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Rayleigh's method, sphere terminal velocity

Short answer

Using Rayleigh's method, V = K da ρsf ρlb μc ge. Dimensional analysis yields V/√(gd) = φ( μ/(ρl d3/2 g1/2), ρs/ρl ), where the function φ is found from experiments.

Sphere of diameter d and density ρs\rho_s settles at a terminal velocity V in a liquid of density ρl\rho_l and dynamic viscosity μ\mu. Determine an expression of velocity in which velocity also depends on acceleration due to gravity g. Use Rayleigh's Method.

Rayleigh's method writes the dependent variable as a product of the independent variables, each raised to an unknown power. The powers are found by equating the dimensions on both sides.

Functional relation

V=K da ρs f ρl b μc geV = K\, d^{a}\,\rho_s^{\,f}\,\rho_l^{\,b}\,\mu^{c}\,g^{e}

Dimensions

V=LT−1V = LT^{-1}, d=Ld = L, ρ=ML−3\rho = ML^{-3}, μ=ML−1T−1\mu = ML^{-1}T^{-1}, g=LT−2g = LT^{-2}.

DimensionEquation
M0=f+b+c0 = f + b + c
L1=a−3f−3b−c+e1 = a - 3f - 3b - c + e
T−1=−c−2e-1 = -c - 2e

There are 5 unknown powers and only 3 equations, so express three of them in terms of cc and ff:

e=1−c2,b=−c−f,a=12−3c2e = \frac{1-c}{2},\qquad b = -c - f,\qquad a = \frac12 - \frac{3c}{2}

Substitute

V=K d 12−3c2 ρs f ρl −c−f μc g 1−c2V = K\, d^{\,\frac12-\frac{3c}{2}}\,\rho_s^{\,f}\,\rho_l^{\,-c-f}\,\mu^{c}\,g^{\,\frac{1-c}{2}}

Group terms with the same power:

V=Kgd[μρl d3/2g1/2]c[ρsρl]fV = K\sqrt{gd}\left[\frac{\mu}{\rho_l\, d^{3/2}g^{1/2}}\right]^{c}\left[\frac{\rho_s}{\rho_l}\right]^{f}

Answer:

Vgd=ϕ[μρl d3/2g1/2, ρsρl]\frac{V}{\sqrt{gd}} = \phi\left[\frac{\mu}{\rho_l\,d^{3/2}g^{1/2}},\ \frac{\rho_s}{\rho_l}\right]

This says the dimensionless velocity (Froude number) depends on a viscous term (a form of inverse Reynolds number) and on the density ratio. The functions of these groups are found by experiment.

✓ Answer checked by Claude Opus 5.5 (high effort)

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