← 2015 Paper 2
UPSC 2015 Maths Optional Paper 2 Q5d — Step-by-Step Solution
10 marks · Section B
Sources, sinks, doublets · Mechanics & Fluid Dynamics · asked 8× in 13 yrs · Read the full method →
Question
Consider a uniform flow U0 in the positive x-direction. A cylinder of radius a is located at the origin. Find the stream function and the velocity potential. Find also the stagnation points.
Technique
Standard “uniform flow + image doublet” complex potential; separate real and imaginary parts; impose ψ=0 on cylinder surface (r=a); stagnation points where ∇ϕ=0.
Solution
Setup. 2D incompressible irrotational flow past a circular cylinder. Use polar coordinates (r,θ) with origin at cylinder centre.
For uniform flow U0 in the +x direction superposed with a doublet at the origin (so the cylinder surface r=a is a streamline), the standard result is:
Step 1 — Complex potential
w(z)=U0(z+za2).
(Uniform flow U0z + doublet U0a2/z.)
Step 2 — Velocity potential ϕ and stream function ψ
w=ϕ+iψ with z=reiθ:
U0(reiθ+ra2e−iθ)=U0[(r+ra2)cosθ+i(r−ra2)sinθ].
So
ϕ(r,θ)=U0(r+ra2)cosθ,ψ(r,θ)=U0(r−ra2)sinθ.
Step 3 — Verify cylinder is a streamline
At r=a: ψ=U0(a−a)sinθ=0. So the cylinder surface r=a is the streamline ψ=0 ✓.
Step 4 — Stagnation points
Velocity components:
vr=r1∂θ∂ψ=U0(1−r2a2)cosθ,
vθ=−∂r∂ψ=−U0(1+r2a2)sinθ.
Stagnation: vr=vθ=0.
vr=0⇒ either r=a or cosθ=0 (i.e. θ=±π/2).
vθ=0⇒sinθ=0 (i.e. θ=0 or π); 1+a2/r2>0 always.
Combining: vθ=0 requires sinθ=0, then vr=U0(1−a2/r2)cosθ=0 requires r=a (since cosθ=±1=0).
Stagnation points: r=a, θ=0 or π — i.e. (a,0) (front of cylinder) and (−a,0) (rear).
In Cartesian: (a,0) and (−a,0).
Answer
Stagnation points: (±a,0).