← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

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

Question

A sphere of radius RR, whose centre is at rest, vibrates radially in an infinite incompressible fluid of density ρ\rho, which is at rest at infinity. If the pressure at infinity is Π\Pi, so that the pressure at the surface of the sphere at time tt is

Π+12ρ{d2R2dt2+(dRdt)2}.\Pi+\frac12\rho\left\{\frac{d^2R^2}{dt^2}+\left(\frac{dR}{dt}\right)^2\right\}.

(The printed statement is truncated — it gives the target expression but omits the “show that”. We derive the rigorous Rayleigh result and reconcile it with the printed form below.)

Technique

Radial incompressible potential ϕ=−R2R˙/r\phi=-R^2\dot R/r (from u(R)=R˙u(R)=\dot R); unsteady Bernoulli with conditions at infinity; evaluate at r=Rr=R.

Solution

Setup. Incompressible irrotational radial flow about a sphere of (time-dependent) radius R(t)R(t) centred at the origin. By spherical symmetry the velocity is purely radial, u=u(r,t)u=u(r,t), with velocity potential ϕ(r,t)\phi(r,t).

Step 1 — Velocity potential (mass conservation)

Incompressibility ∇2ϕ=0\nabla^2\phi=0 with radial symmetry gives ϕ=−F(t)r\phi=-\dfrac{F(t)}{r} (the regular-at-infinity harmonic), so u=∂ϕ∂r=F(t)r2u=\dfrac{\partial\phi}{\partial r}=\dfrac{F(t)}{r^2}. At the sphere surface r=Rr=R the fluid moves with the boundary: u(R,t)=R˙u(R,t)=\dot R. Hence

F(t)R2=R˙ ⟹ F(t)=R2R˙,ϕ=−R2R˙r,u=R2R˙r2.\frac{F(t)}{R^2}=\dot R\ \Longrightarrow\ F(t)=R^2\dot R,\qquad \phi=-\frac{R^2\dot R}{r},\qquad u=\frac{R^2\dot R}{r^2}.

Step 2 — Unsteady Bernoulli equation

For unsteady irrotational flow,

pρ+∂ϕ∂t+12u2=f(t).\frac{p}{\rho}+\frac{\partial\phi}{\partial t}+\frac12 u^2=\text{f}(t).

At infinity ϕ→0, u→0, p→Π\phi\to0,\ u\to0,\ p\to\Pi, so f(t)=Π/ρ\text{f}(t)=\Pi/\rho. Thus

p=Π−ρ(∂ϕ∂t+12u2).p=\Pi-\rho\Big(\frac{\partial\phi}{\partial t}+\tfrac12 u^2\Big).

Step 3 — Evaluate ∂ϕ/∂t\partial\phi/\partial t and u2u^2 at the surface

∂ϕ∂t=−1rddt(R2R˙)=−1r(2RR˙2+R2R¨),u2=R4R˙2r4.\frac{\partial\phi}{\partial t}=-\frac{1}{r}\frac{d}{dt}\big(R^2\dot R\big)=-\frac{1}{r}\big(2R\dot R^2+R^2\ddot R\big),\qquad u^2=\frac{R^4\dot R^2}{r^4}.

At r=Rr=R:

∂ϕ∂t∣R=−(2R˙2+RR¨),u2∣R=R˙2.\left.\frac{\partial\phi}{\partial t}\right|_{R}=-\big(2\dot R^2+R\ddot R\big),\qquad u^2\big|_R=\dot R^2.

Therefore the surface pressure is

pR=Π−ρ[−(2R˙2+RR¨)+12R˙2]=Π+ρ(RR¨+32R˙2).p_R=\Pi-\rho\Big[-(2\dot R^2+R\ddot R)+\tfrac12\dot R^2\Big]=\Pi+\rho\Big(R\ddot R+\tfrac32\dot R^2\Big).

Answer

  pR=Π+ρ(R R¨+32R˙2).  \boxed{\;p_R=\Pi+\rho\Big(R\,\ddot R+\tfrac32\dot R^2\Big).\;}
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