1. Drilling Calculations Course

2. Common Units and symbols
parameter
Unit
Symbol
length
Feet,inches or Mtr
Ft ,inor M
Area
Square feet, Square inches or square meters
Volume
Cubic feett, cubic centimeters cubic meters or
Cuft. Cu In or Meters cube
Capacity
US barrels,US gallons, Cubic ft
bbl, gal(US), cuft
Mass
Pounds,Short tons
lbs, Sh tn
Force (weight)
Pounds-force
lbf
Presure
Ppounds –force per square inch
psi
Density (mudweight)
Pounds per gallon, pounds per cubic ft
Ppg ,pcf
torque
Foot pounds
ftlbs

3. AREAS

4. Areas Area of a square Area of a Square = length X breadth
This same expression is applicable to a rectangle and parallelogram
Length
breadth

5. Area of a Triangle Area of a triangle = Base x Height 2 Height Height

6. Area of a Circle Area of a circle = Π x radius x radius
Diameter
Radius
Area of a circle = Π x radius x radius
Where r= radius of the circle
Π = 3.142
Since diameter= 2 x radius
Area of circle= Π x Diameter x Diamater 4

7. VOLUMES

8. Annular area Annular or cross sectional area = Π {(D x D)-(d x d)} 4
This can be used to calculate annular areas of the wellbore or
Cross sectional areas of tubulars

9. Volumes Volume of a cube Volume of a cube = length x breadth x height
hieght
breadth
length
Volume of a cube = length x breadth x height

10. Cylindrical Capacities
Height
Area
Volume of a cylinder = Area x Height
= Π x Diameter x Diameter x Height 4
This is the basis for all volume the capaciy calculations in well control

11. Capacity Calculations
Rule of thumb:
Capacity of a cylinder in bbls/ft= Diamater x Diameter
1029.4
Multiply this by the Height of the cylinder ( or length of the hole) gives the volumetric capacity in barrels
HVolume = Diameter x Diameter x height(or length)
This expression is generally used on the rig for volume and capacity calculations

12. Annular volume calculations
Top view d D
Annular Volume in bbls
{(D x D)-(d x d)}xlenght
1029.4

13. Hole volume calculations
ANNULAR VOLUME AROUND DP =
[( D3² - D1² ) / ] x L1
ANNULAR VOLUME AROUND DC =
[( D3² - D2² ) / ] x L2

14. Hole volume calculations example
Using the following data, calculate the total annular volume.
L1 = 4000 ft
L2 = 800 ft
D1 = 5 in
D2 = 6.5 in
D3 = 8.5 in

15. Hole volume calculations
ANNULAR VOLUME AROUND DP =
[( 8.5² - 5² ) / ] x 4000 = bbls
ANNULAR VOLUME AROUND DC =
[( 8.5² - 6.5² ) / ] x 800 = bbls
Total Annular Volume = = bbl

16. Pump displacement volumes
The same expression for calculation od hole volumes can
also be used for the calculation of mud pump displacements
Triplex Pump displacement in bbls/ft = D² x L x e x3
x 12 x100
Where D= liner size in inches
L= Stroke length in inches
e= Volumetric efficiency in %

17. Pump displacement volumes example
Where D= 7 inches
L= 11 inches
e = 97%
Pump displacement= 7 x 7 x 97 x 11 x 3
x 12 x 100
= bbls/ stroke

18. Weight on bit
Drill collars and/or heavy wall drill pipes are uses to put weight on the bit
Drill collars are also used to keep the string in tension and prevent it from buckling
There is a point in the drill collars where compression stops and tension starts this point is called the neutral point
Weight on bit is usually about 75-90%of the buoyed drill collar weight

19. Tension
Neutral point
compression

20. Bouyancy factor The apparent loss of weight when an object
Is immersed in a fluid
Is calculated using the expression:
65.44-mw or mw
This factor must be multiplied by the weight in air
to calculate the buoyed weight
Buoyed weight= weight in air x buoyancy factor

21. Apparent or buoyed weight
WB = WA x fluid density
Steel density
Where:
WB = weight of tubulars in fluid
WA = Weight of tubulars in air
Fluid density = Density of fluid in ppg
Steel Density =Density of Steel in PPG

22. Example
A drill string weighs lbs in air how much will it weigh in 11.5ppg mud?
Buoyed weight = weight in air X buoyancy factor
Bouyancy factor= mw
65.44
= =
Buoyed weight = 180,000 X
= lbs

23. A practical application of this is expression is
when calculating the number of drill collars to
give a desired weight on bit

24. Example
Find the maximum weight on bit if a drill string has 20 drill collars each 8 inch by 2.5 inch and 30ft in length and weight 154 lbs/ft
The density of the drilling fluid is 13.4ppg and 90% of the buoyed weight will be utilized

25. Answer Weight of drill collars in air=20x30x154= 92400 lbs
Bouyancy factor= = 0.795
65.44
Effective drill collar wt in fluid
= x =0.795 =73458lbs
90% of available drill collar weight =0.9x73458=66112 lbs

26. Example
Find the maximum weight on bit if a drill string has 30 drill collars each 6.5 inch by 3 inch and 30ft in length and has a nominal wt of pounds per foot
The density of the drilling fluid is 12.4 ppg and 80% of the buoyed weight will be utilized

27. Example
Find the maximum weight on bit available if a drill string has 15 drill collars each 9.5 inch by 2.75 inch and 30ft in length
The density of the drilling fluid is 13 ppg and 80% of the buoyed weight will be utilized

28. WET AND DRY TRIPPING

29. Tripping Dry
When a length of pipe is pulled from the hole, the mud level will fall.

30. The trip tank is then used to fill up the hole.
Tripping Dry
The volume of fall is equal to the volume of steel pulled from the hole.
The trip tank is then used to fill up the hole.
If 1 barrel of steel is removed from the hole, then using the trip tank, we have to add 1 barrel of mud.

31. 1- Calculate the volume of steel pulled: Length x Metal Displacement
Tripping Dry
1- Calculate the volume of steel pulled:
Length x Metal Displacement
Example:
DP Metal Disp = bbls/ft
Length Pulled 93 feet
Volume Of Steel Pulled:
93 x = bbls

32. Tripping Dry 2- Fill up the hole:
You must pump barrel of mud from the trip tank.
You must investigate ( flow check) if more mud or less mud is needed.

33. It will drop inside the pipe and in the annulus.
Tripping Dry
3- NO FILL UP:
If you fail to fill up the hole, the mud level will drop by the volume of steel pulled.
It will drop inside the pipe and in the annulus.

34. Tripping Dry 3- NO FILL UP: Example: DP Capacity: 0.01776 bbl/ft
Volume Of Steel Pulled:
93 x = bbls
DP Capacity: bbl/ft
Annular Capacity: bbl/ft
The mud will drop inside the pipe and the annular:
= bbl/ft

35. Tripping Dry 3- NO FILL UP: Example Cont’d:
The volume of drop is bbls and will drop in a volume of
bbl / ft,
then the length of drop will be:
0.711 / = feet.
If 93 feet are pulled with no fill up, the mud level will drop by feet.

36. Tripping Wet
When a length of pipe is pulled from the hole, the mud level will fall.

37. The trip tank is then used to fill up the hole.
Tripping Wet
The volume of fall is equal to the volume of steel pulled from the hole plus the volume of mud inside this pipe.
The trip tank is then used to fill up the hole.
If 3 barrels of steel and mud are removed from the hole, then using the trip tank, we have to add 3 barrels of mud.

38. 1- Calculate the volume of steel pulled: Length x Metal Displacement
Tripping Wet
1- Calculate the volume of steel pulled:
Length x Metal Displacement
Example:
DP Metal Disp = bbls/ft
Length Pulled 93 feet
Volume Of Steel Pulled:
93 x = bbls

39. 2- Calculate the volume of mud pulled:
Tripping Wet
2- Calculate the volume of mud pulled:
Length x DP Capacity
Example:
DP Capacity = bbls/ft
Length Pulled 93 feet
Volume Of Mud Pulled:
93 x = 1.65 bbls

40. 3- Calculate the total volume of steel and mud pulled:
Tripping Wet
3- Calculate the total volume of steel and mud pulled:
= 2.36 barrels

41. Tripping Wet 4- Fill up the hole:
You must pump 2.36 barrels of mud from the trip tank.
You must investigate ( flow check) if more mud or less mud is needed.

42. It will drop inside the annulus.
Tripping Wet
5- NO FILL UP:
If you fail to fill up the hole, the mud level will drop by the volume of steel and mud pulled.
It will drop inside the annulus.

43. Tripping Wet 5- NO FILL UP: Example: Annular Capacity: 0.0504 bbl/ft
Volume Of Steel and Mud Pulled:
93 x ( ) = 2.36 bbls
Annular Capacity: bbl/ft
The mud will drop inside the annular by:
2.36 / = feet

44. Mud Weight
The Mud weight is the density ( mass per unit volume ) of a drilling fluid.
Mud Weigth can be expressed in:
Pound per Gallon ( ppg )
Pound per Cubic Feet ( pcf )
Specific Gravity ( sg )

45. Pressure can be expressed in:
The pressure is a force or weight per unit area exerted by a gas or liquid against the walls of a vessel, hole, or object inserted into said gas or liquid.
Pressure can be expressed in:
Pounds per square inches ( psi ) OR
Kilograms per square centimeters ( kg / cm²)

46. A cubic foot contain 7.48 US gallons.
Derivation of 0.052
A cubic foot contain 7.48 US gallons.
If the fluid used to fill the cube weights 1 ppg, then the cube would weight 7.48 pounds.
The base of the cube has: 12x12=144 square inches.
7.48/144=
A 1 foot column of 1ppg fluid
exerts a pressure of psi

47. Pressure or Mud Gradient
Since the pressure is measured in psi and depth is measured in feet, it is convenient to convert mud weights from ppg to a pressure gradient in psi/ft.
The conversion factor is 0.052
Pressure gradient (psi/ft) = Fluid Density (ppg) x 0.052
Example:
A fluid having a mud weight of 10 ppg will have:
10 x = 0.52 psi ft as a pressure gradient.
A 1 foot column of 10 ppg mud will have a pressure of 0.52 psi

48. Hyd Pressure(psi) = Fluid gradient (psi/ft) x TVD
Hydrostatic Pressure
Hydrostatic Pressure is the pressure exerted by a column of fluid ( at rest )and is calculated by multiplying the gradient of the fluid by the True Vertical Depth at which the pressure is being measured:
Hyd Pressure(psi) = Fluid gradient (psi/ft) x TVD
Example:
What is the hydrostatic pressure at 10,000 ft for a mud gradient of 0.52 psi /ft ?
Hyd Pressure(psi) = 0.52 x 10,000
Hyd Pressure = 5,200 psi

49. Hydrostatic Pressure
________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
It is the vertical height/depth of the fluid column that matters, its shape is unimportant

50. Hyd Pressure(psi) = Fluid Density (ppg) x 0.052 x TVD
Hydrostatic Pressure
Using the formula for the pressure gradient and for the hydrostatic pressure:
Pressure gradient (psi/ft) = Fluid Density (ppg) x 0.052
Hyd Pressure(psi) = Fluid gradient (psi/ft) x TVD
We can write:
Hyd Pressure(psi) = Fluid Density (ppg) x x TVD

51. It is usefull to visualize the well as a U-tube.
Hydrostatic Pressure
It is usefull to visualize the well as a U-tube.
One column is for the pipe in the well and the other column is for the annulus.
If the same fluid is used is both columns, hydrostatic would be equal and the fluid would be static on both sides of the tube.
Mud Hydrostatic
in the string
Mud Hydrostatic
in the annulus

52. Well bore hydraulics

53. Well bore pressure losses
2600 psi
2500 psi
0 psi
100 psi
80 SPM
400 psi
300 psi
2100 psi
500 psi
1600 psi
1300 psi
300 psi

54. Hydraulic horsepower = PxQ
1714=
Where P= Pressure loss
V= Quantity of flow in Gallons per minute
Annular velocity= Pump output (bbls/min)
Annular capacity (bbls/ft)

55. Example
What is the hydraulic horsepower delivered by a pump at 400 gallons per minute at 3000psi
3000 x 400
1714
=700HP

56. Annular Velocity
Annular velocity is the speed with which the fluid travels up the annulus.
It is an indication of the amount of HHP available in the annulus

57. Annular Velocity Annular velocity = pump output (bbls/min)
Annular capacity (bbls/str)

58. Pump pressure vs pump rate
Pump pressure increased as pump speed increases
P2 =P1x SPM2 2
SPM1
Where:
P1=old pump pressure
P2=new pump pressure
SPM1=Old pump rate
SPM2= New pump rate

59. Pump pressure vs fluid density
Pump pressure also increases with fluid density
P2= P1 x MW2
MW1
Where P1=old pump pressure
P2=new pump pressure
MW1= Old mud wdensity
MW2 new mud density

60. Example At 100 spm the pumping pressure is 2850 psi with 13 ppg
What is the new pressure if the pumps are decreased to 65 spm?
What is the new pressure if the mud weight is then decreased to 10.5 ppg?

61. Solution:
2850 x 65 x x
100 x
= 973 psi
The new pump pressure will be 973 psi approximately

62. Work Done by the Drilling line

63. Each time the block are raised and lowered during the drilling process a certain amount of work is done by the drilling line
Work done = Force X Distance
= Ton X Miles
Tin mile calculations are used to determine the actual slip and cut off programs for the rig

64. Varieties of Ton-Miles Calculations
Round Trip Ton Miles
Drilling Ton miles
Casing Ton miles
Coring Ton miles

65. Drilling Ton Miles Dtm=3X(RTtm2-RTtm1) Where:
Dtm =Drilling Ton Miles between Trips
RTtm2 =Ton miles at depth when bit is run
RTtm1 =Ton miles at depth when bit is pulled

66. Round trip Ton Mile calculations
RTtm = Wp X (Lp+D) +2XDX(2Wb+Wc)
5280 X 2000
Where:
RTtm =Round trip Ton Miles
Wp= Buoyant weight of drill pipe in mud
D=Depth of the hole
Lp=average stand length
Wb=Weight of travelling block assembly
Wc=Buoyant weight of drill collars minus the buoyant
weight of the same length of drill pipe

67. Example Depth 7000 ft. Drilling fluid density 10 ppg
Drill pipe 5 inch diameter x 19.5 lb/ft.
Collars 500 ft of 8 inch OD x 3 inch ID x 147 lb/ft.
Travelling block assembly lbs.
Calculate the Round Trip ton miles at 7000 ft.

68. Answer:
From Drilling Data Handbook table the buoyancy factor = 0.848
The WP Buoyant weight of drill pipe = 19.5 x = lb/ft.
Wc,the Buoyant weight of drill collars =
(500 x 147 x 0,848) ‑ (500 x 16,53)= ‑ 8265 = lbs
Take average length of 1 stand = 90 ft.
RTtm = WP x D x ( LP + D ) + 2 x D x ( 2 Wb + Wc )
5280 x 2000
and with the numbers entered into the formula -
RTt= 16,53 x 7000 x ( ) + 2 x 7000 x ( 2 x x ) x 2000
= 189 TM

69. Ctm= D xWcx(Lc +D)+4 x D xWb 5280 x2000x2
Casing Ton Miles
Ctm= D xWcx(Lc +D)+4 x D xWb
5280 x2000x2
Where:Ctm= casing ton miles
Wc=Submerged wt of casing in mud (lbs/ft)
Lc=Average lenght of 1 joint f casing
Wb=weight of travelling block assembly

70. Ton miles for coring Coringtm=2x(RTtm2-RTtm1)
Where:RTtm1=Ton miles for trip where core barrel was run
:RTtm2=Ton miles for trip where core barrel was pulled

71. Stuck pipe calculations

72. Free point determination
L=M xexA
12 xP
Where:
L= lenght of free pipe
M=Modulus of elasticity of steel
e =Strech in drill pipe
A=cross sectional area of drill åpipe in sq inches
P= difference betweem maximum and minimum pull on pipe

73. Example A 10000 ft string of 4½" and 16,6 lb/ft drill pipe is stuck.
Maximum pull of lbs and a minimum pull of lbs gave a stretch figure of 45 inch. The cross sec­tional area of 4½" drill pipe is 4.4 sq.in.
Calculate the position of the free point.

74. M =
e = 45 inch
A = 4.4 sq.in
P = = lbs
Length of free pipe =
L = M x e x A = x 45 x 4.4
12 x P x 70000
= 7071 ft below the rotary table

75. Accummulator Volumetric requirements

76. Definitions Stored Hydraulic Fluid.
The fluid volume recoverable from the accumulator system between the maximum designed accumulator operating pressure and the precharge pressure.
Usable Hydraulic Fluid. The hydraulic fluid recoverable from the accumulator system between the maximum accumu­lator operating pressure and minimum calculated operating pressure or 200 psi above pre­charge pressure whichever is greatest.
Minimum Calculated Operating Pressure. The minimum calculated pressure to effectively close and seal a ram-type BOP against a wellbore pressure equal to the maximum rated working pressure of the BOP divided by the closing ratio specified for that BOP.
Component Minimum Operating Pressure Recom­mended by the Manufacturer. The minimum operating pres­sure to effectively close and seal ram-type or annular-type preventers under normal operating conditions, as prescribed by the manufacturer.

78. Boyles Law Pressure X Volume is constant Or P1V1=P2V2
At constant temparature If the volume of an enclosed gas bubble is doubled them the presure is reduced by half and if the pressure of a gas bubble is doubled the volume is halved

79. Example Accumulator bottle size 10 gallons.
Precharge pressure psi
Initial condition with only gas :
Pressure x Volume = Constant
1.000 x 10 =
The pump system is started and hydraulic fluid is pumped into the accumulator bottle until maximum operating pressure is reached at psi :
P1 x V1 = P2 x V  x 10 = x V2
V2= X = gallon gas
3000
Stored hydraulic fluid = = gal

81. The pump system is isolated and the BOP’s functioned until accumulator pressure reach precharge pressure psi:
P1 x V1 = P2 x V  x 10 = x V2
V2= 10 X =8.33 gallons gas
1200
Usable hydraulic fluid = = 5 gal
If the minimum operating pressure recommended by the manufacture is psi as for Shaffer Annular Preventer with pipe size smaller than 7” the usable hydraulic fluid would be :
P1 x V1 = P2 x V  x 10 = x V2
V2= 10 X =6.66 gallons gas
1500
Usable hydraulic fluid = = gal

82. Well Control

83. Mud Weight
The Mud weight is the density ( mass per unit volume ) of a drilling fluid.
Mud Weight can be expressed in:
Pound per Gallon ( ppg )
Pound per Cubic Feet ( pcf )
Specific Gravity ( sg )

84. Pressure can be expressed in:
The pressure is a force or weight per unit area exerted by a gas or liquid against the walls of a vessel, hole, or object inserted into said gas or liquid.
Pressure can be expressed in:
Pounds per square inches ( psi )
Kilograms per square centimeters ( kg / cm²)

85. Hyd Pressure(psi) = Fluid Density (ppg) x 0.052 x TVD
Hydrostatic Pressure
Using the formula for the pressure gradient and for the hydrostatic pressure:
Pressure gradient (psi/ft) = Fluid Density (ppg) x 0.052
Hyd Pressure(psi) = Fluid gradient (psi/ft) x TVD
We can write:
Hyd Pressure(psi) = Fluid Density (ppg) x x TVD

86. It is usefull to visualize the well as a U-tube.
Hydrostatic Pressure
It is usefull to visualize the well as a U-tube.
One column is for the pipe in the well and the other column is for the annulus.
If the same fluid is used is both columns, hydrostatic would be equal and the fluid would be static on both sides of the tube.
Mud Hydrostatic
in the string
Mud Hydrostatic
in the annulus

87. Hydrostatic Pressure
If the pipe or the annulus are not kept full, then the hydrostatic pressure in reduced.
Pipe is not full
HP is reduced
Annulus is not full
HP is reduced

88. Balanced when Hydrostatic Pressure = Formation Pressure
________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

89. Under Balanced when HP < FP

90. Over Balanced when HP > FP

91. The first section to fill is the well data.
Kill Sheet
The kill sheet is a tool used to kill the well. It gathers all the informations needed and help you to perform the calculations.
The first section to fill is the well data.

92. Kill Sheet
The pump displacement must be entered and the Slow Circulating Rate must be recorded.

93. Kill Sheet
This section is used to calculate all the hole volumes, to convert them in strokes and in time.

95. Drillpipe Internal volume

96. Heavy weight dril pipe Internal Volume

97. Dril Collars Internal Volume

98. Drill Collar open hole annular volume

99. Drill pipe- Open hole annular volume

100. Drill pipe-Casing annular volume

101. Kill Sheet
The formulas are used to calculate the kill mud, Initial Circulating Pressure, Final Circulating Pressure and to prepare the graph.

102. Kill Sheet
The graph will be followed to maintain a constant Bottom Hole Pressure.

103. It is useful to pump a slug before tripping.
Pumping a Slug
It is useful to pump a slug before tripping.
The slug weight being heavier than the mud, a length of pipe will be empty.
The HP is not reduced because the heavier mud will compensate for the empty pipe.

104. The total HP is the same on both sides of the pipe.
Pumping a Slug
The total HP is the same on both sides of the pipe.
HP kmw
HP mud
HP mud

105. Pumping a Slug Example:
If 20 bbls of 12 ppg slug are pumped in a 10,000 ft hole containing 10 ppg mud, what will be the height of empty pipe?
DP capacity = bbl/ft
1- Calculate the height of the slug:
20 / = 1126 ft

106. Pumping a Slug 2- Calculate the HP of the slug:
1126 x 12 x = psi
702.6 psi

107. Pumping a Slug 2- Calculate the HP of the mud in the annulus:
10,000 x 10 x = 5,200 psi
702.6 psi
5,200 psi

108. Pumping a Slug
3- The total hydrostatic beeing the same on both sides, calculate the HP of the mud below the slug:
5, = psi
702.6 psi
5,200 psi
psi

109. Pumping a Slug
4- Calculate the height of mud needed to give psi as a HP:
TVD = / ( 10 x ) = feet
1,126 ft
10,000 ft ft

110. Pumping a Slug 4- Calculate the height of empty pipe
10, ,126 = ft
225.2 ft
1,126 ft
10,000 ft ft

111. Practical Exercise
3- If 28 bbls of 14 ppg slug are pumped in a 12,000 ft hole containing 11 ppg mud, what will be the height of empty pipe?
DP capacity = bbl/ft

112. MUD WEIGHT CHANGE
2600 psi
A well is being drilled using 10 ppg mud. At 80 spm the total circulating system pressure losses are 2600 psi.
It is decided to increase the mud weight to 11 ppg.
80 spm
Mud wt 10 ppg

113. MUD WEIGHT CHANGE 2860 psi 80 spm Mud wt 11 ppg
It is a good drilling practice to calculate the new circulating pressure before changing the mud weight.
The way we calculate this change in pressure is to use the following formula;
New Mud ppg
Old Mud ppg x Old psi.
11 ppg ppg x 2600 = 2860 psi
The new pump pressure would be approximately 2860 psi.
2860 psi
80 spm
Mud wt 11 ppg

114. Final Circulating Pressure
MUD WEIGHT CHANGE
Final Circulating Pressure
The formula that was just used to calculate the pressure change due to a change in mud weight, is also the formula used to calculate the Final Circulating Pressure.
Kill Mud ppg
Old Mud ppg x Slow Pump psi.

115. PUMP STROKE CHANGE
A well is being drilled using 10 ppg mud. At 80 spm the total circulating system pressure losses are 2600 psi.
It is decided to increase the pump speed from 80 spm to 100 spm.
2600 psi
80 spm
Mud wt 10 ppg

116. PUMP STROKE CHANGE 4063 psi 100 spm Mud wt 10 ppg
It is a good drilling practice to calculate the new circulating pressure before changing the pump speed.
The way we calculate this change in pressure is to use the following formula; New SPM Old psi x Old SPM
2600 x 100 spm spm = 4063 psi
The new pump pressure would be approximately 4063 psi.
4063 psi
100 spm
Mud wt 10 ppg

117. Height of Influx

118. Influx gradient Where: Gi=Influx Gradient Gm= Mud Gradient
hi= Hieght of influx