Question

A uniform flow, a point source of strength \( +\sigma \) at \( (a, 0) \) and a point sink of strength \( -\sigma \) at \( (-a, 0) \) are shown in the figure. The velocity potential \( \phi \) resulting from the superposition of these flow fields is given by 


 

• A. \( \phi = -U_\infty x + \frac{\sigma}{2\pi} \ln \sqrt{(x+a)^2 + y^2} - \frac{\sigma}{2\pi} \ln \sqrt{(x-a)^2 + y^2} \)
• B. \( \phi = -U_\infty x + \frac{\sigma}{2\pi} \ln \sqrt{( (x+a)^2 + y^2 } - \frac{\sigma}{2\pi} \ln \sqrt{( (x-a)^2 + y^2} \)
• C. \( \phi = U_\infty x + \frac{\sigma}{2\pi} \ln \sqrt{(x+a)^2 + y^2} - \frac{\sigma}{2\pi} \ln \sqrt{(x-a)^2 + y^2} \)
• D. \( \phi = U_\infty x + \frac{\sigma}{2\pi} \ln \sqrt{(x+a)^2 + y^2} - \frac{\sigma}{2\pi} \ln \sqrt{( (x-a)^2 + y^2} \)

Hint:

To determine the velocity potential in superimposed flow fields, ensure the correct signs and positions of sources and sinks and verify the direction of uniform flow.

Answer 1

The Correct Option is B

Step 1: Understand the components of the velocity potential.
The velocity potential \( \phi \) for the given setup is obtained by superimposing the potentials of the uniform flow, the source, and the sink: 1. Uniform flow potential: The velocity potential for uniform flow with velocity \( U_\infty \) in the positive \( x \)-direction is: \[ \phi_{\text{uniform}} = -U_\infty x. \] 2. Point source at \( (a, 0) \): The velocity potential for a point source of strength \( +\sigma \) located at \( (a, 0) \) is: \[ \phi_{\text{source}} = \frac{\sigma}{2\pi} \ln \sqrt{(x-a)^2 + y^2}. \] 3. Point sink at \( (-a, 0) \): The velocity potential for a point sink of strength \( -\sigma \) located at \( (-a, 0) \) is: \[ \phi_{\text{sink}} = -\frac{\sigma}{2\pi} \ln \sqrt{(x+a)^2 + y^2}. \] Step 2: Superpose the potentials.
The total velocity potential \( \phi \) is the sum of the three components: \[ \phi = \phi_{\text{uniform}} + \phi_{\text{source}} + \phi_{\text{sink}}. \] Substituting the expressions for each term: \[ \phi = -U_\infty x + \frac{\sigma}{2\pi} \ln \sqrt{(x-a)^2 + y^2} - \frac{\sigma}{2\pi} \ln \sqrt{(x+a)^2 + y^2}. \] Step 3: Verify the options.
Option (A): Incorrect, as it reverses the roles of the source and sink terms.
Option (B): Correct, as it matches the derived expression for \( \phi \).
Option (C): Incorrect, as it uses \( U_\infty x \) instead of \( -U_\infty x \).
Option (D): Incorrect, as it uses \( U_\infty x \) and reverses the source and sink terms.
Conclusion: The velocity potential resulting from the superposition of the flow fields is: \[ \phi = -U_\infty x + \frac{\sigma}{2\pi} \ln \sqrt{(x-a)^2 + y^2} - \frac{\sigma}{2\pi} \ln \sqrt{(x+a)^2 + y^2}. \]