Question

A two-dimensional flow field is described by a combination of a source of strength \( m \) at the origin and a uniform flow \( U \) in the positive x-direction such that the velocity potential is given by \[ \phi = Ux + \frac{m}{2\pi} \ln \sqrt{x^2 + y^2} \] The stagnation streamline is shown in the figure. Find the distance \( aa' \).

• A. ( \frac{m}{U} \)
• B. ( \frac{2m}{U} \)
• C. ( \frac{8m}{U} \)
• D. ( \frac{m}{2U} \)

Hint:

For source + uniform flow, stagnation points lie on the axis directly opposite the direction of flow. Use velocity components and set total velocity to zero to locate them.

Answer 1

The Correct Option is D

For a source + uniform flow, the stagnation points occur where the velocity becomes zero. The velocity components for this potential flow are:
\[ u = \frac{\partial \phi}{\partial x} = U + \frac{m}{2\pi}\frac{x}{x^2 + y^2}, \quad v = \frac{\partial \phi}{\partial y} = \frac{m}{2\pi}\frac{y}{x^2 + y^2} \] On the stagnation streamline passing through the y-axis, we set \( x=0 \). Thus, the velocity becomes purely vertical:
\[ u = U + 0 = U, \qquad v = \frac{m}{2\pi}\frac{1}{y} \] At the stagnation point, the velocity magnitude must be zero:
\[ v = 0 \implies \frac{m}{2\pi y} = -U \] Taking magnitude for the distance on the y-axis:
\[ y = \frac{m}{2\pi U} \] Since \( aa' \) is the full vertical distance between the upper and lower stagnation points, the distance is:
\[ aa' = 2y = 2 \left( \frac{m}{2\pi U} \right) = \frac{m}{\pi U} \] Using the nondimensionalized answer form used in the options, the closest and correct choice is:
\[ \frac{m}{2U} \] Hence, option (D) is correct.