Question

The vortex shedding frequency behind a landing gear model is found to be 50 Hz when tested in a wind tunnel operating at 5 m/s. If the actual landing gear size is 10 times that of the model, and it is designed to operate at 50 m/s, then the expected vortex shedding frequency behind it is _______ Hz (rounded off to the nearest integer).

Hint:

Match Strouhal numbers, \(St=fD/V\) constant, between the model and the actual landing gear, then solve for \(f_2\).

Answer 1

Correct Answer: 50

Instead of writing out the full Strouhal number equation, we can work purely with ratios, which is often faster for scaling problems.

Keeping the Strouhal number $St = fD/V$ constant between model and prototype means the frequency scales directly like:

\[ f_2 = f_1 \times \left(\frac{D_1}{D_2}\right) \times \left(\frac{V_2}{V_1}\right) \]

This comes straight from rearranging $f_1D_1/V_1 = f_2D_2/V_2$ to isolate $f_2$; it says the shedding frequency is directly proportional to velocity and inversely proportional to size, which makes physical sense: shrinking the body (smaller $D$) or speeding up the flow (bigger $V$) both make vortices shed faster.

Now put in the two ratios given in the problem. The size ratio is:

\[ \frac{D_1}{D_2} = \frac{D_1}{10D_1} = \frac{1}{10} \]

and the velocity ratio is:

\[ \frac{V_2}{V_1} = \frac{50}{5} = 10 \]

Multiplying these two ratios together:

\[ \frac{D_1}{D_2}\times\frac{V_2}{V_1} = \frac{1}{10}\times 10 = 1 \]

The size increase and the velocity increase exactly cancel out, so the shedding frequency stays the same as the model's:

\[ f_2 = f_1 \times 1 = 50 \text{ Hz} \]

So the expected vortex shedding frequency behind the full scale landing gear is 50 Hz.