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UPSC 2017 Maths Optional Paper 2 Q8c — Step-by-Step Solution

15 marks · Section B

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

Question

If the velocity of an incompressible fluid at the point $(x,y,z)$ is given by $\left(\dfrac{3xz}{r^5},\dfrac{3yz}{r^5},\dfrac{3z^2-r^2}{r^5}\right),\ r^2=x^2+y^2+z^2$, then prove that the liquid motion is possible and that the velocity potential is $\dfrac{z}{r^3}$. Further, determine the streamlines.

Technique

Continuity $\nabla\cdot\mathbf q=0$ for “motion possible”; verify $\mathbf q=-\nabla\phi$ for the stated potential; streamlines from $dx/u=dy/v=dz/w$ — the $x,y$ pair gives meridian planes, the Stokes stream function $\psi=\rho^2/r^3$ gives the in-plane curves.

Solution

Setup. Velocity components, with $r=\sqrt{x^2+y^2+z^2}$:

Step 1 — Motion is possible (incompressibility: $\nabla\cdot\mathbf q=0$)

For an incompressible fluid the motion is possible iff the equation of continuity $\nabla\cdot\mathbf q=u_x+v_y+w_z=0$ holds. Using $\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$ etc.:

Adding and using $x^2+y^2+z^2=r^2$:

The numerator of the second term is $15zr^2-15z^3+15z^3-5zr^2=10zr^2$, so the second term is $\dfrac{10zr^2}{r^7}=\dfrac{10z}{r^5}$, giving

The continuity equation is satisfied, so the liquid motion is possible.

Step 2 — Velocity potential $\phi=\dfrac{z}{r^3}$

Flow has a velocity potential if it is irrotational; with the convention $\mathbf q=-\nabla\phi$ we verify $\phi=\dfrac{z}{r^3}$ reproduces the velocity. With $\dfrac{\partial}{\partial x}\Big(\dfrac{z}{r^3}\Big)=z(-3)r^{-4}\dfrac{x}{r}=-\dfrac{3xz}{r^5}$:

All three match, so

(Since $\phi$ exists, the flow is irrotational; one checks $\nabla^2\phi=0$, consistent with Step 1 — this is a doublet along the $z$-axis.)

Step 3 — Streamlines

Streamlines satisfy $\dfrac{dx}{u}=\dfrac{dy}{v}=\dfrac{dz}{w}$, i.e.

First relation (from the $x$- and $y$-ratios):

So each streamline lies in a meridian plane $y=(\tan\beta)\,x$ through the $z$-axis (axisymmetric flow).

Second relation. Introduce the cylindrical radius $\rho=\sqrt{x^2+y^2}$. In a meridian plane the in-plane components are $u_\rho=\dfrac{3\rho z}{r^5}$, $w=\dfrac{3z^2-r^2}{r^5}$ ($r^2=\rho^2+z^2$). The Stokes stream function $\psi$ with $u_\rho=-\dfrac1\rho\psi_z,\ w=\dfrac1\rho\psi_\rho$ is

which one checks reproduces $u_\rho,w$. Streamlines are $\psi=$const:

Answer