## Procedure

In an oil mud, diesel oil can be used instead of water. 2. Set mud balance at 8.33 ppg. 3. Fill the mud balance with cuttings until a balance is obtained with the lid in place. 4. Remove lid, fill cup with water (cuttings included), replace lid, and dry outside of mud balance. 5. Move counterweight to obtain new balance. The specific gravity of the cuttings is calculated as follows where SG specific gravity of cuttings bulk density Rw resulting weight with...

## Changing oilwater ratio

NOTE If the oil water ratio is to be increased, add oil if it is to be decreased, add water. Retort analysis by volume oil 51 by volume water 17 by volume solids 32 Example 1 Increase the oil water ratio to 80 20 In 100 bbl of this mud, there are 68 bbl of liquid (oil plus water). To increase the oil water ratio, add oil. The total liquid volume will be increased by the volume of the oil added, but the water volume will not change. The 17 bbl of water now in the mud represents 25 of the liquid...

## Estimated Type of Influx

Influx weight, ppg mud wt, ppg ((SICP SIDPP) height of influx, ft x 0.052) 4 6 ppg oil kick or combination 7 9 ppg saltwater kick Example Determine the type of the influx using the following data Shut-in casing pressure 1044 psi Height of influx 400 ft Shut-in drill pipe pressure 780 psi Mud weight 15.0 ppg Influx weight, ppg 15.0 ppg ((1044 780) -- 400 x 0.052) Influx weight, ppg 15.0 ppg 264 Influx weight 2.31 ppg Therefore, the influx is probably gas.

## Weight of slug required for a desired length of dry pipe with a set volume of slug

Step 1 Length of slug in drill pipe, ft Slug length, ft slug vol., bbl - drill pipe capacity, bbl ft Step 2 Hydrostatic pressure required to give desired drop inside drill pipe HP, psi mud wt, ppg x 0.052 x ft of dry pipe 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 12.2 ppg Drill pipe capacity 4-1 2 in. 16.6 lb ft 0.0 1422 bbl ft Volume of slug 25 bbl Step 1...

## 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, psi height of slug in annulus, ft X difference in gradient, psi ft between Drill pipe capacity 4-1 2 in. 16.6 lb ft 0.01422 bbl ft Annulus volume (8-1 2 in. by 4-1 2 in.) 19.8 ft bbl a) Volume gained in mud pits...

## Oilwater ratio from retort data

Obtain the percent-by-volume oil and percent-by-volume water from retort analysis or mud still analysis. From the data obtained, the oil water ratio is calculated as follows a) oil in liquid phase by vol oil_ x 100 b) water in liquid phase by vol water_ x 100 c) Result The oil water ratio is reported as the percent oil and the percent water. Example Retort analysis by volume oil 51 by volume water 17 by volume solids 32 a) oil in liquid phase 51 x 100 b) water in liquid phase 17 x 100 c) Result...

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

Step 1 Hydrostatic pressure required to give desired drop inside drill pipe HP, psi mud wt, ppg x 0.052 x ft of dry pipe Step 2 Difference in pressure gradient between slug weight and mud weight psi ft (slug wt, ppg mud wt, ppg) x 0.052 Step 3 Length of slug in drill pipe Slug length, ft pressure, psi - difference in pressure gradient, psi ft Slug vol., bbl slug length, ft x drill pipe capacity, bbl ft Example Determine the barrels of slug required for the following Desired length of dry pipe...

## Differential Hydrostatic Pressure Between Cement in the Annulus and Mud Inside the Casing

Determine the hydrostatic pressure exerted by the cement and any mud remaining in the annulus. 2. Determine the hydrostatic pressure exerted by the mud and cement remaining in the casing. 3. Determine the differential pressure. Example 9-5 8 in. casing 43.5 lb ft in 12-1 4 in. hole Well depth 8000 ft Cementing program LEAD slurry 2000 ft 13.8 lb gal TAIL slurry 1000 ft 15.8 lb gal Mud weight 10.0 lb gal Float collar No. of feet above shoe 44 ft

## Strokes

A Surface to bit strokes Strokes drill string volume, bbl - pump output, bbl stk Surface to bit strokes 172.16 bbl - 0.136 bbl stk Surface to bit strokes 1266 b Bit to surface or bottoms-up strokes Strokes annular volume, bbl - pump output, bbl stk Bit to surface strokes 1231.84 bbl - 0.136 bbl stk Bit to surface strokes 9058 c Total strokes required to pump from the Kelly to the shale shaker Strokes drill string vol., bbl annular vol., bbl - pump output, bbl stk Total strokes 172.16 1231.84 -...

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

Displacement, bbl ft OD, in.2 ID, in.2 Weight, lb ft displacement, bbl ft x 2747 lb bbl Example Determine the capacity, bbl ft, displacement, bbl ft, and weight, lb ft, for the following Drill collar OD 8.0 in. Drill collar ID 2-13 16 in. Convert 13 16 to decimal equivalent 13 h- 16 0.8125 b Displacement, bbl ft 8.02 2.81252 c Weight, lb ft 0.0544879 bbl ft x 2747 lb bbl Weight 149.678 lb ft

## Mixing Fluids of Different Densities

Where Vi volume of fluid 1 bbl, gal, etc. Di density of fluid 1 ppg,lb ft3, etc. V2 volume of fluid 2 bbl, gal, etc. D2 density of fluid 2 ppg,lb ft3, etc. VF volume of final fluid mix DF density of final fluid mix Example 1 A limit is placed on the desired volume Determine the volume of 11.0 ppg mud and 14.0 ppg mud required to build 300 bbl of 11.5 ppg mud Given 400 bbl of 11.0 ppg mud on hand, and 400 bbl of 14.0 ppg mud on hand Solution let V1 bbl of 11.0 ppg mud V2 bbl of 14.0 ppg mud b...

## Determine the feet of free pipe and the free point constant Method

The depth at which the pipe is stuck and the number of feet of free pipe can be estimated by the drill pipe stretch table below and the following formula. Feet of stretch, in. x free point constant free pipe pull force in thousands of pounds Example 3-1 2 in. 13.30 lb ft drill pipe 20 in. of stretch with 35,000 lb of pull force From drill pipe stretch table Free point constant 9052.5 for 3-1 2 in. drill pipe 13.30 lb ft

## Pressure Chart for Bringing Well on Choke

Pressure stroke relationship is not a straight line effect. While bringing the well on choke, to maintain a constant bottomhole pressure, the following chart should be used Line 1 Reset stroke counter to 0 Line 3 3 4 stroke rate 50 x 0.75 Line 4 7 8 stroke rate 50 x 0.875 Strokes side Example kill rate speed 50 spm Pressure side Example. Shut-in casing pressure SICP 800 psi Choke line pressure loss CLPL 300 psi Divide choke line pressure loss CLPL by 4, because there are 4 steps on the chart...

## Basic Formulas

Converting Pressure into Mud Weight 5. Equivalent Circulating Density 6. Maximum Allowable Mud Weight 11. Buoyancy Factor 12. Hydrostatic Pressure Decrease POOH 12. Loss of Overbalance Due to Falling Mud Level 15. Drill Pipe Drill Collar Calculations 19. Temperature Conversion Formulas

## Feet of pipe pulled WET to lose overbalance

Feet overbalance, psi x casing cap. pipe cap. pipe disp. mud wt., ppg x 0.052 x pipe cap. h- 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 capacity 0.01776 bbl ft Pipe displacement 0.0075 bbl ft Feet 150 psi x 0.0773 0.01776 0.0075 bbl ft 11.5 ppg x 0.052 0.01776 0.0075 bbl ft

## Tonmiles while making short trip

The ton-miles of work performed in short trip operations, as for drilling and coring 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...

## 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

## 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

## Oil And Gas Nomenclature

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

## 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...

## 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...

## 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...

## 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

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...

## 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

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 Bubble Rising

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 ....

## Equivalent Circulation Density ECD

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...

## 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...

## 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...

## Deviation Departure 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...

## 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 Pressure MASICP

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

## Petex Practical Well

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...

## 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...

## 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...

## Feet of drill pipe that can be used with a specific 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 BHA. Lengthmax r T x f MOP Wbhal x BF Wdp 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...

## Method Ben Eaton Method

F S D - Pf D x y 1 - y Pf D where S D overburden gradient, psi ft Pf D formation pressure gradient at depth of interest, psi ft y Poisson's ratio 1. Obtain overburden gradient from Overburden Stress Gradient Chart. 2. Obtain formation pressure gradient from electric logs, density measurements, or from logging operations. 3. Obtain Poisson's ratio from Poisson's Ratio Chart. 4. Determine fracture gradient using above equation. 5. Determine fracture pressure, psi psi F x D 6. Determine maximum...

## Casing Pressure Decrease when Bringing Well on Choke

When bringing the well on choke with a subsea stack, the casing pressure annulus pressure must be allowed to decrease by the amount of choke line pressure loss friction pressure Reduced casing pressure, psi shut-in casing pressure, psi choke line pressure loss, psi Example Shut-in casing annulus pressure SICP 800 psi Choke line pressure loss CLPL 300 psi Reduced casing pressure, psi 800 psi 300 psi Reduced casing pressure 500 psi

## Casing Burst Pressure Subsea Stack

Step 1 Determine the internal yield pressure of the casing from the Dimensions and Strengths section of cement company's service handbook. Step 2 Correct internal yield pressure for safety factor. Some operators use 80 some use 75 , and others use70 Correct internal yield pressure, psi internal yield pressure, psi x SF Step 3 Determine the hydrostatic pressure of the mud in use NOTE The depth is from the rotary Kelly bushing RKB to the mud line and includes the air gap plus the depth of...

## Formation Pressure Tests

Two methods of testing Equivalent mud weight test Precautions to be undertaken before testing 1. Circulate and condition the mud to ensure the mud weight is consistent throughout the system. 2. Change the pressure gauge if possible to a smaller increment gauge so a more accurate measure can be determined. 4. Begin pumping at a very slow rate 1 4 to 1 2 bbl min. 5. Monitor pressure, time, and barrels pumped. 6. Some operators may have different procedures in running this test, others may include...

## Breakover Point Between Stripping and Snubbing

Example Use the following data to determine the breakover point DATA Mud weight 12.5 ppg Drill collars 6-1 4 in. 2-13 16 in. 83 lb ft Length of drill collars 276 ft Shut-in casing pressure 2400 psi Determine the force, lb, created by wellbore pressure on 6-1 4 in. drill collars Force, lb pipe or collar OD, In 2 x 0.7854 x wellbore pressure, psi Force, lb 6.252 x 0.7854 x 2400 psi Force 73,631 lb Determine the weight, lb, of the drill collars Wt, lb drill collar weight, lb ft x drill collar...

## Calculations Required for Spotting Pills

A Barrels of spotting fluid pill required b Pump strokes required to spot the pill Step 1 Determine the annular capacity, bbl ft, for drill pipe and drill collars in the annulus Annular capacity, bbl ft Dh in.2 Dp in.2 Step 2 Determine the volume of pill required in the annulus Vopl bbl annular capacity, bbl ft x section length, ft x washout factor Step 3 Determine total volume, bbl, of spotting fluid pill required Barrels Barrels required in annulus plus barrels to be left in drill string Step...

## Gas Migration in a Shutin Well

Estimating the rate of gas migration, ft hr Example Determine the estimated rate of gas migration using a mud weight of 11.0 ppg Determining the actual rate of gas migration after a well has been shut-in on a kick Rate of gas migration, ft hr increase in casing pressure, psi hr pressure gradient of mud weight in use, psi ft Example Determine the rate of gas migration with the following data Stabilised shut-in casing pressure 500 psi SICP after one hour 700 psi Pressure gradient for 12.0 ppg mud...

## Basic solids analysis calculations

NOTE Steps 1 4 are performed on high salt content muds. For low chloride muds begin with Step 5. Step 1 Percent by volume saltwater SW SW 5.88 x 10-8 x ppm Cl 12 1 x by vol water Step 2 Percent by volume suspended solids SS SS 100 by vol oil by vol SW Step 3 Average specific gravity of saltwater ASGsw ASGsw ppm Cl 095 x 1.94 x 10-6 1 Step 4 Average specific gravity of solids ASG ASG 12 x MW by vol SW x ASGsw 0.84 x by vol oil Step 5 Average specific gravity of solids ASG ASG 12 x MW by vol...

## Fracture Gradient Determination Subsea Applications

In offshore drilling operations, it is necessary to correct the calculated fracture gradient for the effect of water depth and flow-line height air gap above mean sea level. The following procedure can be used Example Air gap 100 ft Density of seawater 8.9 ppg Water depth 2000 ft Feet of casing below mud-line 4000 ft 1. Convert water to equivalent land area, ft a Determine the hydrostatic pressure of the seawater HPsw 8.9 ppg x 0.052 x 2000 ft b From Eaton's Overburden Stress Chart, determine...

## Evaluation of Centrifuge

QU QM x MW PO l TQW x PO PW 1 PU PO b Fraction of old mud in Underflow e Water flow rate into mixing pit QP TQM x 35 MW 1 TQU x 35 PU 1 0.6129 x QC 0.6129 x QD QB QM QU QP QC_ QD_ x 35 MW mud density into centrifuge, ppg PU QM mud volume into centrifuge, gal m PW QW dilution water volume, gal mm PO CD additive content in mud, lb bbl CC QU Underflow mud volume, gal mm QC FU fraction of old mud in Underflow QD QB mass rate of API barite, lb mm QP water flow rate into mixing pit, gal mm Underflow...

## Lubricate and Bleed

The lubricate and bleed method involves alternately pumping a kill fluid into the tubing or into the casing if there is no tubing in the well, allowing the kill fluid to fall, then bleeding off a volume of gas until kill fluid reaches the choke. As each volume of kill fluid is pumped into the tubing, the SITP should decrease by a calculated value until the well is eventually killed. This method is often used for two reasons 1 shut-in pressures approach the rated working pressure of the wellhead...

## Kill Sheet for a Highly Deviated Well

Whenever a kick is taken in a highly deviated well, the circulating pressure can be excessive when the kill weight mud gets to the kick-off point KOP . If the pressure is excessive, the pressure schedule should be divided into two sections 1 from surface to KOP, and 2 from KOP to TD. The following calculations are used Determine strokes from surface to KOP Strokes drill pipe capacity, bbl ft x measured depth to KOP, ft x pump output, bbl stk Strokes drill string capacity, bbl ft x measured...

## Determine the height ft of unweighted spotting fluid that will balance formation pressure in the ann

A Determine the difference in pressure gradient, psi ft, between the mud weight and the spotting fluid psi ft mud wt, ppg spotting fluid wt, ppg x 0.052 b Determine the height, ft, of unweighted spotting fluid that will balance formation pressure in the annulus Height ft amount of overbalance, psi - difference in pressure gradient, psi ft Example. Use the following data to determine the height, ft, of spotting fluid that will balance formation pressure in the annulus Data Mud weight 11.2 ppg...

## Determine free point constant FPC

The free point constant can be determined for any type of steel drill pipe if the outside diameter, in., and inside diameter, in., are known where As pipe wall cross sectional area, sq in. Example 1 From the drill pipe stretch table 4-1 2 in. drill pipe 16.6 lb ft ID 3.826 in. FPC 452 3.8262 x 0.7854 x 2500 FPC 4.407 x 2500 FPC 11,017.5 Example 2 Determine the free point constant and the depth the pipe is stuck using the following data 2-3 8 in. tubing 6.5 lb ft ID 2.441 in. 25 in. of stretch...

## Bit Nozzle Selection Optimised Hydraulics

These series of formulas will determine the correct jet sizes when optimising for jet impact or hydraulic horsepower and optimum flow rate for two or three nozzles. 1. Nozzle area, sq in. Nozzle area, sq in. N12 N22 N32 2. Bit nozzle pressure loss, psi Pb Pb gpm2 x MW, ppg 3. Total pressure losses except bit nozzle pressure loss, psi Pc Pci amp Pc2 circulating pressure, psi bit nozzle pressure Loss. 4. Determine slope of line M M log Pcr- Pc 5. Optimum pressure losses Popt a For impact force...

## Directional Survey Calculations

The following are the two most commonly used methods to calculate directional surveys North MD x sin. I1 I2 x cos. Al A2 2 2 East MD x sin. I1 I2 x sin. Al A2 2 2 North MD cos. I1 cos. I2 sin. A2 sin. Al I2 I1 A2 Al East MD cos. I1 cos. I2 cos. A2 cos. Al I2 I1 A2 Al Vert. MD sin. I2 sin. I1 I2 I1 where MD course length between surveys in measured depth, ft I1, I2 inclination angle at upper and lower surveys, degrees A1, A2 direction at upper and lower surveys Example Use the Angle Averaging...

## Evaluation of Hydrocyclone

Determine the mass of solids for an unweighted mud and the volume of water discarded by one cone of a hydrocyclone desander or desilter Volume fraction of solids SF SF MW 8.22 Mass rate of solids MS MS 19,530 x SF x V Volume rate of water WR WR 900 1 SF V where SF fraction percentage of solids MW average density of discarded mud, ppg MS mass rate of solids removed by one cone of a hydrocyclone, lb hr V volume of slurry sample collected, quarts T time to collect slurry sample, seconds WR volume...