IWCF P & P

MAERSK TRAINING CENTRE

Drilling Section

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Well Control

Training Manual

Table of content:

01 Pressure in the earth crust Page 007

01.01 Sedimentation 02.01 Compression 03.01 Pressure 04.01 Pressure in fluids 05.01 Pressure gradient 06.01 Abnormal/subnormal pressure

02 Pressure balance in the well bore Page 019

01.02 Pressure balance

02.02 Overbalance and underbalance 03.02 Lost circulation

04.02 Rate of penetration versus overbalance 05.02 Drilling break

06.02 Necessary overbalance 07.02 Trip margin

08.02 Riser margin 09.02 Relationship

10.02 Equivalent drilling fluid density

03 Dynamic pressure regime when circulating Page 028

01.03 Circulation of drilling fluid

02.03 Dynamic pressure in the well bore

04 Consideration with a closed in well Page 033

01.04 Closed in well 02.04 U-tube

05 Properties of gasses and gas laws Page 036

01.05 Drilling with underbalance 02.05 Properties of gas and gas laws 03.05 Expansion of gas

04.05 Formation strength 05.05 Leak-off test

06.05 Maximum allowable annular surface pressure

06 Drilling fluid volume and capacities Page 044

01.06 Calculating drilling fluid volume – capacities 02.06 Drilling fluid volume and capacities from tables 03.06 Surface to bit strokes & bit to surface strokes 04.06 Use of barite to increase drilling fluid volume 05.06 Volume increase due to barite addition

07 Wellbore kicks Page 053

01.07 Kick occurrences 02.07 Warning signals

03.07 Warning signals while drilling

04.07 Warning signals while tripping or making connection 05.07 Procedure for shutting in the well

06.07 Pressure after shut in

08 Circulating a kick out of the well bore Page 069

01.08 General points

02.08 Circulating out an influx using Driller’s Method 03.08 Wait and Weight Method or Engineer’s Method 04.08 The Concurrent Method

05.08 Advantages and disadvantages of the three methods 06.08 Pressure control schemes

09 Calculations of density and pressure gradient of an influx Page 094

01.09 General points 02.09 Example

10 Lost circulation Page 097

01.10 General

02.10 Causes of lost circulation

03.10 Well control with partly lost circulation 04.10 Well control with total lost circulation

11 Volumetric wellcontrol and other Page 102

01.11 General

02.11 Volumetric Method – Specification required 03.11 Volumetric Method – Handling

04.11 Lubrication Technique

05.11 Volumetric Method – Example 06.11 Low Choke Method – Dynamic Kill 07.11 Bullheading

12 Kick with bit off bottom Page 113

01.12 Introduction 02.12 Stripping

03.12 Closing Procedures

04.12 Rig layout for combined stripping and volumetric method 05.12 Procedure

13 Gas cut drilling fluid Page 119

01.13 General

02.13 Causes of gas cut drilling fluid 03.13 Gas kicks in Oil Based Mud 04.13 Influx volume

14 Deviated and Horizontal well control Page 126

01.14 Introduction 02.14 Complications

03.14 Horizontal well control example 04.14 Wait and Weight Method

05.14 Driller’s Method

06.14 Horizontal well kill method

15 Pulling Pipe Page 138

01.15 Introduction 02.15 Pumping slug 03.15 Inadequate hole fill

04.15 Hole not taking correct amount of fluid 05.15 Hole not giving correct amount of fluid

Abbreviations:

A Area

Atm Atmosphere

BHA Bottom hole assembly

BHP Bottom hole pressure

BOP Blow out preventer

Cap Capacity

DC Drill collar

DP Drill pipe

EDC Equivalent circulating density

EFD Equivalent formation density

EOB End of build

FCP Final circulating pressure

FOSV Full opening safety valve

Ft Feet

G Pressure gradient psi/ft

Gal Gallons

GMD Gas migration distance

GMR Gas migration rate

GPM Gallons per minute

HCR High closing ratio

HPHT High Pressure/High Temperature

H2S Hydrogen sulfide

ICP Initial circulating pressure

ID Inside diameter

KMW Kill mud weight

KOP Kick off point

Lb Pounds

Lbs/ft Pounds per feet

LOT Leak off test

MAASP Maximum allowable annular surface pressure

MD Measured depth

MW Mud weight

MWF Final mud weight

OBM Oil base mud

OD Outside diameter

OH Open hole

OMW Original mud weight

P Pressure

PA Pressure annulus

Pc Pressure dynamic

PDP Pressure drill pipe

Pf Pressure formation (pore pressure)

Ph Pressure hydrostatic

PL Pressure loss

PPM Part per million

PSI Pound per inch²

PWD Pressure while drilling

ROP Rate of penetration

RPM Rotation per minute

RRCP Reduced rate circulating pressure

SF Safety factor

SICP Shut in casing pressure

SIDPP Shut in drill pipe pressure

SPM Strokes per minute

SX Sacks

T Temperature TVD True vertical depth

V Volume

WBM Water base mud

01

PRESSURE IN THE EARTHS CRUST

The theory behind the pressure being present in the different depths in the earth rock formations are based on the historic development during millions of years where settling of particles has taken place in the ocean.

Large and small rock particles are transported by rivers and streams, ice and wind and deposited on the seabed offshore. In the sea several different chemical substances are present which also separates from the water and sink to the seabed. Amongst others carbonates, sulphates and chlorides are known to be dissolved in the seawater. Small organisms which live in the sea has a life cyclus and when they die their solid remains also sink to the seabed.

When this process continues during millions of years the layers of settling will obtain a considerable thickness on the sea floor.

02.01 Compression:

The rock particles and solid matter will eventually become more and more compacted as they bear more and more weight from the overlaying deposits. As this process continues the water that is found between the rock particles will usually escape. However there will usually be small cavities left between the particles, which contain the remaining water. These cavities or void spaces make the rock formations more or less porous. A porous formation can contain fluids, gas or hydrocarbons.

As compression and compaction continue during time, combined with thermal and chemical processes the unconsolidated particles will eventually become rock formations within the earth crust.

These sedimentary rock formations are generally porous, and the pores are filled with a fluid or gas. Fig 01

SHALE

SANDSTONE

SALT

Porous/

impermeable

Porous/

permeable

Tight and

without pores

If communication between the cavities or pores in the formation is present this allows the fluid to flow away and escape. Under certain conditions the formation fluid can become trapped. If a porous fluid-bearing formation becomes covered with an impermeable layer of rock such as a clay stone, the fluid becomes trapped.

03.01 Pressure

Before describing the conditions in which the formation fluids are found at different depths in the rock formations the terms mass, density, force, energy and pressure will be considered.

Mass

Mass is defined as the term for a quantity of matter. The unit of measurement that is used is the pound.

Density

Density is an expression giving the mass of gas, fluid or solid matter in relationship to its volume, E.I. mass per unit volume. Other means to express density is the term relative density. By relative density is understood, the mass of a particular volume of substance divided by the mass of an equal volume of fresh water. Due to the definition of the relative density it remains dimensionless.

In this lecture mass in pounds, and volume in gallons is used, therefore the density is given in pounds per gallon (ppg).

Force

When a mass hangs by a string, a force will keep the string in tension. The product of gravitational acceleration and the mass causes the force itself.

Fig 02

Mass

This force can be measured by a dynamometer, Fig. 03. This instrument consists of a spring. One end is fixed and the other end shows on a scale how much gravity force is exerted.

Fig 03

Force is expressed in the unit pounds-force, which is defined as follows.

One pound-force is the force, which will influence a body with a one pound mass when subjected to a gravitational acceleration of 9.80665 m/s2.

The gravitational acceleration of 9.80665 m/s2 is present at latitude 45° North on the earth's globe. Gravitational acceleration differs in various parts of the globe. This means that one pound-force is not an equal value everywhere on the globe.

As an example the gravitational acceleration at the North Pole is equal to 9.831 m/s2, which gives a force influence on a mass of one pound according to the following

-G = 1 x 9,831

9,80665 = 1.0025 [ pounds ]

At the equator the gravitational acceleration = 9,781 m/s2

The force influence on one pound mass becomes

G = 1 x 9,781

9,80665 = 0,9974 [ pounds ]

In practice this variation in gravitational acceleration is ignored and a one pound mass is considered to exert a one pound-force influence.

Pointer Scale

Pressure

Pressure is defined in physics as force per area unit.

Pressure = Force Area unit

The total force, which acts on a plane, is divided by the area of the plane. The result is called pressure.

The unit for force is pounds-force and the unit for area is square inch. Therefore the unit for pressure will be:

Pressure = Pounds

Square inch [pounds per square inch ]

M = 1 pound

G = 1 pound ( 45° latitude North ) g = 9,80665 m/s2 A = 1 inch2 Pressure (P) = P x G A = 1 1 = 1 Fig 04 04.01 Pressure in fluids

Considering a vertical cylindrical volume of static fresh water with a cross-sectional area of one inch2 and height of 10 ft, the pressure at the bottom of this cylinder can be calculated

-The fluids total volume is 1 in2 x 10 x 12 = 120 in3

The density of fresh water is 8.34 ppg

8.34 pounds per gallon = 8.34 x 7,48

1728 pounds / inch 3 Fig 05

M

G

A

The mass of the fluid column will be M = 8.34 pounds gallons x 7,48 gallon ft x 1 1728 ft

inch x 120 inch = 4,33 pounds 3

3 3

3

The pressure at the base of the fluid column is caused by gravitational acceleration that acts on the fluid column divided by the fluid columns' cross sectional area.

psi 4.33 = inch pound 1 4.33 = Ph ₂

It is important to realise that the pressure at the bottom of a static fluid column is only depending on the vertical height of the column and the density of the fluid.

05.01 Pressure gradient

Considering a porous and permeable rock formation in which the pores are filled with fresh water (density 8.34 ppg).

It is now possible to calculate the pressure at 5000 feet depth

-psi = 10 5000 x 4,33 = Ph 2165

It is also possible to calculate the pressure increase that every foot of depth will represent.

ft pr psi 0.433 = 5000 2165 = ft per increase Pressure

This quantity which represents pressure increase in psi/ft is named Pressure Gradient 8G.

When the pressure gradient for a fluid or gas is known it is easy to calculate the pressure at any given depth.

From the shown example of freshwater (8.34 ppg) and pressure gradient (0.433 psi/ft) it is possible to calculate the pressure gradient for a fluid or for a gas with a density of 1 ppg.

ft psi 0,052 = 8,34 0,433 = ppg 1 for gradient Pressure /

With this new figure it is now possible to calculate the pressure gradient for any fluid or gas. Pressure gradient = 0.052 x density in ppg

Example:

Answer: 0.052 x 10.4 = 0.541 psi/ft

Calculate the pressure exerted from this fluid at a depth of 4000 ft -Answer: 0.541 x 4000 = 2164 psi

Fig 06 shows different pressure gradients and illustrates how pressure increases with depth-Fig 06

DEPTH

PRESSURE

1 Gas grad. 0.07 psi/ft

2 Oil grad. 0.30 psi/ft

3 Fresh W. grad 0.433 psi/ft

4 Salt W. grad 0.465 psi/ft

5 10 ppg grad. 0.52 psi/ft

6 15 ppg grad. 0.7785 psi/ft

06.01 Abnormal / Subnormal Pressure

So far it has been assumed that there is a direct proportional relation between formation pressure and fluid density and true vertical depth from the surface.

That means that the formation fluid pressure is only affected by the fluid density and from the true vertical depth.

The influence of the overlying rock formations has so far not been considered.

The reason is that in case of a permeable and porous formation system every single rock particle rests upon or leans up against other particles just below and to the side of it. Therefore the rock structure supports its own weight, and regardless of depth does not affect the formation fluid pressure.

Artesian Well

When talking about artesian wells, we are normally talking about water wells where we have a porous sandstone witch has communication to higher laying areas creating abnormal pressure below a cap rock.

Fig 07

Under compaction

Let us consider that at a particular period in a rock formations' development it was not possible for the formation fluids to escape since an impermeable formation type placed on top prevents this from happening. Therefore the rock particles can not be compacted and consolidated sufficiently to carry the weight of the overlying rock. Since the fluid trapped in between the particles could not escape the fluid will be exposed to compressing forces. These forces result in an increased formation fluid pressure, which is abnormal at the given depth. It can be realised that the trapped formation fluid has to carry the weight of the overlaying formation, along with the formation rock in which it is trapped. In a situation such as this the formation pressure will be greatly different from a calculated normal pressure/depth forecast.

Example:

A formation at 5000 ft depth contains formation fluid. The formation fluid has communication to the surface through porous and permeable formation rock. See fig. 08

POROUS SANDSTONE BELOW CAP ROCK

HYDROSTATIC PRESSURE FROM FORMATION WATER COLUMN LAKE NORMAL FORMATION

PRESSURE AT THE WELL UNTILL BELOW THE CAP

Formation pressure at 5000 ft will be the fluid column pressure Density for formation fluid = 8.95 ppg

Pressure gradient for formation fluid = 8.95 x O.052 = 0.465 psi/ft Pf (Pressure of Formation) = 5000 x O.465 = 2325 psi

Fig 08

If it is considered that this formation fluid was trapped in an earlier period in the sedimentary process and therefore could not escape the later compaction process, it is possible that the fluid may be exposed to the weight of the overlying rock mass.

Assuming formation fluid is 10% and an equivalent formation density of 21 ppg this results in the following formation pressure

-Pf = 5146.7 psi

This formation fluid is over-pressured or abnormal. Over-pressured formations are often encountered with thick salt sediments and salt domes. Salt does not have the same structure as normal rock formations. Salt is termed a "plastic" formation, which means that it is not self-supporting, it can move and deform under pressure, and (this is not necessarily a rapid process). When pressure is applied to a salt formation it behaves more as fluids rather than as solid matter. The relative strength of salt is very low compared to other rock types.

Because of the salt's qualities the weight from the overlying formation including the weight of the salt layers themselves will be transferred to the formation below the salt. The pressure in the salt and in the formation below it will often have a pressure gradient of 1 psi/ft instead of the normal pressure gradient for formation fluid, which is 0.465 psi/ft.

f P = ( 0.1 x 5000 x 8.95 x 0.052) + (0.9 x 5000 x 21 x 0.052) 5000 ft 5000 ft 2325 psi 5147 psi Impermeable zone

Abnormal pressures can also occur when an encapsulated and normal pressured formation for the particular depth at a later stage in history with movements or surface erosion is brought closer to the surface.

The particular formation in question can be found deeper or shallower in relation to its original position. If it is the case that the formation pressure cannot adjust to its new depth it will hold its original pressure.

Example:

If a sandstone formation at 4000 ft depth is considered it will have a normal pressure of 1860 psi. On account of geological processes the area of sandstone becomes isolated by impermeable rock. Through earth movements the formation moves to a shallower depth of 2500 ft. In this situation the sandstone will retain it's original 1860 psi pore pressure but he surrounding formation has a pore pressure of 1160 psi. Such an isolated zone is called a high-pressure zone or abnormal pressured zone.

It may as well be the case that the isolated sandstone by earth movements was brought down to 5000 ft depth. The normal pressure for 5000 ft would be 2325 psi and the isolated sandstone area with its 1860 psi would become a low-pressure or subnormal-pressured

zone.

Fig 09

Abnormal pressured formations can also develop because of differences in the contained formation fluid and gas densities.

Figure 10 shows an anticline. An anticline is the geological term for an area of formations which, due to earth movements has been pushed upwards to take a shape like a dome. In the figure the anticline consists of porous sandstone which contains gas. A layer of impermeable shale that prevents the gas from escaping caps the sandstone. The formation surrounding the anticline has a pore content of salt water and a base depth of 5000 ft. The formation pressure is considered to be normal. Formation pressure of the salt water bearing rock at 5000 ft will therefore be:

4000 ft 2500 ft 5000 ft 1860 psi 1860 psi 1860 psi 1160 psi 2325 psi 1860 psi

Fig 10

If the sandstone in the anticline contained salt water instead of gas, the formation pressure at the very top of the anticline would be exactly the same as the formation just above.

Example:

Pf = 3000 x 0.465 = 1395 psi

The sandstone however is containing gas, which has a pressure gradient of 0.1 psi/ft. This results in the pressure at top of the anticline to be substantially higher than the calculated 1395 psi for a salt-water formation.

The reason is that the hydrostatic pressure of gas within the anticline is much lower than the corresponding hydrostatic pressure of salt water on the outside.

Pressure from the 2000 ft high gas column will be: Ph = 2000 x 0.1 = 200 psi

Therefore the formation pressure at the very top of the anticline below the cap rock will be: Pf = 2325 - 200 = 2125 psi

Formation structures of this type give a real problem if the formations above and/or below will not withstand the 12.45 ppg hydrostatic pressure from the drilling fluid that is required to

psi 2325 = 0.465 x 5000 = Pf Porous with water Sandstone with gas 5000 ft 2325 psi 3000 ft 1395 psi Anticline Tight Shale 2125 psi

balance the zone at 2000 ft. It may be necessary to set several casing strings in order to isolate the pressure.

High-permeability limestone formations have small formation strength gradients, and lost circulation may be the result when the bottom well pressure exceeds formation pressure by as little as 200 psi. This value may be less than the dynamic pressure drop in the annulus or less than a safe trip margin. Such conditions can be risky if insufficient information is available.

Transition zones and under compacted shale

Wherever massive shale formations are found the risk for transition zones and high pressure is present. This is caused by thick impermeable shale restricting the disposal of formation fluid. Due to new sediments are settled on the seabed increasing weight load is exerted on the shale from the formation above. The water, gas or oil trapped within the shale cannot escape. The result is the development of abnormal pore pressures. The terminology under compacted shales is used to indicate these circumstances.

A seal of harder rock often caps the top of the abnormal pressured shale. Just after the cap is penetrated the Rate of Penetration (ROP) increases. The reason is that the shale is easier to drill since the differential pressure between drilling fluid hydrostatic pressure and the formation pressure decreases. A reduction in overbalance results in a faster drilling rate. When the Driller maintains his drilling

parameters constant t.i. constant rotary speed, constant weight on bit and constant pump rate, the Rate of Penetration should be constant as well, unless changes in the drilled formation takes place.

The indication of changes in the formation can therefore be observed by the Driller by means of changes in Rate of Penetration. To confirm whether the well is still in balance, the Driller must stop and observe if the well is static. The terminology for this operation is "flow checking the well".

Fig 11

Whenever thick shales are encountered it is important to be careful and expect abnormal pressure in the formation. Shale related abnormal pressures can occur at any depth from surface to very deep and is the most common reason for abnormal formation pressure.

Because the formation fluid in under compacted shale is unable to escape, a typical trend will indicate that the cuttings density decrease with depth. The density decrease with depth can indicate that abnormal pressure is encountered.

ENCLOSED SAND LENS WITH FORMATION FLUID UNCONSOLIDATED

SHALE-DENSITY DECREASES WITH DEPTH-WATER ENCLOSED

SAND WITH COMMUNICATION TO SURFACE SHALE-DENSITY INCREASES WITH DEPTH - WATER ESCAPES

Surcharged formations by underground blowouts

A different reason for abnormal formation pressures are the result of previous blowouts underground. Shallower sands can become charged as the result of an uncontrolled underground blow out from an adjacent well or from a bad cement job. Even the well has successfully been closed in on surface the pressure from the deeper zone can communicate to the shallower sand reservoir.

When the next well is drilled the abnormal pressure is encountered at the much shallower depth.

See Fig 12

Fig 12 Fig 13

Surcharged formations by natural causes

Shallow formations may also be surcharged by natural causes. This can be the result of a fault in the formations. A fault gives a means of communication between deeper formations with high pressure and shallower formations. The higher pressure escapes into the shallower formation where an abnormal pressure will be the result.

See Fig 13. UNDERGROUND BLOWOUT Pf FAULT ZONE Pf

02

PRESSURE BALANCE IN THE WELL BORE

During drilling of a well the formation pressure must always be counter balanced by an equal amount of pressure exerted from within the well. This is achieved by using a drilling fluid having a sufficient density.

Drilling fluid which is a combination of different fluids and chemicals has several important functions in the drilling process but a main function is the ability to give pressure balance in the well.

The density of the fluid can be adjusted by adding high density material or by diluting by water. It is in this way that balance and control of the formation pressure can be achieved.

02.02 Overbalance and underbalance

Underbalance is the term used when at a particular depth the formation pressure exceeds the hydrostatic pressure exerted by the drilling fluid column. In this situation there is a risk that fluid from the formation will intrude into the wellbore and begin to displace the drilling fluid. On surface the drilling fluid returns rate will increase and later the active drilling fluid pits will show a gain of fluid. If this sequence of events takes place in a well a kick is said to have occurred.

The rate of influx is dependent on the degree of underbalance and on the formation's permeability. To drill a well underbalanced is dangerous in most parts of the world and is therefore usually not practised in oil well drilling.

However in certain areas where sufficient data are available it is practised anyway mainly because drilling can take place with a high penetration rate.

03.02 Lost circulation

Overbalance in the well is present when the drilling fluid hydrostatic pressure exerts a higher pressure than the formation pressure. In this situation formation fluids cannot intrude into the wellbore. The reverse can occur whereby drilling fluid will seep into the formation, and lost circulation may be the result. This is not a desirable situation.

If drilling fluid seeps into the formation the formations' permeability becomes reduced. When the overbalance becomes too large the formation will break allowing a large amount of the drilling fluid to flow into the formation. This situation is called lost circulation.

When lost circulation has been the result a dangerous situation is created. The drilling fluid level drops and hydrostatic pressure is lost. When hydrostatic pressure is lost the formation pressure higher up becomes underbalanced which can result in a blow out.

04.02 Rate Of Penetration versus overbalance

The difference between the hydrostatic pressure exerted by the drilling fluid at the bottom of the wellbore and the formation pressure is called the differential pressure. When the

hydrostatic pressure exerted by the drilling fluid is higher than the formation pressure the bottom hole pressure is in overbalance.

The relationship between differential pressure and Rate of Penetration shows that Rate of Penetration increases when the differential pressure decreases. Penetration is given in feet per minute and differential pressure in psi.

Fig 14

The graph is interesting in several ways. Assume drilling with a differential pressure of 2430 psi in a particular formation with constant drilling parameters E.I. :

- Constant Weight on Bit - Constant drilling fluid density - Constant rotary RPM and - Constant pump rate

it can be seen that the penetration rate is 4 ft per minute.

Ft/min psi Ra te of P en etra ti o n Differential Pressure

P =

Without changing any other parameters imagine that the formation pressure increases by 980 psi. This results in a new differential pressure of 1450 psi and a corresponding increased penetration rate to 6 ft per minute.

It is realised that when the differential pressure decreases the penetration rate will increase.

05.02 Drilling break

An increase in Rate of Penetration (ROP) with constant drilling parameters is called a drilling break.

It should be known that a drilling break is an early warning of a kick. If the Driller reacts on the observation by making a flow check the well may still be overbalanced with the pumps stopped.

Even that an increase in Rate of Penetration may be caused by other factors than a change in differential pressure, the Driller should always play safe and perform a flow check in order to confirm that the well is in balance. A natural reaction must also be to inform the supervisors of any drilling breaks.

06.02 Necessary overbalance

By means of the graph it is seen that to obtain the highest possible penetration rate the degree of overbalance has to be as small as possible. In practice a sufficient overbalance must be used to avoid kicks from taking place.

07.02 Trip margin

A situation that can bring the well in underbalance is when the drill string is pulled upwards during a connection and when tripping the string out of the well. The lower part of the drill string acts as a piston that results in reducing the pressure below the string when pulling upwards.

When the pressure in the wellbore is reduced the formation fluids can enter the well underneath the bit.

To what extent this occurs is dependent on:

- How quickly the drill string is pulled upwards - The dimension of the wellbore

- Dimensions of the drill string

- The rheological characteristics of the drilling fluid

- Other factors like degree of balling of the Bottom Hole Assembly etc.

To prevent formation fluids from being swabbed into the wellbore caused by any of these reasons in combination a necessary overbalance is used. This small degree of overbalance is called a trip margin.

Fig 15

Fig. 15 shows the conditions when drilling in normal pressure conditions. The tolerance area (given by the area between the formation strength pressure and the formation pressure) is relatively large.

When the drilling fluid density is adjusted to be in the centre of the area, there is only a small risk for swabbing in connection with a trip. There is also allowance for a relatively large surge pressure in excess of the hydrostatic pressure without risk for exceeding the formation strength.

Surge pressure in the well is the result of lowering the drill string too quickly. The piston effect results in increasing the pressure below the drill string.

Fig 16 and 17 shows different measurements taken with a PWD tool under “normal” tripping conditions. Fig 16 Fig 17 Formation Strength Formation Pressure Fluid Density Swab Pressure Surge Pressure Pulling Speed (mins/stand Start EMW (G) End EMW (G) Pressure Drop psi 4 5 7 8 0.965 0.964 0.962 0.962 0.956 0.956 0.958 0.960 140 124 62 31 SWAB PRESSURE SWAB PRESSURE Running Speed (mins/stand Pump Rate 0 gpm Pump Rate 180 gpm Pump Rate 250 gpm 1 2 3 4 295 psi 124 psi 93 psi 651 psi 434 psi 356 psi 837 psi 636 psi 527 psi SURGE PRESSURE SURGE PRESSURE

08.02 Riser margin

When drilling takes place from floating rigs (semi-submersible and drill ship), there can be several hundred feet of distance between the rig and the sea floor. The marine riser connects the rig to the sea floor amongst other to allow returns to be taken to the rig. The drilling fluid that is contained in the marine riser is contributing to balancing the formation pressure in the well.

If a marine riser by accident or on purpose is disconnected from the wellhead at the seabed the bottom hole pressure will be reduced. The reason is that the drilling fluid in the marine riser from the well head to the bell nipple is removed and replaced by a shorter column of seawater. An important factor is that the seawater has a lower density than the drilling fluid. To prevent that the reduction in hydrostatic pressure leads to a kick and a blowout a preparation must be made that will ensure that a sufficient overbalance in the well, even with the marine riser disconnected. This overbalance is called a riser margin.

It is realised that there are many precautions to take into consideration, when deciding the drilling fluid density to be used in a particular situation.

09.02 Relationship between hydrostatic pressure, drilling fluid density and true vertical depth

Example:

Well depth TVD 6000 ft Drilling fluid density 10.5 ppg What is the hydrostatic bottom hole pressure?

Answer: Ph = 10.5 x 0.052 x 6000 = 3276 psi

It is required to increase the hydrostatic bottom hole pressure by 500 psi. Which new drilling fluid density shall be used?

Answer: Ph = 3276 + 500 = 3776 psi

The new drilling fluid density will therefore be:

3776

MW = --- = 12.1 ppg 6000 x 0.052

The increase in drilling fluid density will be:

∆MW = 12.1 - 10.5 = 1.6 ppg

With the new drilling fluid density drill to 9000 ft TVD and calculate the bottom hole pressure at this depth?

Answer: Ph = 12.1 x 0.052 x 9000 = 5663 psi

What is the pressure gradient of this drilling fluid column?

Answer: G drilling fluid = 12.1 x 0.052 = 0.629 psi per foot

This can also be calculated a different way:

5665

Gmud = --- = 0.629 psi per foot

9000

All results comes from utilising the formula:

Ph = TVDft x Drilling Fluid Densityppg x 0.052

0.052 is a constant, which represents the pressure gradient in psi per foot for a fluid density equal to 1 ppg.

Pressure Gradient

10.02 Equivalent drilling fluid Density.

Considering a well with a true vertical depth of 6000 ft full of drilling fluid that has a density of 11 ppg.

The well is closed-in at the surface with the Blow Out Preventer ( BOP ) and drilling fluid is pumped slowly into the wellbore. Pressure at the top of the well will now increase to 900 psi. See Fig 18

Find what the bottom hole pressure in the well will be?

It is seen that the pressure now consists of two components.

- The hydrostatic pressure from the drilling fluid (which changes with depth)

- The static pressure at the surface (which gives a constant extra pressure at all depths in the well).

Fig 18

Hydrostatic pressure 11 x 0.052 x 6000 = 3430 psi

Closed-in pressure = 900 psi

Bottom hole pressure = 4330 psi

Which drilling fluid density must be used if the above bottom hole pressure shall be maintained by using only hydrostatic pressure?

The calculated drilling fluid density is called the equivalent drilling fluid density.

MW = 4330 6000 x 0.052 = 13.9 [ppg] 900 psi 900 psi 900 psi 6000 ft MW 11 ppg

This means that the original 11.0 ppg drilling fluid must be replaced by a drilling fluid which has a density of 13.9 ppg if the same bottom hole pressure shall be present without extra pressure being applied at the top of the well.

Pressure in all depths in the well will be different in the two examples. This can be confirmed by simple calculation.

What is the pressure at 3000 ft in the two examples? Example with closed-in pressure on surface:

Example without closed-in pressure on surface:

It must be realised that pressures throughout the well will be lower, if a particular bottom hole pressure is achieved only by drilling fluid density, rather than using a lower drilling fluid density combined with a static pressure applied at the surface.

psi 2616 = Pressure Total psi 900 = Pressure Static Applied psi 1716 = 3000 0.052x x 11 = P 1. h psi 2168 = 3000 x 0.052 x 13.9 = P 2. h

03

Dynamic pressure regime when circulating

Whilst drilling the drilling fluid is continuously circulated to clean out the rock fragments (cuttings) from underneath the bit whilst removing them up to the surface where they are separated from the drilling fluid by the mud cleaning equipment.

To establish the circulation in the system it is required to have a dynamic fluid differential pressure between certain areas in the system. This pressure difference represents a certain energy that is used to overcome the resistance against fluid movement, resistance that is existing in the system.

This resistance against fluid flow or friction as it is generally called in a hydraulic system is largely dependent upon:

- The fluids' rheology (viscosity, density etc.) - The fluids' velocity

- Type of flow regime ( laminar or turbulent)

If a fluid is pumped through an enclosed pipe system with a constant velocity the resistance in the system depends on the flow area. Where the fluid flow meets diameter reductions, a local increase in velocity is the result and therefore a greater friction. Conversely where the flow meets a larger diameter the velocity will decrease and the friction will consequently also decrease.

Fig 19

Fig. 19. shows a circulating fluid system where the initial pressure at the pump is 1400 psi and the final pressure is 0 psi at the flow line. It is seen that the 1400 psi represents the energy required to overcome the friction that is present against the flow of the fluid in the system. Large obstructions to flow give large pressure losses. By means of pressure gauges placed in the system the pressure losses in the different parts of the system can be monitored.

1400 1320 1280 1220 1170 800 0

80 40 60 50 370 800

Recorded Pressure (psi)

Applying these considerations to the circulation of drilling fluid the Fig. 20. shows a pipe system in which the drilling fluid pump ( mud pump ) shall pump drilling fluid through. This simplified pipe system consists of drill pipe, drill collars, bit nozzles and annulus. The drilling fluid enters the top of the drill string with a pressure of 2200 psi. On the way down through the string some of this pressure is lost depending on

- The dimensions of the drill pipe (Internal diameter) - The characteristics of the drilling fluid.

Fig 20

P1 = Pressure as drilling fluids enters the drill pipe (2200 psi)

P2 = Pressure as drilling fluid enters the drill collars (1900 psi)

P3 = Pressure as drilling fluid enters the bit nozzles (1700 psi)

P4 = Pressure as drilling fluids enters annulus (130 psi)

P5 = Pressure as drilling fluid enters the flow line ( 0 psi)

The largest pressure loss in the well system takes place when fluid flows through the bit nozzles that have a relatively small flow-through area.

On the way towards the surface through the annulus, the pressure loss will be the lowest in the system, because the friction is not at all large on account of the large cross-sectional area of the annulus.

The pressure figures used in Fig. 20. are based on average calculations for a simple rotary assembly, and they show that 94% of the total pressure loss occurs in the drill string and bit nozzles.

NATIONAL DRILL PIPE

DRILL COLLARS ANNULUS BIT P1 P2 P3 P4 P5 PSI

The figures show that to circulate the drilling fluid from the bottom of the well up to the surface it is only necessary to use approximately 6% of the total pump pressure. This dynamic pressure will be transmitted to the bottom hole pressure.

When the pump is running and circulation takes place there will be a higher bottom hole pressure than when the pump is stopped.

With the pumps stopped only hydrostatic pressure is present in the well to balance the formation pressure.

02.03 Dynamic pressure in the wellbore ( Circulating Pressure)

Dynamic Pressure ( PC ) is dependent on three factors:

- Components in the flow system

(Flow area, length of drill string, nozzles size etc) - The fluid characteristics ( Rheology )

- The flow rate

(SPM, liner size, pump efficiency etc)

Change in drilling fluid characteristics ( such as viscosity and gel-strength ) can change the friction against flow in a system.

A fluid's flow resistance is largely depending on the drilling fluid density. In well control calculations it is accepted that dynamic pressure loss is proportionally depending on drilling fluid density.

PC1 = Circulation pressure when drilling fluid density is MW1 PC2 = Circulation pressure when drilling fluid density is MW2

The expression for the relationship between circulation pressure and drilling fluid density has proved to be realistic in most practical cases. See fig. 21.

[psi] MW MW x 1 P = 2 P 1 2 C C PSI PC2 PC1

Low fluid density High fluid density

Example:

At 100 SPM the pump pressure is 1000 psi with a drilling fluid density of 10 ppg. What would the pump pressure be at 100 SPM if the drilling fluid density was increased to 12 ppg??

New pump pressure:

To calculate the new pump pressure it is required to know the original pump pressure, which is read just after the pump ( standpipe pressure ).

The third factor that affects the circulation pressure is the speed of the flow of drilling fluid. This velocity of flow is directly related on the pump speed ( SPM = strokes per minute).

The relationship between pump speed and dynamic pressure can be expressed as:

Where SPM is the number of strokes per minute in the two cases. Example:

Circulation pressure is 1200 psi with 40 SPM.

What will the circulation pressure be if the pump speed was increased to 80 SPM? Answer:

It is realised that if the pump speed is increased to twice its original value the dynamic pressure will be increased almost fourfold. The graph in Fig. 22 illustrates this fact.

The power 1.86 is an experience figure, which is obtained from experiments. However in well control calculations it is generally accepted to use the power 2 in stead of 1.86.

[psi] 1200 = 10 12 x 1000 = 2 PC C C 1.86 2 1 P 2 = P 1 x SPM SPM æ èç ö ø÷ psi 4356 = 40 80 x 1200 = 2 P 1.86 C ÷ ø ö ç è æ

For well control calculations use the formula below:

Fig 22

Fig. 23. shows circulation pressures and pressure losses between the drill string and annulus with three different pump rates.

Fig 23 C C 2 2 1 P 2 = P 1 x SPM SPM æ èç ö ø÷ 10 20 30 40 50 60 70 80 90 SPM 1000 2000 3000 4000 Pc 1000 2000 3000 4000 Pc 80 spm 60 spm 40 spm

04

CONSIDERATIONS WITH A CLOSED-IN WELL

Fig. 24 illustrates a wellbore with pressure gauges. The drill string consists of pipe connected to each other, right down to the bottom of the well. Through the bit nozzles, the string is in communication with the annulus. In principle this can represent two pipes, one inside the other but there is only communication at the bottom of the well. On top of the well the BOP equipment is installed. This equipment makes it possible to contain and close off the annulus and its contents. Under the BOP a pressure gauge is installed which measures the surface annulus/casing pressure.

On the top of the drill pipe after the pumps another gauge which measures drill pipe pressure is installed. The two gauges are necessary to get an indication of down hole conditions.

Fig 24 Fig 25

02.04 U-tube

A simplified and equal system can be represented by two tubes standing upright side-by-side and connected at the bottom. The example is called a U-tube. See. Fig. 25.

The pressure in the same horizontal levels in the connected system is always the same if fluid density is the same, when no circulation is taken place and no pressures are closed in on the top on any of the two legs.

It is seen that the hydrostatic pressure at the bottom of such a U-tube system, irrespective of which leg of the U-tube column is considered will be equal. This is easily confirmed by a simple calculation: NATIONAL ANNULAR DRILLSTRING DRILLCOLLAR BOP PDP PA PA PDP A DRI LL S T RI N G ANNULUS

Example:

True vertical depth = 10000 ft Drilling fluid Density = 10 ppg

Both drill pipe and annulus are open at the surface and the U-tube is in balance. Bottom hole pressure Ph at the point A can be found either by the drill pipe or by the annulus when drilling

fluid density is uniform :

If the BOP is closed on the annulus and the drilling fluid in the annulus is replaced with saltwater

( 8.34 ppg ) the following can be calculated:

The internal contents of the string ( drill pipe and drill collar ) has not changed so PH at A is

still 5200, but the hydrostatic pressure in the annulus is only (Fig 26):

Fig 26 Fig 27 [psi] 5200 10000 x 0.052 x 10 = Ph = [psi] Depth Vertical True x 0.052 x Density Fluid Drilling = Ph ppg ft psi 860 = 4340 -5200 = P psi 4337 = 10000 x 0.052 x 8.34 = P a ha PA PDP A DR IL L S T RI NG AN NUL US 10 ppg 8.34 ppg 10000 ft PA PDP A DRI LL S T RI N G ANNULUS 9 ppg 8.34 ppg 10000 ft

Example:

Considering the same well with the same bottom hole pressure, but now with 9 ppg drilling fluid in the drill string and upper 7000 ft of annulus, while there remains saltwater in the lower part of the annulus (Fig 27).

When PSIDP = Pressure ( Shut in drill pipe )

When PSIA = Pressure ( Shut in annulus )

The example represents a typical kick situation, where insufficient drilling fluid density has resulted in a saltwater influx into the annulus. The influx has replaced a quantity of drilling fluid. Notice that the drill pipe bottom hole pressure consists of two parts, first the PSIDP value

and secondly the hydrostatic pressure of the drilling fluid. The annulus bottom hole pressure consists of three parts. The PSIA value, the hydrostatic pressure of drilling fluid and the

hydrostatic pressure of the saltwater.

[psi] 520 = ) 10000 x 0.052 x 9 ( - 5200 = PSIDP [psi] 3 62 ) 0.052 x 8.34 x 3000 0.052 x 9 x 7000 ( - 5200 = PSIA + =

05

PROPERTIES OF GASSES AND GAS LAWS

If drilling takes placed being underbalanced the risk of taking a kick is always present. The influx resulting from a kick can be water, oil or gas.

When dealing with gas the drill crew must be aware that gas behaves differently than fluid.

02.05 Properties of gas and Gas Laws

A given mass of gas can be compressed or expanded, and as the volume changes the pressure will do the same.

Boyles Law states that:

P1 x V1 = P2 x V2

or

Pressure x Volume = Constant → See Fig 28

Fig 28 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 11000 12000 13000 14000 15000 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 VOLUME PRESSURE

This means that when a given volume V1 with an absolute pressure P1 is changed in

pressure or volume we get a new pressure P2 with a new volume V2.

Example: V1 = 5 gal P1 = 170 psi V2 = 3 gal Calculate P2 [psi] 283 = 3 5 x 170 = V V x P = P _ V x P = V x P 2 1 1 2 2 2 1 1

It is important to know that gas expands if pressure is reduced.

Boyle’s Law is only true when the temperature is constant. If the temperature changes the formula given below is used where → T = temperature

1 1 1 2 2 2 P x V T = P x V T

It must be noted that the temperature to use is an absolute temperature which is given in Kelvin degrees, (ºK ) for the Centigrade system.

If the Fahrenheit system is used the absolute temperature must be given in Rankin (ºR ) degrees.

ºK is obtained by addition of 273º to the temperature given in Centigrade ºC.

K C

T° = t + 273°

ºR is obtained by addition of 460º to the temperature given in Fahrenheit ºF.

R F T° = t + 460° Example: V1 = 12 gal P1 = 90 psi T1 = 20ºC V2 = 12 gal T2 = 80ºC [psi] 108 = 12 x 20) + (273 80) + (273 x 12 x 90 = V x T T x V x P = P 2 1 2 1 1 2

Example:

The formula, which relates to the properties of gasses, is here used in a practical example. The well has a depth of 10000 ft and there is a swabbed gas bubble on bottom. The drilling fluid density is 12.5. The well is open and in balance. Consequently no closed-in pressure at the surface. The pressure in the gas is therefore equal to the hydrostatic pressure at 10000 ft → Ph.

Hydrostatic pressure Ph is 6500 psi.

If the BOP is closed and the gas is allowed to rise upwards ( migrate ), the gas volume will not change and in accordance with the gas law the pressure will also remain unchanged. Assuming the temperature is constant the gas would retain its original volume and pressure all the way to the surface.

Fig 29

Considering that the gas has migrated halfway up the wellbore it will still have a pressure of 6500 psi. The pressure at surface ( annulus ) at this stage would be: (See Fig 29.)

[psi] 3250 = 5000 x 0.052 x 12.5 -6500 = PSIA

Bottom hole pressure:

[psi] 9750 = 5000 x 0.052 x 12.5 + 6500 = Pbottom PA P_A PA 10000 ft 12.5 ppg 0 psi 6500 psi 6500 psi 5000 ft 6500 psi 9750 psi 3250 psi 5000 ft 12.5 ppg 12.5 ppg 6500 psi 6500 psi 12.5 ppg 13000 psi 10000 ft

When the gas is allowed to rise all the way to the surface without expanding, the pressure at the surface would be 6500 psi.

Bottom hole pressure would be:

[psi] 13000 = 10000 x 0.052 x 12.5 + 6500 = Pbottom

This extreme pressure throughout the wellbore cannot be controlled, and it is not reasonable to assume that the situation would develop all the way as described. The weakest point in the wellbore is normally believed to be at the casing shoe level. When the pressure increases above the strength at the weakest point the formation at that point will fracture. The risk for an underground blow out is high.

A gas kick can never be allowed to migrate up through the annulus without expanding. A skilled drill crew must take proper and timely action to avoid the dangerous situation that is likely to occur.

In the given example the temperature influence neither the changed height due to annulus geometry was taken into account since these factors only have a small influence in practice.

03.05 Expansion of Gas

Although some kicks are predominantly salt water or oil, at least some gas is usually present. Because salt water and oil do not expand as pressure decreases, they are not as troublesome as gas. It is important for the persons who control blowouts to understand the behaviour of gas in a well.

The gas volume change as a result of pressure change is predictable, and this allows calculation under illustrative conditions of changes in bottom well pressure as gas rises through the drilling fluid. When the pressure of a given mass of gas is doubled, the volume is halved. When the pressure is halved the volume is doubled. This relationship between pressure and volume results in the greatest expansion of the gas in the upper part of the well. See Fig 28.

Although cut drilling fluid is one of the early indicators of abnormal pressure, minor gas-cutting results in only a small reduction in the hydrostatic head. In a gas-cut column of drilling fluid, the pressure increases rapidly with depth, so that the volume of gas scattered through the well bore is very small, and reduces the overall drilling fluid density in the well very little. A slug of gas in the bottom of a well is potentially dangerous, because it will expand greatly when it rises or is pumped up. Under low pressure near the surface, it will displace a large amount of drilling fluid from the well and consequently greatly reduce bottom hole pressure giving risk for a blowout.

The safe handling of a gas kick requires knowledge about the principle of gas expansion and consequently lowering the pressure in the gas bubble as it is circulated up through the annulus in order to maintain the correct and constant bottom hole pressure. The theoretical knowledge requires practice as well as knowledge about well control equipment.

When the gas in a well control situation is circulated to the surface and expanding, more drilling fluid must be allowed to flow out of the annulus than is pumped into the drill pipe. Thus, the pit level will increase.

The expected drilling fluid volume increase should be known prior to circulating out the kick. This detail is discussed in the kick control section.

To control a correct and constant bottom hole pressure, the surface pressures are used as a parameter for control. This is done by means of the choke in connection with a stroke counter for the mud pumps, and simple recognised procedures.

04.05 Formation strength

From the previous examples it is realised that pressure throughout the wellbore increases when gas rises up the annulus in a closed-in well. Gas must be circulated out of a well under control. One of the most important limitations that should be known is the maximum pressure the formation ( or weak point ) can withstand before it fractures and allows the drilling fluid to flow into the formation.

If the formation strength is exceeded in a kick situation there is a high risk for an underground blow-out and perhaps complete loss of control of the well.

The formation strength is recorded by means of a leak-off test.

05.05 Leak-off test

A leak-off test can be carried out in various ways. The aim is to find the surface annulus pressure value for when the drilling fluid begins to seep into the formation, without at the same time to cause an actual fracture of the formation. The less drilling fluid volume that is pumped into the formation, the less damage there is caused to the formation. After the test the formation should easily heal again as a result of the drilling fluid's wall building effect. A leak-off test is carried out just after casing has been set and cemented.

A leak-off test may be conducted as follows:

Between 10 and 30 feet is drilled below the casing shoe to expose virgin hole.

The well is circulated to obtain a representative and accurately known drilling fluid density in the well.

The well is closed-in and drilling fluid is pumped into the well at a very slow rate.

The cement pump is generally used since they have a smaller displacement and thus are easier to control and are fitted with very accurate low pressure gauges.

Accurately measured volumes are pumped into the well, one barrel in this example, until an increase in casing pressure is registered. At this point pumping is stopped for about one minute, until the surface annulus pressure has stabilised. When no pressure decrease is observed the pressure is plotted on a graph paper. The pumping is resumed and the same volume is again pumped. The pumps are stopped and the new pressure is plotted after it has stabilised. This procedure is repeated until it is observed that the pressure increase per volume portion is no longer proportional.

This is easy seen on the plotted graph at the point where the straight line begins to bend. The pressure on the graph where this happens is the annulus surface leak-off pressure. See Fig. 30.

Fig 30

The leak-off test should be interrupted at this point. If the pumping is continued the pressure will decrease as a result of an increasing amount of drilling fluid which is injected into the formation. Furthermore the formation strength will be reduced. It has been proven that a leak-off test performed too far has damaged the formation. In that case a second leak-off test will indicate a lower formation strength.

Fig. 30 shows the results from a leak-off test carried out after casing has been cemented at 3000 ft. Drilling fluid density was 9.6 ppg. The leak-off pressure is seen to be 720 psi. The combined pressure the casing shoe is exposed to is the hydrostatic pressure of the drilling fluid and the surface leak off pressure and this combined pressure becomes the maximum allowable shoe pressure at any given time.

Calculate the maximum allowable pressure at the casing shoe: Answer: [psi] 2218 = 720 + 3000 x 0.052 x 9.6 = Pshoe

When we know the maximum allowable shoe pressure, we are able to calculate the

equivalent drilling fluid density or maximum allowable drilling fluid density

PRESSURE V O L U M E DEPTH S H O E P R E S S U R E 100 200 300 400 500 600 700 800 900 1000 1100 1 2 3 4 5 6 7 8 9 101112131415 * * * * * * * * * * * * * * Ph Max Pshoe PRESSURE ANNULUS

[ppg] 14.22 = 0.052 x 3000 2218 = density fluid drilling Equivalent

06.05 Maximum Allowable Annular Surface Pressure ( MAASP )

MAASP means the highest surface pressure that can be allowed at the top of the casing in excess of hydrostatic pressure that is likely to causes losses at the shoe formation if exceeded. There are three factors that decide the Initial MAASP.

- The maximum pressure that the surface equipment can handle - The maximum pressure the casing can handle

- The maximum pressure that the formation at the casing shoe ( or weak point ) can handle.

In most cases it is the formation strength at the casing shoe that is the deciding factor, and Initial MAASP is then given from the leak-off test which has previously been described.

As the maximum allowable shoe pressure remains constant the hydrostatic pressure inside the casing is the determine factor for the MAASP at any given time-See Fig 31

MAASP = Maximum Allowable Shoe Pressure – Pressure Hydrostatic Inside Casing

Fig 31

As illustrated in Fig 31 the MAASP will increase if pressure hydrostatic inside the casing decrease for whatever reason and visa versa.

P

h

MAASP

=

P

h

=

MAASP

One important issue when circulating out a kick is to monitor the Initial MAASP value. If the Initial MAASP is approached before the kick is circulated into the casing the responsible rig management must take safe action. It may be impossible to avoid exceeding the Initial MAASP, but the competent and responsible management may decide to evacuate the rig for non-essential personnel until the situation has proven to be safe.

Once the influx is inside the casing the initial value can be exceeded.

06

DRILLING FLUID VOLUME AND CAPACITIES

In routine operations as well as during well control operations it is necessary to know the total drilling fluid volume and the volume for the individual sections in the circulating system.

How much volume does the drill string contain and what is the volume in the different parts of the annulus?

These questions can easily be answered if the dimensions of the different components in the drill string and annulus are known. There are two ways to find the different capacities and volumes:

- By calculating the volumes - By reading tables

01.06 Calculating drilling fluid Volume - Capacities

The internal capacity of drill pipe and drill collars is calculated based on formulas for cylinders.

For a cylinder with a diameter d (inches) and a height of 1 foot the volume will be:

h x A = V /ft] ft [ 144 x 4 h x d x = V p 2 3 1 ft2 = 144 in2 1 ft3 = 0.1781 bbl Then: V = x d x 0.1781 4 x 144 = d 1029.4 bbl / ft 2 2 p Fig 32

Calculations of annular capacities are basically calculations of a hollow cylinder, or the difference between two cylinders, - one inside the other.

For a hollow cylinder with an outside diameter OD in and inside diameter ID in and a height of 1 ft the following formula can be derived

1 ft

D

ann 2 2 2 2 V = OD 1029.4 - ID1029.4 = OD - ID 1029.4 [ bbl / ft ]

The total inside or outside capacity for a certain length of pipe can be worked out by multiplying the capacity in bbl/ft by the length in ft. The result is the capacity in bbl.

Example:

Wellbore inside diameter = casing id = 9-7/8 in

Vertical depth = 5.000 ft

Drill pipe 5"OD & 4-1/4"ID = 4.600 ft Drill collars 7"OD & 2-13/16”ID = 400 ft

Fig 33

Internal capacities drill pipe:

Total Volume of drill pipe = 0.01754 x 4600 = 80.68 bbl Internal capacities drill collars:

Total Volume drill collars = 0.00768 x 400 = 3.07 bbl Annulus Capacities:

Total Volume between casing and drill pipe = 0.0704 x 4600 = 323.84 bbl

Total Volume between casing and drill collars = 0.04713 x 400 = 18.85 bbl

V drill pipe = (4 1 / 4) 1029,4 = 0,01754 bbl / ft 2 V drill collar = (2 13 / 16) 1029,4 = 0,00768 bbl / ft 2 V drill pipe = (9 7 / 8 ) - 5 1029,4 = 0,0704 bbl / ft 2 2 V drill collar = (9 7 / 8 ) - 7 1029,4 = 0,04713 bbl / ft 2 2 1 ft O D I D

02.06 Drilling fluid Volume and Capacities from Tables

It is common practice to use tables that give capacities in bbl/ft or litre/meter for different sizes of pipe and casing. These tables are made taking into consideration the physical outline of the pipes

( tool-joint etc. ).

Tables of this kind can be found in different Data Handbooks or in the Drilling Data

Handbook

( DDH ) sixth edition 1991. Section D “Capacities and Annular Volumes” and section G “Pumping and Pressure Losses”.

All the tables in DDH is in SI-units, but at the bottom of each table a conversion factor is found in order to convert to oil-field units.

Fig 34

Fig. 34 shows an example of a well and drill string.

Internal capacity of drill pipe: ( table D7 )

10000 ft of 5" Drill-pipe , 19.5 lbs/ft, Grade G-105. Reading in table: 9.05 l/m,(9.05 x 0.00192 = bbl/ft)

Internal capacity of drill collars: ( table D8 ) 500 ft of 7"OD x 2 13/16"ID Drill collar

Reading in table: 4.01 l/m,(4.01 x 0.00192 = bbl/ft)

Total internal capacity of drill string: 173.76 + 3.85 = 177.61 bbl

Volume between drill pipe and casing: ( table D14 ) 7500 ft of Casing 9-5/8”, 47 lbs/ft

Reading in table: 24.9 l/m, (24.9 x 0.00192 = bbl/ft)

Volume between drill pipe and Open-Hole: ( table D12 ) 2500 ft of 8-5/8” Open-Hole

Reading in table: 24.4 l/m, (24.4 x 0.00192 = bbl/ft)

Volume between drill collars and Open-Hole: (table D11 ) 500 ft of 8-5/8” Open-Hole bbl 6 7 . 3 17 = 10000ft x 0.00192 x 9.05 bbl/ft bbl / ft ( 4.01 x 0.00192 ) x 500 ft = 3.85 bbl bbl / ft (24.9 x 0.00192) x 7500 ft = 358.56 bbl bbb/ ft (24.4 x 0.00192) x 2500 ft = 117.12 bbl Drill Pipe 5” - 19.5 lbs/ft 10000 ft Casing 9-7/8” - 47 lbs/ft 7500 ft Open Hole - 8-5/8” 3000 ft Drill Collar 7” x 2-13/16” 500 ft

Reading in table: 12.9 l/m, (12.9 x 0.00192 = bbl/ft)

Total capacity of annulus: 558.56 + 117.12 + 12.38 = 688.06 bbl By making the above calculations the exact quantities of drilling fluid contained in the different parts of the well is known.

03.06 Surface-to-Bit Strokes & Bit-to-Surface Strokes

The exact number of strokes required to pump from the surface through the drill string to the bit, is known as surface-to-bit strokes.

The number of pump strokes required to pump from the bottom of the well to the surface, is known as bit-to-surface strokes.

These values can be calculated when the pump displacement per stroke is known. Pump displacement can be found in the DDH Section G table G6.

Given:

National pump 12-P-160. w/ 6" liners. The number 12 represents the stroke length in inches. Volumetric efficiency 97 %.

From the table is read 16.68 l/stroke with volumetric efficiency of 100 %. At the bottom of the table a conversion factor is found to convert into bbl.

With 97 % efficiency the pump output would be:

By using the capacity figures in fig. 34 we can now calculate surface-to-bit strokes as follows:

Surface-to-bit strokes

Respectively bit-to-surface strokes is now calculated. Bit-to-surface strokes bbl / ft (12.9 x 0.00192) x 500 ft = 12.38 bbl 16.68 x 0.264 42 = 0.1048 bbl / stroke 0.1048 x 97 100 = 0.1017 bbl / stroke

Total inside volume of drillstring

Mud pump output per stroke = Strokes

strokes 46 17 = 1017 . 0 61 . 7 17 = strokes bit Surface®

The circulating time required is controlled by the speed of the drilling fluid pump. In case of a pump speed of 30 strokes per minute ( SPM ) the result would be:

Another volume that is often necessary to know is the bit-to-shoe time and the corresponding pump strokes.

Considering fig. 34 it can be seen that:

Therefore:

Total annulus capacity

Mud pump output per stroke = Strokes

Bit surface strokes: 688.06

0.1017 = 6765 strokes ® minutes 225.5 = 30 5 676 = time surface Bit® minutes 2 58. = 30 46 17 = time bit Surface® strokes 1273 = 1017 . 0 38 . 12 + 117.12 = strokes shoe Bit®

Bit shoe time (at 30 SPM) = 1273

30 = 42.4 minutes ®