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Settling velocity of silica particle 0.05 cm at 10°C

Short answer

Stokes' law gives 24.7 cm/s but Re=135.7 (>1) invalidates it. Using the transition-zone drag equation iteratively (Fair and Geyer) yields a settling velocity of about 9.40 cm/s (94.0 mm/s).

Find the settling velocity of silica particle of specific gravity 2.65 at 10°C if the diameter of the particle is 0.05 cm. Assume kinematic viscosity at 10°C is 0.91 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=0.5d=0.5 mm =0.05=0.05 cm, Ss=2.65S_s=2.65, ν=0.0091\nu=0.0091 cm²/s (given: 0.91 centistokes), g=981g=981 cm/s².

Step 1: Stokes' law (trial)

vs=g(Ss−1)d218ν=981×1.65×0.05218×0.0091=24.705 cm/sv_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.05^2}{18\times0.0091}=24.705\ \text{cm/s} Re=vsdν=24.705×0.050.0091=135.74Re=\frac{v_s d}{\nu}=\frac{24.705\times0.05}{0.0091}=135.74

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)
124.705135.740.77411.805
211.80564.861.0829.984
39.98454.861.1839.553
49.55352.491.2119.438
59.43851.861.2199.407
69.40751.691.2229.399

The values converge to vs=9.40v_s=9.40 cm/s, Re=51.6Re=51.6, Cd=1.22C_d=1.22.

Answer: settling velocity vs≈9.40v_s\approx9.40 cm/s (94.094.0 mm/s), with Re=51.6Re=51.6. (Stokes' law would overestimate it as 24.70 cm/s.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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