ENGINEERING CALCULATIONS
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INTRODUCTION
This section is a guide for calculations regarding the drilling fluid such as capacity of mud,
tanks volumes, tubulars and holes, circulation times, velocity of the mud in the annular and
in the drill pipes and other important calculations. The capacity of calculating muds
formulations and various situations by means of solids and liquids additives is necessary in
fluids engineering.
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FIELD UNIT SYSTEM
The unit of measure most commonly used is the Field Unit.
However, the decimal measuring system is more and more used in drilling.
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FIELD UNIT SYSTEM
Metric System
Field Unit
Mass
Kilogramme (kg)
Pounds (lb)
Length
Meters (m)
Feet (ft) and inches (in.)
Volume, Capacity
and Displacement
Cubic meters (m3) and litres
(l)
Barrels (bbl) and gallons
(gal)
Density
grams/ cubic centimetres
(g/cm3) e (kg/l) – Either
equal at specific weight
Pounds/gallon (lb/gal)
and pounds/foot3 (lb/ft3)
Pressure
kiloPascals (kPa), bar or
atmospheres
pounds/inch2 (lb/in.2 o
psi)
Concentration
kilogram/cubic meter
(kg/m3)
pounds /barrel (lb/bbl)
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CONVERSION FACTORS
Multiply
by
To obtain
Volume
Barrels(bbl)
5.615
Feet3 (ft3)
Barrels(bbl)
0.159
Meters3 (m3)
Barrels(bbl)
42
Cubic feet (ft3)
0.0283
Cubic feet(ft3)
7.48
Gallons, U.S. (gal)
Meters3 (m3)
Gallons, U.S. (gal)
Meters3 (m3)
Gallons, U.S. (gal)
0.00379
Gallons, U.S. (gal)
3.785
Litre(l)
Cubic Meters (m3)
6.289
Barrels(bbl)
Cubic Meters (m3)
1,000
Litres(l)
Pounds (lb)
453.6
Grams (g)
Pounds (lb)
0.4536
Kilograms(kg)
Kilograms (kg)
2.204
Pound (lb)
Metric tons (mt)
1,000
Kilograms (kg)
0.3048
Metres (m)
Mass or Weight
Length
Feet (ft)
Inches (in.)
2.54
Centimetre(cm)
Inches (in.)
25.4
Milliletre (mm)
Meters (m)
3.281
Feet (ft)
Miles (mi)
1.609
Kilometres (km)
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CONVERSION FACTORS
Multiply
To obtain
by
Pressure
lb/in.2 (psi)
6.8948
lb/in.2 (psi)
0.068948
lb/in.2 (psi)
0.0703
kg/cm2
kiloPascal (kPa)
0.145
lb/in.2 (psi)
bar (bar)
100
kiloPascal (kPa)
bar (bar)
kiloPascal (kPa)
Concentration
pounds/barrels (lb/bbl)
2.853
kg/m3
kilograms/cubic metre(kg/m3)
0.3505
lb/bbl
119.83
kg/m3
kilogram/cubic meter (kg/m3)
0.008345
lb/gal
pound/gallon (lb/gal)
0.11983
g/cm3, kg/l or SG
Density
pounds /gallons (lb/gal)
pound/cubic feet (lb/ft3)
16.02
kg/m3 and g/l
g/cm3, kg/l or SG
8.345
lb/gal
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HOLE CALCULATIONS
¾ Capacity
¾ Volume
¾ Displacement
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HOLE CALCULATIONS
Mud pits and tanks – Capacity and Volume
Rectangular Tanks
V = Volume or Tank Capacity
L = Length
W = Width
H = Height
M = Fluid Level
Volume = L x W x H
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HOLE CALCULATIONS
Vertical cylindrical tanks
Vcyl = Cylindrical tank capacity
D = Cylinder diameter
Acqua distillata
H = Cylinder height
M = Fluid level
π= 3.1416
To calculate the diameter, measure the circumference and divide by 3.1416:
D = Circumference/π
The formula to calculate the capacity of a vertical cylindrical tank is:
Vcyl
=
2
π D
H
4
The volume of the fluid (VMud) in a vertical cylindircal tank is calculated as
follows :
π x D2 M
3
3
VMud (ft or m ) =
4
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HOLE CALCULATIONS
Horizontal Cylindrical Tanks
VCyl = Horizontal cylinder capacity
D = Cylinder diameter
L = Cylinder length
M = Material level
π = 3.1416
VCyl
Agip KCO
L
=
2
⎡
⎢ (2 M − D ) MD − M
⎢⎣
2
2
D2
⎞ πD
−1 ⎛ 2 M
+
− 1⎟ +
sin ⎜
2
4
⎝ D
⎠
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HOLE CALCULATIONS
Hole Volume
The volume of each interval is calculated as follows:
V
Hole
π D
=
section
2
L
4
Where: DW = Internal diameter (ID) of the casing, liner or open hole
L = Interval Length
If the diameter (DW) is in inches:
VHole
D2 W ( in.)
section (bbl / ft ) =
1029
On the contrary, with the decimal measuring system:
V Hole
Agip KCO
section
D 2 W ( in. )
m /m =
1974
(
3
)
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HOLE CALCULATIONS
Drill Pipe or Drill Collars capacity (volume of the internal diameter)
The volume, with the drill string in the hole, is the sum of the internal capacity of
the drill pipes plus the volume of the annular space.
V
Pipe
(bbl
/ ft
)=
ID
)=
ID
2
P
(in . )
1029
With the metric system:
V Pipe
Agip KCO
(l
/ m
2
P
(in . )
1974
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HOLE CALCULATIONS
Annular volume
The volume of the annular space is determined subtracting the external volume of the drill
string from the hole capacity(or casing).
V Annulus
ID 2 W (in .) − OD 2 P (in .)
(bbl/f t ) =
1029
Where: IDW = Internal diameter hole or casing
ODP = External diameter drill pipes or drill collars
With the metric system:
ID 2 W (in.) − OD 2 P (in.)
VAnnulus (l/m ) =
1974
Or:
VAnnulus = CapacityWell – DisplacementDrillstring – CapacityDrillstring
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HOLE CALCULATIONS
Displacement
The drill string displacement can be estimated (VPipe Displ.) using the OD and the ID
of the drill pipes and the drill collars.
V Pipe
Displaceme
nt
(bbl/ft ) =
OD
2
P
(in. ) − ID 2 P (in. )
1029
Where: ODP = External diameter drill pipe or drill collars
IDP = Internal diameter drill pipe or drill collars
With the metric system:
VPipe Displaceme
Agip KCO
OD 2 P (in .) − ID 2 P (in .)
nt (l/m ) =
1974
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PUMPS CAPACITY
TRIPLEX PUMPS
In the triplex pump, pistons are three and operate in the same direction. Generally, they
are short stroke, (from 6-in. to 12-in.) and operate at rates from 60- to 120-stk/min.
Triplex pump performance calculation formula:
V Pump
Agip KCO
Output
=
3 x 3 . 1416 xID
2
Liner
xLxP Eff
4
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PUMPS CAPACITY
Where:
VPump Output = Pumps Performance
IDLiner = Liner Internal Diameter
L = Stroke Length
PEff = Volumetric efficiency (dimensionless)
V Pump
Output
(bbl
/ stk ) =
ID
2
Liner
(in .)xL (in .)xP Eff
4117 . 7
With the metric system:
V Pump
Agip KCO
Output
(l
/ stk
)=
ID
2
Liner
(in . )xL (in . )xP Eff
25 . 90
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PUMPS CAPACITY
DUPLEX PUMPS
The pistons of a duplex pump are two and operate in both directions. The difference
between the calculations for a duplex and for a triplex is that the volume of the rod pistons
must be subtracted form the volume of one of the cylinders plus the difference in number of
cylinders, 4 for a duplex and 3 for a triplex. Generally, the duplex pumps have a longer
stroke (from 10 to 18 in.) and operate at lower rate from 40 to 80-stk/min.
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PUMPS CAPACITY
The general equation for a duplex pump is :
VPump Output
=
[
(
) ]
2π
x ID 2 Liner xL + ID 2 Liner − OD 2 Rod xL xPEff
4
Where:
IDLiner = Internal diameter liner
VPump Output = Pump rate
ODRod = Rode external diameter
L = Stroke length
PEff = Volumetric efficiency (dimensionless)
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PUMPS CAPACITY
Pump rate bbl/stroke for a duplex pump with ID liner, OD pipe and stroke length are in
inches:
V Pump
Output
⎡ 2 xID
(bbl / stk ) = ⎢
⎢⎣
2
Liner
(in .) − OD
2
Rod
6174
(in . ) ⎤ xL (in .)xP
⎥
⎥⎦
Eff
With the metric system:
V Pump
Agip KCO
Output
⎡
(l / stk ) = ⎢ 2 xID
⎣
2
Liner
(in .) − OD
38 . 85
2
Rod
(in . )⎤ xL (in . )xP
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Eff
ANNULAR VELOCITY
The annular velocity (AV) is the average velocity of the fluid which flows in the
annular space. For a correct well cleaning, a minimum velocity is needed. The latter
depends on a certain number of factors such as ROP, cuttings size, hole inclination,
mud density and rheology.
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ANNULAR VELOCITY
¾ Equations for the calculation of the anular velocity based on pumps rate and the
annular volume:
AV
=
AV (ft/min
V
V
)=
AV (m/min ) =
Agip KCO
→ Output
Pump
Ann
V Pump
Output
(bbl
/ min
V Ann (bbl / ft )
)
(l / min )
V Ann (l / m )
VPump
Output
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CIRCULATION TIME
The total circulation time is the time (or number of strokes) to make a complete loop,
from the tank, following the path down to the bit, go up from the annular and come
back in the circulating tank.
Total circulatio n time (min) =
Agip KCO
VSystem
VPump Output
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CIRCULATION TIME
Where:
VSystem = Total attive volume (bbl o m3)
V Pump Output = Pumps capacity (bbl/min o m3/min)
Total circulation (strokes) = Total circulation time (min) x pumps velocity (stks/min)
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CIRCULATION TIME
Bottoms-up is the time (or number of strokes) the mud circulates from bottom hole
throught the annular up on the surface.
The bottoms-up is calculated as follows:
Bottom
− up time(min)
=
V Annulus
V Pump
Output
Where:
VAnnulus = Annular volume (bbl or m3)
VPump Output = Pumps capacity (bbl/min or m3/min)
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CIRCULATION TIME
Bottoms-up (strokes) = Bottoms-up (min) x pumps velocity (stk/min)
The total circulation time (or number of strokes) to make a complete circulation, starting
from the tank down to the bit and up to the annular returning back in the tanks. The
circulation total time is calculated as follows:
Hole cycle time(min)=
Agip KCO
VHole − VDrillsringDispl
VPump Output
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CIRCULATION TIME
Where:
V Hole = Total well volume (bbl o m3)
V Drillstring Displ = Pipes displacement (bbl o m3)
V Pump Output = Pumps capacity (bbl/min o m3/min)
Complete circulation (strokes) = time for a complete circulation (min) x pumps
velocity(stk/min)
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HYDROSTATIC PRESSURE
The hydrostatic pressure (Ph) is the pressure exerted by a column of liquid and
depends on the density of the fluid and the vertical depth or True Vertical Depth (TVD).
In a well, it is the pressure exerted on the wall to avoid cavings and controls the
formation pressure as well.
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HYDROSTATIC PRESSURE
Hydrostatic pressure is calculated as follows:
P H (bar
)=
Mud Weight
(kg/l )x TVD (m )
10.2
Hydrostatic pressure = Mud density x TVD x conversion factor (0.0981)
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HYDROSTATIC PRESSURE
Field unit system:
P H (lb/in.2) = Mud density (lb/gal) x TVD (ft) x 0.052
Conversion
Agip KCO
factor 0.052 =
12 in./ft
231 in. 3 /gal
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EXAMPLES
• Data:
• Surface Casing: 1,600 m - 133/8-in.
48-lb/ft, (323-mm ID)
• Bit diameter: 121/4 in
• T.D.: 3,400 m
• Drillstring:
Drill Pipe: 5-in. 19.50-lb/ft,
(127-mm OD, 108.6-mm ID),
Drill collars 200 m of 71/4-in. x 23/4-in (185-mm OD x72-mm ID)
• Surface systems:
2 tanks: depth 4-m, width 3-m, length 10-m. Both tanks have 2.5 m of mud with
pipes in well.
• Mud density: SG 1.50 o 1.5 kg/l
• Mud pumps: Triplex: 6 in x 12 in
(152.4 mm x 304.8 mm)
110 stk/min, with 90% efficiency
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EXAMPLES
13.3/8 in. casing
1600 m
12.1/4 in. Open Hole
Drill Pipe 5 in. 19.5-lb/ft
DC 71/4-in. x 23/4-in
3400 m
Well Diagram
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EXAMPLES
Part I: Determine the capacity of the surface system in m3, m3/m e m3/cm.
VPit (m3) 1 tank = 4 m x 3 m x 10 m = 120 m3
VPit (m3) 2 tanks = 120 m3 x 2 = 240 m3
VPit (m3/m) 2 tanks = 240 m3 ÷ 4 = 60 m3/m
VPit (m3/cm) 2 tanks = 60 m3/m ÷ 100 cm/m = 0.60 m3/cm
Part II: Determine the total volume of the mud on the surface, in m3.
VSurface (m3)= VMud (m3) 2 tanks = 60 m3/m x 2.5 m = 150 m3
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EXAMPLES
Part III: Determine the total volume of the well without drill string.
Calculate the volume of the mud in each interval and sum the volumes.
V Well
(m )
3
ID 2 Well
=
1,273,000
( )
VCsgl m
3
x L (m
)
323 2 mm 2
=
x 1,600 (m ) = 131.1 m 3
1,273,000
( )
250.82 mm2
VOH m =
x1,800 (m) = 88.9 m3
1,273,000
3
Total without the drill string:
V System = V Csg + V OH = 131.1 m3 + 88.9 m3 = 220 m3
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EXAMPLES
Part IV: Determine the total volume with the string in the well .
Volume in the string:
V Drillstrin
( )
VDP m
3
(m )
3
g
)x
L (m )
108.6 2 (mm )
=
x 3200 (m ) = 29.6 m 3
1,273,000
( )
V DC m
ID 2 DS (mm
=
1,273,000
3
72 2 (mm
=
1,273,000
)
x 200 (m ) = 0.8 m 3
Total volume in the string:
V Drillstring = V DP + V DC = 29.6 m 3 + 0.8 m 3 = 30.4 m 3
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EXAMPLES
Volume in the annulus:
VAnnulus
ID 2 Well (mm) − OD 2 DS (mm)
x L (m )
m =
1,273,000
( )
3
3232 (mm) − 1272 (mm)
VAnn(Csg DP) m =
x 1,600 (m) = 0.06927 x 1,600 = 110.8 m 3
1,273,000
( )
3
250.82 (mm) − 127 2 (mm)
VAnn(OH DP) m =
x 1,600 (m ) = 0.03673 x 1,600 = 58.8 m 3
1,273,000
( )
3
250.82 (mm) − 1852 (mm)
VAnn (OH DC ) m =
x 200 (m ) = 0.02252 x 200 = 4.5 m 3
1,273,000
( )
3
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EXAMPLES
VAnnulus
(m ) = V
3
Total
Ann (Csg DP )
+ VAnn (OH DP ) + VAnn (OH DC )
V Annulus Total = 110.8 + 58.8 + 4.5 = 174.1 m3
Total volume of the hole with a string:
V Well/DS = V Annulus Total + V Drillstring = 174.1 + 30.4 = 204.5 m 3
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EXAMPLES
Part V: Determine the circulation total volume.
V Total = V Well/DS + V Surface = 204.5 m3 + 150 m3 = 354.5 m3
Part VI: Determine the time needed for a complete circulation and bottoms-up.
ID2 Liner (mm) x L (mm) x Eff (decimal)
VP ump Output (l/stk) =
424,333
152.4 2 Liner (mm) x 304.8 (mm) x 0.9 (decimal)
VP ump Output (l/stk ) =
= 15.01 (l/stk )
424,333
(
VP ump Output (l/min ) = 15.01 (l/stk ) x 110 stk/min = 1,651 (l/min ) = 1.651 m 3 / min
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)
EXAMPLES
354.5 m 3
Total circulatio n time (min ) =
= 215 min
3
1.651 m /min
(
)
Total circulation (stk) = 215 min x 110 stk/min = 23,650 stk
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EXAMPLES
Hole cycle time (min ) =
( )
V Well/DS m 3
(
V P ump Output m 3 / min
)
=
204 .5
= 124 min
1.651
Hole cycle (stk) = 124 min x 110 stk = 13,640 stk
Bottoms - up time (min ) =
V Annulus
(m )
3
(
Total
V P ump Output m 3 / min
)
=
174 . 1
= 106 min
1.651
Bottoms-up (stk) = 106 min x 110 stk = 11,660 stk
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EXAMPLES
Part VII: Determine the velocity in the annular per each single interval.
AV =
V Pump
AV(OH/DC) =
AV(OH/DP) =
AV(Csg/DP) =
Agip KCO
V
Output
Ann
(
(m
) = 1.651(m / min) = 73 m/min
/ m) 0.02252(m / m)
VPumpOutput m3 / min
VAnn OH/DC
3
(
(m
3
3
) = 1.651 (m / min) = 45 m/min
0.03673 (m / m )
/ m)
VPump Output m 3 / min
VAnn OH/DP
(
(m
3
3
3
) = 1.651 (m / min) = 24 m/min
/ m)
0.06927 (m / m )
VPump Output m 3 / min
VAnn Csg/DP
3
3
3
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EXAMPLES
Part VIII: Determine the hydrostatic pressure at the TD
P
H
(bar ) =
P H (bar
Agip KCO
)=
Mud
(kg/l
Weight
)x TVD
(m )
10.2
1.5
(kg/l ) x
10.2
3,400
(m ) =
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41
DENSITY INCREASE THROUGH ADDITIVES
Weight increase (specific weight d = 4.2)
(
W kg/m
with:
3
(
d 2 − d1 )
) = 4200 (4.2 - d )
2
W = weight of the barite to be added in kg/m3
d1 = mud initial denisty (specific weight)
d2 = desired mud density (specific weight)
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DENSITY INCREASE THROUGH ADDITIVES
Weight increase using Calcium Carbonate
(
W M kg/m
with:
3
) = 2650
(d 2 − d 1 )
(2.65 - d 2 )
W = weight of the calcium carbonate to be added in kg/m3
d1 = mud initial density (specific weight)
d2 = desired mud density (specific weight)
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-43
RPW2021A
43
DENSITY REDUCTION THROUGH WATER OR OIL
Water needed to reduce density (density H2O d = 1)
(
VWater liters/m
with:
3
) = 1000 x
(d1 − d 2 )
(d 2 − 1)
VWater = water volume (litres) to be added at 1 m3 of mud
d1 = mud initial density (specific weight)
d2 = desired mud density (specific weight)
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-44
RPW2021A
44
DENSITY REDUCTION THROUGH WATER OR OIL
Oil needed to reduce the density (oil density d = 0.85)
(
Voil liters/m
3
) = 1000 x
(d 1 − d 2 )
(d 2 − 0 .85 )
with: VOil = oil volume (litres) to be added 1 m3 of mud
d1 = mud initial density (specific weight)
d2 = desired mud density (specific weight)
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-45
RPW2021A
45
VOLUME INCREASE WITH ADDITIVES
Final volume after the addition of additives
V
F
with:
(l ) =
V
I
+
M
d
a
a
VF = final volume (litres)
VI = initial volume of 1,000 l (1 m3)
Ma = additives weight
da = additives specific weight
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-46
RPW2021A
46
MIXING OF LIQUIDS WITH DIFFERENT DENSITY
Final volume after mixing
VA + VB = VF
(VA x dA)+ (VB x dB) = (VF x dF)
with: VA = fluid volume A (m3)
VB = fluid volume B (m3)
VF = final volume(m3)
dA = fluid density A (kg/l)
dB = fluid density B (kg/l)
dF = final density of the mixed fluids (kg/l)
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-47
RPW2021A
47
MUD REPORT
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-48
RPW2021A
48
MUD REPORT
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-49
RPW2021A
49
MUD REPORT
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-50
RPW2021A
50
Agip KCO
Well Area Operations
Drilling Muds
ENGINEERING
Drilling Supervisors Training
Course CALCULATIONS-MAP-51
RPW2021A
51