## 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 µ1=µ2 (the same fluid,

therefore the same viscosity):

*D²1(ρ1-ρL) = D²2(ρ2-ρL)*

*ρ1 = 2.65*

*ρ2 = 4.25*

*ρL = 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-ρL)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 µ1=µ2 (the same fluid,

therefore the same viscosity):

*D²1(ρ1-ρL) = D²2(ρ2-ρL)*

*ρ1 = 2.65*

*ρ2 = 4.25*

*ρL = 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*

*µ1 = 1 cP*

*µ2 = 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.*

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