ENGINEERING CALCULATIONS
INTRODUCTION
This section is a guide for calculations regarding the drilling fluid such as capacity of mud, tanks volumes, volumes, tubulars tubulars and holes, holes, circulation circulation times, velocity velocity of the mud in the annular and 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.
FIELD FIEL D UNIT SYSTEM SYSTEM
The unit unit of measure measure most most commonly commonly used used is the Field Unit. Unit. However, the decimal measuring system is more and more used in drilling.
FIELD FIEL D UNIT SYSTEM SYSTEM
Metric tr ic Sys ystem tem
Field Unit
Mass
Kilog Kilogra ramm mme e (kg) (kg)
Poun Pound ds (lb) (lb)
Length
Meter ters (m) (m)
Feet Feet (ft) (ft) and and inc inche hes s (in. (in.))
Volume, Capacity and Displacement
Cubic meters (m3) and litres (l)
Barrel Barrels s (bbl) (bbl) and and gallo gallons ns (gal)
Density
grams/ grams/ cubic cubic centimet centimetres res (g/cm3) e (kg/l) (kg/l) – Eithe Eitherr equal at specific weight
Pounds Pounds/gal /gallon lon (lb/gal (lb/gal)) and pounds/foot 3 (lb/ft3)
Pressure
kiloPas kiloPascals cals (kPa), (kPa), bar bar or atmospheres
pounds/inch 2 (lb/in.2 o psi)
Concentration
kilog kilogra ram/c m/cub ubic ic meter meter (kg/m3)
poun pounds ds /bar /barre rell (lb/ (lb/bb bbl) l)
CONVERSION CONVERSION FACTORS FA CTORS Multiply
by
To obtain
Volume Barrels(bbl)
5.615
Feet3 (ft3)
Barrels(bbl)
0.159
Meters3 (m3)
Barrels(bbl)
42
Cubi Cubic c feet feet (ft (ft3)
0.0283
Cubic Cubic feet(f feet(ftt3)
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)
Cubi Cubic c Mete Meters rs (m )
6.289
Barrels(bbl)
Cubi Cubic c Mete Meters rs (m3)
1,000
Litres(l)
Pounds Pounds (lb)
453.6
Grams Grams (g)
Pound Pounds s (lb) (lb)
0.4536
Kilograms(kg)
Kilogra Kilograms ms (kg) (kg)
2.204
Poun Pound d (lb) (lb)
Metr Metric ic tons tons (mt) (mt)
1,000
Kilogra Kilograms ms (kg) (kg)
0.3048
Metr Metres es (m) (m)
3
Mass Mass or Weight
Length Feet Feet (ft) (ft) Inches Inches (in.) (in.)
2.54
Centimetre(cm)
Inches Inches (in.) (in.)
25.4
Millil Milliletr etre e (mm) (mm)
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
HOLE CALCULATIONS
Capacity
Volume
Displacement
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
HOLE CALCULATIONS
Vertical cylindrical tanks Vcyl = Cylindrical tank capacity D = Cylinder diameter H = Cylinder height
Acqua distillata
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
=
π
D
2
H
4 The volume of the fluid (VMud) in a vertical cylindircal tank is calculated as
follows :
2
HOLE CALCULATIONS
Horizontal Cylindrical Tanks VCyl = Horizontal cylinder capacity D = Cylinder diameter L = Cylinder length M = Material level
π = 3.1416
VCyl =
L ⎡
D 2
sin ⎢ (2 M − D ) MD − M + 2 ⎢⎣ 2 2
⎞ π D 2 − 1⎟ + ⎜ 4 ⎝ D ⎠
−1 ⎛ 2 M
⎤ ⎥ ⎥⎦
HOLE CALCULATIONS Hole Volume The volume of each interval is calculated as follows:
V
Hole
π
=
section
2
D
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
(
3
/
)
D 2W
( in. )
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
(l /
m
2
P
1974
(in . )
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 (bbl/f t ) =
ID 2 W (in .) − OD
2
P
(in .)
1029
Where: IDW = Internal diameter hole or casing ODP = External diameter drill pipes or drill collars With the metric system:
VAnnulus (l/m ) =
ID 2 W (in.) − OD 2 P (in.)
1974 Or: V Annulus = CapacityWell – DisplacementDrillstring – CapacityDrillstring
Annular Volume
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
nt
(l/m ) =
OD 2 P (in .) − ID 2 P (in .)
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
Output
=
3 x 3 . 1416 xID 4
2
xLxP Eff
Liner
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
Output
(l / stk ) =
ID
2
Liner
(in . ) xL (in . )xP Eff 25 90
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.
PUMPS CAPACITY
The general equation for a duplex pump is :
VPump Output
=
2π 4
[
(
) ]
x ID 2 Liner xL + ID 2 Liner − OD 2 Rod xL xP Eff
Where:
IDLiner = Internal diameter liner VPump Output = Pump rate ODRod = Rode external diameter L = Stroke length PEff = Volumetric efficiency (dimensionless)
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
(in . )⎤
Rod
(in . )⎤
6174
⎥ xL (in . ) xP Eff ⎥⎦
With the metric system:
V Pump
Output
⎡ 2 xID (l / stk ) = ⎢ ⎣
2
Liner
(in . ) − OD 38 . 85
2
⎥ xL (in . ) xP 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.
ANNULAR VELOCITY
Equations for the calculation of the anular velocity based on pumps rate and the annular volume:
AV
=
AV (ft/min
V
→ Output
Pump
V
)=
AV (m/min ) =
Ann
V Pump
Output
(bbl
/ min
V Ann (bbl / ft )
(l / min ) V Ann (l / m )
VPump
Output
)
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) =
VSystem VPump Output
CIRCULATION TIME
Where: VSystem = Total attive volume (bbl o m 3) V Pump Output = Pumps capacity (bbl/min o m3/min) Total circulation (strokes) = Total circulation time (min) x pumps velocity (stks/min)
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
Where: V Annulus = Annular volume (bbl or m3) VPump Output = Pumps capacity (bbl/min or m3/min)
Output
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)=
VHole − VDrillsringDispl VPump Output
CIRCULATION TIME
Where: V Hole = Total well volume (bbl o m3) V Drillstring Displ = Pipes displacement (bbl o m 3) V Pump Output = Pumps capacity (bbl/min o m3/min) Complete circulation (strokes) = time for a complete circulation (min) x pumps velocity(stk/min)
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.
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)
HYDROSTATIC PRESSURE
Field unit system:
P H (lb/in.2) = Mud density (lb/gal) x TVD (ft) x 0.052
Conversion
12 in./ft factor 0.052 = 231 in. 3 /gal
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 7 1/4-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)
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
EXAMPLES
Part I: Determine the capacity of the surface system in m 3, m 3/m e m 3/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 m 3/cm Part II: Determine the total vo lume of the mud on the surf ace, in m 3. VSurface (m3)= VMud (m3) 2 tanks = 60 m3/m x 2.5 m = 150 m3
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
(m ) = 3
Well
ID
2
Well
1,273,000 323 2 mm 2
( ) = 1,273,000
VCsgl m
( )
3
VOH m = 3
x L (m
x 1,600 (m ) = 131.1 m 3
250.82 mm2
x1,800 (m) = 88.9 m3 1,273,000
Total without the drill string: V
= V Csg + V
)
= 131.1 m3 + 88.9 m3 = 220 m3
EXAMPLES
Part IV: Determine the total volume with the string in the well . Volume in the string:
V Drillstrin
(m ) = 3
g
( )=
VDP m
3
2
DS
1,273,000
72
(mm )
1,273,000
108.6 2 (mm )
( )=
V DC m
3
ID
2
x L (m
)
x 3200 (m ) = 29.6 m 3
(mm )
1,273,000
x 200 (m
)=
0.8 m
Total volume in the string : V
=V
+V
= 29.6 m 3 + 0.8 m 3 = 30.4 m 3
3
EXAMPLES Volume in the annulus: ID 2 Well (mm) − OD 2 DS (mm)
( )=
VAnnulus m
3
1,273,000
( )
VAnn(Csg DP) m = 3
3232 (mm) − 1272 (mm) 1,273,000
x L (m )
x 1,600 (m) = 0.06927 x 1,600 = 110.8 m 3
( )=
250.82 (mm) − 127 2 (mm)
( )=
250.82 (mm) − 1852 (mm)
VAnn (OH DP) m
VAnn (OH DC ) m
3
3
1,273,000
1,273,000
x 1,600 (m ) = 0.03673 x 1,600 = 58.8 m 3
x 200 (m ) = 0.02252 x 200 = 4.5 m 3
EXAMPLES
VAnnulus
Total
m
3
= VAnn (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
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 bot toms-up.
VP ump Output (l/stk ) =
VP ump Output (l/stk ) = V
(l/min )
ID2 Liner (mm) x L (mm) x Eff (decimal) 424,333
152.4 2 Liner (mm) x 304.8 (mm) x 0.9 (decimal) 424,333
= 15.01 (l/stk )
(
15.01 (l/stk ) x 110 stk/min 1,651 (l/min ) 1.651 m 3 / min
)
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
EXAMPLES
Hole cycle time (min ) =
V Well/DS m 3
(
)
3
V P ump Output m / min
)
=
204 . 5 1.651
= 124 min
Hole cycle (stk) = 124 min x 110 stk = 13,640 stk
Bottoms - up time (min ) =
V Annulus V P ump
Output
(m ) 3
Total
(m
3
/ min
Bottoms-up (stk) = 106 min x 110 stk = 11,660 stk
)
=
174 . 1 1.651
= 106 min
EXAMPLES
Part VII: Determine the velocity in the annular per each single int erval.
AV
=
V Pump
AV(OH/DC) =
AV(OH/DP) =
AV(Csg/DP) =
V
Output Ann
) = 73 m/min (m / m) 0.02252(m / m)
VPumpOutput m3 / min VAnn OH/DC
3
3
3
=
(
1.651 m 3 / min 3
) = 24 m/min (m / m) 0.06927 (m / m)
VPump Output m 3 / min VAnn Csg/DP
(
) = 45 m/min (m / m) 0.03673 (m / m)
VPump Output m 3 / min VAnn OH/DP
=
1.651 m3 / min
3
=
(
1.651 m 3 / min 3
EXAMPLES
Part VIII: Determine the hydrostatic pressure at the TD
P
P
H
H
(bar ) =
(bar ) =
Mud
(kg/l
Weight
) x TVD
(m )
10.2
1.5
(kg/l ) x 10.2
3,400
(m )
= 500
bars
DENSITY INCREASE THROUGH ADDITIVES
Weight increase (specific weight d = 4.2)
(
W kg/m
with:
3
(d 2 − d 1 ) ) = 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)
DENSITY INCREASE THROUGH ADDITIVES
Weight increase using Calcium Carbonate
WM
with:
(kg/m ) = 2650 3
(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)
DENSITY REDUCTION THROUGH WATER OR OIL
Water needed to reduce density (densit y H 2O d = 1)
(
VWater liters/m
with:
3
) = 1000 x
(d 1 − 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)
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)
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
MIXING OF LIQUIDS WITH DIFFERENT DENSITY Final volume after mixing
V A + VB = VF (V A x d A)+ (VB x dB) = (VF x dF)
with: VA = fluid volume A (m3) VB = fluid volume B (m3) VF = final volume(m3) d A = fluid density A (kg/l) dB = fluid density B (kg/l) dF = final density of the mixed fluids (kg/l)
MUD REPORT
MUD REPORT
MUD REPORT
