← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

Sources, sinks, doublets · Mechanics & Fluid Dynamics · asked 8× in 13 yrs · Read the full method →

Question

If fluid fills the region of space on the positive side of the $x$-axis, which is a rigid boundary and if there be a source $m$ at the point $(0,a)$ and an equal sink at $(0,b)$ and if the pressure on the negative side be the same as the pressure at infinity, show that the resultant pressure on the boundary is

where $\rho$ is the density of the fluid.

Technique

Image method for rigid wall (same-strength reflection); Bernoulli; residue-theorem integral.

Solution

Setup. Fluid fills the upper half-plane $y>0$; the boundary is the $x$-axis ($y=0$), a rigid wall. Source of strength $m$ at $(0,a)$, sink of strength $m$ at $(0,b)$, both in the fluid ($a,b>0$).

Strategy. Image method to enforce zero normal velocity on $y=0$; Bernoulli to get pressure; integrate $(p_\infty-p)$ over the boundary.

Step 1 — Image system

For a rigid boundary $y=0$, each singularity has a same-strength image at the mirror point:

• Source $m$ at $(0,a)$ → image source $m$ at $(0,-a)$.
• Sink $-m$ at $(0,b)$ → image sink $-m$ at $(0,-b)$.

Complex potential (using $z=x+iy$, source at $z_0$ contributing $m\log(z-z_0)$):

Step 2 — Velocity on the boundary

On $z=x$ (real, $y=0$):

So $|V|^{2}=u^{2}=\dfrac{4m^{2}x^{2}(b^{2}-a^{2})^{2}}{(x^{2}+a^{2})^{2}(x^{2}+b^{2})^{2}}$.

Step 3 — Pressure via Bernoulli

$p+\tfrac{1}{2}\rho|V|^{2}=p_\infty$ at infinity (where $|V|\to 0$). So

The net force per unit length on the wall (upward, since pressure above is less than $p_\infty$):

where

Step 4 — Evaluate $J$ via residues

Poles in upper half-plane: $z=ia$ (order 2), $z=ib$ (order 2). Residue computations (carried out via logarithmic differentiation) give

(Method: compute residues of $\dfrac{z^{2}}{(z^{2}+a^{2})^{2}(z^{2}+b^{2})^{2}}$ at $ia$ and $ib$; combine; the symmetric simplification yields $-(a-b)^{3}/[4ab(a^{2}-b^{2})^{3}]\cdot i$, and times $2\pi i$ gives the stated $J$.)

Step 5 — Assemble the force

Using $(a^{2}-b^{2})^{2}=(a-b)^{2}(a+b)^{2}$:

Comparison with the stated answer. The question states the result as $\dfrac{\pi\rho m^{2}(a-b)^{2}}{2ab(a+b)}$, which differs from the derivation above by a factor of $2$. The discrepancy may stem from a different source-strength convention (some texts define “strength $m$” so that the complex potential is $\tfrac{m}{2}\log(z-z_0)$, halving the velocity and quartering $|V|^{2}$) or from integrating over only the half-boundary $x\in[0,\infty)$ (which would halve the integral due to the even integrand). For the problem as stated and standard convention $w=m\log(z-z_0)$:

Answer