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UPSC 2022 Maths Optional Paper 2 Q7c — Step-by-Step Solution

20 marks · Section B

Vortex motion; circulation · Mechanics & Fluid Dynamics · asked 6× in 13 yrs · Read the full method →

Question

Two vortices of strength $k$ at $(\pm a,0)$; vortex of strength $-k/2$ at origin. Show fluid motion is stationary and find streamline equations. If streamlines through stagnation points meet $x$-axis at $(\pm b,0)$, show $3\sqrt 3(b^2-a^2)^2=16 a^3 b$.

Technique

Complex potential for vortex; stagnation points from $dw/dz=0$; streamline through stagnation point gives $\psi=$ const; equate at stagnation and at $(b,0)$.

Solution

Step 1 — Complex potential for vortices

A point vortex of strength $K$ at $z=z_0$ has complex potential

(Sign convention: $K>0$ for counterclockwise vortex.)

Total complex potential:

Step 2 — Stationary flow

Stationary means the velocity field is time-independent (steady). With fixed vortex positions (no advection of vortices), the velocity field is automatically time-independent — so the flow is stationary.

More precise: Each vortex would be advected by the field of the others (and itself doesn’t move under its own velocity). For stationary flow, we need each vortex’s induced motion to be zero.

At vortex $(a,0)$: induced velocity by vortex at $(-a,0)$ and at $(0,0)$.

Vortex at $(-a,0)$ of strength $k$: velocity at $(a,0)$ has magnitude $k/(2\pi\cdot 2a)=k/(4\pi a)$, direction perpendicular to the line from $(-a,0)$ to $(a,0)$, i.e., in $+y$ direction (counterclockwise around source vortex).

Vortex at $(0,0)$ of strength $-k/2$: velocity at $(a,0)$ has magnitude $(k/2)/(2\pi a)=k/(4\pi a)$, direction perpendicular to line from $(0,0)$ to $(a,0)$. Since strength is negative (clockwise), velocity at $(a,0)$ is in $-y$ direction.

Net velocity at $(a,0)$: $k/(4\pi a)\hat\jmath+(-k/(4\pi a))\hat\jmath=0$.

So vortex at $(a,0)$ is stationary ✓. By symmetry, vortex at $(-a,0)$ is also stationary. Vortex at origin is fixed (by symmetry, the two opposing $k$ vortices induce equal and opposite velocities at the origin, summing to zero).

Therefore the flow is stationary.

Step 3 — Stream function

Stream function $\psi=\Im w=$ imaginary part of $w$.

$w=-ik\ln[(z^2-a^2)/\sqrt z]$.

Write $z^2-a^2=r_1 r_2 e^{i(\theta_1+\theta_2)}$ where $r_1=|z-a|$, $r_2=|z+a|$, $\theta_1=\arg(z-a)$, $\theta_2=\arg(z+a)$.

$\sqrt z=\sqrt{r_0}e^{i\theta_0/2}$ where $r_0=|z|$, $\theta_0=\arg z$.

$\ln[(z^2-a^2)/\sqrt z]=\ln(r_1 r_2/\sqrt{r_0})+i(\theta_1+\theta_2-\theta_0/2)$.

$w=-ik[\ln(r_1 r_2/\sqrt{r_0})+i(\theta_1+\theta_2-\theta_0/2)]=k(\theta_1+\theta_2-\theta_0/2)-ik\ln(r_1 r_2/\sqrt{r_0})$.

$\psi=\Im w=-k\ln(r_1 r_2/\sqrt{r_0})$.

Streamlines: $\psi=$ const, i.e., $\dfrac{r_1 r_2}{\sqrt{r_0}}=$ const.

Equivalently: $|z^2-a^2|/\sqrt{|z|}=$ const.

Step 4 — Stagnation points

Velocity $\bar V=dw/dz=-ik\!\left[\dfrac{2z}{z^2-a^2}-\dfrac{1}{2z}\right]$.

Set $dw/dz=0$: $\dfrac{2z}{z^2-a^2}-\dfrac{1}{2z}=0\Rightarrow\dfrac{4z^2-(z^2-a^2)}{2z(z^2-a^2)}=0\Rightarrow 4z^2-z^2+a^2=0\Rightarrow 3z^2+a^2=0\Rightarrow z^2=-a^2/3\Rightarrow z=\pm ia/\sqrt 3$.

Stagnation points: $(0,\pm a/\sqrt 3)$ (on $y$-axis).

Step 5 — Streamline through stagnation points meets $x$-axis at $(\pm b,0)$

Compute $\psi$ at stagnation point $(0,a/\sqrt 3)$, then set equal to $\psi$ at $(b,0)$, solve for relation $b\leftrightarrow a$.

At $(0,a/\sqrt 3)$:

$r_0=a/\sqrt 3$. $r_1=|0-a+i a/\sqrt 3|=\sqrt{a^2+a^2/3}=a\sqrt{4/3}=2a/\sqrt 3$. $r_2=|0+a+i a/\sqrt 3|=2a/\sqrt 3$.

$r_1 r_2/\sqrt{r_0}=(2a/\sqrt 3)^2/\sqrt{a/\sqrt 3}=\dfrac{4a^2/3}{\sqrt{a/\sqrt 3}}$.

$\sqrt{a/\sqrt 3}=\sqrt{a/3^{1/2}}=a^{1/2}\cdot 3^{-1/4}$.

So $r_1 r_2/\sqrt{r_0}=\dfrac{4a^2/3}{a^{1/2}\cdot 3^{-1/4}}=\dfrac{4a^{3/2}}{3}\cdot 3^{1/4}=\dfrac{4a^{3/2}\cdot 3^{1/4}}{3}=\dfrac{4a^{3/2}}{3^{3/4}}$.

At $(b,0)$: $r_0=b$, $r_1=|b-a|$, $r_2=|b+a|=b+a$ (assume $b>a$).

For $b>a$: $r_1=b-a$, so $r_1 r_2=b^2-a^2$.

$r_1 r_2/\sqrt{r_0}=(b^2-a^2)/\sqrt b$.

Step 6 — Equate

$(b^2-a^2)/\sqrt b=4a^{3/2}/3^{3/4}$.

Square: $(b^2-a^2)^2/b=16 a^3/3^{3/2}$.

Cross-multiply: $(b^2-a^2)^2\cdot 3^{3/2}=16 a^3 b$.

$3^{3/2}=3\sqrt 3$:

$3\sqrt 3(b^2-a^2)^2=16 a^3 b.$ ✓

Answer