## Basic Calculations

Slug Calculations 33. Accumulator Capacity 37. Bulk Density of Cuttings 41. Drill String Design (Limitations) 42. Ton-Mile Calculations 44. Cementing Calculations 47. Weighted Cement Calculations 53. Calculations for the Number of Sacks of Cement Required 54. Calculations for the Number of Feet to Be Cemented 57. Setting a Balanced Cement Plug 61. Differential Hydrostatic Pressure Between Cement in the Annulus and Mud Inside the Casing 65. Hydraulicing Casing 66. Depth...

## Round trip tonmiles RTtm

RT _ Wp x D x (Lp + D) + (2 x D) (2 x Wb + Wc) 5280 x 2000 Wp buoyed weight of drill pipe, lb ft D depth of hole, ft Lp length of one stand of drill pipe, (ave), ft Wb weight of traveling block assembly, lb Wc buoyed weight of drill collars in mud minus the buoyed weight of the same length of drill pipe, lb 2000 number of pounds in one ton 5280 number of feet in one mile Traveling block assembly 15,0001b Average length of one stand 60 ft (double) BF 65.5 - 9.6ppg. + 65.5 BF 0.8534 b) Buoyed...

## Tonmiles during coring operations

The ton-miles of work performed in coring operations, as for drilling operations, is expressed in terms of work performed in making round trips. To determine ton-miles while coring, take 2 times ton-miles for one round trip at the depth where coring stopped minus ton-miles for one round trip at the depth where coring began T4 ton-miles for one round trip depth where coring stopped before coming out of hole T3 ton-miles for one round trip depth where coring started after going in hole

## English units calculation

bbl slug pumped x slug wt, ppg mud wt, ppg - bbl slug Example Determine the number of barrels of mud gained due to pumping the slug and determine the feet of dry pipe. Barrels of slug pumped 25 barrels Drill pipe capacity 0.01776 bbl ft Barrels gained 25bbl x 14.2ppg 12.6ppg - 25bbl 28.175 - 25 3.175 bbl Determine the feet of dry pipe after pumping the slug. Feet of dry pipe 3.175bbl - 0.01776bbl ft 179 feet

## Length of bottomhole assembly necessary for a desired weight on bit

Where WOB desired weight to be used while drilling f safety factor to place neutral point in drill collars Wdc drill collar weight, lb ft BF buoyancy factor Example Desired WOB while drilling 50,0001b Safety factor 15 Drill collar weight 147 lb ft b Length of bottomhole assembly necessary 50,000 x 1.15

## Basic Formulas

Hydrostatic Pressure 3. Converting Pressure into Mud Weight 4. Specific Gravity 5. Equivalent Circulating Density 6. Maximum Allowable Mud Weight 7. Pump Output 7. Annular Velocity 9. Capacity Formulas 12. Control Drilling 19. Buoyancy Factor 20. Hydrostatic Pressure Decrease When Pulling Pipe out of the Hole 20. Loss of Overbalance Due to Falling Mud Level 22. Formation Temperature 24. Hydraulic Horsepower 25. Drill Pipe Drill Collar Calculations 25. Pump Pressure Pump...

## Feet of drill pipe that can be used with a specific bottomhole assembly BHA

NOTE Obtain tensile strength for new pipe from cementing handbook or other source. b Determine maximum length of drill pipe that can be run into the hole with a specific bottomhole assembly where T tensile strength, lb for new pipe f safety factor to correct new pipe to no. 2 pipe MOP margin of overpull Wbha BHA weight in air, lb ft Wdp drill pipe weight in air, lb ft, including tool joint BF buoyancy factor c Determine total depth that can be reached with a specific bottomhole assembly Total...

## Basic formula

New circulating _ x f new pump rate, spm pressure, psi p I old pump rate, spm Example Determine the new circulating pressure, psi using the following data Present circulating pressure 1800 psi Old pump rate 60 spm New circulating 18 30spm V pressure, psi V 60 spm New circulating 1800 ix0 25 pressure, psi

## Formation Temperature FT

Vincrease F per ft of depth x TVD, ft Example If the temperature increase in a specific area is 0.012 F ft of depth and the ambient surface temperature is 70 F, determine the estimated formation temperature at a TVD of 15,000ft FT, F 70 F 0.012 F ft x 15,000ft FT, F 70 F 180 F FT 250 F estimated formation temperature where HHP hydraulic horsepower P circulating pressure, psi Q circulating rate, gpm Example circulating pressure 2950 psi circulating rate 520 gpm Drill Pipe Drill Collar...

## When pulling DRY pipe Step

Arre s of stands x length per x displacement displaced pu ed stand ft bbl ft HP, psi barrels displaced x 0 052 x mud decrease casing pipe capacity, - displacement, bbl ft bbl ft Example Determine the hydrostatic pressure decrease when pulling DRY pipe out of the hole Number of stands pulled 5 Average length per stand 92 ft Pipe displacement 0.0075 bbl ft Casing capacity 0.0773 bbl ft

## Drilling or connection tonmiles

The ton-miles of work performed in drilling operations is expressed in terms of work performed in making round trips. These are the actual ton-miles of work involved in drilling down the length of a section of drill pipe usually approximately 30 ft plus picking up, connecting, and starting to drill with the next section. To determine connection or drilling ton-miles, take 3 times ton-miles for current round trip minus ton-miles for previous round trip where Td drilling or connection ton-miles...

## Barrels of slug required for a desired length of dry pipe Step

Hydrostatic pressure required to give desired drop inside drill pipe HP, psi mud wt, ppg x 0.052 x ft of dry pipe Difference in pressure gradient between slug weight and mud weight psi ft slug wt, ppg - mud wt, ppg x 0.052 Slug length, ft pressure, psi ,. Slug vol, bbl slug length, ft x P. 6 Example Determine the barrels of slug required for the following Desired length of dry pipe 2 stands 184 ft Mud weight 12.2 ppg Drill pipe capacity 0.01422 bbl ft Hydrostatic pressure required HP, psi 12.2...

## Volume of slug

Slug length, ft slug vol, bbl drill pipe capacity, bbl ft Hydrostatic pressure required to give desired drop inside drill pipe HP, psi mud wt, ppg x 0.052 x ft of dry pipe Weight of slug, ppg Slug wt, ppg HP, psi 0.052 slug length, ft mud wt, ppg Example Determine the weight of slug required for the following Desired length of dry pipe 2 stands 184 ft Mud weight 12.2ppg Drill pipe capacity 0.01422 bbl ft Slug length, ft 25 bbl 0.01422bbl ft Slug length 1758 ft Hydrostatic pressure required HP,...

## Feet of pipe pulled WET to lose overbalance

Overbalance, psi x casing cap. - pipe cap. - pipe disp. -------- -- mud wt., ppg x 0.052 x pipe cap. pipe disp., bbl ft Example Determine the feet of WET pipe that must be pulled to lose the overbalance using the following data Amount of overbalance 150 psi Casing capacity 0.0773 bbl ft Pipe displacement 0.0075 bbl ft Mud weight 11.5 ppg 150psi x 0.0073 - 0.01776 - 0.0075bbl ft 11.5ppg x 0.052 0.01776 0.0075bbl ft drilling fluid _ metal displacement, n nQ8, Pressure drop per density kg ym_X...

## Volume height and pressure gained because of slug

A Volume gained in mud pits after slug is pumped, due to U-tubing Vol, bbl ft of dry pipe x drill pipe capacity, bbl ft b Height, ft, that the slug would occupy in annulus Height, ft annulus vol, ft bbl x slug vol, bbl c Hydrostatic pressure gained in annulus because of slug HP si - sluS x difference in gradient, psi ft 'pS1 in annulus, ft between slug wt and mud wt Example Feet of dry pipe 2 stands 184 ft Drill pipe capacity 0.01422 bbl ft Annulus volume 8-1 2in. by 4-l 2in. 19.8ft bbl a...

## Capacities bblft displacement bblft and weight lbft can be calculated from the following formulas

Weight, lb ft displacement, bbl ft x 27471b bbl Example Determine the capacity, bbl ft, displacement, bbl ft, and weight, lb ft, for the following Drill collar OD 8.0in. Drill collar ID 2-13 16in. Convert 13 16 to decimal equivalent a Capacity, bbl ft 7 y 3 1029.4 b Displacement, bbl ft Displacement, bbl ft c Weight, lb ft 0.0544879 bbl ft x 27471b bbl Weight 149.678 lb ft

## Capacity of tubulars and open hole drill pipe drill collars tubing casing hole and any cylindrical o

Example Determine the capacity, bbi ft, of a 12-l 4in. hole 12.252 Example Determine the capacity, ft bbl, of 12-l 4in. hole Example Determine the capacity, gal ft, of 8-1 2in. hole Capacity, gal ft Capacity 2.9477764gal ft Example Determine the capacity, ft gal, of 8-1 2 in. hole r, 24.51 Capacity, ft gal - Example Determine the capacity, ft3 linft, for a 6.0 in. hole 6 02 f Capacity, linft ft3 183,35 ' y ID, in.2 Example Determine the capacity, linft ft3, for a 6.0in. hole Capacity, linft ft3...

## Amount of cuttings drilled per foot of hole drilled

A BARRELS of cuttings drilled per foot of hole drilled Example Determine the number of barrels of cuttings drilled for one foot of 12-l 4in.-hole drilled with 20 0.20 porosity Barrels 0.1457766 x 0.80 Barrels 0.1166213 b CUBIC FEET of cuttings drilled per foot of hole drilled Cubic feet x 0.7854 1 - porosity Example Determine the cubic feet of cuttings drilled for one foot of 12-1 4 in. hole with 20 0.20 porosity Cubic feet x 0.7854 1 - 0.20 Cubic feet 150 0626 x 0.7854 x 0.80 144 where Wcg...

## Rule of thumb formulas

Weight, lb ft, for REGULAR DRILL COLLARS can be approximated using the following formula Weight, lb ft OD, in.2 - ID, in.2 2.66 Drill collar OD 8.0in. Drill collar ID 2-13 16in. Decimal equivalent 2.8125 in. Weight, lb ft 8.02 - 2.81252 2.66 Weight, lb ft 56.089844 x 2.66 Weight 149.19898 lb ft Weight, lb ft, for SPIRAL DRILL COLLARS can be approximated using the following formula Weight, lb ft OD, in.2 - ID, in.2 2.56 Drill collar OD 8.0in. Drill collar ID 2-13 16in. Decimal equivalent 2.8125...

## Cement additive calculations

A Weight of additive per sack of cement Weight, lb percent of additive x 941b sk b Total water requirement, gal sk, of cement , Cement water Additive water Water, gal sk . . ,,, requirement, gal sk requirement, gal sk SG of cement x8.331b gal weight of additive, lb SG of additive x 8.331b gal water volume, gal Yield, ft3 sk vol of slurry, gal sk 7.48 gal ft3 e Slurry density, lb gal 94 wtof additive 8.33 x vol of water sk Example Class A cement plus 4 bentonite using normal mixing water Amount...

## Water requirements

Weight, lb sk 94 8.33 x vol of water, gal of additive x 94 Vol, gal sk --1---1- water vol, gal c Water requirement using material balance equation D,V, D2V2 Example Class H cement plus 6 bentonite to be mixed at 14.01b gal. Specific gravity of bentonite 2.65. Bentonite requirement, lb sk Water requirement, gal sk Slurry yield, ft3 sk Check slurry weight, lb gal Weight, lb sk 94 0.06 x 94 8.33 x y Weight, lb sk 94 5.64 8.33y Weight 99.64 8.33y Vol, gaVsk - - - - y Vol, gal sk 3.6 0.26 y 3 Water...

## Subsea applications

In subsea applications the hydrostatic pressure exerted by the hydraulic fluid must be compensated for in the calculations Example Same guidelines as in surface applications Hydrostatic pressure of hydraulic fluid 445 psi Adjust all pressures for the hydrostatic pressure of the hydraulic fluid Pre-charge pressure 1000 psi 445 psi 1445 psi Minimum pressure 1200 psi 445 psi 1645 psi Maximum pressure 3000 psi 445 psi 3445 psi Determine hydraulic fluid necessary to increase pressure from pre-charge...

## Weighted Cement Calculations

Amount of high density additive required per sack of cement to achieve a required cement slurry density wt x 11.207983, . ,, n - wt x CW - 94 - 8.33 x CW where x additive required, pounds per sack of cement Wt required slurry density, lb gal SGc specific gravity of cement CW water requirement of cement AW water requirement of additive SGa specific gravity of additive Additive gal 94 lb sk Gravity Example Determine how much hematite, lb sk of cement would be required to increase the density of...

## SI units calculations

Drilling fluid metal disp-, density, kg m3 m3 m_ dry pipe, kPa m cafng capaclty lt - mfal disP x 102 m ' m m- m drilling fluid Pressure drop per density, kg m3 meter tripping wet pipe, kPa m Level drop for POOH _ length of drill collars, m x metal disp., m3 m drill collars, m casing capacity, m3 m

## Tonmiles while making short trip

The ton-miles of work performed in short trip operations is also expressed in terms of round trips. Analysis shows that the ton-miles of work done in making a short trip is equal to the difference in round trip ton-miles for the two depths in question. where Tst ton-miles for short trip T6 - ton-miles for one round trip at the deeper depth, the depth of the bit before starting the short trip. T5 ton-miles for one round trip at the shallower depth, the depth that the bit is pulled up to.