The size distribution of a powder is measured by sedimentation in a vessel havin

Question

The size distribution of a powder is measured by sedimentation in a vessel having the sampling
point 180 mm below the liquid surface. If the viscosity of the liquid is 1.2 mNs/m2,
and the densities of the powder and liquid are 2650 and 1000 kg/m3 respectively, determine
the time which must elapse before any sample will exclude particles larger than
20 µm.
If Stokes’ law applies when the Reynolds number is less than 0.2, what is the approximate
maximum size of particle to which Stokes’ Law may be applied under these
conditions?

Explanation / Answer

The problem involves determining the time taken for a 20 µm particle to fall below the sampling point, that is 180 mm. Assuming that Stokes’ law is applicable, equation 3.88 may be used, taking the initial velocity as v = 0. Thus: y = (bt/a) - (b/a2)(1 - e-at ) where: b = g(1 - ?/?s) = 9.81[1 - (1000/2650)] = 6.108 m/s2 and: a = 18µ/d2?s = (18 × 1.2 × 10-3)/[(20 × 10-6)2 × 2650] = 20,377 s-1 In this case: y = 180 mm or 0.180 m Thus: 0.180 = (6.108/20,377)t - (6.108/20,3772)(1 - e-20,377t ) = 0.0003t + (1.4071 × 10-8e-20,377t ) Ignoring the exponential term as being negligible, then: t = (0.180/0.0003) = 600 s The velocity is given by differentiating equation 3.88 giving: ?y = (b/a)(1 - e-at ) When t = 600 s: ?y = [(6.108d2 × 2650)/(18 × 0.0012)]{1 - exp[-(18 × 0.0012 × 600)/d2 × 2650]} = 7.49 × 105d2[1 - exp(-4.89 × 10-3d -2)] For Re = 0.2, then d(7.49 × 105d2)[1 - exp(-4.89 × 10-3d -2)] × 2650/0.0012 = 0.2 or: 1.65 × 1012d3[1 - exp(-4.89 × 10-3d -2)] = 0.2 When d is small, the exponential term may be neglected and: d3 = 1.212 × 10-13 or: d = 5.46 × 10-5m or 54.6 µm

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