# Three Examples About Stokes’ Law

## Example 1

A viscosified seawater fluid (SG 1.1, plastic viscosity (PV) 2.0 centipoise
(cP), and yield point (YP) 12.0 lb/100 sq ft) is circulated to clean out
a cased well bore. What size of low-gravity solids will settle out with
5-micron barite particles? What is the settling velocity of 10-micron
barite particles in rig tanks? From Stokes’ law, settling velocity is:

Vs=Cg(ρs-ρL)D²E/µ

For equivalent settling rates, Vs=Vs0. And for µ1=µ2 (the same fluid,
therefore the same viscosity):

D²1(ρ1-ρL) = D²2(ρ2-ρL)
ρ1 = 2.65
ρ2 = 4.25
ρL = 1.1
(D²1×1.55) = (D²2×3.15)
D²1⁄D²2=3.15⁄1.55= 2.03
D1⁄D2=1.42

Thus a 5-micron barite particle will settle at the same rate as a 7-micron
low-gravity particle, and a 10-micron barite particle will settle at the
same rate as a 14-micron low-gravity particle.

Again, Stokes’ law applied to settling velocity is:

Vs=Cg(ρs-ρL)D²E⁄µ

Settling velocity, for 5 micron barite (or 7 micron drilled solid) particle is: ## Example 2

What are the equivalent diameters of barite (SG 4.25) and drilled solids
(SG 2.65) in an 11.5-ppg drilling fluid with PV = 20 cP and YP = 12 lb/
100 sq ft?

From Stokes’ law, settling velocity is:

Vs=Cg(ρs-ρL)D²E/µ

For equivalent settling rates, Vs=Vs0. And for µ1=µ2 (the same fluid,
therefore the same viscosity):

D²1(ρ1-ρL) = D²2(ρ2-ρL)
ρ1 = 2.65
ρ2 = 4.25
ρL = 1.38
(D²1×1.27) = (D²2×2.87)
D²1⁄D²2=2.87⁄1.27= 2.26
D1⁄D2=1.5

Thus a 10-micron barite particle will settle at the same rate as a
15-micron low-gravity particle; a 50-micron barite particle will settle at
the same rate as a 75-micron low-gravity particle, and so forth.
Stokes’ law shows that as fluid viscosity and density increase,
separation efficiency decreases. If the drilling-fluid weight is 14.0 ppg
(SG 1.68), In drilling fluid weighing 14 ppg, a 10-micron barite particle will settle at
the same rate as a 16-micron drilled-solid particle, and a 50-micron
barite particle will settle at the same rate as an 80 (or 81.4)-micron drilledsolid particle.

Frequently disregarded is the fact that the efficiency of a separator is
viscosity dependent. The D50 cut point increases with viscosity, as shown
by Stokes’ law:

Vs=Cg(ρs-ρL)D²E/µ

## Example 3

A 4-inch cone will separate half of the 12-micron low-gravity (SG 2.6)
particles in water, (that is, the D50 cut point is 12 microns). What is the
D50 cut point in a fluid of 50 cP viscosity of the same density?
For constant settling velocity (if fluid density is unchanged) and other
constant parameters:

For ρ1−ρ2

D²1⁄µ1=D²2⁄µ2
D1 = 12 micron
µ1 = 1 cP
µ2 = 50 cP
D²2 = 7200
D2 = 84.8 microns

Thus, if fluid SG remains 1.1, and viscosity is 50 cP, the D50 is raised to
85 microns. Similarly, for a 4-inch hydrocyclone:

 Fluid Viscosity, cP D50, microns (SG 2.6) 1.0 12.0 10.0 37.9 20.0 53.7 30.0 65.8 40.0 75.8 50.0 84.8

Cut-point performance can be further projected by dividing by the projected SG at various viscosities. Thus, in terms of the preceding example, for 20-cP viscosity and 1.4 (11.7 ppg)-density fluid, the D50 would be: 53.7 microns× 1.1⁄1.4 = 42.1 microns.

The D50 cut points for 6-inch and 8-inch hydrocyclones (common desilter and desander sizes) are usually given as 25 microns and 60 microns.

The variation of D50 cut points with viscosity for a 6-inch desilter
hydrocyclone (SG 1.1 fluid) can be seen to be:

 Fluid Viscosity, cP D50, microns (SG 2.6) 1.0 25.0 10.0 79.0 20.0 112 30.0 137 40.0 158 50.0 176

Again, cut-point performance can be further projected by dividing by
the projected SG at various viscosities. Thus, in the preceding situation,
for a 4-inch hydrocyclone, with 20-cP viscosity and 1.4 (11.7 ppg)-
density fluid, the D50 would be:

67.1 microns ×1.1⁄1.4= 52:7 microns.