Drilling Mud Calculations

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Chapter 2: Drilling Hydraulics

DRILLING MUD CALCULATIONS

The most common mud engineering calculations are those concerned with the changes of mud volume and density caused by the addition of various solids or liquids to the system. The first step is to compute the system volume, which is the sum of the mud in the hole and surface pits. While the surface volume is readily obtained from the pit size, the downhole volume is difficult to determine. Boreholes are not always cut to gauge (the same size as the bit) and unless a caliper log is available, which is unusual at the time of drilling, the true hole size must be estimated. In hard rock areas, little error may result from assuming bit size to exist; in salt or sloughing sections, however, this will be a gross error. With experience in the area, the mud engineer is able to make reasonable approximations. Those lacking this experience may compute hole volume (borehole less drill string volume) and apply any correction factor deemed applicable. Consider then the volume and density change of a mud (or water) resulting from the addition of solids. Two basic assumptions must be made: (1)

The volumes of each material are additive. This may immediately raise a question concerning bentonite and water mixtures since it is known that bentonite swells when wet. This expansion is due, however, to the adsorption of water; hence the clay volume increase is at the expense of water volume, and the total volume (clay plus water) is, for practical purposes, unchanged.

(2)

The weights of each material are additive.

Writing expressions for these assumptions: Vs + Vm1 = Vm 2

……………….. (1)

ρ sVs + ρ m1Vm1 = ρ m 2Vm 2

……………….. (2)

where: Vs = volume of solid Vm1 = volume of initial mud (or any liquid) Vm2 = final volume of mixture

ρs = density of solid ρm1 = density of initial mud ρm2 = density of final mud Solving for Vs: Vs =

Vm 2 ( ρ m 2 − ρ m1 ) ρ s − ρ m1

……………….. (3)

As to units, the densities may be in any consistent set; lb/gal or gm/cc is commonly used in field or laboratory problems, respectively. Vs will then be in the same units chosen for Vm2; bbl, cc, etc. Equation (3) is not particularly useful as it stands, since the net volume of a powdered solid is not readily measurable. However, the corresponding weight to add is

ρ sVs =

ARIFF OTHMAN

ρ sVm 2 ( ρ m 2 − ρ m1 ) ρ s − ρ m1

Drilling Engineering – SKM3413

……………….. (4)

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Chapter 2: Drilling Hydraulics

Example 1 A 9.5 lb/gal mud contains clay (SG = 2.5) and fresh water. Compute (a) the volume % and (b) the weight % clay in this mud. Solution 1 (a)

(b)

Altering Equation (3): Volume % solids =

Vs ρ − ρ m1 9.5 − 8.33 X 100 = m 2 X 100 = X 100 = 9.4% Vm 2 ρ s − ρ m1 (2.5)(8.33) − 8.33

Weight % solids =

ρ sVs ρ ( ρ − ρ m1 ) 20.8(9.5 − 8.33) X 100 = s m 2 X 100 = X 100 = 20.6% ρ m 2Vm 2 ρ m 2 ( ρ s − ρ m1 ) 9.5(20.8 − 8.33)

Example 2 For laboratory purposes, it is desired to mix one liter of bentonite fresh water mud having a viscosity of 30 cp (3.0% bentonite powder is needed to produce 30 cp bentonite fresh water mud). (a) (b)

What will be the resulting mud density? How much of each material should be used?

Solution 2 Altering Equation (3) (a)

0.03 =

ρ m 2 − ρ m1 ρ m 2 − 1.0 = ρ s − ρ m1 2.5 − 1.0

from which ρ m 2 = 1.045 gm/cc = 8.7 lb/gal (b)

Vs =

1000(1.045 − 1.0) = 30 cc = 2.5 X 30 = 75 gm 2.5 − 1.0

Also: Vm1 = Vm 2 − Vs = 1000 − 30 = 970 cc water For certain types of problems it is convenient to express Equation (3) in a different form. Suppose that the quantity of solids (Vs) necessary to increase (or decrease) the density of an initial mud is desired. Then: Vs =

(Vm1 + Vs )( ρ m 2 − ρ m1 ) ρ s − ρ m1

……………….. (3a)

where Vm1 + Vs = Vm 2 (volumes additive) Solving for Vs gives Vs =

ARIFF OTHMAN

Vm1 ( ρ m 2 − ρ m1 ) ρs − ρm2

Drilling Engineering – SKM3413

……………….. (5)

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Chapter 2: Drilling Hydraulics

Example 3

(a) (b)

How much weighting material (BaSO4, the mineral barite, SG = 4.3) should be added to the mud of Example 2 to increase its density to 10 lb/gal? What will the resulting volume be?

Solution 3

(a)

(b)

Vs =

1000(10 − 8.7) = 50.4 cc or 4.3 X 50.4 = 217 gm 35.8 − 10

Vm 2 = 1000 + 50.4 = 1050 cc

Since barite is so universally used as a weighting material, it is useful to express Equation (5) in field units. Barite is sold in 100 lb bags or sacks. Such a sack contains 100/(4.3)(62.4) = 0.373 cuft, or 0.373/5.61 = 0.0665 barrels of net material. Therefore 1 barrel (net) of barite = 1/0.0665 ≅ 15 sacks. Let SB = sacks of barite necessary to increase the density of 100 bbl of mud from ρm1 to ρm2. Substituting these special conditions into Equation (5): S B 100( ρ m 2 − ρ m1 ) = 15 35.8 − ρ m 2 SB =

or

1500( ρ m 2 − ρ m1 ) 35.8 − ρ m 2

……………….. (5a)

Example 4

(a) (b)

How many sacks of barite are necessary to increase the density of 1000 bbl of mud from 10 to 14 lb/gal? What will be the final mud volume?

Solution 4

(a)

Using Equation (5a)

SB = (b)

1500(14 − 10) sacks sacks = 275 = 2750 35.8 − 14 100 bbl 1000 bbl

Vm 2 = 1000 +

2750 = 1180 bbl 15

Example 5

(a) (b)

How much fresh water must be added to 1000 bbl of 12 lb/gal mud to reduce its density to 10 lb/gal? What will the resulting volume be?

Solution 5

(a)

(1000 bbl)(42 gal/bbl)(12 lb/gal) + (Vw )(42)(8.33) = (1000 + Vw )(42)(10) from which Vw = 1200 bbl

ARIFF OTHMAN

Drilling Engineering – SKM3413

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(b)

Chapter 2: Drilling Hydraulics

Vw = 1000 + 1200 = 2200 bbl

Equations (6) and (7) are other commonly used forms of Equation (3). Vw =

Vm1 ( ρ m1 − ρ m 2 ) ρ m 2 − 8.33

……………….. (6)

Sc =

875( ρ m 2 − ρ m1 ) 20.8 − ρ m 2

……………….. (7)

where:

Vw = barrels of water necessary to reduce density of Vm1 barrels initial mud from ρm1 to ρm2. Sc = sacks (100lb) of clay (SG = 2.5) necessary to change density of 100 bbl initial mud from ρm1 to ρm2. In working with laboratory size samples, it is convenient to measure quantities in grams or cubic centimeters. For field use, it is necessary to express these results in pounds per barrel. It is then useful to realize: 1 lb/bbl X

454 gm/lb = gm/cc 3785 cc/gal X 42 gal/bbl

or

gm/350 cc = lb/bbl

……………….. (8)

For laboratory or pilot testing purposes, it is convenient to work with a 350 cc quantity so that treating agent additions in gm/per 350 cc of mud will be equivalent to field additions in lb/bbl. Example 6

A mud engineer finds from pilot tests that 2.0 gm of CMC is required to obtain the desired water loss reduction for a one liter mud sample. How much CHIC should be added to the actual 1000 barrel system? Solution 6

CMC needed =

350 X 2.0 X 1000 = 700 lb 1000

In making recommendations for mud treating, it is necessary to know the time required for the entire mud system to make a complete cycle. This is called the cycle time and is computed from a knowledge of pumping rate and system volume. Normally in field operation the types of pump used is a duplex mud pump. The displacement of a duplex mud pump can be computed from:

q = 0.00679 SN (2 D 2 − d 2 )e

……………….. (9)

where:

q S

= pump discharge rate, gal/min = stroke length, in.

ARIFF OTHMAN

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N D d e

Chapter 2: Drilling Hydraulics

= = = =

complete strokes per minute piston (liner) diameter, in. piston rod diameter, in. pump volumetric efficiency, commonly used as 90% for power pumps and 85% for steam.

Cycle time is then expressed as tc =

6180 Vm SN (2 D 2 − d 2 )e

……………….. (10)

where:

tc = cycle time, min Vm = system volume, bbl Example 7

What is the cycle time for the following conditions?

Vm = 1000 bbl Pump liners = 75 in. diameter Stroke length = 16 in. Piston rod diameter = 2¼ in. N = 40 strokes per minute Power pump is used Solution 7 tc =

(

(6180)(1000)

)

(16)(40) (2)(7.5) 2 − (2.25) 2 (0.90)

= 100 min.

Treating materials would then be added at a rate allowing their uniform distribution in the system. In Example (7), if 20 sacks of material were needed, they could be added at the approximate rate of one sack per five minutes.

ARIFF OTHMAN

Drilling Engineering – SKM3413

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