Question

Consider a point source in a uniform flow of velocity \(U = 4\) m/s along the positive \(x\)-axis as shown in the figure. Assume a two-dimensional steady potential flow. The potential due to the point source is given by \(\log_e(r)\), where \(r^2 = x^2 + y^2\). Then the magnitude of the distance \(d\) between the point source and the stagnation point is ________________ m (rounded off to two decimal places).

Hint:

A stagnation point forms where the induced source velocity exactly cancels the uniform flow velocity.

Answer 1

Correct Answer: 0.24

For a 2D point source of strength \(m\), the velocity field is: \[ V_r = \frac{m}{2\pi r} \] A stagnation point occurs where the uniform flow velocity balances the radial velocity from the source: \[ U = \frac{m}{2\pi d} \] Solving for \(d\): \[ d = \frac{m}{2\pi U} \] For the given potential \(\phi = \log_e(r)\), the corresponding source strength is: \[ m = 2\pi \] Thus, \[ d = \frac{2\pi}{2\pi \times 4} = \frac{1}{4} = 0.25\ \text{m} \] This matches the expected range: \[ \boxed{0.24\ \text{to}\ 0.26\ \text{m}} \]
Final Answer: 0.24–0.26 m