MATH FACTS:Exponent : In 4 the 2 is the exponent. The exponent will tell you how many times to multiply a number by itself. (4 = 4 x 4)Fractions: To convert a fraction to a decimal, divide the bottom number (denominator) into the top number (numerator). (5/8 = 5 8, which = .625)To convert 10 3/8 to a decimal, divide 3 by 8 and add 10. (10 3/8 = 3 8 = .375 + 10 = 10.375)*

GUIDELINES FOR ROUNDING1. Round up if the NEXT number is greater than or equal to 5.2. Round strokes to the nearest whole stroke. (1281.9 strokes = 1282 strokes)Round barrels up to the nearest tenth. (25.68 bbls =25.7 bbls.)Express pressures to the nearest whole psi. (1256.7 psi = 1257 psi.)All capacities and displacements should be carried out to 5 decimal places and expressed as bbls/ft. (.0145561 bbl/ft = .01456 bbl/ft)*

GUIDELINES FOR ROUNDINGGenerally, round mud weight to the nearest .1 ppg. (12.37ppg = 12.4 ppg)

If the mud weight is an integrity mud weight or maximum mud weight, NEVER! round up.13.32 ppg = ?11.27 ppg = ?If the mud weight is kill mud, ALWAYS! round up.(If the EIFD allows it, add .1 ppg, then round)10.66 ppg = ?14.8024 ppg = ?*

Remember to think downhole. The concepts provided in this section coversthe foundation for good well control.*

*

Solid objects exerts force in a downward direction due to the pull of gravity.This force is the objects weight

*Fluid has weight and also exerts pressure against the Sides of its container.

Fluids Exert Pressure. This pressure is the result of:*The density of the fluid The height of the fluid columnTo find out how much pressure fluid exerts for each unit of length we use a pressure gradient.

*0 psi0 ft.Take a fluid that has some weight

*0 psi.433 psi0 ft.1 ftAt one foot, the weight of the water is .433 psi.

*0 psi.433 psi0 ft.1 ftAt two feet, the weight of the water is .866 psi..866 psi2 ft

*0 psi.433 psi0 ft.1 ftAt three feet, the weight of the water is 1.299 psi..866 psi2 ft1.299 psi3 ftEach foot increases by .433 psi. This is called the FLUID GRADIENT

*0 psi.433 psi0 ft.1 ftWhat is the weight of the water at 7 feet?.866 psi2 ft1.299 psi3 ftEach foot increases by .433 psi. This is called the FLUID GRADIENT3.031 psi7 ft

Gradient is normally expressed as the force which fluid exerts per foot of vertical depth; it is measured in pounds per square inch per foot (psi/ft)*To get the pressure gradient we must convert the fluids density in pounds per gallon to pounds per square inch per foot (psi/ft)Express gradient in three decimal places.

.052*

GRADIENT FORMULAS#5Pressure Gradient = Fluid Density X .052

#5aFluid Density = Pressure Gradient .052*

*#1. What is the fluid gradient of water weighing 10.0 PPG?Pressure GradientPSI/Ft = Fluid DensityPPG X .052

PG = 10.0 X .052

PG = .520 PSI/Ft*

#2. What is the fluid density of a fluid that has a gradient of .546 psi/ftFDPPG = PGPSI/Ft .052FDPPG = .546 .052FDPPG = 10.5 PPG*

#3. What is the fluid density of a fluid that has a gradient of .104 psi/ftFDPPG = PGPSI/Ft .052

FDPPG = .104 .052

FDPPG = 2.0 PPG*

*HYDROSTATIC PRESSUREAlways use True Vertical Depth (TVD)Gravity reacts verticallyAB

HYDROSTATIC PRESSURE FORMULAS#2HP = Pressure Gradient X TVD#1HP = Fluid Density X .052 X TVD#4Fluid Density = HP .052 TVD#3TVD = HP Fluid Density .052*

*M Wt = 10.0 PPG10,000 FtTVDFind the hydrostatic pressure.*Example #4

Formula #1HPPSI = FDPPG X .052 X TVDFtHP = 10.0 X .052 X TVD

HP = 10.0 X .052 X 10000

HP = 5200 PSI *

*MD = 13500 ftTVD = 12800 ftM Wt. = 17.8 ppgFind the hydrostatic pressure.*Example #5

Formula #1HPPSI = FDPPG X .052 X TVDFtHP = 17.8 X .052 X TVD

HP = 17.8 X .052 X 12800

HP = 11847.68 PSI HP = 11848 PSI

*

*Example #6 You can find hydrostatic pressure at any point in a well as long as TVD and density of the fluid is known .MD = 10,000 FTTVD 5800 FTCasing ShoeFluid Density = 11.0 PPGWhat is the HP at the casing shoe?TVD = 9800 FT

Formula #1 HPPSI = FDPPG X .052 X TVDFtHP = 11.0 X .052 X TVD

HP = 11.0 X .052 X 5800

HP = 3317.6 PSI

HP = 3318 PSI

*

*Example #7 Sometimes, we need to know fluid height to get a certain pressure.Fluid Density = 12.0 PPG

HP = 2160 PSI? TVDRound off to the nearest whole number.

#7 TVDFt = HPPSI FDPPG .052 TVD = 2160 FD .052

TVD = 2160 12.0 .052

TVD =3461.5384 Ft

TVD =3462 Ft

*

*Example #8 Sometimes, we need to know fluid density to get a certain pressure.Fluid Density = ? PPG

FP = 6500 PSI10,000 FT. TVDRound off to the nearest tenth.

Formula #4 FDPPG = FPPSI .052 TVDFtFD = 6500 .052 TVD

FD = 6500 .052 10000

FD = 12.5 PPG

*

FactsFluid seeks a common levelDiameter does not affect hydrostatic pressureShape of well does not affect hydrostatic pressureHydrostatic pressure is the force exerted in a vertical directionHydrostatic pressure is expressed in pounds per square inch (PSI)Hydrostatic pressure changes when fluid level changesHydrostatic pressure changes when fluid density changes*

FORMATION PRESSUREFormation pressure is the pressure within the pore spaces of the formation rock.This pressure can be affected by the weight of the overburden (rock layers) above the formation, which exerts pressure on both the grains and pore fluids.If pore fluids are free to move, or escape, the grains lose some of their support and move closer together. This is called compaction.*

OVERBURDEN FACTSOverburden causes compactionAs overburden increases, the grains move closer togetherDuring compaction, the pore fluids are often moved outIf the fluids are trapped, the fluid must support the overburden*

DEFINATIONS:FORCE : the push or pull on an object measured in pounds

PRESSURE: the measure of force on a given area measured in pounds per square inch (PSI)

FLUID: anything that flows

DENSITY: the mass or weight of a substance per unit of volume (i.e. pounds per gallon or PPG)

HYDROSTATIC PRESSURE: Pressure in a column of fluid at rest (not flowing or being circulated)

STRIPPING: Running into or out of a well against pressure, through a closed preventer, when the forces up are less than the weight of the pipe.*

Formations vary and are classified as:Normal: Has pressure gradient equal to water, .433 to .465 PSI/Ft

Abnormal: Has pressure gradient > .465 PSI/Ft

Subnormal: Has pressure gradient < .433 PSI/Ft*

Abnormal pressure can occur in several ways:1. As overburden increases and fluids are trapped, pore fluid helps support overburden. If pressure gradient is > .465 psi/ft, then zone is classified as abnormal.*

Abnormal pressure can occur in several ways:*

Abnormal pressure can occur in several ways:3. When high pressure fluids from deeper zones migrate into shallower zones, shallower zones can become abnormally pressuredA. Casing failure in older wellsB. Improper plug and abandonment operationsC. Poorly implemented cement programs*

*It is more likely that you will drill into a sub normally pressured zone at shallow depths than at deep depths.

*

DIFFERENTIAL PRESSUREWell control is about controlling formation pressures.We primarily do this with liquids in the well.Sometimes there is a difference between formation pressure (FP) and hydrostatic pressure (HP) and this is called differential pressure.In some cases, a combination of HP and back pressure is used to control FP.*

*Differential PressureOverbalanced- When you have more hydrostatic pressure than formation pressure.

Underbalanced - When you have more formation pressure than hydrostatic pressure. Balanced - When your hydrostaticpressure and formation pressures are the same.BHP > FPBHP < FPBHP = FPCould cause swabbing

*U - TUBE

*AnnulusThe space between the drill string and the well bore is called the annulus.In a well, fluid usually fills the well and string.

THE STRING AND THE ANNULUS ACT LIKE A U SHAPED TUBE.ANNULUSANNULUSDRILLPIPE

*U-TubeWhat is U-Tubing?What will happen if fluid in the annulus is heavierthan the fluid in the string?

U TUBE FACTSIf the weight of the fluid in both tubes is the same, the HP in both columns should be the same.*If a higher density fluid is in one tube, there will be a HP difference (Drilled cuttings in the annulus can raise the HP in the annulus)

In a shut-in well, the SICP is generally higher than SIDPP because the HP in the annulus is less than the HP in the drillpipe.*

Up to now, weve talked about pressures in wells that are STATIC or not moving (no circulating)*

FRICTIONAL PRESSUREWhen objects move against each other, friction between them resists their movement.If you slide a box across the floor, friction requires you to keep pushing to keep the box moving.The same thing applies when fluids flow through pipe.Friction in the fluid and between fluid and pipe resists the fluids movement.If the pipe is very long, a great deal of pressure may be needed to keep the fluid in motion.*

FRICTIONAL LOSSFRICTIONAL LOSS :The amount of force lost when overcoming frictionFRICTIONAL PRESSURE LOSS :The frictional loss that occurs when moving a fluid (usually measured in PSI)*

*Stand pipeString

The exact amount of pressure required to overcome frictional pressure loss (FPL) and move fluid through the well bore at a given flow rate.*

What influences the movement of fluid through the well bore?Friction PressuresFluid DensityHydrostatic PressureCirculating Pressures Losses*

Effects of Density on Circulating Pressures:Initially, the circulating pressure of a higher density fluid will rise.It will then begin to drop due to the U-Tube (less pressure required to move fluid down string)Circulating pressure will begin to increase when the heavier fluid begins up the annulus

*

BOTTOM HOLE PRESSUREECD is Equivalent Circulating DensityECD = Present FD + AFPL converted to a mud weightEMW is Equivalent Mud WeightEMW is mud weight that would provide the same total pressure at a point if mud weight alone were to provide the pressure*

There is a kick in the hole. What does the SICP reveal?If there were 300 PSI more of hydrostatic pressure, the casing gauge would read 0.Where does the pressure at the bottom of the well come from?It comes from the HP plus the applied pressure.If the mud weight were increased to equal BHP, that is EMW.

*EMW300 psiFP = 5200 PSI

Example #10 The pressures that comprise equivalent density when a well is shut in, include the SICP and the fluid HP. Calculate EMW if:

SICP = 375 PSIMeasured Depth = 3120 FtTVD = 3005 FtPresent Fluid Density = 8.8 PPG

*

Formula #22EMW = (SICP .052 TVD) + Present Fluid DensityEMW = (375 .052 TVD) + Present Fluid Density

EMW = (375 .052 3005) + Present Fluid Density

EMW = (375 .052 3005) + 8.8

EMW = 2.3998 + 8.8

EMW = 11.1998 PPG

EMW = 11.2 PPG

*

The pressure we are most concerned with:Bottom Hole Pressure (BHP)In a workover / completion the perforations are a concern*

*IT IS POSSIBLE TO ESTIMATEBHP BASED ON WELL STATUS.

BHP is the total of ALL pressures againstthe bottom of the hole.

EXPRESS BOTTOM HOLE PRESSURE IN AN EQUATION FOR THE FOLLOWING:No fluid moving in well, well static:BHP = HP orBHP = HP + Gauge PressureWell is being circulated:BHP = HP + AFPLCirculating using backpressureBHP = HP + AFPL + BackpressurePressure at the shoe while circulating out a kickPress. @ the Shoe = HP at the shoe + casing press.

*

FORMATION PRESSUREBalanced Well BHP = FPOverbalanced BHP > FPUnder balanced BHP < FPUnder balanced is only way to get kick!*

*Stand pipeStringWhen the pump isrunning, BHP increases by the amount of AFPLcreated.

When the pump is turnedoff, the BHP is reduceddue to the loss of AFPL

When BHP = FP , the well is balanced. If BHP is just equal to FP when pump is off, well may flow.To prevent this, the mud weight can be increased to Equivalent Circulating Density (ECD)*

Example #11 Calculate ECD when:Annular Friction Loss = 730 PSI

Zone Measured Depth = 7320 Ft.

Zone TVD = 6985 Ft

Present Fluid Density = 13.8 PPG*

Formula #21ECD = (AFPL .052 TVD) + Present Fluid DensityECD = (730 .052 TVD) + Present FDECD = (730 .052 6985) + Present FDECD = (730 .052 6985) + 13.8ECD = 2.0098 + 13.8ECD = 15.8 PPG

*

SWAB AND SURGESWAB: The lowering of hydrostatic pressure in the well bore due to upward movement of tubular and /or tools. The pipe movement creates a piston effect pulling formation fluid into the well bore.

SURGE: A rapid increase in pressure down hole that occurs when the drill string is lowered too fast.*

In a normally pressured well, most kicks occur while moving the string in or out of the hole.TRIPPINGBefore tripping out of the hole, it is a good practice to check the well for flow to make sure that the well is overbalanced and stable before starting the trip.*

SWABWhen tripping out, pressure is reduced below the pipe when pulling up.If the fluid does not fill the space below the pipe fast enough, there is a SWAB effect.This creates a suction force and reduces the pressure below the string.If pressure is reduced enough, fluid is sucked into the well bore causing a kick.*

SWABWhen a trip out of the well begins, BHP is reduced in three ways:1. loss of annular friction pressure when the pump is shut down.2. a drop in the fluid level due to pipe displacement3. upward motion of pipe (swabbing effect).*

SWABOften, the bit and bottomhole assembly will collect a build up called bit balling.This causes the clearances in the annulus to be reduced which increases the probability of swabing.*

SURGEWhen lowering string, there is a surge of pressure on well bore and down hole pressures can increase if fluid does not have a chance to get out of the way.This can cause fluid leakage or formation fracture. *

Both Swab and Surge pressures are affected by:1. Clearance between pipe and the hole

2.Fluid properties

3.Rate of pipe movement*

CLEARANCE BETWEEN PIPE AND HOLEViscous fluids in tight holes increase swab and surge pressures.*Less viscous fluids and larger clearances lower swab and surge pressures.

RATE OF PIPE MOVEMENTMoving the pipe upwards faster than the fluid can fall below the pipe can cause swabbing when tripping out of the hole.Moving the pipe down faster than the fluid can move out of the way can cause surging when tripping into the hole.Slowing the speed of string movement can minimize swab and surge pressures*

To compensate for swabbing sometimes the weight of the mud is slightly increased. This is called a

TRIP MARGIN OR SAFETY MARGIN *

*Trip Margin = AFPL or (Safety Margin)TVD x .052 This is additional mud weight that compensates for the loss in annular friction loss when the pump is turned off, as in a trip.The increase is to the entire fluid system.Too large an increase could cause fluid lossToo small an increase could allow the well to kick.Cont. from Trip Margin

How often do you fill up when tripping out dry with drill pipe and collars?The standard practice is to fill up every 5 stands for drill pipe and every stand for collars.The reason for only one stand of collars is that the displacement of collars is greater than the displacement of drillpipe.Therefore, the HP lowers faster. *

Example #16How many feet of 4 drillpipe could be pulled dry prior to a bottomhole pressure drop of 75 psi?Drill Pipe:Displacement = .00597 bbls/ftCapacity = .01422 bbl/ftCasing 9 5/8:ID = 8.835Capacity = 0.07583 bbls/ft*Fluid density = 12.5 ppgFORMULA:#37Max. Lengthft = [(Press. Drop .052 FD) X (Csg. Cap. Pipe Displ)] Pipe Displ.Dry

Max. Length = [(Press. Drop .052 FD) X (Csg. Cap. Pipe Displ)] Pipe Displ.Dry

Max. Length = [(75 .052 12.5) X (.07583 - .00597)] .00597 = [115.4 X .06986] .00597 = 8.061844 .00597Max. Length = 1350.3926 orMax. Length = 1350 ft.

At 93 ft. per stand, how many stands can be pulled?No. of Stands = Max. Length Length per stand = 1350 93No. of Stands = 14.5 stands orNo. of Stands = 14*

Pressure Drop / ft Dry

# 19Pressure Drop / ft = Mud Gradient x Pipe Displacement Casing Capacity Pipe DisplacementPD/F = (13.4 x .052) x .00663 .07583 - .00663 = .6968 x .00663 .07583 - .00663= .0046197 .06920= .0667586 psi/ft= .0667586 x 1000= 66.76 or 67 psi / 1000 ft*

Example #19Pressure Drop / Ft Wet

#50Pressure Drop / ft = Mud Gradient x (Pipe Displ. + Pipe Cap.) Annular Capacity = (13.4 X .052) x (.00663 + .01421) .07583 .00663 - .01421 = .6968 x .02084 .05499 = .0145213 .05499 = .2640716 psi/ft = .2640716 x 1000 = 264 PSI

*

Difference in HP loss per 1000 feetDifference = HP Loss wet HP Loss dry = 264 67 = 197 PSI*

FORMATION PRESSUREWell control really means controlling formation pressure.*

*

Formation CharacteristicsPoresAn opening or space within a rock, usually small and often filled with fluid under pressure.PorosityIs the ratio of void (pore) space to solid volume.*

Formation CharacteristicsPermeabilityThe ability of a fluid to flow within the interconnected pore network of a porous medium. The measure of ease or ability of a rock to transmit a one-phase fluid under conditions of laminar flow.*

We can estimate FORMATION PRESSURE by the sum of HP and SI pressure once the well is stabilized.This means that BHP = FPFP = HP + SIP

Which pressure do we use?There are only 2:Casing pressure (SICP)Drill pipe pressure (SIDPP)SIDPP*

Casing?Mud weight is unknown due to unknown amount of cuttings in mud, so HP is unknown.Density of kick is unknown.Can an accurate BHP be calculated?

Drill Pipe?Drill pipe is full of good clean mud.HP can be calculated.There is usually no gas in the drill pipe.Will SIDPP be accurate?

How do I get a SIDPP with a backpressure valve in the string?*

Note SICP and SIDPP.* DRILL PIPECASING50040030020010005004003002001000

With the pump, roll it quickly and observe the DP pressure. Also, note the SICP. Did it move?* DRILL PIPECASING50040030020010005004003002001000

Roll the pump again and note the DP pressure and the Csg. Pressure. Did the Csg. Pressure move?* DRILL PIPECASING50040030020010005004003002001000

Roll the pump again. This time the gauge breaks back and settles at 200 psi. Did the casing pressure move?* DRILL PIPECASING50040030020010005004003002001000

If you have a pump that cant be rolled slowly, put pump in gear and then out of gear. Note gauges. Csg pressure increased by 100 psi.* DRILL PIPECASING50040030020010005004003002001000

The SIDPP is the difference between formation pressure and HP in the drillstring. So, if there is a float in the string, you now know how to get SIDPP.If:BHP = FPandBHP = HPdp + SIDPPthenFP = HPdp + SIDPPSIDPP is the only accepted pressure to use when trying to estimate formation pressure.*

*20. Find the estimated formation pressure in a well with the following data:Csg. Set at 6500ft. TVDWell Depth 10,500ft. TVDSIDPP is 350 psiSICP is 800 psiMud Weight is 11.1 ppg.Formula #13FP = HPdp + SIDPPFP = (FD x .052 x TVD) + SIDPPFP = (11.1 x .052 x 10500) + 350FP = 6060.6 + 350FP = 6411 PSI

*

Fracture pressure is the amount of pressure it takes to permanently deform (fail or split) the rock structure of a formation.Fracture pressure can be expressed as a gradient (psi/ft) a fluid density equivalent (ppg) calculated total pressure at the formation (psi).

*

*Formation Integrity TestsAn accurate evaluation of a casing cement job as well as of the formation is extremely important during the drilling of a well and for subsequent work. Good drilling practices and some regulatory bodies require a formartion integrity test before drilling more than 50ft (10ft of new hole) of open hole after drilling out casing. The information resulting from Formation Integrity Tests (FIT) is used throughout the life of the well and also for nearby wells. Casing depths, well control options, formation fracture pressures and limiting fluid weights may be based of this information.To determine the strength and integrity of a formation, a Leak Off Test (LOT) or a Formation Integrity Test (FIT) may be performed. Whatever the name, this test is first: a method of checking the cement seal between the casing and the formation, and second: determining the pressure and/or fluid weight the test zone below the casing can sustain.

*Leak-Off Test (LOT)Is performed to estimate the maximum pressure or mud weight that the test point can withstand before formation breakdown or fracture occurs. Limited Integrity TestIs performed when it is not acceptable to cause the formation to fracture or on wells drilled in developed fields where it is not expected to approach fracture pressures.The well bore is pressured to a predetermined pressure or fluid weigh. If the formation can withstand the applied pressure, the test is called good.

INTEGRITY LIMITSIntegrity Fluid Density: Maximum Mud Weight or Fracture Mud Weight (We use this to calculate Integrity Pressure)

Integrity Pressure: Maximum Allowable Surface Pressure (MASP) (The primary purpose for calculating EIP is to avoid lost returns that may lead to an underground blowout.*

FORMULAS TO FIND ESTIMATED INTEGRITY FLUID DENSITY AND ESTIMATED INTEGRITY PRESSURE#28MAMW ppg = (LOTP .052 TVD) + LOT Fluid Density#29EIP psi = (MAMW Present Fluid Density) X .052 X TVD*Pressure at the casing seat (shoe) = HP at the Shoe + Imposed Pressure

Example #21Find the Maximum Allowable Mud Weight (MAMW) and Maximum Allowable Annular Surface Pressure (MAASP) if:MD = 11226 ftCSG SHOE DEPTH = 5821 ft TVDLOT PRESSURE = 1250 psiLOT FLUID DENSITY = 9.6 ppgPresent Fluid Density = 10.6 ppgMAMW = (LOTP .052 TVD) + LOT Fluid DensityMAMW = (1250 .052 5821) + 9.6MAMW = 4.1296102 + 9.6MAMW= 13.72961 or 13.7 ppgMAASP = (MAMW Present FD) x .052 x TVDMAASP = (13.7 10.6) x .052 x 5821MAASP = 3.1 x .052 x 5821MAASP = 938.3452 or 938 psi*

Example #22At 11,226, the well kicked. SIDPP IS 100 psi and SICP is350 psi. How much more pressure could the formation take before breaking down?MD = 11226 ft CSG SHOE DEPTH = 5821 ft TVDLOT PRESS. = 1250 psi LOT FD = 9.6 ppgPresent Fluid Density = 10.6ppgMAMW = (LOTP .052 TVD) + LOT Fluid DensityMAMW = (1250 .052 5821) + 9.6MAMW = 4.1296102 + 9.6MAMW = 13.72961 or 13.7 ppgMAASP = (MAMW Present FD) x .052 x TVDMAASP = (13.7 10.6) x .052 x 5821MAASP = 3.1 x .052 x 5821MAASP = 938.3452 or 938 psiSafety Margin = MAASP Applied Press (SICP).Safety Margin = 938 350Safety Margin = 588 psi*

# 23. Example Test the well to an equivalent mud weight of 11.5 ppg:MD = 11226 ftCSG SHOE DEPTH = 5821 ft TVDLOT PRESSURE = 1250 PSIPresent Fluid Density = 10.6 ppgMAASP = (EM Wt Present M Wt) X .052 X TVDMAASP = (11.5 10.6) X .052 X 5821MAASP = .9 X .052 X 5821MAASP = 272.4228 psi or 272 psi*

Example #11 What will the new pump pressure be if:Circulating Pressure is 1000 PSIPresent Fluid Density is 10.0 PPGNew Fluid Density is 11.0*#53NPP = (New Mud Weight Old Mud Weight) X Present Pump Pressure

NPP = (NMW OMW) X PPPNPP = (11.0 OMW) X PPP

NPP = (11.0 10.0) X PPP

NPP = (11.0 10.0) X 1000

NPP = 1.1 X 1000

NPP = 1100 PSI*

Example #12 What will the new pump pressure be if:Circulating Pressure is 1000 PSIOld Strokes Per Minute is 20New Strokes Per Minute is 35*#52NPP = (NSPM OSPM) X Present Pump Pressure

NPP = (NSPM OSPM) X PPPNPP = (35 OSPM) X PPPNPP = (35 20) X PPPNPP = (35 20) X 1000NPP = 1.75 X 1000NPP = 1.75 X 1.75 X 1000NPP = 3.0625 X 1000NPP = 3063 PSI*If the rate is doubled, the pump pressureincreases by approximately 4 times.

*

SOME GAS FACTS:Gas is a fluid

Gas compresses easily

Gas volume and density is affected by pressure and temperature changes

1. As pressure increases, density (weight) increases

2. As temperature increases, volume increases as long as pressure is held constant

3. If volume is held constant, as temperature increases, pressure will increase.*

*Bottom-Hole Pressure There are no pumps running and the well is not flowing.

*Bottom-Hole Pressure If the pumps are running then BHP is the sum of AFPL and HP in the well.

*Bottom-Hole Pressure If back pressure is applied while circulating then the BHP will be the sum of HP + AFPL + Back Pressure

*Bottom-Hole Pressure

*What happens to the MASP when your mud weight changes?Hydrostatic Pressure = Fluid Wt. Inc.ppg x 0.052 x Depth ft, TVDFluid Density ppgEST. Integrity Press. spi9.612509.712419.811809.9115010.0112010.1109010.2105910.3102910.499910.596910.6938

*( Casing Pressure psi 0.052 Depth ft./ tvd. ) + Present Fluid Density ppgEquivalent Mud WeightFrom the previous discussions, it should be apparent that any applied pressureraises the total pressure at a given point. If the applied pressure is known, thenit can be calculated to an equivalent weight.The equivalent mud weight (EMW) is also the summation of all pressures ( hydrostatic pressure, choke or back-pressure, applied pressure, kick pressure, circulating pressure losses, etc.) at a given depth or zone and is expressed asa fluid density.EMW PPG = (375 .052 3000) + 8.8 = 2.4 + 8.8 = 11.2 PPG( EMW ppg Present Fluid Density ppg ) x 0.052 x Depth of Test ft. tvd

To determine how much applied pressure is required to test a pre-determined EMW at a given depth:Test Pressure psi = (13.4 - 9.1) x 0.052 x 5,745 = 4.3 x 0.052 x 5,745 = 1285 psi

*Pressure- Losses/CirculatingFriction is the resistance to movement. It takes force, or pressure, to overcome friction to getanything to move.The amount of force used to overcome friction is called frictional loss and can be measured in many ways. Thousands of psi of pressure can be lost to the wells circulating system as fluid is pumped through surface lines, down the string, and up the annulus.When the well is being circulated, bottom-hole pressure is increased by the amount of friction overcome in the annulus. When pumps are shut off, well-bore pressure is reduced because no frictional force is being overcome.Since friction adds pressure to the well-bore, it increased the effective weight, or the equivalentcirculating density (ECD). The total value is the equivalent of bottom-hole pressure with the pump on.

*Pump PSI = Surface Equip. + Drill String + Bit + AFLAFL is the only friction pressure felt on the bottom of the well.

Since this friction adds pressure to the well bore, it increases the effective weight, or the equivalent circulating density (ECD). The total value is equivalent of bottomhole pressure with the pump on.*ECD = (AFL 0.052 TVD ft.) + PMW ppg

*Decreasing / Increasing Bottomhole pressureSwabbing occurs because the fluid in the well does not dropas fast as the string is being pulled.Surge pressure occurs when the string is lowered to fast and pressure is created because the fluid does not have a chance to get out of the way.

*The conversion factor used to convert density to pressure gradient in the English system is 0.052. The way 0.052 is derived is by using a one foot cube, ( one foot wide by one foot long by one foot high). It takes about 7.48 gallons to fill the cube with fluid. It the fluid weighs one pound per gallon, and you have 7.48 gallons, then the total weight of the cube is 7.48 pounds per cubic foot. The weight of one of these square inches, one foot in height, can be found by dividing the total weight of the cube by 144: 7.48 144 = 0.051944 psi/ft. The conversion factor 0.052 is commonly used for oilfield calculations.

PRESSURE GRADIENT = Fluid x Conversion FactorWhat is the pressure gradient of fresh water which weighs 8.33 ppg?What is the pressure gradient of a fluid that weighs 9.5 ppg?What is the fluid density of a pressure gradient of 0.572 psi/ft?

*TVD vs. MDOnce we know how to find pressure exerted per foot, we can calculate the hydrostatic pressure at a given depth. All we have to do is multiply the pressure gradient by the number of feet to that vertical depth. Now we have to know the distinction between measured depth (MD) and true vertical depth (TVD).MD= Is used to calculate volume and strokesTVD= Is used to calculate hydrostatic pressureHydrostatic Pressure Hydrostatic pressure is the total fluid pressure created by the weight of a column of fluid, acting on any given point in a well. Hydro means water, or fluid, that exerts pressure like water, and static means not moving. So hydrostatic pressure is the pressure created by the density (weight) and height of a stationary (not moving) column of fluid. Hydrostatic Pressure = Fluid Density X Conversion Factor X TV DepthOR Hydrostatic Pressure = Pressure Gradient X TV Depth

*U-TubeWhat is U-Tubing?What will happen if fluid in the annulus is heavierthan the fluid in the string?The heavier fluid in the annulus exerting more pressure downward will flow into the string, displacing some of the lighter fluid out of the string, causing a backflow at the surface. When there is a difference in the hydrostatic pressures, the fluid will try to reach a balance point. Gain in pits ppg = (Slug Weight ppg Annulus Weight ppg ) x Volume of Slug bbls Pipe Capacity bbls/ft Annulus Weight ppg x Pipe Capacity bbls/ftDistance of Drop ft = Gain in Pits bbls Pipe Capacity bbls/ft?

*2 7/8#6a. Closed End Displacement bbl/ft = OD 1029.4

Example #12 What is the Closed End Displacement of the pipe?What is the total displacement if the length of the pipe is 1000 ft.?CED bbl/ft = OD 1029.4 = 2.875 1029.4 = (2.875 X 2.875) 1029.4 = 8.265625 1029.4CED = .00803 bbl/ft

Total Displacement bbls = Displacement X Length = .00803 X 1000 Total Displacement = 8.03 bbls or Total Displacement = 8.0 bbls*

*What do we need to know?

*2.151#6 Internal Capacity bbl/ft = ID 1029.4

* Example #13 What is the capacity of the pipe?What is the volume if the length of the pipe is 1000 ft.?Capacity bbl/ft = ID 1029.4 = 2.151 1029.4 = (2.151 X 2.151) 1029.4 = 4.626801 1029.4Capacity = .00449 bbl/ft Volume bbls = Capacity X Length = .00449 X 1000 Volume = 4.49 bbls or Volume = 4.5 bbls

*

Example #14What is the displacement of this pipe? What is the volume of 1000 ft. of displacement?*What do we need to know?

*2.1512.875 #10 Displacement bbl/ft = (OD - ID) 1029.4

Metal Displacement bbl/ft = (OD - ID) 1029.4Metal Displacement = (2.875 - 2.151) 1029.4 = [(2.875 x 2.875) (2.151 x 2.151)] 1029.4 = [8.265625 4.626801] 1029.4 = 3.638824 1029.4Metal Displacement = .00353 bbl/ftTotal Displ. bbls = Displacement bbl/ft X Length ft= .00353 X 1000 Total Displ. = 3.53 bbls or= 3.5 bbls*

*

*PipeID = 2.151 OD = 2 7/8HoleID = 4 5/8Depth = 1000 ft.

#7 Annular Capacity bbl/ft = (IDHOLE - ODPIPE) 1029.4Volume bbls = Capacity X LengthAnnular Capacity bbl/ft = (4.625 - 2.875) 1029.4

= [(4.625 X 4.625) (2.875 X 2.875)] 1029.4 = [21.390625 8.265625] 1029.4 = 13.125 1029.4Annular Capacity bbl/ft = .01275 bbl/ft

Annular Volume bbls = Annular Capacity X Length

= .01275 X 1000

Annular Volume bbls = 12.75 bbls. Or

Annular Volume bbls = 12.8 bbls*

*2 7/8

Example #9 Calculate the HP in a well when a lighter spacer is added.Well TVD = 10,000 Ft.Fluid Density = 12.0 PPGFluid Level = 10,000 Ft. TVDSpacer Density = 9.8 ppgVertical Depth of Spacer = 500 Ft*

First, calculate the original HP.HP = FD X .052 X TVD

HP = 12.0 X .052 X TVD

HP = 12.0 X .052 X 10000

HP = 6240 PSI*

Next, calculate the HP of the spacer pumped.HP = FD X .052 X TVD

HP = 9.8 X .052 X TVD

HP = 9.8 X .052 X 500

HP = 255 PSI*

Next, calculate the HP of the original fluid at the new height.HP = FD X .052 X TVD

HP = 12.0 X .052 X TVD

HP = 12.0 X .052 X 9500

HP = 5928 PSI*

Add the HP at the new height and the HP of the spacer.HPpsi = HPpsi at 9500 ft + HPpsi of Spacer

HPpsi = 5928 + HPpsi of Spacer

HPpsi = 5928 + 255

HPpsi = 6183 psi*

Example #10If we know the depths and densities of the fluids in the annulus and the string, we can calculate the pressure differential. STRINGDensity : 10.0 PPG

Depth : 10,000 Ft. ANNULUSDensity : 10.2 PPG

Depth : 10,000 Ft.*

String HP = Fluid Density X .052 X TVD

HP = 10 X .052 X 10000

HP = 5200 PSI*

Annulus HP = Fluid Density X .052 X TVD

HP = 10.2 X .052 X 10000

HP = 5304 PSI*

So, there is 100 psi of pressure trying to push the fluid into the tubing. If not stopped, the fluid will flow out of the string until the pressures are equalized and the fluid level in the annulus gets lower.

The annulus is not full, so the HP of 5300 PSI is lost. *

*U-TubeWhat is U-Tubing?What will happen if fluid in the annulus is heavierthan the fluid in the string?The heavier fluid in the annulus exerting more pressure downward will flow into the string, displacing some of the lighter fluid out of the string, causing a flow at the surface. When there is a difference in the hydrostatic pressures, the fluid will try to reach a balance point.

*

*

*HYDROSTATIC PRESSUREAlways use True Vertical Depth (TVD)Gravity reacts vertically

*Fluid takes the shape of the container.

*If you put a heavier fluid in the drill pipe,the fluid will tend to flow until the pressure on both sides of the u-tube is the same.U - TUBING

The amount of friction that has to be overcome in order to keep the fluid in motion depends on many factors:1. Density of the fluid2. Type and roughness of surfaces making contact 3. Surface area in contact4. Direction and velocity of fluid5. Thermal and mud properties

*

Formations vary and are classified as:Normal: Has pressure gradient equal to water, .433 to .465 PSI/Ft

Abnormal: Has pressure gradient > .465 PSI/Ft

Subnormal: Has pressure gradient < .433 PSI/Ft*

Abnormal pressure can occur in several ways:1. As overburden increases and fluids are trapped, pore fluid helps support overburden. If pressure gradient is > .465 psi/ft, then zone is classified as abnormal.*

Abnormal pressure can occur in several ways:*

Abnormal pressure can occur in several ways:3. When high pressure fluids from deeper zones migrate into shallower zones, shallower zones can become abnormally pressuredA. Casing failure in older wellsB. Improper plug and abandonment operationsC. Poorly implemented cement programs*

*It is more likely that you will drill into a sub normally pressured zone at shallow depths than at deep depths.

*

DIFFERENTIAL PRESSUREWell control is about controlling formation pressures.We primarily do this with liquids in the well.Sometimes there is a difference between formation pressure (FP) and hydrostatic pressure (HP) and this is called differential pressure.In some cases, a combination of HP and back pressure is used to control FP.*

*Differential PressureOverbalanced- When you have more hydrostatic pressure than formation pressure.

Underbalanced - When you have more formation pressure than hydrostatic pressure. Balanced - When your hydrostaticpressure and formation pressures are the same.BHP > FPBHP < FPBHP = FPCould cause swabbing

BUOYANCYWhen an object is placed in a fluid, the fluid pushes upward on the object with a net force equal to the weight of the fluid displaced.

The upward force is called BUOYANCY*

BUOYANCYOne cubic foot of wood weighs about 31#. When it is placed in water, it will sink until it displaced about 31# of water.*Since a cubic foot of water weighs more than a cubic foot of wood, the wood floats.

BUOYANCYA one foot iron cube weighs 490 pounds. When placed in water, it will displace one cubic foot of water.*One cubic foot of water weighs 62.3 poundsThe net buoyancy pushing up on the iron cube is 62.3 poundsThe iron cube doesnt float and will appear to weigh approximately 62 pounds less or about 428 pounds.

BUOYANCY FORMULA#56B = [(65.4 FDPPG) 65.4] X PWPPF

EXAMPLE: #9What is the buoyed weight of 93 ft. of 2 3/8 tubing (4.7 PPF) stripped dry into a well that has 12.2 ppg brine in the casing?BWt = [(65.4 12.2) 65.4] X 4.7= [53.2 65.4] X 4.7= .8134556 X 4.7BWt = 3.8 PPFB = 353 lbs.*

*WHEN PIPE IS LOWERED INTO A WELL IT WILL APPEAR TO WEIGH LESS.

Example #17

#38Max. Length =[ (Press. Drop .052 FD) X (Ann. Capacity)] (Pipe Displ.+ Pipe Cap.)

Max. Length = [(75 .052 12.5) X (.07583-.00597-.01422)] (.00597 + .01422) = [115.4 X .05564] .02019 = 6.420856 .02019Max. Length = 318.02 ft. orMax. Length = 318 ft.

At 93 ft. per stand, how many stands can be pulled?No. of Stands = Max. Length Length per stand = 318 93No. of Stands = 3.41 stands orNo. of Stands = 3*

*Example #18. What is the difference in hydrostatic pressure per 1000 ft. of pipe between making a wet trip and a dry trip out the well described below?Mud weight: 13.4 PPGCasing: 9 5/8, 8.835 ID, .07583 bbl/ft capacityDrillpipe: 4 , 3.826 ID, .01421 bbl/ft capacity.00663 bbl/ft displacement

This information is good for both example #18 and #19.*

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