Increase Mud Density
Mud weight, ppg, increase with barite (average specific gravity of barite—4.2)
Barite, sk/100 bbl =
Example: Determine the number of sacks of barite required to increase the density of 100bbl of 12.0ppg (W,) mud to 14.0ppg (W2):
Barite, sk/1 OObbl =
21.0
Barite = 140 sk/100 bbl
Metric calculation
Barite, kg/m
3 _ (kill fluid density, kg/1 - original fluid density, kg/1) x 4.2
4.2 - kill fluid density, kg/1
iVill flnirl Hf^nci
Barite, kg/m
Barite, kg/m3 = original fluid density kg/mj
4200 - kill fluid density, kg/m3
Volume increase, bbl, due to mud weight increase with barite
Volume increase, per 100 bbl
Example: Determine the volume increase when increasing the density from 12.0ppg (W,) to 14.0ppg (W2):
Volume increase, per 100 bbl =-
Volume increase = 9.52bbl per lOObbl
Starting volume, bbl, of original mud weight required to yield a predetermined final volume of desired mud weight with barite
Example: Determine the starting volume, bbl, of 12.0ppg (W,) mud required to achieve 100 bbl (VF) of 14.0 ppg (W2) mud with barite:
Starting volume, bbl =
Starting volume, bbl =
Starting volume = 91.3 bbl
Mud weight increase with calcium carbonate (SG—2.7)
NOTE: The maximum practical mud weight attainable with calcium carbonate is 14.0 ppg.
Sacks/100 bbl =
Example: Determine the number of sacks of calcium carbonate/100 bbl required to increase the density from 12.0ppg (W,) to 13.0ppg (W2):
Sacks/100 bbl = 99.5
Volume increase, bbl, due to mud weight increase with calcium carbonate
Example; Determine the volume increase, bbl/100 bbl, when increasing the density from 12.0ppg (W,) to 13.0ppg (W2):
Volume increase, per 100 bbl =-
Volume increase = 10.53 bbl per 100bbl
Starting volume, bbl, of original mud weight required to yield a predetermined final volume of desired mud weight with calcium carbonate o • , u , VF 22.5 - W,
Example: Determine the starting volume, bbl, of 12.0ppg (W,) mud required to achieve 100bbl (VF) of 13.0ppg (W2) mud with calcium carbonate:
Starting volume, bbl =-
10.5
Starting volume = 90.5 bbl
Mud weight increase with hematite (SG—4.8)
Example: Determine the hematite, sk/100 bbl, required to increase the density of 100bbl of 12.0ppg (W,) to 14.0ppg (W2):
Hematite, _ 3360 sk/100 bbl 26
Hematite = 129.2sk/100bbl
Volume increase, bbl, due to mud weight increase with hematite
Volume increase, = 100(W2 - Wt) per 100 bbl 40 - W2
Example: Determine the volume increase, bbl/100 bbl, when increasing the density from 12.0ppg (W,) to 14.0ppg (W2):
Volume increase, = 100(14.0 - 12.0) per 100 bbl 40-14.0
Volume increase, _ 200 per 100 bbl ~ 26
Volume increase = 7.7 bbl per 100 bbl
Starting volume, bbl, of original mud weight required to yield a predetermined final volume of desired mud weight with hematite
Example: Determine the starting volume, bbl, of 12.0ppg (W,) mud required to achieve lOObbl (VF) of 14.0ppg (W2) mud with hematite:
Starting volume, bbl =-
vu, 2600
Starting volume, bbl = ^ Starting volume = 92.9 bbl
Dilution
Mud weight reduction with water
W2-Dw
Example: Determine the number of barrels of water weighing 8.33 ppg (Dw) required to reduce lOObbl (V,) of 14.0ppg (W,) to 12.0 ppg (W,):
Water, bbl
3.67
Water = 54.5 bbl Mud weight reduction with diesel oil
Diesel, bbl
Example: Determine the number of barrels of diesel weighing 7.0ppg (Dw) required to reduce 100bbl (V,) of 14.0ppg (W,) mud to 12.0 ppg (W2):
Diesel = 40 bbl
where Vi = volume of fluid 1 (bbl, gal, etc.) D, = density of fluid 1 (ppg, lb/ft3, etc.) V2 = volume of fluid 2 (bbl, gal, etc.) D2 = density of fluid 2 (ppg, lb/ft3, etc.) VF = volume of final fluid mix Dh = density of final fluid mix
Example 1: A limit is placed on the desired volume:
Determine the volume of 11.0 ppg mud and 14.0 ppg mud required to build 300 bbl of 11.5 ppg mud:
Given: 400 bbl of 11.0 ppg mud on hand, and 400 bbl of 14.0 ppg mud on hand
Solution: let V, = bbl of ll.Oppg mud V2 = bbl of 14.0 ppg mud then a) V, + V2 = 300 bbl b) (11.0) V, + (14.0) V2 = (11.5) (300)
Multiply Equation A by the density of the lowest mud weight (D, = ll.Oppg) and subtract the result from Equation B:
b) (11.0XV,) + (14.0)(V2) = 3450 -a) (11.0)^) + (11.0)(V2) = 3300
V2 = 50 bbl of 14.0 ppg mud V, + V2 = 300 bbl V, = 300 - 50 V, = 250 bbl of 11.0 ppg mud
Therefore:
Df = final density, ppg
(50X14.0) + (250)(11.0) = 300Df 700 + 2750 = 300Df 3450 = 300Df 3450 300 = Df 11.5 ppg = Dp
Example 2: No limit is placed on volume:
Determine the density and volume when the following two muds are mixed together:
Given: 400bbl of ll.Oppg mud, and 400bbl of 14.0ppg mud
Solution: let V, = bbl of ll.Oppg mud
D] = density of 11.0 ppg mud V2 = bbl of 14.0ppg mud D2 = density of 14.0 ppg mud VF = final volume, bbl Dp = final density, ppg
(400)(11.0) + (400)(14.0) = 800 Df 400 + 5600 = 800 DF 10,000 = 800 Dp 10,000- 800 = Dp 12.5ppg = DF
Therefore: final volume = 800 bbl final density = 12.5 ppg
Oil-Based Mud Calculations
Density of oil/water mixture being used
Example: If the oil/water (o/w) ratio is 75/25 (75% oil, Vh and 25% water, V2), the following material balance is set up:
NOTE: The weight of diesel oil, D, = 7.0ppg The weight of water, D2 = 8.33 ppg
(0.75X7.0) + (0.25)(8.33) = (0.75 + 0.25) DF 5.25 + 2.0825 = 1.0 DF 7.33 = DF
Therefore: The density of the oil/water mixture = 7.33ppg
Starting volume of liquid (oil plus water) required to prepare a desired volume of mud
Wi = initial density of oil/water mixture, ppg W2 = desired density, ppg DV = desired volume, bbl
Example: Wi = 7.33 ppg (o/w ratio - 75/25) W2 = 16.0 ppg Dv = 100 bbl
Oil/water ratio from retort data
Obtain the percent-by-volume oil and percent-by-volume water from retort analysis or mud still analysis. Using the data obtained, the oil/water ratio is calculated as follows:
liquid phase % by vol oil + % by vol water b) % water in __% by vol water_^
liquid phase % by vol oil + % by vol water c) Result: The oil/water ratio is reported as the percent oil and the percent water.
Example: Retort annalysis: % by volume oil =51
51 x 17
% water in liquid phase = 25
c) Result: Therefore, the oil/water ratio is reported as 75/25:
Changing oil/water ratio
NOTE: If the oil/water ratio is to be increased, add oil; if it is to be decreased, add water.
Retort analysis: % by volume oil =51 %j by volume water = 17 % by volume solids = 32
The oil/water ratio is 75/25.
Example I: Increase the oil/water ratio to 80/20:
In 100bbl of this mud, there are 68 bbl of liquid (oil plus water). To increase the oil/water ratio, add oil. The total liquid volume will be increased by the volume of the oil added, but the water volume will not change. The 17 bbl of water now in the mud represents 25°/) of the liquid volume, but it will represent only 20% of the new liquid volume.
Therefore: let x = final liquid volume then, 0.20x = 17
The new liquid volume = 85 bbl
Barrels of oil to be added:
Oil, bbl = new liquid vol - original liquid vol Oil, bbl = 85-68
Oil = 17 bbl oil per 100 bbl of mud
liquid phase original liquid oil + new oil added
liquid phase 68 + 17
Therefore: The new oil/water ratio would be 80/20.
Example 2: Change the oil/water ratio to 70/30:
As in Example 1, there are 68 bbl of liquid in 100 bbl of this mud. In this case, however, water will be added and the volume of oil will remain constant. The 51 bbl of oil represents 75% of the original liquid volume and 70% of the final volume:
Therefore: let x = final liquid volume then, 0.70x = 51
The new liquid volume = 73 bbl
Barrels of water to be added:
Water, bbl - new liquid vol - original liquid vol Water, bbl = 73-68
Water = 5 bbl of water per 100 bbl of mud
Therefore, the new oil/water ratio would be 70/30.
Basic solids analysis calculations
NOTE: Steps 1-4 are performed on high salt content muds. For low chloride muds begin with Step 5.
Step 1
Percent-by-volume saltwater (SW)
SW = (5.88 x 10"8) x [(ppm CI)12 + l] x % by vol water
Step 2
Percent-by-volume suspended solids (SS) SS = 100 - % by vol oil - % by vol SW
Step 3
Average specific gravity of saltwater (ASGsw) ASGsw = (ppm CI)095 x (1.94 x 10"6) + 1
92 Formulas and Calculations Step 4
Average specific gravity of solids (ASG)
ASG = (12 x MW) - (% by vol SW x ASGsw) - (0.84 x % by vol oil)
Step 5
Average specific gravity of solids (ASG)
% by vol solids
Step 6
Percent-by-volume low gravity solids (LGS)
Step 7
Percent-by-volume barite
Step 8
Pounds-per-barrel barite Barite, lb/bbl = % by vol barite x 14.71
Step 9
Bentonite determination
If cation exchange capacity (CEC)/methylene blue test (MBT) of shale and mud are KNOWN:
a) Bentonite, lb/bbl:
where S = CEC of shale M = CEC of mud b) Bentonite, % by volume:
If the cation exchange capacity (CEC)/methylene blue (MBT) of SHALE is UNKNOWN:
where M = CEC of mud b) Bentonite, lb/bbl = bentonite, % by vol x 9.1
Step 10
Drilled solids, % by volume Drilled solids, % by vol = LGS, % by vol - bent, % by vol
Step 11
Drilled solids, lb/bbl Drilled solids, lb/bbl = drilled solids, % by vol x 9.1
Example: Mud weight = 16.0 ppg
Chlorides = 73,OOOppm
CEC of shale = 7 lb/bbl Rector Analysis:
water = 57.0% by volume oil = 7.5% by volume solids = 35.5% by volume
1. Percent by volume saltwater (SW)
SW = [(5.88—8 x 685468.39) + 1] x 57 SW = (0.0403055 + 1) x 57 SW = 59.2974 percent by volume
2. Percent by volume suspended solids (SS)
SS = 100 - 7.5 - 59.2974 SS = 33.2026 percent by volume
3. Average specific gravity of saltwater (ASGsw)
ASGsw = (41,701.984 x 1.94"6) +1 ASGsw = 0.0809018 + 1 ASGsw = 1.0809
4. Average specific gravity of solids (ASG) (12 x 16) - (59.2974 x 1.0809) - (0.84 x 7.5)
33.2026
LGS = 11.154 percent by volume
7. Percent by volume barite
Barite, % by volume = 33.2026 - 11.154 Barite = 22.0486 % by volume
8. Barite, lb/bbl
Barite, lb/bbl = 22.0486 x 14.71 Barite = 324.3349 lb/bbl
9. Bentonite determination a) lb/bbl =-x (30 - 9 x —) x 11.154
-a1 6sJ
lb/bbl = 1.1206897 x 2.2615385 x 11.154 Bent = 28.26965 Ib/bb b) Bentonite, % by volume
Bent, % by vol = 28.2696 + 9.1 Bent = 3.10655% by vol
10. Drilled solids, percent by volume
Drilled solids, % by vol = 11.154 - 3.10655 Drilled solids = 8.047% by vol
11. Drilled solids, pounds per barrel
Drilled solids, lb/bbl = 8.047 x 9.1 Drilled solids = 73.2277lb/bbl
_ __Solids Fractions
Maximum recommended solids fractions (SF)
Maximum recommended low gravity solids (LGS)
x 200
where SF = maximum recommended solids fractions, % by vol MW = mud weight, ppg
LGS = maximum recommended low gravity solids, % by vol
Example: Mud weight = 14.0 ppg
Determine: Maximum recommended solids, % by volume Low gravity solids fraction, % by volume
Maximum recommended solids fractions (SF), % by volume:
SF = (2.917 x 14.0) - 14.17 SF = 40.838 - 14.17 SF = 26.67% by volume
Low gravity solids (LGS), % by volume:
Low gravity solids (LGS), % by volume:
x 200
x 200
LGS = 0.2667 - (0.3125 x 0.6807) x 200 LGS = (0.2667 - 0.2127) x 200 LGS = 0.054 x 200 LGS = 10.8% by volume
Vm (Fct - Fcop) Fcop - Fca where Vwm = barrels of dilution water or mud required Vm = barrels of mud in circulating system Fct = percent low gravity solids in system Fcop = percent total optimum low gravity solids desired Fca = percent low gravity solids (bentonite and/or chemicals
Example: lOOObbl of mud in system. Total LGS = 6%. Reduce solids to
If dilution is done with a 2% bentonite slurry, the total would be:
added)
4 Dilute with water:
Vwm = lOOObbl
Displacement—Barrels of Water/Slurry Required__
Fct - Fca where Vwm = barrels of mud to be jetted and water or slurry to be added to maintain constant circulating volume:
Example: lOOObbl in mud system. Total LGS = 6%. Reduce solids to 4%:
2000
If displacement is done by adding 2% bentonite slurry, the total volume would be:
6-2 2000
Evaluation of Hydrocyclone
Determine the mass of solids (for an unweighted mud) and the volume of water discarded by one cone of a hydrocyclone (desander or desilter):
Volume fraction of solids (SF):
13.37
Mass rate of solids (MS):
where SF = fraction percentage of solids
MW = average density of discarded mud, ppg MS = mass rate of solids removed by one cone of a hydrocyclone, lb/hr
V = volume of slurry sample collected, quarts T = time to collect slurry sample, seconds WR = volume of water ejected by one cone of a hydrocyclone, gal/hr
Example: Average weight of slurry sample collected = 16.0 ppg Sample collected in 45 seconds Volume of slurry sample collected = 2 quarts a) Volume fraction of solids:
13.37
b) Mass rate of solids:
MS = 11,204.36 x 0.0444 MS = 497.97 lb/hr c) Volume rate of water:
Evaluation of Centrifuge a) Underflow mud volume:
b) Fraction of old mud in underflow:
c) Mass rate of clay:
d) Mass rate of additives:
e) Water flow rate into mixing pit:
[QM x (35 - MW)] - [QU x (35 - PU)] = - (0.6129 x QC) - (0.6129 x QD) 35 - PW
f) Mass rate for API barite:
where MW = mud density into centrifuge, ppg
QM = mud volume into centrifuge, gal/min PW = dilution water density, ppg QW = dilution water volume, gal/min PU = underflow mud density, ppg PO = overflow mud density, ppg CC = clay content in mud, lb/bbl CD = additive content in mud, lb/bbl QU = underflow mud volume, gal/min
FU = fraction of old mud in underflow
QC = mass rate of clay, lb/min QD = mass rate of additives, lb/min
QP = water flow rate into mixing pit, gal/min
QB = mass rate of API barite, lb/min
Example: Mud density into centrifuge (MW) = 16.2ppg
Mud volume into centrifuge (QM) = 16.5 gal/min Dilution water density (PW) = 8.34ppg Dilution water volume (QW) = 10.5 gal/min Underflow mud density (PU) = 23.4ppg Overflow mud density (PO) = 9.3 ppg
Clay content of mud (CC) = 22.51b/bbl
Additive content of mud (CD) = 6 lb/bbl
Determine: Flow rate of underflow
Volume fraction of old mud in the underflow Mass rate of clay into mixing pit Mass rate of additives into mixing pit Water flow rate into mixing pit Mass rate of API barite into mixing pit a) Underflow mud volume, gal/min:
14.1
QU = 7.4 gal/min b) Volume fraction of old mud in the underflow: 35 - 23.4
11.6
QC = 7.55 lb/min d) Mass rate of additives into mixing pit, lb/min: 6 x [16.5 - (7.4 x 0.324)]
QD = 2.01 lb/min e) Water flow into mixing pit, gal/min:
QB = 16.5 - 7.4 - 8.20 - 0.348 - 0.0926 x 35 QB = 0.4594 x 35 QB = 16.079 lb/min
References
Chenevert, Martin E., and Reuven Hollo, TI-59 Drilling Engineering Manual, PennWell Publishing Company, Tulsa, 1981.
Crammer Jr., John L. Basic Drilling Engineering Manual, PennWell Publishing Company, Tulsa, 1982.
Manual of Drilling Fluids Technology, Baroid Division, N.L. Petroleum Services, Houston, Texas, 1979.
Mud Facts Engineering Handbook, Milchem Incorporated, Houston, Texas, 1984.
CHAPTER FOUR
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