Question

An inviscid steady incompressible flow is formed by combining a uniform flow with velocity $U_\infty$ and a clockwise vortex of strength $K$ at the origin, as shown in the figure. Velocity potential ($\phi$) for the combined flow in polar coordinates $(r,\theta)$ is

• A. $\displaystyle \phi=\frac{K\theta}{2\pi}-U_\infty r\cos\theta$
• B. $\displaystyle \phi=\frac{K\theta}{2\pi}-U_\infty r\sin\theta$
• C. $\displaystyle \phi=K\ln r+U_\infty r\cos\theta$
• D. $\displaystyle \phi=-K\ln r+U_\infty r\sin\theta$

Hint:

Elementary potentials: uniform flow $\phi=U_\infty r\cos\theta$; source/sink $\phi= \pm Q\,\ln r/(2\pi)$; vortex $\phi=\Gamma\theta/(2\pi)$. Superpose and pick the angular dependence: $\cos\theta$ for flow along $x$ (not $\sin\theta$).

Answer 1

The Correct Option is A

Step 1: Write the potentials of the elementary flows.
- Uniform flow of speed $U_\infty$ along the $+x$–direction has velocity potential \[ \phi_{\text{uni}}=U_\infty x=U_\infty r\cos\theta. \] - A point vortex of circulation $\Gamma$ at the origin has potential \[ \phi_{\text{vortex}}=\frac{\Gamma}{2\pi}\,\theta, \] since $u_\theta=\dfrac{1}{r}\dfrac{\partial \phi}{\partial \theta}=\dfrac{\Gamma}{2\pi r}$. 
Step 2: Account for clockwise sense.
By convention, {counterclockwise} circulation is positive. A {clockwise} vortex therefore has $\Gamma<0$. If its (signed) strength is denoted by $K$, then the vortex potential is $\phi_{\text{vortex}}=\dfrac{K\theta}{2\pi}$. 
Step 3: Superpose the potentials.
The flow is the linear superposition of the two potential flows, so \[ \phi=\phi_{\text{vortex}}+\phi_{\text{uni}} =\frac{K\theta}{2\pi}+U_\infty r\cos\theta. \] Among the choices, the only expression with the correct $\theta$–term and the correct angular dependence ($\cos\theta$ for a uniform flow along $x$) is option (A), noting that the sign of $K$ encodes the clockwise sense. Final Answer: \[ \boxed{\text{(A)}} \]