The resistance force on a spherical particle settling in a fluid is given by Stokes’ Law. Obtain an expression for the terminal falling velocity of the particle. It is convenient to express experimental results in the form of a dimensionless group which may be plotted against a Reynolds group with

Question

The resistance force on a spherical particle settling in a fluid is given by Stokes’ Law.
Obtain an expression for the terminal falling velocity of the particle. It is convenient to express experimental results in the form of a dimensionless group which may be plotted against a Reynolds group with respect to the particle. Suggest a suitable form for this dimensionless group.
Force on particle from Stokes’ Law [latex]=3 \pi \mu d u[/latex]; where μ is the fluid viscosity, d is the particle diameter and u is the velocity of the particle relative to the fluid.

What will be the terminal falling velocity of a particle of diameter 10 μm and of density 1600 kg/m³ settling in a liquid of density 1000 kg/m³ and of viscosity 0.001 Ns/m²?
If Stokes’ Law applies for particle Reynolds numbers up to 0.2, what is the diameter of the largest particle whose behaviour is governed by Stokes’ Law for this solid and liquid?

Accepted Answer

The accelerating force due to gravity = mass of particle - mass of liquid displacedg.
For a particle of radius r, volume [latex]4 \pi r^3 / 3[/latex], or, in terms of diameter, d, volume = [latex]4 \pi\left(d^3 / 2^3 ight) / 3=\pi d^3 / 6[/latex]. Mass of particle [latex]=\pi d^3 ho_s / 6[/latex], where [latex] ho_s[/latex] is the density of the solid.
Mass of liquid displaced [latex]=\pi d^3 ho / 6[/latex], where ρ is the density of the liquid, and accelerating force due to gravity [latex]=\left(\pi d^3 ho_s / 6-\pi d^3 ho / 6 ight) g=\left(\pi d^3 / 6 ight)\left( ho_s- ho ight) g[/latex].

At steady state, that is when the terminal velocity is attained, the accelerating force due to gravity must equal the drag force on the particle F, or: [latex]\left(\pi d^3 / 6 ight)\left( ho_s- ho ight) g=3 \pi \mu d u_0[/latex] where [latex]u_0[/latex] is the terminal velocity of the particle.

Thus: [latex]\underline{\underline{u_0=\left(d^2 g / 18 \mu ight)\left( ho_s- ho ight)}}[/latex] (i)

It is assumed that the resistance per unit projected area of the particle, R′, is a function of particle diameter, d; liquid density, ρ; liquid viscosity, μ, and particle velocity, u or [latex] R^{\prime}=f(d, ho, \mu, u)[/latex]. The dimensions of each variable are [latex]R^{\prime}= M / L T T ^2, d= L , ho= M / L ^3, \mu= M / L T[/latex], and u = L/T. With 5 variables and 3 fundamental dimensions, there will be [latex](5-3)=2[/latex] dimensionless groups. Taking d, ρ and u as the recurring set, then:

[latex]d \equiv L , \quad L =d[/latex]

[latex] ho \equiv M / L ^3, \quad M = ho L ^3= ho d^3[/latex]

[latex]u \equiv L / T , \quad T = L / u=d / u[/latex]

Thus: dimensionless group 1: [latex]R^{\prime} L T T ^2 / M =R^{\prime} d\left(d^2 / u^2 ight) /\left( ho d^3 ight)=R^{\prime} / ho u^2[/latex]

dimensionless group 2: [latex]\mu L T / M =\mu d(d / u) /\left( ho d^3 ight)=\mu /(d u ho)[/latex]

and: [latex]R^{\prime} / ho u^2=f(\mu / d u ho)[/latex]

or: [latex]R^{\prime} / ho u^2=K(d u ho / \mu)^n=K R e^n[/latex] (ii)

[latex]\underline{ ext{In this way the experimental data should be plotted as the group (R/ρu²) against Re }}[/latex].

For this particular example, [latex]d=10 \mu m =\left(10 imes 10^{-6} ight)=10^{-5} m ; ho_s=1600 kg / m ^3; ho=1000 kg / m ^3[/latex] and [latex]\mu=0.001 Ns / m ^2[/latex].

Thus, in equation (i): [latex]u_0=\left(\left(10^{-5} ight)^2 imes 9.81 /(18 imes 0.001) ight)(1600-1000)[/latex]

[latex]=3.27 imes 10^{-5} m / s[/latex] or [latex]\underline{\underline{0.033 mm / s}}[/latex]

When Re [latex]R e=0.2, d u ho / \mu=0.2[/latex] or when the terminal velocity is reached:

[latex]d u_0=0.2 \mu / ho=(0.2 imes 0.001) / 1000=2 imes 10^{-7}[/latex]

or: [latex]u_0=\left(2 imes 10^{-7} ight) / d[/latex]

In equation (i): [latex]u_0=\left(d^2 g / 18 \mu ight)\left( ho_s- ho ight)[/latex]

[latex]\left(2 imes 10^{-7} ight) / d=\left(d^2 imes 9.81 /(18 imes 0.001) ight)(1600-1000)[/latex]

∴ [latex]d^3=6.12 imes 10^{-13}[/latex]

and: [latex]d=8.5 imes 10^{-5}[/latex] m or [latex]\underline{\underline{85 \mu m}}[/latex]

Question 1.P.17: The resistance force on a spherical particle settling in a f...

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