Calculations For Mud Engineers

Solids Analysis For Unweighted Freshwater Muds

Material Balance

Conversion Factors

SG of water = 1.0

SG of Drill Solids = 2.6

SG of Barite = 4.2

1 gallon water = 8.34 lbs

1 barrel water = 42 gallons = 350 lbs/bbl

1 gallon barite = (4.2) (8.34) = 35 lbs/gal

1 barrel barite = (4.2) (350) = 1470 lbs/bbl

1 gallon bentonite = (2.6) (8.34) = 21.7 lbs/gal

1 barrel bentonite = (2.6) (350) = 910 lbs/bbl

1 barrel of water = 350 lbs

Material Balance

Calculate the requirements of water and barite to make 1.0 bbl final volume. Mud will contain 20 lbs/bbl bentonite which will weigh 8.8 lbs/gal. Let x = barrels of bentonite slurry and (1-x) equal the volume of barite. Using the material balance equation, perform the calculations.

(V1) (D1) + (V2) (D2) = (VF) (DF)

(8.8)(X) + (35)(1-X) = 12

8.8X + 35 - 35X = 12

23 = 26.2 X

X = 0.88 bbl bentonite slurry

1 - X = 0.12 bbl barite

(0.12) ( 1470) = 176 lbs barite

Material Balance

Starting with 1.0 barrel of bentonite slurry, calculate the requirements of water and barite required to increase the density to 12.0 lb/gal. The bentonite slurry will weigh 8.8 lbs/gal.

Let X = barrels of barite.

(V1) (D1) + (V2) (D2) = (VF) (DF)

(1.0)(8.8) + (35) (X) = (1+X) (12)

8.8 + 35 X = 12 + 12X

23X = 3.2

X = 0.14 barrel barite

= (0.14)(1470) = 204 lbs barite

The volume increase of adding barite to 1.0 barrel of bentonite slurry ( 8.8 lb/gal) to reach a final density of 12.0 lb/gal is 0.14 bbl. The final volume will then be 1.14 barrels.

Material Balance

Prepare a mud that contains 12 lb/bbl bentonite starting with a mixture of prehydrated bentonite that contains 35 lb/bbl bentonite. Let X equal the volume of prehydrated bentonite to begin with.

(X) (35) = 12

X = (12/35)

X = 0.34 bbl of 35 lb/bbl prehydrated bentonite

1 - X = 0.64 bbl water

Dilution

Reduce the density of a 14.5 lb/gal drilling mud to 13.0 lb/gal using freshwater (8.34 lb/gal). Let X = volume of freshwater required and (1-X) equal to the volume of original mud.

(V1) (D1) + (V2) (D2) = (VF) (DF)

(X) (8.34) + (1-X) (14.5) = 13.0 (1)

8.34 X + 14.5 - 14.5 X = 13.0 (1.0)

6.16 X = 1.5

X = 0.24 bbl water

1 - X = 0.76 bbl original mud

Dilution

Reduce the density of a 14.5 lb/gal drilling mud to 13.0 lb/gal using freshwater (8.34 lb/gal). Start with 1.0 bbl of the 14.5 lb/gal mud. Let X = volume of freshwater required.

(V1) (D1) + (V2) (D2) = (VF) (DF)

14.5 (1.0) + 8.34 (X) = 13 ( 1 + X)

14.5 + 8.34 X = 13 + 13X

4.66 X = 1.5

X = 0.32 bbl freshwater

Final volume = 1.32 bbl.

Solids Analysis

Given : Mud Weight : 11.9 lb/gal

Retort Solids : 18 %

Retort Oil : 3 %

Chlorides : 20,000 mg/l

MBT : 27 lb/bbl

SG LGS : 2.6

SG Oil : 0.84

SG HGS : 4.2

Determine:

a). The corrected solids content

b). The ASG

c). The vol % LGS and % HGS

d). The lb/bbl LGS and HGS

e). The volume % bentonite

f). The volume % drill solids

g). Lb/bbl bentonite and drill solids

Solids Analysis

a). Correction for salinity

Vol % salt water = vol % retort water * [1+(5.88 * 10-8)(chlorides)1.2)]

=0.79 * [1 +(5.88 * 10-8)(20,0001.2)]

=0.79 * 1.01

=0.80

SG Saltwater = 1 + (1.9 * 10 -6) * (chlorides)0.95

= 1 + (1.9 * 10 -6) *(20,000)0.95

= 1.02

Corrected Solids = 1.0 - 0.80 - 0.03 = 0.17 or 17 %

Solids Analysis

b). Calculate the ASG by material balance:

V1D1 + V2D2 + V3D3 = VFDF

(8.33)(1.023)(0.80) + (8.33)(.84)(0.03) + (8.33)(0.17)(ASG) = 1(11.9)

6.79 + 0.21 + 1.44 (ASG) = 11.9

ASG = 3.40

c). HGS + LGS = Total Solids

If X = LGS, then (0.17 - X) = HGS

(4.2)(0.17 - X) + (2.6)(X) = (ASG)(0.17)

0.71 - (4.2)(X) + (2.6)(X) = (3.44)(0.17)

1.6(X) = 0.13

X = 0.086 or 9 % LGS

0.17 - 0.084 = 8 % HGS

Solids Analysis

d). Lb/bbl LGS = (% LGS)(2.6)(8.33)(42)

= (0.086)(2.6)(8.33)(42)

= 78.23 lb/bbl

Lb/bbl HGS = (%HGS)(4.2)(8.33)(42)

= (0.084)(4.2)(8.33)(42)

= 123.4 lb/bbl

e). % Bentonite = (MBT - % LGS) / 8

= (27 - 9)/8

= 2.25 %

f). % DS = % LGS - % Bentonite

= 9 - 2.25

= 6.75 %

Solids Analysis

g). lb/bbl Bentonite = (9)(% Bentonite)

= (9)(2.25)

= 20.3 lb/bbl Bentonite

lb/bbl DS = (9)(% DS)

= (9)(6.75)

= 60.7 lb/bbl DS

Loss of Overbalance - Lost Returns

Hydrocyclone Evaluation