← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q7b — Step-by-Step Solution

20 marks · Section B

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

Question

The space between two concentric spherical shells of radii $a,b\ (a<b)$ is filled with a liquid of density $\rho$. If the shells are set in motion, the inner one with velocity $U$ in the $x$-direction and the outer one with velocity $V$ in the $y$-direction, then show that the initial motion of the liquid is given by velocity potential

where $r^2=x^2+y^2+z^2$, the coordinates being rectangular. Evaluate the velocity at any point of the liquid.

Technique

Initial irrotational incompressible motion ⇒ harmonic $\phi$; build from the harmonics $x,\ x/r^3$ (and $y,\ y/r^3$); fix four coefficients from the four normal-velocity boundary conditions on $r=a$ (inner $U\hat i$) and $r=b$ (outer $V\hat j$); velocity $\mathbf q=-\nabla\phi$.

Solution

Setup. The liquid is incompressible and the initial motion (from rest) is irrotational, so $\phi$ is harmonic, $\nabla^2\phi=0$, with velocity $\mathbf q=-\nabla\phi$. The natural harmonic building blocks that are linear in a direction are

both solutions of Laplace’s equation, and similarly with $y$. We seek $\phi$ as a combination matching the boundary velocities on $r=a$ and $r=b$.

Step 1 — Form of the potential

Take

each bracket harmonic. The $x$-group must carry the inner sphere’s $U\hat i$ condition, the $y$-group the outer sphere’s $V\hat j$ condition.

Step 2 — Boundary conditions

On a rigid sphere $r=R$ moving with velocity $\mathbf W$, the normal velocity of the liquid equals that of the sphere: $\mathbf q\cdot\hat n=\mathbf W\cdot\hat n$, with $\hat n=\mathbf r/r$. Writing $\mathbf q=-\nabla\phi$:

• Inner $r=a$, $\mathbf W=U\hat i$: $-\partial_r\phi\big|_{a}=U\,\dfrac{x}{a}$ for all surface points.
• Outer $r=b$, $\mathbf W=V\hat j$: $-\partial_r\phi\big|_{b}=V\,\dfrac{y}{b}$ for all surface points.

For a term $Ax+Bx/r^3$, the radial derivative is $\partial_r(Ax+Bxr^{-3})=(A-2Br^{-3})\dfrac{x}{r}\cdot r/r\ldots$; carrying this out and imposing the two $x$-conditions ($-\partial_r\phi=U x/a$ at $a$ and $=0$ at $b$, since the outer sphere has no $x$-motion) gives two linear equations for $A,B$. Likewise the $y$-conditions ($0$ at $a$, $Vy/b$ at $b$) give $C,E$.

Step 3 — Solve for the coefficients

Solving the two pairs (standard concentric-sphere algebra) yields

Substituting,

which is exactly the stated potential. $\blacksquare$

Step 4 — Velocity at any point, $\mathbf q=-\nabla\phi$

Using $\partial_x(x\,r^{-3})=r^{-3}-3x^2r^{-5}$, $\partial_y(x r^{-3})=-3xy r^{-5}$, etc., and writing $\Delta=b^3-a^3$:

Equivalently, collecting the inverse-cube (dipole) terms,

Answer