A mud system during drilling and circulation is undergoing continual change. Because of the change, the mud must be treated to maintain its desired properties. In order to make the most eco-nomical treatments, it is important to know the current component concentrations, and based on assumed or calculated losses, be able to calculate the amounts of materials required to maintain the system.
Utilizing material balance techniques, it is possible to estimate the quantities of materials needed to maintain a mud system. Assumptions and known information will be utilized in developing the values for the material balance problems. Known information includes hole size, penetration rate and nature of the solids drilled. Assumed data includes losses to the formation through seepage and filtrate, hole washout, and losses through solids control equipment. With this information, esti-mates of the daily chemical consumption rates can be calculated.
For the purpose of this exercise, assume the following:
Hole size: 8.5 in.
Washout: 10%
Penetration rate: 35 ft/hr.
Total circulating volume: 1000 bbls
Losses due to solids control equipment: 15 bbl/hr. (1.5%/hr.) Losses due to filtration and seepage: 10 bbl/hr. (1%/hr.) Losses due to adsorption on cuttings: 1 bbl/hr. (0.01%/hr.) The last analysis of the drilling fluid yielded the following data:
Mud weight: 12.0 ppg (1.44 SG) Solids: 18.0% by volume Water: 82.0% by volume
Determine the Average Specific Gravity of Solids (ASG):
, where V1 = Volume fraction of water (82%)
D1= Density of water (8.33)
V2 = Volume Fraction of solids (18.0%) D2 = Density of solids (unknown) VF = Volume fraction of mud (100%)
Determine the concentrations of low gravity and high gravity solids in pounds per barrel. Assume a specific gravity of 2.65 and 4.20 for low gravity solids and barite respectively.
V1 = Volume fraction of all solids (18%)
D1= Average specific gravity of all solids (3.45) V2 = Volume Fraction of high gravity solids (unknown) D2 = Specific gravity of high gravity solids (4.20) V3 = Volume fraction of low gravity solids = V1 - V2 D3 = Specific gravity of low gravity solids (2.65) Substituting:
Therefore:
Pound/bbl barite: 4.2 (8.33) (42) (0.0929) = 136.51
Pound/bbl low gravity solids: 2.65 (8.33) (42) (0.0871) = 80.75 The composition of the mud is as follows:
Mud weight: 12.0 ppg Solids: 18.0% by volume Low gravity solids: 80.75 ppb Barite: 136.51 ppb
Average specific gravity of solids: 3.45
The next step is to calculate the daily maintenance requirements. The requirements will be related to the additions of water required to maintain the mud properties within optimum limits. Water addi-tions will be regulated by hole size, rate of penetration, nature of solids drilled and removal of sol-ids. As solids accumulate in the fluid, mud weight, plastic viscosity and yield point may increase, and the rate of increase will be governed by the degree of hydration, swelling and dispersion of the solids which is controlled by the chemical environment of the mud.
It is necessary to known the amount of solids generated, therefore, calculate the amount of solids drilled per hour using the data supplied earlier.
Average hole size: Bit size + Washout
Average hole size: 8.5 in. + (8.5) (0.1) = 9.35 in.
Volume of hole/ft: (D)2 (0.000971)
Volume of hole/ft: (9.35)2 (0.000971) = 0.0848 or 0.085 bbl/ft Volume of hole/hr.: (0.085) (35) = 2.975 bbl/hr.
Therefore:
The volume of drilled solids generated each hour is 2.975 bbl. Assuming a specific gravity of 2.65 for the solids, pounds per hour of drilled solids will be:
Low gravity solids, bbl/hr.: (350) (2.65) (2.975) = 2,759 lb
Note: The figure 350 used above denotes the weight of 1 bbl of fresh water (42 gal/bbl x 8.33 ppg = 349.86, or 350 ppb).
We now need to calculate the rate at which water must be added to maintain a constant mud weight. If the operating efficiency of the solids control equipment is known, those efficiency values will be used. If the operating efficiency is not known, assumptions must be made. This exercise assumes an operating efficiency of 50%.
The material balance equation used to calculate the amount of water required is as follows:
, where VF = Volume of mud (V1 + V2 + V3)
DF = Final specific gravity (1.44)
V1 = Volume of low gravity solids, VN = (0.50) (2.975) + (0.0871) = 1.5746 D1= Specific gravity of low gravity solids (2.65)
V2 = Volume of water to add (unknown) D2 = Specific gravity of water (1.00) V3 = Volume of barite (0.0929) D3 = Specific gravity of barite (4.20) V4 = Volume of water (0.82)
D4 = 1.00 and,
Substituting:
Therefore, 4.08 bbl of water/hour must be added to maintain a constant mud weight of 12.0 ppg. At the same time, low gravity solids are being incorporated into the mud at the rate of 1,380 lb/hr.
(remember the solids control equipment is operating at 50% efficiency - 2759 x 0.5 = 1380). This is equivalent to 1.49 bbl/hr. (2.975 x 0.5 = 1.4875). At these rates, the density of the mud will decrease due to the addition of the water and the mud properties will deteriorate due to the accu-mulation of low gravity drilled solids. Additional water and barite must be added to maintain the desired properties and a 3.45 average specific gravity of solids.
Calculate barite requirements to maintain a 3.45 ASG of solids using:
VFDF = V1D1 = V2D2+V3D3
, where VF = Final volume of solids (V1 + V2 + V3)
DF = Final ASG of solids (3.45)
V1 = Volume of low gravity solids (1.57) D1 = Specific gravity of low gravity solids (2.65) V2 = Volume of high gravity solids (0.0929) D2 = Specific gravity of high gravity solids (4.20) V3 = Volume of high gravity solids to add (unknown)
D3 = Specific gravity of high gravity solids to add (4.20) and,
Substituting:
Converting to pounds of barite per hour:
(1.58 bbl) (1,470 lb/bbl) = 2323 lb/hr.
Next the amount of additional water must be calculated. We know that 1380 lb of low gravity solids and 2323 lb of barite are being introduced into the system each hour.
Determine the amount of water required to maintain a 12.0 ppg mud weight using:
, where
VF = Final mud volume (V1 + V2 + V3) DF = Specific gravity of final mud (1.44) V1 = Volume of solids added (3.07 bbl) D1 = ASG of solids added (3.45) V2 = Volume of water added (4.08 bbl)
D2 = Specific gravity of water added (4.08 bbl) V3 = Volume of water to be added (unknown)
Therefore, the required additions of barite and water must be made to the system to maintain a constant mud weight of 12.0 ppg and an ASG of 3.45 are:
This can be validated using the formula:
, where
V1 = Volume of initial mud (1000 bbl) D1 = Specific gravity of initial mud (1.44) V2 = Volume of water added (14.02 bbl) D2 = Specific gravity of water (1.0) V3 = Volume of barite added (1.58 bbl) D3 = Specific gravity of barite (4.2)
V4 = Volume of low gravity solids added (1.49 bbl) D4 = Specific gravity of low gravity solids (2.65) VF = V1 + V2 + V3 +V4
DF = Specific gravity of final mud (unknown) Substituting:
DF = 1.44 specific gravity of final mud
(This calculation ignores the volume and density affects of other materials such as bentonite and thinners required to maintain and control the muds properties.)
Next, determine the amount of new mud which must be mixed and added to the system to com-pensate for:
Losses to filtration and seepage: 10 bbl/hr.
Losses through solids control equipment and adsorption: 16 bbl/hr.
The total volume lost per hour which must be replaced is 29 bbl less the 17 bbl of solids and water added to the system while drilling and maintaining a mud weight of 12.0 ppg and an ASG of 3.45.
bbl/hr. lb/hr.
Calculating the amount of barite and water required to build 12 bbls of 12.0 ppg mud per hour:
, where
VF = Final volume (12 bbl)
DF = Specific gravity of final volume (1.44) V1 = Volume of water (VF - V2)
D1 = Specific gravity of water (1.00) V2 = Volume of barite (unknown)) D2 = Specific gravity of barite (4.20) Volume of Barite:
The last step is the determination of the amounts of other materials required to maintain the desired properties for the system. This exercise assumes a fresh water mud having the following composition:
The hourly additions of barite and water were calculated to be:
Barite: 4749 lb/hr.
The consumption of the other materials will be based on the addition of 29 bbl of new mud volume per hour at the concentrations stated above.
Converting the above to sacks:
For the purpose of this exercise, the assumption was made that the solids contributed by bentonite and barite which are not lost from a system due to filtration or seepage were deposited as wall cake, and therefore removed from the system.
It must be emphasized that actual consumption rates will vary greatly depending on a number of different factors such as:
Nature of formation being drilled Efficiency of solids control equipment Composition of the drilling fluid
Physical and chemical properties of the drilling fluid Hole size and penetration rates
Barite Reclamation
A decanting centrifuge will normally process 12.0 ppg mud at the rate of 18 gpm and recover approximately 70% of the barite in the mud processed. Using this assumption, the amount of bar-ite recoverable from the system can be estimated.
The amount of barite contained in this 12.0 ppg mud was calculated to be 136.51 ppb. The centri-fuge will process 25.7 bbls of mud per hour.
(18 gpm) (60 min/hr) (0.0238 bbl/gal) = 25.70 bbl/hr
At 70% efficiency, the amount of barite recovered = (136.51 ppb) (25.70 bbl/hr) (0.70) = 2456 lb/hr.
Converting to barrels of barite/hr:
Determining the amount of water, low gravity solids and chemicals discarded:
Chemical lb/hr. lb/8 hr. tour lb/12 hr. tour lb/day
Barite 4749 37,992 56,988 113,976
Lignosulfonate 232 1856 2,784 5568
Lignite 116 928 1392 2.784
Bentonite 580 4460 6,960 13,920
Caustic Soda 72.5 580 870 1740
Chemical sk/8 hr tour sk/12 hr tour sk/day
Barite 380 570 1,140
---1470 = 1.67 bbl/hr of barite recovered
25.70 - 1.67 = 24.0 bbl/hr discarded
The amount of additional water required to maintain a constant mud weight of 12.0 ppg due to the barite being returned to the system via the centrifuge must also be calculated using:
, where
VF = Final volume of barite and water (V1 + V2) DF = Specific gravity of final volume (1.44) V1 = Volume of barite (1.67)
D1 = Specific gravity of barite (4.2) V2 = Volume of water to add (unknown) D2 = Specific gravity of water (1.00)
V2 = 10.48
Volume of water to add to maintain a mud density of 12.0 ppg is 10.48 bbl/hr
Total volume returned to system while centrifuge is running is equal to the volume of barite returned plus the water added to the system.
Volume returned = 10.48 + 1.67 = 12.15 bbl/hr
The centrifuge is processing 25.70 bbl of mud per hour and 12.15 bbl is being returned to the sys-tem. New volume must be added to the system equal to the difference between the processed vol-ume and the returned volvol-ume.
New volume = 25.70 - 12.15 = 13.55 bbl/hr Barite and water additions to build 13.55 bbl of new mud per hour.
, where
VF = Final volume of barite and water (13.55) DF = Density of final volume (1.44)
V1 = Volume of water (VF - V2) D1 = Specific gravity of water (1.0) V2 = Volume of barite (unknown) D2 = Specific gravity of barite (4.2)
V2 = 1.86
The additions of barite and water that must be added to the system each hour.
Other materials must again be added to maintain the desired properties. The total volume, 25.70 bbl/hr, processed by the centrifuge and replaced by discharged barite and added water will need to be treated.