Nomenclature Pko
Di drill pipe or drill collar ID, in. Dp drill pipe or drill collar OD, in Example 1 Determine surge pressure for plugged pipe Data Well depth 15,000 ft Drill pipe OD Hole size 7-7 8 in. Drill pipe ID Drill collar length 700 ft Mud weight Average pipe running speed 270 ft mm Drill collar 6-1 4 OD x 2-3 4 ID Viscometer readings f600 140 3. Determine velocity, ft mm v 0.45 4.52_ 270 v 0.45 0.484 270 v 252 ft min 4. Determine maximum pipe velocity, ft min Vm 1.5 x 252 5. Determine pressure losses,...
Hydrostatic Pressure Exerts by Each Barrel of Mud in the Casing
psi bbl 1029.4 x 0.052 x mud wt, ppg Dh2 Dp2 Example Dh 9-5 8 in, casing 43.5 lb ft 8.755 in. ID Dp 5.0 in. OD Mud weight 10.5 ppg psi bbl 1029.4 x 0.052 x 10.5 ppg 8.7552 5.02 psi bbl 19.93029 x 0.052 x 10.5 ppg psi bbl 10.88 psi bbl 1029.4 x 0.052 x mud wt ppg ID2 Example Dh 9-5 8 in. casing 43.5 lb ft 8.755 in. ID Mud weight 10.5 ppg psi bbl 1029.4 x 0.052 x 10.5 ppg 8.7552 psi bbl 13.429872 x 0.052 x 10.5 ppg psi bbl 7.33
Hydrostatic Pressure Decrease at TD Caused by Gas Cut Mud Method
HP decrease, psi 100 weight of uncut mud, ppg weight of gas cut mud, ppg Example Determine the hydrostatic pressure decrease mud using the following data Weight of uncut mud 18.0 ppg Weight of gas cut mud 9.0 ppg HP decrease, psi 100 x 18.0 ppg 9.0 ppg HP Decrease 100 psi Method 2 P MG - C V where P reduction in bottomhole pressure, psi MG mud gradient, psi ft C annular volume, bbl ft V pit gain, bbl C 0.0459 bbl ft Dh 8.5 in. Dp 5.0 in. V 20 bbl Solution P 0.624 psi ft - 0.0459 bbl ft 20 P...
Equivalent Circulating Density ECD ppg
ECD, ppg annular pressure, loss, psi - 0.052 - TVD, ft mud weight, in use, ppg Example annular pressure loss 200 psi true vertical depth 10,000 ft ECD, ppg 200 psi - 0.052 - 10,000 ft 9.6 ppg ECD 10.0 ppg 6. Maximum Allowable Mud Weight from Leak-off Test Data ppg Leak-off Pressure, psi - 0.052 - Casing Shoe TVD, ft mud weight, ppg Example leak-off test pressure 1140 psi casing shoe TVD 4000 ft Mud weight 10.0 ppg ppg 1140 psi - 0.052 - 4000 ft 10.0 ppg ppg 15.48
Psistroke
ICP psi FCP_psi - strokes to bit_ _psi stk Example Use the following data and fill out a kill sheet Example Use the following data and fill out a kill sheet drill pipe 5.0 in. 19.5 lb ft capacity drill collars 8.0 in. OD 3.0 in. ID Mud pump 7 in. x 12 in. triplex 95 eff.
Drill Stem Bore Pressure Losses
where P drill stem bore pressure losses, psi MW mud weight, ppg L length of pipe, ft Q circulation rate, gpm Example Mud weight 10.9 ppg Length of pipe 6500 ft Circulation rate 350 gpm Drill pipe ID 4.276 in. P 0.000061 x 10.9 x 6500 x 350 186 4.276486
Adjusting Maximum Allowable Shutin Casing Pressure For an Increase in Mud Weight
MASICP Pl D x mud wt2 mud wtx 0.052 where MASICP maximum allowable shut-in casing annulus pressure, psi PL leak-off pressure, psi D true vertical depth to casing shoe, ft Mud wt2 new mud wt, ppg Mud wt1 original mud wt, ppg Example Leak-off pressure at casing setting depth TVD of 4000 ft was 1040 psi with 10.0 ppg in use. Determine the maximum allowable shut-in casing pressure with a mud weight of 12.5 ppg MASICP 1040 psi 4000 x 12.5 10.0 0.052 MASICP 1040 psi 520 MASICP 520 psi
Nomenclature Iou
n dimensionless K dimensionless X dimensionless f 600 600 viscometer dial reading f300 300 viscometer dial reading Example Mud weight 14.0 ppg f 600 64 f300 37 Avc critical annular velocity, ft mm Hole diameter 8.5 in. Pipe OD 7.0 in. 4. Determine critical annular velocity X 81600 0.2684 079 0.387 8.5 70.79 x 14.0 AVc 1035 k 2 079 AVc 103 5 08264 AVc 310 ft mm
Critical RPM RPM to Avoid Due to Excessive Vibration Accurate to Approximately
Critical RPM 33055 x V OD, in.2 ID, in.2 L, ft2 Example L length of one joint of drill pipe 31 ft OD drill pipe outside diameter 5.0 in. ID drill pipe inside diameter 4.276 in. Critical RPM 33055 x V 5.02 4.2762 312 Critical RPM 34.3965 x 6.579 Critical RPM 226.296 NOTE As a rule of thumb, for 5.0 in. drill pipe, do not exceed 200 RPM at any depth.
Annular Pressure Losses
P 1.4327 x 10 7 x MW x Lx V2 Dh Dp where P annular pressure losses, psi L length, ft Dh hole or casing ID, in. Example Mud weight 12.5 ppg Circulation rate 350 gpm Drill pipe OD 5.0 in. Determine annular velocity, ft mm v MW mud weight, ppg V annular velocity, ft mm Dp drill pipe or drill collar OD, in. Length 6500 ft Hole size 8.5 in. Determine annular pressure losses, psi P 1.4327 x 10 7 x 12.5 x 6500 x 1812
Reduction in Bottomhole Pressure if Riser is Disconnected
Example Use the following data and determine the reduction in bottom-hole pressure if the riser is disconnected Data Air gap 75 ft Water depth 700 ft Seawater gradient 0.445 psi ft Well depth 2020 ft RKB Mud weight 9.0 ppg Step 1 Determine bottomhole pressure BHP 9.0 ppg x 0.052 x 2020 ft BHP 945.4 psi Step 2 Determine bottomhole pressure with riser disconnected BHP 0.445 x 700 9.0 x 0.052 x 2020 700 75 BHP 311.5 582.7 BHP 894.2 psi Step 3 Determine bottomhole pressure reduction BHP reduction...
Nomenclature
pressure at top of gas bubble, psi gradient of influx, psi ft total depth, ft feet in annulus occupied by drill string volume mud weight, ppg pressure at depth of interest, psi pressure exerted by influx, psi pressure gradient of mud weight in use, ppg height of influx, ft formation pressure, psi pressure gradient of kill weight mud, ppg pressure at surface, psi shut-in drill pipe pressure, psi shut-in casing pressure, temperature, degrees Fahrenheit, at depth of interest temperature, degrees...
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 Water depth 1000 ft Hydrostatic pressure of hydraulic fluid 445 psi Step 1 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 Step 2 Determine hydraulic fluid necessary to...
Calculations for the Number of Feet to Be Cemented
If the number of sacks of cement is known, use the following Step 1 Determine the following capacities Annular capacity, ft 3 ft Dh, in.2 Dp, in.2 Step 2 Determine the slurry volume, ft3 Slurry vol, ft3 number of sacks of cement to be used x slurry yield, ft3 sk Step 3 Determine the amount of cement, ft3, to be left in casing Cement in feet of setting depth of x casing capacity, ft3 ft h- excess casing, ft3 casing cementing tool, ft Step 4 Determine the height of cement in the annulus feet of...
Surface Equipment Pressure Losses
where SEpl surface equipment pressure loss, psi Q circulation rate, gpm C friction factor for type of surface equipment W mud weight, ppg
Cement additive calculations
a Weight of additive per sack of cement Weight, lb percent of additive x 94 lb sk b Total water requirement, gal sk, of cement Water, gal sk Cement water requirement, gal sk Additive water requirement, gal sk Vol gal sk 94 lb_ weight of additive, lb_ water volume, gal SG of cement x 8.33 lb gal SG of additive x 8.33 lb gal Yield, ft3 sk vol. of slurry, gal sk 7.48 gal ft3 Density, lb gal 94 wt of additive 8.33 x vol. of water sk Example Class A cement plus 4 bentonite using normal mixing water...
Capacity of tubulars and open hole drill pipe drill collars tubing casing hole
a Capacity, bbl ft ID in.2 Example Determine the capacity, bbl ft, of a 12-1 4 in. hole b Capacity, ft bbl 1029.4 Example Determine the capacity, ft bbl, of 12-1 4 in. hole c Capacity, gal ft ID in.2 Example Determine the capacity, gal ft, of 8-1 2 in. hole d Capacity, ft gal ID in 2 Example Determine the capacity, ft gal, of 8-1 2 in. hole e Capacity, ft3 linft ID2 Example Determine the capacity, ft3 linft, for a 6.0 in. hole f Capacity, linftlft3 183.35 Example Determine the capacity, linft...
Gas Flow Into the Wellbore
Flow rate into the wellbore increases as wellbore depth through a gas sand increases Q 0.007 x md x Dp x L - U x ln Re Rw 1,440 where Q flow rate, bbl min md permeability, millidarcys Dp pressure differential, psi L length of section open to wellbore, ft U viscosity of intruding gas, centipoise Re radius of drainage, ft Rw radius of wellbore, ft Example md 200 md Dp 624 psi L 2Oft U 0.3cp ln Re - Rw 2,0 Q 0.007 x 200 x 624 x 20 - 0.3 x 2.0 x 1440 Q 20 bbl min Therefore If one minute is required...
Maximum Influx Barrels to Equal Maximum Allowable Shutin Casing Pressure MASICP
Example Maximum influx height to equal MASICP from above example 2185 ft Annular capacity drill collars open hole 12-1 4 in. x 8.0 in. 0.0826 bbl ft Annular capacity drill pipe open hole 12-1 4 in. x 5.0 in. 0.1215 bbl ft Drill collar length 500 ft Step 1 Determine the number of barrels opposite drill collars Barrels 0.0836 bbl ft x 500 ft Barrels 41.8 Step 2 Determine the number of barrels opposite drill pipe Influx height, ft, opposite drill pipe ft 2185 ft 500 ft Barrels opposite drill pipe...
d Exponent
The d exponent is derived from the general drilling equation R N a Wd - D where R penetration rate d exponent in general drilling equation, dimensionless N rotary speed, rpm a a constant, dimensionless d exponent equation d log R - 60N - log 12W - 1000D where d d exponent, dimensionless R penetration rate, ft hr N rotary speed, rpm W weight on bit, 1,000 lb Example R 30 ft hr N 120 rpm W 35,000 lb D 8.5 in. Solution d log 30 - 60 x 120 - log 12 x 35 1000 x 8.5 d log 30 - 7200 - log 420 - 8500 d...
Basic Calculations
3. Accumulator Capacity Usable Volume Per Bottle 4. Bulk Density of Cuttings Using Mud Balance 5. Drill String Design Limitations 8. Weighted Cement Calculations 9. Calculations for the Number of Sacks of Cement Required 10. Calculations for the Number of Feet to Be Cemented 11. Setting a Balanced Cement Plug 12. Differential Hydrostatic Pressure Between Cement in the Annulus and Mud Inside the Casing 15. Lost Returns Loss of Overbalance 17. Calculations Required for Spotting Pills 18. Pressure...
Method
Pump soft line or other plugging material down the drill pipe and notice how many strokes are required before the pump pressure increases. Depth of washout, ft strokes required x pump output, bbl stk - drill pipe capacity, bbl ft Example Drill pipe 3-1 2 in. 13.3 lb ft Capacity 0.00742 bbl ft Pump output 0.112 bbl stk 5-1 2 in. by 14 in. duplex 90 efficiency NOTE A pressure increase was noticed after 360 strokes. Depth of washout, ft 360 stk x 0.112 bbl stk - 0.00742 bbl ft Depth of washout...
Water Requirements and Specific Gravity of Common Cement Additives
Water Requirement gal 94 lb sk Specific Gravity Table 2-1 continued Water Requirements and Specific Gravity of Common Cement Additives Water Requirement gal 94 lb sk Specific Gravity Table 2-1 continued Water Requirements and Specific Gravity of Common Cement Additives Water Requirement gal 94 lb sk Specific Gravity
Pressure required to overcome the muds gel strength in the annulus
Pgs y - 300 Dh, in. Dp, in. x L where Pgs pressure required to break gel strength, psi L length of drill string, ft y 10 mm. gel strength of drilling fluid, lb 100 sq ft Dh hole diameter, in. Dp pipe diameter, in. Example L 12,000 ft y 10 lb 100 sq ft Pgs 10 - 300 x 12.25 5.0 x 12,000 ft Pgs 10 - 2175 x 12,000 ft Pgs 55.2 psi Therefore, approximately 55 psi would be required to break circulation.
Height Gain From Stripping into Influx
where L length of pipe stripped, ft Cdp capacity of drill pipe, drill collars, or tubing, bbl ft Ddp displacement of drill pipe, drill collars or tubing, bbl ft Ca annular capacity, bbl ft Example If 300 ft of 5.0 in. drill pipe 19.5 lb ft is stripped into an influx in a 12-1 4 in. hole, determine the height, ft, gained DATA Drill pipe capacity 0.01776 bbl ft Length drill pipe stripped 300 ft Drill pipe displacement 0.00755 bbl ft Annular capacity 0.1215 bbl ft Solution Height, ft 300 0.01776...
Annular capacity between casing or hole and drill pipe tubing or casing
a Annular capacity, bbl ft Dh2 Dp2 Example Hole size Dh 12-1 4 in. Drill pipe OD Dp 5.0 in. Annular capacity, bbl ft 12.252 5.02 b Annular capacity, ft bbl 1029.4 Example Hole size Dh 12-1 4 in. Drill pipe OD Dp 5.0 in. c Annular capacity, gal ft Dh2 Dp2 Example Hole size Dh 12-1 4 in. Drill pipe OD Dp 5.0 in. Annular capacity, gal ft 12.252 5.02 d Annular capacity, ft gal 24.51 Example Hole size Dh 12-1 4 in. Drill pipe OD Dp 5.0 in. Annular capacity, ft gal 0.19598 ft gal e Annular capacity,...
Volume of Mud to Bleed to Maintain Constant Bottomhole Pressure with a Gas
With pipe in the hole Vmud Dp x Ca . where Vmud volume of mud, bbl, that must be bled to maintain constant bottomhole pressure with a gas bubble rising. Dp incremental pressure steps that the casing pressure will be allowed to increase. Ca annular capacity, bbllft Example Casing pressure increase per step 100 psi Gradient of mud 13.5 ppg x 0.052 0.70 psi ft Annular capacity Dh 12-1 4 in. Dp 5.0 in. 0.1215 bbl ft Vmud 100 psi x 0.1215 bbl ft 0.702 psi ft With no pipe in hole Vmud Dp x Ch ....
Maximum recommended low gravity solids LGS
LGS SF -r- 100 0.3125 x MW -- 8.33 1 x 200 where SF maximum recommended solids fractions, by vol LGS maximum recommended low gravity solids, by vol MW mud weight, ppg Determine Maximum recommended solids, by volume Low gravity solids fraction, by volume Maximum recommended solids fractions SF , by volume SF 2.917 x 14.0 14.17 SF 40.838 14.17 SF 26.67 by volume Low gravity solids LOS , by volume LGS 26.67 -r- 100 0.3125 x 14.0 -r- 8.33 1 x 200 LGS 0.2667 0.3125 x 0.6807 x 200
Hydraulics Analysis
This sequence of calculations is designed to quickly and accurately analyse various parameters of existing bit hydraulics. 1. Annular velocity, ft mm AV AV 24.5 x Q 2. Jet nozzle pressure loss, psi Pb Pb 156.5 x Q2 x MW 3. System hydraulic horsepower available Sys HHP SysHHP surface, psi x Q 4. Hydraulic horsepower at bit HHPb HHPb Q x Pb 5. Hydraulic horsepower per square inch of bit diameter HHPb sq in. HHPb x 1.27 6. Percent pressure loss at bit psib psib Pb x 100 7. Jet velocity, ft sec Vn...
Equivalent Circulation Density ECD
3. Determine annular velocity v , ft mm v 24.5 x Q 4. Determine critical velocity Vc , ft mm Vc 3.878 x 104 x K K 2 n x 2.4 x 2n 1 2 n MW Dh Dp 3n 5. Pressure loss for laminar flow Ps , psi Ps 2.4v x 2n 1 n x_KL 6. Pressure loss for turbulent flow Ps , psi Ps 7.7 x 10 5 x MW08 x Q18 x PV02 x L 7. Determine equivalent circulating density ECD , ppg ECD, ppg Ps 0.052 TVD, ft 0MW, ppg Example Equivalent circulating density ECD , ppg Data Mud weight 12.5 ppg Plastic viscosity 24 cps Yield point 12...
Accumulator Precharge Pressure
The following is a method of measuring the average accumulator pre-charge pressure by operating the unit with the charge pumps switched off P,psi vol. removed, bbl - total acc. vol., bbl x Pf x Ps - Ps Pf where P average pre-charge pressure, psi Pf final accumulator pressure, psi Ps starting accumulator pressure, psi Example Determine the average accumulator pre-charge pressure using the following data Starting accumulator pressure Ps 3000 psi Final accumulator pressure Pf 2200 psi Volume of...
Nomenclature 1
IF sq in. impact force lb sq in of bit diameter Dp pipe or collar OD, in. N1 N2 N3 jet nozzle sizes, 32nd in. HHP hydraulic horsepower at bit IF impact force, lb Nozzle size 1 12-32nd in. Nozzle size 2 12-32nd in. Nozzle size 3 12-32nd in. Circulation rate 520 gpm Surface pressure 3000 psi Hole size 12-1 4 in. Drill pipe OD 5.0 in. Pb 156.5 x 5202 x 12.0 122 122 122 2 3. System hydraulic horsepower available Sys HHP 3000 x 520 5. Hydraulic horsepower per square inch of bit area HHP sq in. 826 x...
Maximum Pressures When Circulating Out a Kick Moore Equations
The following equations will be used 1. Determine formation pressure, psi Pb SIDP mud wt, ppg x 0.052 x TVD, ft 2. Determine the height of the influx, ft hi pit gain, bbl - annular capacity, bbl ft 3. Determine pressure exerted by the influx, psi Pi Pb Pm D X SICP 4. Determine gradient of influx, psi ft Ci Pi - hi 5. Determine Temperature, R, at depth of interest Tdi 70 F 0.012 F ft. x Di 460 6. Determine A for unweighted mud A Pb Pm D X Pi 7. Determine pressure at depth of interest Pdi A A2 pm...
Data from kill sheet
Initial drill pipe circulating pressure ICP 1780 psi Final drill pipe circulating pressure FCP 1067 psi Step 2 Determine interim pressure for the 5.0 in. drill pipe at 1063 strokes psi 1063 strokes 1780 7000 - 15,000 x 1780 1067 1780 0.4666 x 713 1780 333 1447 psi Step 3 Determine interim pressure for 5.0 in. plus 3-1 2 in. drill pipe 1063 381 1444 strokes psi 1444 strokes 1780 11,300 - 15,000 x 1780 1067 1780 0.86666 x 713 1780 618 1162 psi Note. After pumping 1062 strokes, if a straight line...
Weighted Cement Calculations
Amount of high density additive required per sack of cement to achieve a required cement slurry density x Wt x 11.207983 - SGc wt x CW - 94 - 8.33 x CW 1 AW - 100 - wt - SGa x 8.33 - wt AW - 100 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 Example Determine how much hematite, lb sk of cement, would be required to increase the...
Minimum Conductor Casing Setting Depth
Example Using the following data, determine the minimum setting depth of the conductor casing below the seabed Data Maximum mud weight to be used while drilling this interval 9.0 ppg Water depth 450 ft Gradient of seawater 0.445 psi ft Air gap 60 ft Formation fracture gradient 0.68 psi ft Step 1 Determine formation fracture pressure psi 450 x 0.445 0.68 x y psi 200.25 O.68y Step 2 Determine hydrostatic pressure of mud column psi 9.0 ppg x 0.052 x 60 450 y psi 9.0 x 0.052 x 60 450 9.0 x 0.052 x...
DeviationDeparture Calculation
Deviation is defined as departure of the wellbore from the vertical, measured by the horizontal distance from the rotary table to the target. The amount of deviation is a function of the drift angle inclination and hole depth. The following diagram illustrates how to determine the deviation departure AB distance from the surface location to the KOP BC distance from KOP to the true vertical depth TVD BD distance from KOP to the bottom of the hole MD CD Deviation departure departure of the...
References Ndd
Adams, Neal and Tommy Charrier, Drilling Engineering A Complete Well Planning Approach, PennWell Publishing Company, Tulsa, 1985. Chenevert, Martin E., and Reuven Hollo, TI-59 Drilling Engineering Manual, PennWell Publishing Company, Tulsa, 1981. Christmafl, Stan A., Offshore Fracture Gradients, JPT, August 1973. Craig, J. T. and B. V. Randall, Directional Survey Calculations, Petroleum Engineer, March Crammer Jr., John L., Basic Drilling Engineering Manual, PennWell Publishing Company, Tulsa,...
Bottomhole Pressure When Circulating Out a Kick
Example Use the following data and determine the bottomhole pressure when circulating out a kick Height of gas kick in casing 1200 ft Original mud weight 12.0 ppg Choke line pressure loss 220 psi Annulus casing pressure 631 psi Original mud in casing below gas 5500 ft Step 1 Hydrostatic pressure in choke line psi 12.0 ppg x 0.052 x 1500 75 psi 982.8 Step 2 Hydrostatic pressure exerted by gas influx psi 0.12 psi ft x 1200 ft psi 144 Step 3 Hydrostatic pressure of original mud below gas influx...
Dilution of Mud System
Vwm Vm Fct Fcop Fcop Fca where Vwm barrels of dilution water or mud required Vm barrels of mud in circulating system Fct percent low gravity solids in system Fcop percent total optimum low gravity solids desired Fca percent low gravity solids bentonite and or chemicals added Example 1000 bbl of mud in system. Total LOS 6 . Reduce solids to 4 . Dilute with water If dilution is done with a 2 bentonite slurry, the total would be
Surge And Swab Calculations
2 Laminar flow around drill pipe 3 Turbulent flow around drill collars These calculations outline the procedure and calculations necessary to determine the increase or decrease in equivalent mud weight bottomhole pressure due to pressure surges caused by pulling or running pipe. These calculations assume that the end of the pipe is plugged as in running casing with a float shoe or drill pipe with bit and jet nozzles in place , not open ended. A. Surge pressure around drill pipe 1. Estimated...
Annular capacity between casing and multiple strings of tubing
a Annular capacity between casing and multiple strings of tubing, bbl ft Annular capacity, bbl ft Dh2 TV 2 Tz 2 Example Using two strings of tubing of same size Dh casing 7.0 in. 29 lb ft ID 6.184 in. T1 tubing No. 1 2-3 8 in. OD 2.375 in. T2 tubing No. 2 2-3 8 in. OD 2.375 in. Annular capacity, bbl ft 6.1842 2.3752 2.3752 Annular capacity, bbl ft 38.24 11.28 Annular capacity 0.02619 bbl ft b Annular capacity between casing and multiple strings of tubing, ft bbl Example Using two strings of...
Maximum Influx Height Possible to Equal Maximum Allowable Shutin Casing
Influx height MASICP, psi gradient of mud wt in use, psi ft influx gradient, psi ft Example Determine the influx height, ft, necessary to equal the maximum allowable shut-in casing pressure MASICP using the following data Data Maximum allowable shut-in casing pressure 874 psi Mud gradient 10.0 ppg x 0.052 0.52 psi ft Gradient of influx 0.12 psi ft Influx height 874 psi - 0.52 psi ft 0.12 psi fl Influx height 2185 ft
References 1
Adams, Neal, Well Control Problems and Solutions, PennWell Publishing Company, Tulsa, OK, 1980. Adams, Neal, Workover Well Control, PennWell Publishing Company, Tulsa, OK, 1984. Goldsmith, Riley, Why Gas Cut Mud Is Not Always a Serious Problem, World Oil, Oct. 1972. Grayson, Richard and Fred S. Mueller, Pressure Drop Calculations For a Deviated Wellbore, Well Control Trainers Roundtable, April 1991. Petex, Practical Well Control Petroleum Extension Service University of Texas, Austin, Tx, 1982....
Surface Pressure During Drill Stem Tests
psi formation pressure equivalent mud wt, ppg x 0.052 x TVD, ft psi oil specific gravity x 0.052 x TVD, ft Surface pressure, psi formation pressure, psi oil hydrostatic pressure, psi Example Oil bearing sand at 12,500 ft with a formation pressure equivalent to 13.5 ppg. If the specific gravity of the oil is 0.5, what will be the static surface pressure during a drill stem test FP, psi 13.5 ppg x 0.052 x 12,500 ft FP 8775 psi psi 0.5 x 8.33 x 0.052 x 12,500 ft psi 2707 Surface pressure, psi 8775...
Sizing Diverter Lines
Determine diverter line inside diameter, in., equal to the area between the inside diameter of the casing and the outside diameter of drill pipe in use Example Casing 13-3 8 in. J-55 61 Iblft ID 12.515 in. Drill pipe 19.5 lb ft OD 5.0 in. Determine the diverter line inside diameter that will equal the area between the casing and drill pipe Diverter line ID, in. sq. root 12.5152 5.02 Diverter line ID 11.47 in.
Gas Expansion Equations
Basic gas laws Pi Vi Ti P2 V, T2 P2 hydrostatic pressure at the surface or any depth in the wellbore, psi . V1 original pit gain, bbl V2 gas volume at surface or at any depth of interest, bbl T1 temperature of formation fluid, degrees Rankine R F 460 T2 temperature at surface or at any depth of interest, degrees Rankine Basic gas law plus compressibility factor P1 V1 T1 Z1 P2 V2 T2 Z2 where Z1 compressibility factor under pressure in formation, dimensionless Z2 compressibility factor at the...
Calculations Drill string volume
Drill pipe capacity 0.01776 bbl ft x 9925 ft 176.27 bbl HWDP capacity 0.00883 bbl ft x 240 ft 2.12 bbl Drill collar capacity 0.0087 bbl ft x 360 ft 3.13 bbl Total drill string volume 181.5 bbl Drill collar open hole 0.0836 bbl ft x 360 ft 30.10 bbl Drill pipe open hole 0.1215 bbl ft x 6165 ft 749.05 bbl Drill pipe casing 0.1303 bbl ft x 4000 ft 521.20 bbl Total annular volume 1300.35 bbl Strokes to bit Drill string volume 181.5 bbl - 0.136 bbl stk Bit to casing strokes Open hole volume 779.15...
Kill Sheet With a Tapered String
psi _strokes ICP DPL - DSL x ICP FCP Note Whenever a kick is taken with a tapered drill string in the hole, interim pressures should be calculated for a the length of large drill pipe DPL and b the length of large drill pipe plus the length of small drill pipe. Example Drill pipe 1 5.0 in. 19.5 lb ft Capacity 0.01776 bbl ft Length 7000 ft Drill pipe 2 3-1 2 in. 13.3 lb ft Capacity 0.0074 bbl ft Length 6000 ft Drill collars 4 1 2 in. OD x 1-1 2 in. ID Capacity 0.0022 bbl ft Length 2000 ft 7000...
Fracture Gradient Determination Surface Application
Method 1 Matthews and Kelly Method F P D Ki a D where F fracture gradient, psi ft P formation pore pressure, psi a matrix stress at point of interest, psi D depth at point of interest, TVD, ft Ki matrix stress coefficient, dimensionless 1. Obtain formation pore pressure, P, from electric logs, density measurements, or from mud logging personnel. 2. Assume 1.0 psi ft as overburden pressure S and calculate a as follows a S P 3. Determine the depth for determining Ki by D a . 4. From Matrix Stress...