e c t i o n$ and the outer one with velocity $V$ in the $y - d i r e c t i o n$ , then show has the initial motion of the liquid is given by velocity potential $ϕ = \frac{{a^{3} U (1 + \frac{1}{2} b^{3} r^{- 3}) x - b^{3} V (1 + \frac{1}{2} a^{3} r^{- 3}) y}}{(b^{3} - a^{3})}$ , where $r^{2} = x^{2} + y^{2} + z^{2}$ , the coordinate being rectangular. Evaluate the velocity at any point of the liquid.

[2016, 20M]

Since the motion is irrotational, consequently, the corresponding velocity potential $ϕ$ exists such that:

\nabla^{2} ϕ = 0 \dots (1)

The boundary conditions for $ϕ$ are:

{(\frac{- \partial ϕ}{\partial r})}_{r = a} = U \cos θ \dots (2)

{(\frac{- \partial ϕ}{\partial r})}_{r = b} = V \sin θ \dots (3)

2016-7(b)

The above equations suggest that $ϕ$ must involve terms containing $\sin θ$ and $\cos θ$ . So, we take $ϕ$ as:

$ϕ = (A r + \frac{B}{r^{2}}) \cos θ + (C r + \frac{D}{r^{2}}) \sin θ \dots (3)$ $⟹ \frac{- \partial ϕ}{\partial r} = (- A + \frac{2 B}{r^{3}}) \cos θ + (- C + \frac{2 D}{r^{3}}) \sin θ \dots (4)$

Using boundary condiitons for (1) and (2) in (4), we get:

$(- A + \frac{2 B}{a^{3}}) \cos θ + (- C + \frac{2 D}{a^{3}}) \sin θ = U \cos θ \dots (5)$ $(- A + \frac{2 B}{b^{3}}) \cos θ + (- C + \frac{2 D}{b^{3}}) \sin θ = V \sin θ \dots (6)$

Comparing coefficients of $\cos θ$ and $\sin θ$ in (5) and (6), we get:

$\frac{- A + 2 B}{a^{3}} = U, \frac{- C + 2 D}{a^{3}} = 0 \dots (7)$ , and
$\frac{- A + 2 B}{b^{3}} = 0, \frac{- C + 2 D}{b^{3}} = V \dots (8)$

Solving (7) and (8), we get:

$A = \frac{U a^{3}}{b^{3} - a^{3}}$ , $B = \frac{U a^{3} b^{3}}{2 (b^{3} - a^{3})}$ , $C = \frac{- U b^{3}}{b^{3} - a^{3}}$ , $D = \frac{- U a^{3} b^{3}}{2 (b^{3} - a^{3})}$

Putting these values in (3), we get:

$ϕ = (\frac{U a^{3} r}{b^{3} - a^{3}} + \frac{u a^{3} b^{3}}{2 (b^{3} - a^{3}) r^{2}}) \cos θ + (\frac{- U b^{3}}{b^{3} - a^{3}} \cdot r - \frac{U a^{3} b^{3}}{2 (b^{3} - a^{3}) r^{2}}) \sin θ$
$⟹ ϕ = \frac{{U a^{3} (1 + \frac{b^{3}}{2 r^{3}}) r \cos θ - V b^{2} (1 + \frac{a^{3}}{2 r^{3}}) r \sin θ}}{(b^{3} - a^{3})}$ $⟹ ϕ = \frac{{a^{3} U (1 + \frac{b^{3} r^{- 3}}{2}) x - b^{3} V (1 + \frac{a^{3} r^{- 3}}{2}) y}}{b^{3} - a^{3}}$

Now, let velocity at any point= $\vec{q} = u \hat{i} + v \hat{j}$ where, $u = \frac{- \partial ϕ}{\partial x}$ and $v = \frac{- \partial ϕ}{\partial y}$

Therefore,

$u = \frac{- \partial ϕ}{\partial x} = \frac{- a^{3} U (1 + \frac{b^{3} r^{- 3}}{2})}{b^{3} - a^{3}} - \frac{a^{3} U x (\frac{- 3}{2} b^{3} r^{- 4} \frac{x}{r})}{(b^{3} - a^{3})}$ + $\frac{b^{3} V y (- \frac{3}{2} a^{3} r^{- 4}, \frac{x}{r})}{b^{3} - a^{3}}$ $⟹ u = \frac{- a^{3} U (1 + \frac{b^{3} r^{- 3}}{2})}{b^{3} - a^{3}} + \frac{\frac{3}{2} a^{3} b^{3} x^{2} U r^{- 5}}{(b^{3} - a^{3})} - \frac{\frac{3}{2} a^{3} b^{3} x y v r^{- 5}}{b^{3} - a^{3}}$

Also,

⟹ u = \frac{- a^{3} u (1 + b^{3} r^{- 3})}{(b^{3} - a^{3})} + \frac{3 a^{3} b^{3} x^{2} U r^{- 5}}{2 (b^{3} - a^{3})} - \frac{3 a^{3} b^{3} x y V r^{- 5}}{2 (- b^{3} - a^{3})} \dots (9)

v = - \frac{\partial ϕ}{\partial y} = \frac{- a^{3} U x (- \frac{3}{2} b^{2} r^{- 4} \frac{y}{r})}{b^{3} - a^{3}} + \frac{b^{3} V (1 + \frac{1}{2} a^{3} r^{- 3})}{b^{3} - a^{3}} + \frac{b^{3} V y (- \frac{3}{2} a^{3} r^{- 4} \frac{y}{r})}{b^{3} - a^{3}}

⟹ v = \frac{- 3 a^{3} b^{3} x y U r^{- 5}}{2 (b^{3} - a^{3})} + \frac{b^{3} v (1 + \frac{1}{2} a^{3} r^{- 3})}{b^{3} - a^{3}} - \frac{3 a^{3} b^{3} y^{2} V r^{- 5}}{2 (b^{3} - a^{3})} \dots (10)

$∴$ velocity is given by:

$\vec{q} = u \hat{i} + \hat{j}$ , where $u$ and $v$ are given by (9) and (10).

5) Given the velocity potential $ϕ = \frac{1}{2} \log [\frac{(x + a)^{2} + y^{2}}{(x - a)^{2} + y^{2}}]$ , determine the streamlines.

[2014, 20M]

Given, $ϕ = \frac{1}{2} \log {(x + a)^{2} + y^{2}] - \frac{1}{2} \log ((x - a)^{2} + y^{2}) \dots (1)$

To determine stream lines.

$- \frac{\partial ϕ}{\partial x} = u = - \frac{\partial ψ}{\partial y}, - \frac{\partial ϕ}{\partial y} = v = \frac{\partial ψ}{\partial x}$ Hence $\frac{\partial ϕ}{\partial x} = \frac{\partial ψ}{\partial y}, \frac{\partial ϕ}{\partial y} = - \frac{\partial ψ}{\partial x}$

Now, $\frac{\partial ϕ}{\partial y} = \frac{x + a}{(x + a)^{2} + y^{2}} - \frac{x - a}{(x - a)^{2} + y^{2}}$

Integrating wrt $y$ , $ψ = \tan^{- 1} \frac{y}{x + a} - \tan^{- 1} \frac{y}{x - a} + F (x) \dots (2)$

where $F (x)$ is constant of integration. To determine $F (x)$ ,

\frac{\partial ψ}{\partial x} = - \frac{\partial ϕ}{\partial y} = - \frac{y}{(x + a)^{2} + y^{2}} + \frac{y}{(x - a)^{2} + y^{2}}

By $(4),$ $\frac{\partial ψ}{\partial x} = - \frac{y}{(x + a)^{2} + y^{2}} + \frac{y}{(x - a)^{2} + y^{2}} + F^{'} (x)$ Equating (5) to (6), $F^{'} (x) = 0$ . Integrating this $F (x) =$ absolute const and hence neglected. Since it has no effect on the fluid motion.

Now (2) becomes

$ψ = \tan^{- 1} \frac{y}{x + a} - \tan^{- 1} \frac{y}{x - a}$ $= \tan^{- 1} \frac{- 2 a y}{x^{2} - a^{2} + y^{2}}$ $∴$ Stream lines are given by $ψ =$ const. i.e., $\tan^{- 1} [\frac{- 2 a y}{x^{2} - a^{2} + y^{2}}] = c o n s t . o r, \frac{y}{x^{2} - a^{2} + y^{2}} = c o n s t .$ If we take const. $= 0,$ then we get $y = 0,$ i.e., $x$ -axis. If we take conts. $= \infty,$ then we get circle $x^{2} - a^{2} + y^{2} = 0$ .

6) If $n$ rectilinear vortices of the same strength $K$ are symmetrically arranged as generators of a circle cylinder of radius $a$ in an infinite liquid, prove that the vortices will move round the cylinder uniformly in time $\frac{8 π^{2} a^{2}}{(n - 1) K}$ . Find the velocity at any point of the liquid.

[2013, 20M]

From the fig, the $n$ vortices are at $A_{0}$ , $A_{1}$ , $A_{2} \dots$ , $A_{n - 1}$ such that $∠ A_{0} O A_{1} = ∠ A_{1} O A_{2} = \dots = ∠ A_{n - 1} O A_{1} = \frac{2 π}{n}$ The coordinates of the points $A_{r}$ are given by: $z = z_{r} = a e^{(2 π / n) i r}$ where $r = 0, 1, 2 \dots n - 1$ . These are $n$ roots of the equation $z^{n} - a^{n} = 0$

Now, $[z^{n} - a^{n} = 0 \Rightarrow z^{n} = a^{n} e^{2 π r i}]$ Hence, $z^{n} - a^{n} = (z - z_{0}) (z - z_{1}) - \dots (z - z_{n - 1})$ The complex potential due to $n$ vortices at $P$ is given by:

\begin{aligned} w & = \frac{i k}{2 π} [\log (z - z_{0}) + \log (z - z_{1}) + \dots + \log (z - z_{n - 1})] \\ = \frac{i k}{2 π} \log (z - z_{0}) (z - z_{1}) \dots (z - z_{n - 1}) \\ = \frac{i k}{2 π} \log (z^{n} - a^{n}) \dots (1) \end{aligned}

For the point $A_{0}$ , $z = a$ , so that $r = a$ and $θ = 0$ .

If $w^{'}$ is the complex potential at $A_{0}$ , then

\begin{aligned} w^{'} & = w - \frac{i k}{2 π} \log (z - a) \\ = \frac{i k}{2 π} [\log (z^{n} - a^{n}) - \log (z - a)] \end{aligned}

ϕ^{'} + i ψ^{'} = \frac{i k}{2 π} [\log (r^{n} e^{i n θ} - a^{n}) - \log (r e^{i θ} - a)]

ψ^{'} = \frac{k}{4 π} [\log (r^{2 n} + a^{2 n} - 2 r^{n} a^{n} \cos n θ) - \log (r^{2} + a^{2} - 2 r a \cos θ)]

\frac{\partial ψ^{'}}{\partial r} = \frac{k}{4 π} [\frac{2 n r^{2 n - 1} - 2 n r^{n - 1} a^{n} \cos n θ}{r^{2 n} + a^{2 n} - 2 r^{n} a^{n} \cos n θ} - \frac{2 r - 2 a \cos θ}{r^{2} + a^{2} - 2 r a \cos θ}]

\frac{\partial ψ^{'}}{\partial θ} = \frac{k}{4 π} [\frac{2 n r^{n} a^{n} \sin n θ}{r^{2 n} - 2 r^{n} a^{n} \cos n θ + a^{2 n}} - \frac{2 r a \cos θ}{r^{2} + a^{2} - 2 r a \cos θ}]

{(\frac{\partial ψ^{'}}{\partial r})}_{r = a} = \frac{k}{4 π a} [n (\frac{1 - \cos n θ}{1 - \cos u θ}) - (\frac{1 - \cos θ}{1 - \cos θ})] = \frac{k}{4 π a} (n - 1)

\begin{aligned} {(\frac{\partial ψ^{'}}{\partial θ})}_{r = a} & = \frac{k}{4 π} [\frac{n \sin n θ}{1 - \cos n θ} - \frac{\sin θ}{1 - \cos θ}] \\ = \frac{k}{4 π} [\frac{n^{2} \cos n θ}{n \sin n θ} - \frac{\cot θ}{\sin θ}] as θ \to 0 (using L'Hospital's Rule) \\ = \frac{k}{4 π} [\frac{- n^{3} \sin n θ}{n^{2} \cos n θ} - \frac{(- \sin θ)}{\cos θ}] as θ \to 0 (using L'Hospital's Rule) \\ = \frac{k}{4 π} [0 + 0] \\ = 0 \end{aligned}

Finally, $\frac{\partial ψ^{'}}{\partial θ}$ = $\frac{k}{4 π a} (n - 1) = 0$ , as $r \to a, θ \to 0$ .

Consequently, the velocity $q_{0}$ of the vertex $A_{0}$ is given by:

q_{0} = {[{(\frac{\partial ψ^{'}}{\partial r})}^{2} + \frac{1}{r} {(\frac{\partial ψ^{'}}{\partial θ})}^{2}]}^{1 / 2} = \frac{k (n - 1)}{4 π a}

7) Show that $ϕ = x f (r)$ is a possible form for the velocity potential for an incompressible fluid motion. If the fluid velocity $\dot{q} \to 0$ as $r \to \infty$ , find the surfaces of constant speed.

[2012, 30M]

It is given that velocity potential, $ϕ = x f (r)$ .

For motion of fluid to be possible, $\nabla^{2} ϕ = 0$

\Rightarrow \frac{\partial^{2} ϕ}{\partial x^{2}} + \frac{\partial^{2} ϕ}{\partial y^{2}} + \frac{\partial^{2} ϕ}{\partial z^{2}} = 0

$[\frac{x}{r} f^{'} (r) + \frac{x^{3}}{r^{2}} f^{''} (r) + \frac{2 x}{r} f^{'} (r) - \frac{x^{3}}{r^{3}} f^{'} (r)]$ + $[\frac{x}{r} f^{'} (r) - \frac{x y^{2}}{r^{3}} f^{'} (r) + \frac{x y^{2}}{r^{2}} f^{''} (r)]$ + $[\frac{x}{r} f^{'} (r) - \frac{x z^{2}}{r^{3}} f^{'} (r) - \frac{x^{3}}{r^{2}} f^{''} (r)] = 0$

$\Rightarrow x f^{''} (r) + \frac{4 x f^{'} (r)}{r} = 0$ $\frac{f^{''} (r)}{f^{'} (r)} = \frac{- 4}{r}$ $\log f^{'} (r) = - 4 \log r + c_{1}$ $f^{'} (x) = \frac{c_{1}}{r^{4}} \dots (1)$ $f = \frac{- c_{1}}{3 r^{3}} + C_{2} \dots (2)$

Now, $\vec{q} = \nabla ϕ = f (r) \hat{i} + \frac{x}{r} f^{'} (r) \hat{r}$

⟹ \bar{q} = (\frac{- c_{1}}{3 r^{3}} + c_{2}) \hat{ı} + (\frac{x c_{1}}{r^{5}}) \hat{r} [using(1) and (2)]

\begin{aligned} Since \bar{q} & \to 0 as λ \to \infty \Rightarrow c_{2} = 0 \\ \Rightarrow \bar{q} = \frac{c_{1}}{3 r^{3}} (\hat{ı} - \frac{3 x}{r^{2}} \hat{r}) \end{aligned}

$Surfaces of constant speed are given by q^{2} = Constant$ $\begin{aligned} \Rightarrow \vec{q} . \vec{q} & = constant \\ = {(\frac{c_{1}}{3 r^{3}})}^{2} [(\hat{i} - \frac{3 x}{r^{2}} \hat{r}) \cdot (\hat{i} - \frac{3 x}{r^{2}} \hat{r})] \\ = \frac{c_{1}^{2}}{9 r^{6}} (1 + \frac{3 x^{2}}{r^{2}}) \\ = \frac{c_{1}^{2}}{9 r^{8}} (r^{2} + 3 x^{2}) \end{aligned}$

$∴$ Surfaces of constant speed are given by $\frac{(r^{2} + 3 x^{2})}{r^{8}} = Constant$

8) A rigid sphere of radius $a$ is placed in a stream of fluid whose velocity in the undisturbed state is $V$ . Determine the velocity of the fluid at any point of the disturbed stream.

[2012, 12M]

Let $ϕ$ velocity potential of sphere and $U$ be velocity of fluid along $z - a x i s$ . Then $ϕ$ satifies Laplace’s equation, thus $\frac{\partial^{2} ϕ}{\partial r^{2}} + \frac{2}{r} \frac{\partial ϕ}{\partial r} + \frac{1}{r^{2}} \frac{\partial^{2} ϕ}{\partial θ^{2}} + \frac{\cot θ}{r^{2}} \frac{\partial ϕ}{\partial θ} = 0$

$\Rightarrow ϕ$ is of the form $f (r) \cos θ$ Let $ϕ = (A r + \frac{B}{r^{2}}) \cos θ$

At boundary of sphere, i.e, $r = a$ , normal velocity of sphere=velocity of fluid at that point.

\begin{aligned} \Rightarrow - \frac{\partial ϕ}{\partial r} & = 0 \\ \Rightarrow - (A - \frac{2 B}{a^{3}}) \cos θ & = 0 \\ \Rightarrow A & = \frac{2 B}{a^{3}} \dots (1) \\ ⟹ ϕ & = [r (\frac{2 B}{a^{3}}) + \frac{B}{r^{3}}] \cos θ [u s i n g (1)] \end{aligned}

- \frac{\partial ϕ}{\partial r} = - [\frac{2 B}{a^{3}} - \frac{2 B}{r^{3}}] \cos θ

\begin{aligned} As r \to \infty, velocity = U \cos θ (= \frac{\partial ϕ}{\partial x}) \\ \Rightarrow U \cos θ = - [\frac{2 B}{a^{3}} - \frac{2 B}{r^{3}}] \cos θ \end{aligned}

$\Rightarrow B = \frac{- a^{3} U}{2} \Rightarrow A = - U [u s i n g (1)]$ $\Rightarrow ϕ = (- U) (r + \frac{a^{3}}{2 r^{2}}) \cos θ$ Now, velocity components are given by:

q_{r} = - \frac{\partial ϕ}{\partial r} = U (1 - \frac{a^{3}}{r^{3}}) \cos θ

q_{θ} = \frac{- 1}{r} \frac{\partial ϕ}{d θ} = (- u) (1 + \frac{a^{3}}{2 r^{3}})

9) If the velocity potential of a fluid is $ϕ = \frac{1}{r^{3}} z \tan^{1} (\frac{y}{z})$ , $r^{2} = x^{2} + z^{2} + z^{2}$ then show that the stream lines lie on the surfaces $x^{2} + y^{2} + z^{2} = c {(x^{2} + y^{2})}^{2 / 3}$ , $c$ being a constant.

[2008, 12M]

10) Show that the velocity potential $ϕ = \frac{1}{2} a (x^{2} + y^{2} - 2 z^{2})$ satisfies the Laplace equation, and determine the stream lines.

[2002, 12M]

11) If the velocity distribution of an incompressible fluid at the point $(x, y, z)$ is given by $(\frac{3 x z}{r^{5}}, \frac{3 y z}{r^{5}}, \frac{k z^{2} - r^{2}}{r^{5}})$ , then determine the parameter $k$ such that it is a possible motion. Hence find its velocity potential.

[2001, 12M]

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