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Model liquid viscosity, Froude and Reynolds

Short answer

For equal Froude and Reynolds numbers, Vr = √Lr, so νr = Lr3/2 = (1/5)1.5 = 0.0894. The required model kinematic viscosity is νm = 4.645×10−5 × 0.0894 ≈ 4.16×10−6 m²/s.

Oil of kinematic viscosity 4.645×10−54.645\times10^{-5} m²/s is to be used in a prototype in which both viscous and gravity force dominate. A model scale of 1:5 is also desired. What viscosity of model liquid is necessary to make both the Froude number and the Reynolds number same in model and prototype?

If both gravity and viscous forces are to be modelled, the Froude and Reynolds numbers must both be equal in model and prototype. The liquid in the model must then have a different viscosity.

Froude number equal

VmgLm=VpgLp  ⇒  Vr=Lr\frac{V_m}{\sqrt{g L_m}} = \frac{V_p}{\sqrt{g L_p}} \;\Rightarrow\; V_r = \sqrt{L_r}

(the same gg for both).

Reynolds number equal

VmLmνm=VpLpνp  ⇒  νr=VrLr\frac{V_mL_m}{\nu_m} = \frac{V_pL_p}{\nu_p} \;\Rightarrow\; \nu_r = V_rL_r

Combine

νr=Lr1/2Lr=Lr3/2=(15)1.5=0.0894\nu_r = L_r^{1/2}L_r = L_r^{3/2} = \left(\frac15\right)^{1.5} = 0.0894 νm=νp×0.0894=4.645×10−5×0.0894=4.155×10−6 m2/s\nu_m = \nu_p\times 0.0894 = 4.645\times10^{-5}\times0.0894 = 4.155\times10^{-6}\ \text{m}^2/\text{s}

Answer: the model liquid must have a kinematic viscosity of νm≈4.15×10−6\nu_m \approx 4.15\times10^{-6} m²/s (about 0.0415 stokes). The dynamic viscosity then follows as μm=ρmνm\mu_m = \rho_m\nu_m for the chosen liquid; if its density equals that of the oil, μm=0.0894 μp\mu_m = 0.0894\,\mu_p.

✓ Answer checked by Claude Opus 5.5 (high effort)

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