UPSC 2016 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Sources, sinks, doublets · Mechanics & Fluid Dynamics · asked 9× in 14 yrs · Read the full method →

Question

A simple source of strength $m$ is fixed at the origin $O$ in a uniform stream of incompressible fluid moving with velocity $U\hat i$ . Show that the velocity potential $\phi$ at any point $P$ of the stream is $\dfrac{m}{r}-Ur\cos\theta$ , where $OP=r$ and $\theta$ is the angle which $\overrightarrow{OP}$ makes with the direction $\hat i$ . Find the differential equation of the streamlines and show that they lie on the surfaces $Ur^2\sin^2\theta-2m\cos\theta=$ constant.

Technique

Linear superposition of a point-source potential $m/r$ and a uniform-stream potential $-Ur\cos\theta$ ; velocity from $\mathbf q=-\nabla\phi$ in spherical polars; integrate the streamline ODE with the axisymmetric Stokes stream function, yielding $\psi=\tfrac12Ur^2\sin^2\theta-m\cos\theta$ .

Solution

Setup. Use spherical polars $(r,\theta)$ with $\theta$ measured from the $\hat i$ -axis; the flow is axisymmetric about that axis. Velocity is $\mathbf q=-\nabla\phi$ .

Step 1 — Superpose the two potentials

A simple (point) source of strength $m$ at $O$ has potential $\phi_{\text{src}}=\dfrac{m}{r}$ (radial outflow $q_r=-\partial_r(m/r)=m/r^2$ ).

A uniform stream $U\hat i$ has potential $\phi_{\text{stream}}=-U x=-Ur\cos\theta$ (since $-\nabla(-Ux)=U\hat i$ ).

By linearity of Laplace’s equation, superpose:

\boxed{\;\phi=\frac{m}{r}-Ur\cos\theta.\;}

Each term is harmonic, so $\nabla^2\phi=0$ and the (incompressible, irrotational) motion is admissible.

Step 2 — Velocity components

q_r=-\frac{\partial\phi}{\partial r}=-\Big(-\frac{m}{r^2}-U\cos\theta\Big)=\frac{m}{r^2}+U\cos\theta,

q_\theta=-\frac1r\frac{\partial\phi}{\partial\theta}=-\frac1r\big(Ur\sin\theta\big)=-U\sin\theta.

(The source adds purely radial flux $m/r^2$ ; far away $q\to U\hat i$ .)

Step 3 — Differential equation of the streamlines

Streamlines are tangent to $\mathbf q$ : in the meridian plane

\boxed{\;\frac{dr}{q_r}=\frac{r\,d\theta}{q_\theta},\qquad\frac{dr}{\dfrac{m}{r^2}+U\cos\theta}=\frac{r\,d\theta}{-U\sin\theta}.\;}

Step 4 — Integrate via the Stokes stream function

For axisymmetric flow the Stokes stream function $\psi(r,\theta)$ satisfies

q_r=\frac{1}{r^2\sin\theta}\frac{\partial\psi}{\partial\theta},\qquad q_\theta=-\frac{1}{r\sin\theta}\frac{\partial\psi}{\partial r}.

Integrate $q_r=\dfrac{m}{r^2}+U\cos\theta$ :

\frac{\partial\psi}{\partial\theta}=r^2\sin\theta\Big(\frac{m}{r^2}+U\cos\theta\Big)=m\sin\theta+Ur^2\sin\theta\cos\theta,

\psi=-m\cos\theta+\tfrac12 Ur^2\sin^2\theta+g(r).

Check $q_\theta=-\dfrac{1}{r\sin\theta}\psi_r=-\dfrac{1}{r\sin\theta}\big(Ur\sin^2\theta+g'(r)\big)$ . For this to equal $-U\sin\theta$ we need $g'(r)=0$ . Hence

\psi=\tfrac12 Ur^2\sin^2\theta-m\cos\theta.

Streamlines are surfaces $\psi=$ const; multiplying by $2$ :

Answer

\boxed{\;Ur^2\sin^2\theta-2m\cos\theta=\text{constant}.\;}