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