particles exit at the vortex. The D50 cut point of a solids separation device is defined as that particle size at which one-half of the weight of specific size particles go to the underflow and one-half of the weight go to the overflow. For example, a D30 cut point references a particle size which is 30% concentrated in the underflow and 70% in the overflow.
As stated earlier, the cut point is related to the inside diameter of the hydrocyclone. For example, a 12-inch cone has a D50 cut point for low-gravity solids in water of approximately 60 to 80 microns, a 6-inch cone around 30 to 60 microns, and a 4-inch cone around 15 to 20 microns (Table 7-4). However, the cut point will vary with the size and amount of solids in the feed, as well as fluid viscosity.
For comparative purposes, consider a 50-micron equivalent drilled solid diameter. Relatively speaking, the percent discharge is as follows:
If a graph of particle size versus percent of particles recovered to underflow is plotted, the portion of the curve near the D50, or 50%, recovery point (median cut point) is very steep when separations are efficient.
Particle separations in hydrocyclones vary considerably. In addition to proper feed head and the cone apex setting, drilling fluid properties including density, percent solids (and solids distribution) and viscosity, all affect separations. Any increase in these mud properties will increase the cut point of a separation device.
Stokes Law defines the relationship between parameters that control the settling velocity of particles in viscous liquids, not only in settling pits but also in equipment such as hydrocyclones and centrifuges.
Separations in a settling pit are controlled by the force of gravity and the viscosity of the suspending fluid (drilling mud). A large, heavy particle settles faster than a small, lighter particle. This settling process can be increased by reducing the viscosity of the suspending fluid, increasing the gravitational forces on the particles, or by increasing the effective particle(s) size with flocculation or coagulation.
Hydrocyclones and centrifuges increase settling rates by applying increased centrifugal force, which is equivalent to higher gravity force.
Stokes' Law for settling spherical particles in a viscous liquid is expressed as:
vs n where Vs = Settling or terminal velocity, feet/sec C = Units constant, 2.15 x 10"7 g = Acceleration (gravity or apparatus) ft/sec2
De = Particle equivalent diameter, microns ps = Specific gravity of solids (cutting, barite, etc.) p, = Specific gravity of liquid phase |i = Viscosity of media, centipoise
Various size particles with different densities can have the same settling rates. That is, there exists an equivalent diameter for every 2.65 specific gravity drilled solid, be it limestone, sand, or shale, which cannot be separated by gravimetric methods from barite particles of a corresponding equivalent diameter. Presently, it is not possible to separate desirable barite particles from undesirable drilled solid particles that settle at the same rate.
TABLE 7-4. Hydrocyclone Size versus D50 Cut Point
Cone Diameter (inches) Dso Cut Point in Water Dso Cut Point in Drilling Fluid
6 30-35 70-100
Generally, a barite particle (specific gravity = 4.25) will settle at the same rate as a drilled solids particle (specific gravity = 2.65) that is 1 \ times the barite particle's diameter. This may be verified by applying Stokes' Law.
Example #1. A viscosified seawater fluid with a specific gravity of 1.1, PV = 2.0 centipoise, and YP = 12.0 lbs/100 ft2, is circulated to clean out a cased wellbore. What size low-gravity solids will settle out with 5-micron barite particles? With 10-micron barite particles, what is the settling velocity in rig tanks?
Using Stokes' Law, the settling velocity is:
For equivalent settling rates, Vs = Vs. And for ji, = ji2 (the same fluid, therefore, the same viscosity).
(specific gravity = 2.65) in an 11.5 ppg mud with PV = 20 cp and YP = 12 lbs/100 ft2?
Using Stokes' Law, the settling velocity is:
For equivalent settling rates, Vs = Vs. And for ji, = 112 (the same fluid, therefore, the same viscosity).
(D,2 x 1.27) = (D22 x 2.87) D,2/D22 = 2.87/1.27 = 2.26 D/D2 = 1.50
(D,2 x 1.55) = (D22 x 3.15) Dj2/D22 = 3.15/1.55 = 2.03 D,/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.
Settling velocity for a 5-micron barite (or 7-micron drilled solid) particle is:
and for a 10-micron barite (or 14-micron drilled solid) particle:
Thus, a 10-micron barite particle will settle at the same rate as a 15-micron low-gravity particle, and a 50-micron barite particle will settle at the same rate as a 75-micron low-gravity particle.
Stokes' Law shows that as fluid viscosity and density increase, separation efficiency decreases.
If the drilling fluid weight is 14.0 pounds per gallon (specific gravity = 1.68):
d2ds = 4.25-1.68 = 2.57 = 2 65 d2b ~ 2.65-1.68 " 0.97 ~
Therefore, in drilling fluid weighing 14 pounds per gallon, 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-micron (or 81.4) drilled solid particle.
It is important to remember that the efficiency of a separator is viscosity dependent. The median cut, or D50 cut point, increases with viscosity as shown by Stokes' Law:
v _ (2.15 x 1Q-7) x 32.2 ft/sec2 x (4.25 - 1.1) x 100 5 ~ 16 cp
Example #2. What are the equivalent diameters of barite (specific gravity = 4.25) and drilled solids
Example #3. A 4-inch cone will separate half of the 12-micron low-gravity (specific gravity = 2.6) particles in water (that is, the Dso cut point is 12
microns). What is the D50 cut point in a 50-cp viscosity fluid of the same density?
For constant settling velocity, if fluid density is unchanged and other parameters remain constant:
Again, cut point performance can be further projected by dividing by the projected specific gravity at various viscosities. Thus, in the above example, for a 6-inch hydrocyclone, 20 centipoise viscosity, and 1.4 (11.7 ppg) density fluid, the D50 would be:
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