Question 5 of 6: Wind-Tunnel Testing of a Submarine Model

Question 5: Wind-Tunnel Testing of a Submarine Model (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Prototype: $L_p=2.5$ m in water at $15^{\circ}\text{C}$ ($\nu_w=1.139\times10^{-6}\ \text{m}^2/\text{s}$), $V_p=0.5$ m/s. Model: length scale $L_m/L_p=1/8$, in air at $25^{\circ}\text{C}$, 1 atm ($\nu_a=1.562\times10^{-5}\ \text{m}^2/\text{s}$, $a_{\text{air}}=346$ m/s).

Find. The tunnel air speed $V_m$ for dynamic similarity, and whether that similarity is physically valid.

Approach. For a fully submerged body the drag coefficient depends only on Reynolds number, $C_D=f(Re)$, so dynamic similarity requires equal Reynolds numbers in model and prototype.

1. Similarity criterion. Matching Reynolds numbers, $$Re_m=Re_p\ \Rightarrow\ \frac{V_m L_m}{\nu_a}=\frac{V_p L_p}{\nu_w}.$$
2. Solve for the tunnel speed. $$V_m=V_p\,\frac{L_p}{L_m}\,\frac{\nu_a}{\nu_w}=0.5\times 8\times\frac{1.562\times10^{-5}}{1.139\times10^{-6}}=0.5\times8\times13.71.$$ $$\boxed{V_m=54.9\ \text{m/s}.}$$ (Check: $Re=\rho VL/\mu\approx1.10\times10^6$ for both model and prototype.)
3. (b) Validity — compressibility check. The corresponding Mach number of the tunnel air is $$Ma=\frac{V_m}{a_{\text{air}}}=\frac{54.9}{346}=0.16.$$ Since $Ma\lt 0.3$, compressibility effects are negligible and the air behaves as effectively incompressible, so matching Reynolds number alone is sufficient — the assumption of flow similarity is valid. Both flows also sit at $Re\approx1.1\times10^6$, i.e. the same (turbulent) regime, so the drag coefficient measured in air transfers directly to the submarine in water.