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Settling velocity of silica particle 0.02 cm at 20°C

Short answer

Stokes' law gives 3.57 cm/s but Re=7.09. Transition-zone iteration converges to a settling velocity of about 2.63 cm/s (26.3 mm/s) at Re≈5.2.

Find the settling velocity of silica particle of size 0.02 cm with specific gravity 2.65 in water at 20°C. Take kinematic viscosity of water at 20°C as 1.007 centistokes.

Settling velocity of a discrete spherical particle is first tried with Stokes' law, and the Reynolds number is checked. If Re>1Re>1 the transition formula must be used.

Data

d=0.2d=0.2 mm =0.02=0.02 cm, Ss=2.65S_s=2.65, ν=0.01007\nu=0.01007 cm²/s (given: 1.007 centistokes), g=981g=981 cm/s².

Step 1: Stokes' law (trial)

vs=g(Ss−1)d218ν=981×1.65×0.02218×0.01007=3.572 cm/sv_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.02^2}{18\times0.01007}=3.572\ \text{cm/s} Re=vsdν=3.572×0.020.01007=7.09Re=\frac{v_s d}{\nu}=\frac{3.572\times0.02}{0.01007}=7.09

Since Re>1Re>1, Stokes' law is not valid and the flow is in the transition zone (1<Re<10001<Re<1000).

Step 2: Transition zone (Fair and Geyer)

Cd=24Re+3Re+0.34,vs=4g(Ss−1)d3CdC_d=\frac{24}{Re}+\frac{3}{\sqrt{Re}}+0.34,\qquad v_s=\sqrt{\frac{4g(S_s-1)d}{3C_d}}

Trial and error:

Trialvv assumed (cm/s)Re=vd/νRe=vd/\nuCdC_dnew vv (cm/s)
13.5727.094.8492.983
22.9835.935.6232.771
32.7715.505.9802.687
42.6875.346.1372.652
52.6525.276.2032.638
62.6385.246.2322.632

The values converge to vs=2.63v_s=2.63 cm/s, Re=5.2Re=5.2, Cd=6.25C_d=6.25.

Answer: settling velocity vs≈2.63v_s\approx2.63 cm/s (26.326.3 mm/s), with Re=5.2Re=5.2. (Stokes' law would overestimate it as 3.57 cm/s.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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