Download Drilling Mud Calculations PDF

TitleDrilling Mud Calculations
TagsMechanical Engineering Pump Density Applied And Interdisciplinary Physics Barrel (Unit)
File Size155.0 KB
Total Pages5
Document Text Contents
Page 1

2007/2008-1 Chapter 2: Drilling Hydraulics

ARIFF OTHMAN Drilling Engineering – SKM3413 Page 1

2.5 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:

1 2 s m mV V V+ = ……………….. (1)

1 1 2 2 s s m m m mV V Vρ ρ ρ+ = ……………….. (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:

2 2 1
1

( )m m m
s

s m

V
V

Page 2

2007/2008-1 Chapter 2: Drilling Hydraulics

ARIFF OTHMAN Drilling Engineering – SKM3413 Page 2

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) Altering Equation (3):

2 1

2 1

9.5 8.33
Volume % solids X 100 X 100 X 100 9.4%

(2.5)(8.33) 8.33
s m m

m s m

V
V

ρ ρ
ρ ρ

− −
= = = =

− −


(b) 2 1
2 2 2 1

( ) 20.8(9.5 8.33)
Weight % solids X 100 X 100 X 100 20.6%

( ) 9.5(20.8 8.33)
s s s m m

m m m s m

V
V

ρ ρ ρ ρ
ρ ρ ρ ρ

− −
= = = =

− −


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

Solution 2

Altering Equation (3)

(a) 2 1 2
1

1.0
0.03

2.5 1.0
m m m

s m

ρ ρ ρ
ρ ρ

− −
= =

− −


from which 2 1.045 gm/cc 8.7 lb/galmρ = =

(b)
1000(1.045 1.0)

30 cc 2.5 X 30 75 gm
2.5 1.0s

V


= = = =




Also: 1 2 1000 30 970 cc waterm m sV V V= − = − =

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:

1 2 1
1

( )( )m s m m
s

s m

V V
V

ρ ρ
ρ ρ

+ −
=


……………….. (3a)

where 1 2m s mV V V+ = (volumes additive)

Solving for Vs gives

1 2 1
2

( )m m m
s

s m

V
V

ρ ρ
ρ ρ


=


……………….. (5)

Page 3

2007/2008-1 Chapter 2: Drilling Hydraulics

ARIFF OTHMAN Drilling Engineering – SKM3413 Page 3

Example 3

(a) 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?

(b) What will the resulting volume be?

Solution 3

(a)
1000(10 8.7)

50.4 cc
35.8 10s

V


= =


or 4.3 X 50.4 217 gm=

(b) 2 1000 50.4 1050 ccmV = + =

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):

2 1
2

100( )
15 35.8

m mB

m

S ρ ρ
ρ


=


or

2 1
2

1500( )
35.8

m m
B

m

S
ρ ρ

ρ


=


……………….. (5a)

Example 4

(a) How many sacks of barite are necessary to increase the density of 1000 bbl of mud from 10 to
14 lb/gal?

(b) What will be the final mud volume?

Solution 4

(a) Using Equation (5a)

1500(14 10) sacks sacks
275 2750

35.8 14 100 bbl 1000 bblB
S


= = =




(b) 2
2750

1000 1180 bbl
15m

V = + =

Example 5

(a) How much fresh water must be added to 1000 bbl of 12 lb/gal mud to reduce its density to 10
lb/gal?

(b) What will the resulting volume be?

Solution 5

(a) (1000 bbl)(42 gal/bbl)(12 lb/gal) ( )(42)(8.33) (1000 )(42)(10)w wV V+ = +

from which 1200 bblwV =

Page 4

2007/2008-1 Chapter 2: Drilling Hydraulics

ARIFF OTHMAN Drilling Engineering – SKM3413 Page 4

(b) 1000 1200 2200 bblwV = + =

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

1 1 2
2

( )
8.33

m m m
w

m

V
V

ρ ρ
ρ


=


……………….. (6)

2 1
2

875( )
20.8

m m
c

m

S
ρ ρ

ρ


=


……………….. (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:


454 gm/lb

1 lb/bbl X 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:

2 20.00679 (2 )q SN D d e= − ……………….. (9)

where:

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

Similer Documents