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Drilling Calculations Course

Drilling Calculations Course. Common Units and symbols. AREAS. Area of a square. Areas. Length. breadth. Area of a Square = length X breadth This same expression is applicable to a rectangle and parallelogram. Length. breadth. Area of a Triangle. Height. Height. base. base.

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Drilling Calculations Course

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  1. Drilling Calculations Course

  2. Common Units and symbols

  3. AREAS

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

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

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

  9. Volume of a cube Volumes 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) 1029.4 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 ) / 1029.4] x L1 ANNULAR VOLUME AROUND DC = [( D3² - D2² ) / 1029.4] 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 ) / 1029.4] x 4000 = 183.6 bbls ANNULAR VOLUME AROUND DC = [( 8.5² - 6.5² ) / 1029.4] x 800 = 23.31 bbls Total Annular Volume =183.6 + 23.31 = 206.91 bbl

  16. Pump displacement volumes D L 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 1029.4 x 12 x100 Where D= liner size in inches L= Stroke length in inches e= Volumetric efficiency in %

  17. Pump displacement volumes example D L Where D= 7 inches L= 11 inches e = 97% Pump displacement= 7 x 7 x 97 x 11 x 3 1029.4 x 12 x 100 = 0.127 bbls/ stroke

  18. 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 Weight on bit

  19. Tension Neutral point compression

  20. The apparent loss of weight when an object Is immersed in a fluid Is calculated using the expression: 65.44-mw or 1 - mw 65.44 65.44 This factor must be multiplied by the weight in air to calculate the buoyed weight Buoyed weight= weight in air x buoyancy factor Bouyancy factor

  21. 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 Apparent or buoyed weight

  22. A drill string weighs 180000 lbs in air how much will it weigh in 11.5ppg mud? Buoyed weight = weight in air X buoyancy factor Bouyancy factor= 65.44-mw 65.44 = 65.44-11.5 = 0.8242 65.44 Buoyed weight = 180,000 X 0.8242 = 148 356 lbs Example

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

  24. 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 Example

  25. Weight of drill collars in air=20x30x154= 92400 lbs Bouyancy factor= 65.44-13.4 = 0.795 65.44 Effective drill collar wt in fluid = 92400 x =0.795 =73458lbs 90% of available drill collar weight =0.9x73458=66112lbs Answer

  26. 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 88.88 pounds per foot The density of the drilling fluid is 12.4 ppg and 80% of the buoyed weight will be utilized Example

  27. 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 Example

  28. WET AND DRY TRIPPING

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

  30. 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. Tripping Dry 1- Calculate the volume of steel pulled: Length x Metal Displacement Example: DP Metal Disp = 0.00764 bbls/ft Length Pulled 93 feet Volume Of Steel Pulled: 93 x 0.00764 = 0.711 bbls

  32. Tripping Dry 2- Fill up the hole: You must pump 0.711 barrel of mud from the trip tank. You must investigate ( flow check) if more mud or less mud is needed.

  33. 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: Volume Of Steel Pulled: 93 x 0.00764 = 0.711 bbls DP Capacity: 0.01776 bbl/ft Annular Capacity: 0.0504 bbl/ft The mud will drop inside the pipe and the annular: 0.01776 + 0.0504 = 0.06816 bbl/ft

  35. Tripping Dry 3- NO FILL UP: Example Cont’d: The volume of drop is 0.711 bbls and will drop in a volume of 0.06816 bbl / ft, then the length of drop will be: 0.711 / 0.06816 = 10.43 feet. If 93 feet are pulled with no fill up, the mud level will drop by 10.43 feet.

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

  37. 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. Tripping Wet 1- Calculate the volume of steel pulled: Length x Metal Displacement Example: DP Metal Disp = 0.00764 bbls/ft Length Pulled 93 feet Volume Of Steel Pulled: 93 x 0.00764 = 0.711 bbls

  39. Tripping Wet 2- Calculate the volume of mud pulled: Length x DP Capacity Example: DP Capacity = 0.01776 bbls/ft Length Pulled 93 feet Volume Of Mud Pulled: 93 x 0.01776 = 1.65 bbls

  40. Tripping Wet 3- Calculate the total volume of steel and mud pulled: 1.65 + 0.711 = 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. 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: Volume Of Steel and Mud Pulled: 93 x (0.00764+0.01776) = 2.36 bbls Annular Capacity: 0.0504 bbl/ft The mud will drop inside the annular by: 2.36 / 0.0504 = 46.84 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 The pressureis aforce or weight per unit areaexerted 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. 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=0.05194 A 1 foot column of 1ppg fluid exerts a pressure of 0.052 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 0.052 = 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. 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. 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 0.052 x TVD

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