← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q7c — Step-by-Step Solution

20 marks · Section B

Potential flow · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

Given the velocity potential ϕ=12log⁡ ⁣[(x+a)2+y2(x−a)2+y2]\phi=\dfrac{1}{2}\log\!\left[\dfrac{(x+a)^{2}+y^{2}}{(x-a)^{2}+y^{2}}\right], determine the streamlines.

Technique

Recognise source-sink flow; build complex potential ww; ψ=Im⁡(w)\psi=\operatorname{Im}(w); the angle-subtended interpretation gives streamlines as circles via the inscribed-angle theorem.

Solution

Strategy. Recognise the velocity potential as that of a source-sink pair; build the complex potential; extract the stream function; identify its level curves geometrically.

Step 1 — Identify the flow

ϕ=12ln⁡[(x+a)2+y2]−12ln⁡[(x−a)2+y2]=ln⁡r1−ln⁡r2,\phi=\dfrac{1}{2}\ln\bigl[(x+a)^{2}+y^{2}\bigr]-\dfrac{1}{2}\ln\bigl[(x-a)^{2}+y^{2}\bigr]=\ln r_1-\ln r_2, where r1=∣z−(−a)∣=(x+a)2+y2r_1=|z-(-a)|=\sqrt{(x+a)^{2}+y^{2}} and r2=∣z−a∣=(x−a)2+y2r_2=|z-a|=\sqrt{(x-a)^{2}+y^{2}}.

This is the velocity potential of a source at z=−az=-a and a sink at z=+az=+a, both of unit strength (with the convention ϕ=mln⁡r\phi=m\ln r for a source of strength mm).

Step 2 — Complex potential

For a source at z0z_0, the complex potential is w=log⁡(z−z0)w=\log(z-z_0). Combining source at −a-a and sink at +a+a:

w(z)=log⁡(z+a)−log⁡(z−a)=log⁡z+az−a.w(z)=\log(z+a)-\log(z-a)=\log\dfrac{z+a}{z-a}.

Verify: Re⁡(w)=12ln⁡∣z+a∣2∣z−a∣2=ϕ\operatorname{Re}(w)=\dfrac{1}{2}\ln\dfrac{|z+a|^{2}}{|z-a|^{2}}=\phi ✓.

Step 3 — Stream function

ψ=Im⁡(w)=arg⁡z+az−a=arg⁡(z+a)−arg⁡(z−a)=θ1−θ2,\psi=\operatorname{Im}(w)=\arg\dfrac{z+a}{z-a}=\arg(z+a)-\arg(z-a)=\theta_1-\theta_2, where θ1=arctan⁡ ⁣yx+a\theta_1=\arctan\!\dfrac{y}{x+a} (angle to point (x,y)(x,y) from (−a,0)(-a,0)) and θ2=arctan⁡ ⁣yx−a\theta_2=\arctan\!\dfrac{y}{x-a} (from (+a,0)(+a,0)).

Step 4 — Streamlines are ψ=\psi= const

θ1−θ2=\theta_1-\theta_2= const means: the angle subtended at (x,y)(x,y) by the segment from (−a,0)(-a,0) to (a,0)(a,0) is constant.

Geometric fact (inscribed angle theorem): the locus of points from which a fixed segment subtends a constant angle is a circular arc through the segment’s endpoints.

So each streamline is a circle passing through (±a,0)(\pm a, 0).

Step 5 — Algebraic form

Compute tan⁡(θ1−θ2)=tan⁡θ1−tan⁡θ21+tan⁡θ1tan⁡θ2\tan(\theta_1-\theta_2)=\dfrac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2}.

Numerator: yx+a−yx−a=y[(x−a)−(x+a)](x+a)(x−a)=−2ayx2−a2\dfrac{y}{x+a}-\dfrac{y}{x-a}=\dfrac{y[(x-a)-(x+a)]}{(x+a)(x-a)}=\dfrac{-2ay}{x^{2}-a^{2}}.

Denominator: 1+y2(x+a)(x−a)=x2−a2+y2x2−a21+\dfrac{y^{2}}{(x+a)(x-a)}=\dfrac{x^{2}-a^{2}+y^{2}}{x^{2}-a^{2}}.

Therefore

tan⁡(θ1−θ2)=−2ayx2+y2−a2=k(const).\tan(\theta_1-\theta_2)=\dfrac{-2ay}{x^{2}+y^{2}-a^{2}}=k\quad(\text{const}).

Cross-multiplying:

−2ay=k(x2+y2−a2)  ⟹  x2+y2+2aky−a2=0.-2ay=k(x^{2}+y^{2}-a^{2})\;\Longrightarrow\;x^{2}+y^{2}+\dfrac{2a}{k}y-a^{2}=0.

Complete the square in yy:

x2+(y+ak)2=a2+a2k2=a2(k2+1)k2.x^{2}+\bigl(y+\tfrac{a}{k}\bigr)^{2}=a^{2}+\dfrac{a^{2}}{k^{2}}=\dfrac{a^{2}(k^{2}+1)}{k^{2}}.

So the streamlines are

Answer

  x2+(y+ak)2=a2(k2+1)k2,k∈R∖{0}.  \boxed{\;x^{2}+\bigl(y+\tfrac{a}{k}\bigr)^{2}=\dfrac{a^{2}(k^{2}+1)}{k^{2}},\quad k\in\mathbb R\setminus\{0\}.\;}
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