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UPSC 2022 Maths Optional Paper 2 Q8c — Step-by-Step Solution

20 marks · Section B

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

Question

Verify $w=ik\log[(z-ia)/(z+ia)]$ is the complex potential of a steady flow of fluid about a circular cylinder, where $y=0$ is a rigid boundary. Find the force on unit length of the cylinder.

Technique

Verify $\psi=0$ on rigid wall by direct substitution; Blasius theorem for force; residue of $(dw/dz)^2$ at the vortex.

Solution

Setup. Complex potential $w=ik\log[(z-ia)/(z+ia)]$.

This is the potential of two vortices: at $z=ia$ (strength to be determined) and at $z=-ia$ (image strength).

A vortex of strength $\kappa$ at $z_0$ has potential $w=-i\kappa\log(z-z_0)$. Comparing: $w=ik\log(z-ia)-ik\log(z+ia)$, so this is a vortex of strength $-k$ at $ia$ and strength $+k$ at $-ia$.

Step 1 — Verify $y=0$ is a rigid boundary

A streamline of the flow is rigid. On $y=0$, $z=x$ (real), so $z-ia=x-ia$ and $z+ia=x+ia$.

$\dfrac{z-ia}{z+ia}=\dfrac{x-ia}{x+ia}=\dfrac{(x-ia)^2}{x^2+a^2}=\dfrac{x^2-a^2-2iax}{x^2+a^2}$.

Magnitude: $\left|\dfrac{z-ia}{z+ia}\right|=\dfrac{|x-ia|}{|x+ia|}=1$ (both equal $\sqrt{x^2+a^2}$).

So on $y=0$, $\log[(z-ia)/(z+ia)]$ is purely imaginary (real part = $\log 1=0$).

Then $w=ik\cdot(\text{imaginary})=ik\cdot i\cdot(\text{real})=-k\cdot(\text{real})$, i.e., $w$ is purely real on $y=0$.

Stream function $\psi=\Im w=0$ on $y=0$. So $y=0$ is the streamline $\psi=0$ — a rigid boundary ✓.

Step 2 — Identify circular cylinder

Looking for another streamline at $\psi=$ const that’s a circle. The vortex pair structure suggests the cylinder is centred at $z=ia$ (or $-ia$).

Actually the standard setup: vortex at $z=ia$ with image $-ia$ (mirror across $y=0$) creates a flow with $y=0$ as a rigid wall. The vortex at $ia$ is inside a virtual half-plane.

To put a cylinder around the vortex at $ia$: use the Milne-Thomson circle theorem. The cylinder is centred at $ia$ with some radius $\rho$.

Hmm — re-reading the problem: It says “fluid about a circular cylinder, where the plane $y=0$ is a rigid boundary.” So the cylinder is somewhere above $y=0$.

Looking at the structure: vortex at $ia$ (strength $-k$) and image at $-ia$ (strength $+k$). For the flow around an actual cylinder, we’d need additional doublets/vortices to make the cylinder boundary a streamline.

Standard reading: $w=ik\log[(z-ia)/(z+ia)]$ is the potential of two equal-and-opposite vortices forming a vortex pair. The line $y=0$ is the streamline equidistant from both (by symmetry).

The “cylinder” might be infinitesimal — i.e., the vortex itself represents a thin cylinder with circulation. In that case, the “cylinder” is at $z=ia$, of vanishing radius.

Step 3 — Force on the cylinder via Blasius theorem

Blasius theorem. Force on a body inside a region of irrotational flow:

Here $\rho$ = fluid density, $C$ encloses the body.

Take a small circle around the vortex at $ia$, enclosing only that vortex (and the cylinder).

$dw/dz=ik\!\left[\dfrac{1}{z-ia}-\dfrac{1}{z+ia}\right]$.

Near $z=ia$, the dominant term is $ik/(z-ia)$.

$(dw/dz)^2=\!\left(\dfrac{ik}{z-ia}-\dfrac{ik}{z+ia}\right)^2$.

For the contour integral, by residue theorem: only the residue at $z=ia$ contributes (since the contour encloses only this point).

Expand near $z=ia$: $(dw/dz)^2=\!\left[\dfrac{ik}{z-ia}-\dfrac{ik}{2ia}+\cdots\right]^2=\dfrac{(ik)^2}{(z-ia)^2}-2\cdot\dfrac{ik}{z-ia}\cdot\dfrac{ik}{2ia}+\cdots$ $=\dfrac{-k^2}{(z-ia)^2}-\dfrac{(ik)^2}{ia(z-ia)}+\cdots$ $=\dfrac{-k^2}{(z-ia)^2}+\dfrac{k^2}{ia(z-ia)}+\cdots$.

Residue at $z=ia$ of $(dw/dz)^2$ is the coefficient of $1/(z-ia)$, which is $k^2/(ia)=-ik^2/a$.

Wait: $\dfrac{1}{ia}=\dfrac{1}{ia}\cdot\dfrac{-i}{-i}=\dfrac{-i}{-i^2\cdot a}=\dfrac{-i}{a}=-i/a$.

So residue $=k^2\cdot(-i/a)=-ik^2/a$.

By residue theorem: $\oint(dw/dz)^2 dz=2\pi i\cdot(-ik^2/a)=2\pi k^2/a$.

Blasius: $F_x-iF_y=\dfrac{i\rho}{2}\cdot\dfrac{2\pi k^2}{a}=\dfrac{i\pi\rho k^2}{a}$.

Take real and imaginary parts: $F_x=0$, $F_y=-\pi\rho k^2/a$.

So $F_y$ is negative — the cylinder is pulled toward the wall $y=0$.

Answer