A 1:50 ship model is tested in water at a speed chosen to match the prototype's Froude number. What happens to the Reynolds number in this model test?
AReynolds number is also matched because it is dimensionless, just like Froude number
BReynolds number increases by a factor of 50 because the model is smaller and flow is more turbulent
CReynolds number is approximately (1/50)^(3/2) ≈ 0.3% of the full-scale value — severely mismatched
DReynolds number is irrelevant for ship testing because ships operate at the free surface
Froude matching requires V_m = V_p × √(L_m/L_p) = V_p/√50. Reynolds number is Re = VL/ν. At model scale: Re_m = (V_p/√50)(L_p/50)/ν = Re_p × (1/50)^(3/2) ≈ 0.003 Re_p. So Re is only about 0.3% of its full-scale value — viscous effects in the model test are completely different from the prototype. The Froude and Reynolds numbers cannot simultaneously be matched in the same fluid at reduced scale, which is the central challenge of ship model testing.
Question 2 Multiple Choice
An engineer claims that testing a 1:20 aircraft model in a wind tunnel at the same wind speed as the full-scale aircraft automatically matches the Reynolds number. Why is this incorrect?
AReynolds number only applies to internal flows like pipes, not external aerodynamics
BRe = ρVL/μ; with the same fluid and velocity but L scaled by 1/20, Re_model is only 5% of Re_prototype
CThe wind tunnel walls create boundary effects that invalidate Reynolds number matching
DReynolds number is automatically matched whenever geometric similarity is maintained
Re = ρVL/μ scales linearly with characteristic length L. With L_m = L_p/20, V_m = V_p, and the same fluid (same ρ and μ), Re_m = Re_p/20. The model test has only 5% of the full-scale Reynolds number — viscous effects are far more prominent in the model than in the prototype. To recover Re matching at reduced scale, engineers must compensate elsewhere: increasing fluid density (pressurized tunnel), using a denser gas, or increasing velocity. Same fluid at same speed does not give Re matching when size changes.
Question 3 True / False
Pressurizing a wind tunnel increases air density, which raises the Reynolds number at a given velocity and model size, partially compensating for the reduction in Reynolds number caused by scale reduction.
TTrue
FFalse
Answer: True
Re = ρVL/μ, so increasing ρ (by pressurizing the tunnel) directly increases Re at constant V and L. Dynamic viscosity μ is nearly independent of pressure at moderate pressures, so the ratio ρ/μ = 1/ν (kinematic viscosity) increases approximately in proportion to pressure. A tunnel pressurized to 5 atm gives Re roughly 5 times higher than the same tunnel at 1 atm, allowing closer Re matching with a smaller model. This is the principle behind pressurized wind tunnels like those at NASA Langley and the ETW in Europe.
Question 4 True / False
Once geometric similarity between a model and prototype is achieved, dynamic similarity follows automatically — you do not need to separately match any dimensionless numbers.
TTrue
FFalse
Answer: False
Geometric similarity — matching the shape — is necessary but far from sufficient for dynamic similarity. The flow physics depend on force ratios captured in dimensionless numbers: Re (inertia vs. viscosity), Fr (inertia vs. gravity), We (inertia vs. surface tension), Ma (inertia vs. compressibility). Two geometrically identical shapes at different scales can have completely different flow regimes (e.g., laminar vs. turbulent) if Re is not matched. Dynamic similarity requires matching all physically significant dimensionless groups, not just the geometry.
Question 5 Short Answer
Why is it generally impossible to simultaneously match both the Reynolds number and the Froude number when testing a ship model in water, and how do engineers address this limitation in practice?
Think about your answer, then reveal below.
Model answer: Fr = V/√(gL) requires V_m = V_p√(L_m/L_p). Re = VL/ν requires V_m = V_p(L_p/L_m)(ν_m/ν_p). For these to be equal simultaneously: √(L_m/L_p) = (L_p/L_m)(ν_m/ν_p), giving ν_m/ν_p = (L_m/L_p)^(3/2). For a 1:50 model, ν_m must be √50^3 ≈ 1/350 of water's kinematic viscosity — no practical fluid achieves this. Engineers resolve the conflict by matching Fr (which governs wave-making resistance, the dominant physical phenomenon) and using empirical friction correlations (the ITTC line) to separately estimate and correct for the mismatched viscous drag component.
This is the classic 'Froude-Reynolds dilemma' in naval architecture. The fundamental incompatibility arises because the two dimensionless numbers have different velocity scaling laws at fixed scale ratio and fluid. The engineering solution — matching the dominant physics and applying corrections for the rest — is the standard approach across all scale testing. Which number to prioritize depends on what drives the system: Fr for ships and open-channel hydraulics, Re for submerged bodies, Ma for supersonic aerodynamics.