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Settling velocity of discrete particle 0.14 mm at 25°C

Short answer

Stokes' law gives 1.93 cm/s but Re=3.02, so transition formula is applied. After trial and error, settling velocity converges to 1.57 cm/s (15.7 mm/s) with Re=2.5.

Determine the settling velocity of a discrete particle having the diameter 0.14 mm; specific gravity 2.61 in water. The temperature of fluid is 25°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=0.14d=0.14 mm =0.014=0.014 cm, Ss=2.61S_s=2.61, ν=0.00893\nu=0.00893 cm²/s (at 25 °C, about 0.893 centistokes), g=981g=981 cm/s².

Step 1: Stokes' law (trial)

vs=g(Ss−1)d218ν=981×1.61×0.014218×0.00893=1.926 cm/sv_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.61\times0.014^2}{18\times0.00893}=1.926\ \text{cm/s} Re=vsdν=1.926×0.0140.00893=3.02Re=\frac{v_s d}{\nu}=\frac{1.926\times0.014}{0.00893}=3.02

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)
11.9263.0210.0151.716
21.7162.6911.0921.630
31.6302.5611.6061.594
41.5942.5011.8431.578
51.5782.4711.9501.571
61.5712.4611.9981.568

The values converge to vs=1.57v_s=1.57 cm/s, Re=2.5Re=2.5, Cd=12.04C_d=12.04.

Answer: settling velocity vs≈1.57v_s\approx1.57 cm/s (15.715.7 mm/s), with Re=2.5Re=2.5. (Stokes' law would overestimate it as 1.93 cm/s.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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