Question 5 of 6: Wind-tunnel testing of a submarine model — Reynolds similarity

Question 5: Wind-tunnel testing of a submarine model — Reynolds similarity (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.

Find. The tunnel air speed that makes the model flow dynamically similar to the prototype flow, and a judgement on whether that speed keeps the similarity assumption honest.

Approach. Dimensional analysis of drag on a fully submerged body gives $C_D = f(Re)$ alone; equate the model and prototype Reynolds numbers to solve for the tunnel speed; then test the one assumption that equation hides — incompressibility — by computing the model Mach number.

1. Establish which parameter must be matched. For a body deeply submerged in a single fluid, with no free surface to make waves and no cavitation, dimensional analysis of $F = f(\rho, \mu, V, L)$ leaves one independent dimensionless group, so $$C_D = \frac{F}{\tfrac12\rho V^{2}A} = f\!\left(Re\right),\qquad Re = \frac{VL}{\nu}$$ There is no Froude number to match — the submarine runs submerged, not on the surface — and no Weber number, so Reynolds similarity alone gives complete dynamic similarity. That single observation carries most of the 15 marks on part (a).
2. Assemble the fluid properties. Fresh water at 15 °C has $\nu_w = 1.140\times10^{-6}\;\text{m}^2\text{/s}$. For the tunnel air at 25 °C and 101.325 kPa, the ideal-gas law gives $$\rho_a = \frac{p}{RT} = \frac{101\,325}{287 \times 298.15} = 1.184\;\text{kg/m}^{3}, \qquad \nu_a = \frac{\mu_a}{\rho_a} = \frac{1.849\times10^{-5}}{1.184} = 1.562\times10^{-5}\;\text{m}^{2}\text{/s}$$ so air at these conditions is $\nu_a/\nu_w = 13.70$ times more “viscous” kinematically than the lake water. This ratio, not the viscosities themselves, is what drives the answer.
3. Equate the Reynolds numbers (part a). Writing $Re_m = Re_p$ with the model dimensions $L_m = \lambda L_p$, $$\frac{V_m L_m}{\nu_a} = \frac{V_p L_p}{\nu_w} \qquad\Longrightarrow\qquad V_m = V_p \left(\frac{L_p}{L_m}\right)\left(\frac{\nu_a}{\nu_w}\right) = V_p \cdot \frac{1}{\lambda} \cdot \frac{\nu_a}{\nu_w}$$ $$V_m = 0.5 \times 8 \times 13.70 = \boxed{54.8\;\text{m/s}\;(197\;\text{km/h})}$$ Because the scale factor applies to every length equally, it does not matter whether $L$ or $D$ is used as the characteristic length — the ratio $L_p/L_m$ is 8 either way. Using the diameter, $$Re = \frac{0.5 \times 1.3}{1.140\times10^{-6}} = 5.70\times10^{5} = \frac{54.8 \times 0.1625}{1.562\times10^{-5}}$$ which is a free check that the arithmetic closes on both sides.
4. Test the hidden assumption (part b). Reynolds similarity was derived on the assumption that the air behaves as an incompressible fluid, so density does not enter as an independent variable. That is only true while the Mach number stays low. At 25 °C, $$a = \sqrt{\gamma R T} = \sqrt{1.4 \times 287 \times 298.15} = 346\;\text{m/s}, \qquad Ma = \frac{V_m}{a} = \frac{54.8}{346} = 0.158$$ $$\boxed{Ma = 0.158 < 0.3 \;\Rightarrow\; \text{the similarity assumption is valid}}$$
5. Quantify the margin and explain the answer. The conventional $Ma < 0.3$ threshold is not arbitrary: the isentropic density change at the stagnation point is $$\frac{\rho_0}{\rho} - 1 = \left(1+\tfrac{\gamma-1}{2}Ma^{2}\right)^{1/(\gamma-1)} - 1 = 1.3\;\%$$ at $Ma = 0.158$, which is smaller than the repeatability of a student drag balance, so treating the air as incompressible introduces no measurable error. The test is therefore genuinely similar: the model and prototype share the same $Re$, the same $C_D$, and the same flow topology, so the measured drag can be scaled to the prototype through $F_p = C_D\,\tfrac12\rho_w V_p^{2}A_p$. Two practical caveats belong in a complete answer — the tunnel blockage must stay small (the question states the tunnel section is much larger than the model, which secures this) and the model must be geometrically similar including its surface roughness, since at $Re \approx 6\times10^{5}$ the boundary layer is in the transitional range where roughness moves the transition point. Had the required speed come out above roughly 100 m/s, compressibility would have broken the similarity and the team would need a pressurised tunnel, a heavier gas, or a towing tank instead.