## Drilling Fluids

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

### 3 _ (kill fluid density, kg/1 - original fluid density, kg/1) x 4200

1. 2 - kill fluid density, kg/1
2. I. units calculation kill fluid density, kg/m3 - x 4200

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

### Mixing Fluids of Different Densities

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

### Check the calculations. If the calculated amount of liquid is added, what will be the resulting oil/water ratio?

• oil in original vol oil + new vol oil
• x 100

liquid phase original liquid oil + new oil added

liquid phase 68 + 17

### x 100

• oil in _ gQ liquid phase
• water would then be: 100 - 80 = 20

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

### Check the calculations. If the calculated amount of water is added, what will be the resulting oil/water ratio?

• water in 17 + 5
• x 100

### liquid phase 68 + 5

• water in _ ^^ liquid phase
• water in = 100 _ 30 = 7Q liquid phase

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

### 121.60544

1. 2026 ASG = 3.6625
2. Because a high chloride example is being used, Step 5 is omitted.
3. Percent by volume low gravity solids (LGS)

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

### Dilution of Mud System

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:

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:

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

1. 8 + (0.63636 x 26.66) FU = 0.324%
2. Mass rate of clay into mixing pit, lb/min: 22.5 x [16.5 - (7.4 x 0.324)]

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:

1. 66 218.507
2. 66 QP = 8.20 gal/min f) Mass rate of API barite into mixing pit, lb/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.

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### Responses

• chilimanzar
How sacks of barite in 1 barrel of?
9 years ago
• Amanda Took-Brandybuck
What is a LGS centrifuge?
9 years ago
• mark
How much barite is needed to increase the mud weight to kill weight mud?
9 years ago
• Cailyn
What is the average weight of drilling mud?
9 years ago
• jeffrey wesley
How to calculate volume of oil in a 100 bbl tank?
8 years ago
• pirkko
How to calculate solid content unweighted mud?
7 years ago
• SHAWN
How to calculate total fluid needed when drilling?
7 years ago
• jannette
How much Barite per oil well is required?
7 years ago
• sophie
How much barites is required to uncrease the density of 300bbl of mud from 14ppg to15ppg?
4 years ago
• gloria
How to calculate volume of barite and water in mud preparation?
4 years ago
• timba sandyman
How to calculate for mud weight and its volume?
4 years ago
• david
How to calculate final mw when mixing 2 fluids with different densities?
4 years ago
• christina
How much barite is reuired to weight up 0.1 ppg?
3 years ago
• Isaac
How much barite is added to increase mud density?
3 years ago
• silke
How to calculate mud weight given different additives mass?
3 years ago
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What is the foemula to calculate increase in mud weight in oil well drilling?
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How to increase the specific gravity drilling fluid?
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How much barite is required to increase the density of 300 bbls?
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How many sacks of barite to increase 1 bbl?
3 years ago
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How many sacks of drilling fluid?
3 years ago
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How many sacks of bentonite to make mud formula?
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