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Vortex Motion

PYQs

Vortex Motion

1) Does a fluid with velocity $q = [z - \frac{2 x}{r}, 2 y - 3 z - \frac{2 y}{r}, x - 3 y - \frac{2 z}{r}]$ possess vorticity, where $q (u, v, w)$ is the velocity in the Cartesian frame $\vec{r} (x, y, z)$ and $r^{2} = x^{2} + y^{2} + z^{2}$ ? What is the circulation in the circle $x^{2} + y^{2} = 9$ , $z = 0$ ?

[2016, 10M]

Given,

(i) Calculation of vorticity:

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

Now,

\nabla \times \vec{q} = | \begin{matrix} \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{matrix} |

So,

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

$∴$ fluid does not possess vorticity.

(ii) Calculation of circulation:

Given,

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

Let $x = 3 \cos θ$ , $y = 3 \sin θ$ , $z = 0 \dots (2)$

$⟹$ $d x = - 3 \sin θ d θ$ , $d y = - 3 \cos θ d θ$ , $d z = 0 \dots (3)$

Now, the circulation around the circle $C$ is given by:

$Γ = \int_{C} \vec{q} \cdot d \vec{r} = \int_{0}^{2 π} \vec{q} \cdot \vec{d r}$
$⟹ Γ = \int_{0}^{2 π} u d x + v d y + w d z$
$⟹ Γ = \int_{0}^{2 π} (0 - 2 \cos θ) (- 3 \sin θ d θ) + (6 \sin θ - 2 \sin θ - 2) (3 \cos θ d θ) + 0$ $⟹ Γ = \int_{0}^{2 π} 6 \sin θ \cos θ d θ + 12 \sin θ \cos θ d θ$ $⟹ Γ = \int_{0}^{2 π} (2 \sin θ \cos θ \d θ)$
$⟹ Γ = \int_{0}^{2 π} \sin 2 θ d θ$ $⟹ Γ = 0$

Thus, the circulation is 0.

2) Prove that the necessary and sufficient conditions that the vortex lines may be at right angles to the stream lines are $u$ , $v$ , $m = μ (\frac{\partial ϕ}{\partial x}, \frac{\partial ϕ}{\partial y}, \frac{\partial ϕ}{\partial z})$ where $μ$ and $ϕ$ are functions of $x$ , $y$ , $z$ , $t$ .

[2013, 10M]

The differential equations of streamlines and vortex lines are respectively: $\begin{array}{l} \frac{d x}{u} = \frac{d y}{v} = \frac{d z}{w} \end{array} \dots (1)$ $\begin{matrix} \frac{d x}{ξ} = \frac{d y}{η} = \frac{d z}{ζ} \end{matrix} \dots (2)$ (1) and (2) will intersect orthogonally iff $\begin{matrix} ξ + v η + w ζ = 0 \\ ⟹ μ (\frac{\partial w}{\partial y} - \frac{\partial v}{\partial 2}) + v (\frac{\partial u}{\partial z} - \frac{\partial ω}{\partial x}) + ω (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}) = 0 \end{matrix}$ This implies that $u d x + v d y + w d z is$ a perfect differential.

\begin{aligned} ∴ u d x + v d y + w d z & = μ d ψ \\ = μ (\frac{\partial ψ}{\partial x} d x + \frac{\partial ψ}{\partial y} d y + \frac{\partial ψ}{\partial z} d x) . \\ ⟹ u, u, w & = μ (\frac{\partial ϕ}{\partial x}, \frac{\partial ϕ}{\partial y}, \frac{\partial ϕ}{\partial z}) \end{aligned}

3) An infinite row of the equidistant rectilinear vortices are at distance $a$ apart. The vortices are of the same numerical strength $K$ but they are alternately of opposite signs. Find the Complex function that determines the velocity potential and the stream function.

[2011, 30M]

2011- Fluid

Let row of vertices be taken along $x - a x i s$ .

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

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

\begin{aligned} w & = \frac{i k}{2 n} [\log z + \log (z - 2 a) + \log (z + 2 a) + \dots] \\ - \frac{i k}{2 n} [\log (z - a) + \log (z + a) + l o g (z - 3 a) + \dots] \\ = \frac{i k}{2 π} \log (\frac{z [z^{2} - (2 a)^{2}] [z^{2} - (4 a)^{2})] \dots}{[z^{2} - a^{2}] [z^{2} - (3 a^{2})] \dots}) \\ = \frac{i k}{2 π} \log (\frac{\frac{z}{2 a} {[1 - (\frac{z}{2 a})}^{2}] [1 - {(\frac{z}{4 a})}^{2}] c \dots}{[1 - {(\frac{z}{a})}^{2}] [1 - {(\frac{z}{3 a})}^{2}) \dots}) \\ = \frac{i k}{2 π} \log (\frac{\sin π z / 2 a}{\cos (π 2 / 2 a)}) \\ ⟹ w & = \frac{i k}{2 π} \log (\tan (\frac{π z}{2 a})) \end{aligned}

This will determine velocity potential and stream function.

4) In an incompressible fluid the vorticity at every point is constant in magnitude and direction, show that the components of velocity $u$ , $v$ , $w$ are solutions of Laplace’s equation.

[2010, 12M]

Vorticity $(Ω) =$ curl $q$ $\begin{aligned} Ω & = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ u & v & w \end{matrix} | \\ = (\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z}) \hat{ı} + (\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x}) \hat{j} + (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}) \hat{k} \end{aligned}$

Now, since $Ω$ is constant, therefore

\begin{aligned} \frac{\partial w}{\partial y} - \frac{\partial v}{\partial z} & = c_{1} \dots (1) \\ \frac{\partial u}{\partial z} - \frac{\partial w}{\partial x} & = c_{2} \dots (2) \\ \frac{\partial v}{d x} - \frac{\partial u}{\partial y} & = c_{3} \dots (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} [\frac{\partial) v}{\partial y} + \frac{\partial w}{\partial z}] & = 0 \dots (4) \end{aligned}

Acccording to the equation of continuity,

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

$\begin{aligned} \frac{\partial^{2} u}{\partial y^{2}} + \frac{\partial^{2} u}{\partial z^{2}} - \frac{\partial}{\partial x} (- \frac{\partial u}{\partial x}) = 0 \end{aligned}$ $= \frac{\partial^{2} u}{\partial y^{2}} + \frac{\partial^{2} u}{\partial z^{2}} + \frac{\partial^{2} u}{\partial x^{2}} = 0$ $∴ \nabla^{2} u = 0$ Similarly we can show $\nabla^{2} v = 0$ and $\nabla^{2} ω = 0$ .

Hence proved.

5) When a pair of equal and opposite rectilinear vortices are situated in a long circle cylinder at equal distance from its axis, show that the path of each vortex is given by the equation $(r^{2} \sin^{2} θ - b^{2}) {(r^{2} - a^{2})}^{2} = 4 a^{2} b^{2} r^{2} \sin^{2} θ$ , $θ$ being measured from the line through the centre perpendicular to the joint of the vortices.

[2010, 30M]

Let the axis of the cylinder be $x - a x i s$ and vortices are situated at $P (r, θ)$ and $(r, - θ)$ respectively.

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

2010

Image systen due to $Q$ consists: $\to$ Vorex of strength $k$ at $\frac{a^{2}}{r} say Q$ $\to$ 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 π} [ & \log (z - r e^{i θ}) - \log (z - \frac{a^{2}}{r} e^{i θ}) - \log (z - r e^{- i θ}) \\ + \log (z - \frac{a^{2}}{r} e^{- i θ})] \end{aligned}

Complex potential at $P (r, θ)$ will be: $W^{'} = W$ - potential due to $P$ $W^{'} = \frac{i k}{2 π} [- \log (z - \frac{a^{2}}{r} e^{i θ}) - \log (z - i e^{- i θ}) + \log (z - \frac{a^{2}}{r} e^{- i θ})]$ $W^{'} = \frac{i k}{2 π} [\log \frac{(z - \frac{a^{2}}{r} e^{- i θ})}{(z - r e^{- i θ}) (z - \frac{a^{2}}{r} e^{i θ})}]$ At $P$ , $z = r e^{i θ}$ $W^{'} = \frac{i k}{2 π} [\log \frac{(r e^{i θ} - \frac{a^{2}}{r} e^{- i θ})}{(r e^{i θ} - r e^{- i θ}) (e^{i θ} - \frac{a^{2}}{r} e^{i θ})}]$ As $W^{'} = ϕ + i ψ$ $∴ ψ = \frac{k}{2 π} \log \frac{∣ r e^{i θ} - \frac{a^{2} e^{- i θ}}{r} |}{| r e^{i θ} - r e^{- i θ} | | r e^{i θ} - {\frac{a^{2}}{r}}^{i θ} ∣}$

Streamlines are given by $ψ =$ constant $\frac{k}{2 π} [\log \frac{| r^{2} e^{i θ} - a^{2} e^{- i θ} | / r}{| 2 i r \sin θ ∣ | \frac{r^{2} e^{i θ} - a^{2} e^{i θ}}{r} |}] = C_{1}$

$\log \frac{∣ (r^{2} + a^{2}) \sin θ i + (r^{2} - a^{2}) \cos θ |}{| 2 i r \sin θ | | (r^{2} - a^{2}) \cos θ + i (r^{2} - a^{2}) \sin θ |} = C_{2}$ $\Rightarrow \frac{{({(r^{2} + a^{2})}^{2} - {(r^{2} - a^{2})}^{2}) \sin^{2} θ + r^{2} - a^{2})}^{2}}{4 r^{2} \sin^{2} θ {(r^{2} - a^{2})}^{2}} = C_{3}$ $\begin{aligned} \frac{4 r^{2} a^{2} \sin^{2} θ + {(r^{2} - a^{2})}^{2}}{4 r^{2} \sin^{2} θ {(r^{2} - a^{2})}^{2}} & = C_{3} \\ 4 r^{2} a^{2} \sin^{2} θ + {(c^{2} - a^{2})}^{2} = & C_{3} 4 r^{2} \sin^{2} θ {(r^{2} - a^{2})}^{2} \end{aligned}$

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

Hence, proved.

6) Prove that the necessary and sufficient condition for vortex lines and stream to be at right angles to each other is that $u = μ \frac{\partial ϕ}{\partial x}$ , $v = μ \frac{\partial ϕ}{\partial y}$ , $w = μ \frac{\partial ϕ}{\partial z}$ where $μ$ and $ϕ$ are functions of $x$ , $y$ , $z$ and $t$ .

[2005, 12M]

\begin{aligned} ∴ u d x + v d y + w d z & = μ d ψ \\ = μ (\frac{\partial ψ}{\partial x} d x + \frac{\partial ψ}{\partial y} d y + \frac{\partial ψ}{\partial z} d x) . \\ ⟹ u, u, w & = μ (\frac{\partial ϕ}{\partial x}, \frac{\partial ϕ}{\partial y}, \frac{\partial ϕ}{\partial z}) \end{aligned}

7) In an incompressible fluid the vorticity at every point is constant in magnitude and direction. Do the velocity components satisfy the Laplace equation? Justify.

[2004, 12M]

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