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Settling velocity of discrete particle 0.17 mm at 21°C

Short answer

Stokes' law yields 2.66 cm/s, Re=4.62, requiring transition-zone correction. Iterative solution gives a settling velocity of approximately 2.06 cm/s (20.6 mm/s).

Determine the settling velocity of a discrete particle having the diameter 0.17 mm; specific gravity 2.65 in water. The temperature of fluid is 21°C.

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=0.17d=0.17 mm =0.017=0.017 cm, Ss=2.65S_s=2.65, ν=0.00978\nu=0.00978 cm²/s (at 21 °C, about 0.978 centistokes), g=981g=981 cm/s².

Step 1: Stokes' law (trial)

vs=g(Ss−1)d218ν=981×1.65×0.017218×0.00978=2.657 cm/sv_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.017^2}{18\times0.00978}=2.657\ \text{cm/s} Re=vsdν=2.657×0.0170.00978=4.62Re=\frac{v_s d}{\nu}=\frac{2.657\times0.017}{0.00978}=4.62

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)
12.6574.626.9322.301
22.3014.007.8422.163
32.1633.768.2702.106
42.1063.668.4632.082
52.0823.628.5482.072
62.0723.608.5852.067

The values converge to vs=2.06v_s=2.06 cm/s, Re=3.6Re=3.6, Cd=8.61C_d=8.61.

Answer: settling velocity vs≈2.06v_s\approx2.06 cm/s (20.620.6 mm/s), with Re=3.6Re=3.6. (Stokes' law would overestimate it as 2.66 cm/s.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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