drilling engineering ahmed-3

Circulating drilling mud in a well serves several purposes:

(1) The mud removes well cuttings from the bottom of the well to allow drilling to

continue. Without removing the cuttings, drilling would have to stop every few feet

to remove the cuttings that clog up the bottom of the well as has to be done on a cable

tool drilling rig.

(2) As the mud flows across the bit, it cleans the cuttings from the teeth.

(3) The drilling mud also cools and lubricates the bit.

(4) In very soft sediments, the jetting action of the drilling mud squirting out of the bit

helps cut the well.

(5)The drilling mud also controls pressure in the well and prevents blowouts.

At the bottom of the well, there are two fluid pressures. Pressure on fluids in the

pores of the rock (reservoir or fluid pressure) tries to force the fluids to flow through

the rock and into the well. Pressure exerted by the weight of the drilling mud filling

the well tries to force the drilling mud into the surrounding rocks.

If the pressure on the fluid

in the subsurface rock is greater

than the pressure of the drill-ing

mud (underbalance), water, gas,

or oil will flow out of the rock into

the well. This can cause the sides

of the well to cave in, trapping the

equip-ment. In extreme cases, it

can cause a blowout where fluids

flow uncontrolled and often

violently onto the surface.

In order to control

subsurface fluid pressure, the

weight of the drilling mud is

adjusted to exert a greater

pressure on the bottom of the well

than the pressure on the fluid in

the rocks (overbalance).

Some of the drilling mud is then forced into the surrounding rocks during drilling. The rocks act as a filter, and the solid mud particles are plastered to the sides of the well to form a filter or mud cake as the fluids enter the rock. The filter cake is very hard. It stabilizes the sides of the well and prevents subsurface fluids from flowing into the well.

After a well has been drilled, the drilling mud is disposed of and not reused. If it is fresh water drilling mud, it can be spread on the adjacent land to fertilize the crops (land farming). Salt water, oil-based, and emulsion drilling mud, however, usually have to be trucked away to a disposal site. On an offshore drilling rig, a barge can be used to bring the mud ashore to a disposal site.

The kelly cock (Fig. ), a valve, is used either above or below the kelly. It allows drilling mud to be circulated down the drillstring but can be closed by hand with a hexagonal wrench to prevent fluids from flowing up the drillstring. The kelly cock is closed during making a connection to prevent mud from spilling out of the kelly.

Fig. Kelly cock with wrench

In general The concept that a fluid cannot maintain a rigid shape is a basic, but

important characteristic, which means that fluids cannot sustain a shear stress (a tangential force applied to the surface).

Rheology is the study of the deformation of fluids.

The fluid flow behavior is described by an applied shear stress and the resulting shear rate within that fluid.

Two parallel fluid layers are separated by a distance dy. An applied force, F, acting over an area, A, causes the layers to slide past one another. The resistance to this sliding movement, the frictional drag, is called shear stress.

Shear stress, , is defined as the force per unit area required to sustain a constant rate of fluid movement.

Or, an applied force, F, acting along a unit surface area, A, tending to deform the fluid element.

shear rate: Consider that a fluid is placed between two parallel plates that are 1.0 cm apart, the upper plate moving at a velocity of 1.0 cm/sec and the lower plate fixed. The fluid layer at the lower plate is not moving and the layer nearest the top plate is moving at 1.0 cm/sec. Halfway between the plate, a layer is moving at 0.5 cm/sec. The velocity gradient is the rate of change of velocity with distance from the plates. This simple case shows the uniform velocity gradient with shear rate (v1 - v2)/h = shear rate = (cm/sec)/(cm/1) = 1/sec.

Shear rate, , is the velocity gradient, i.e., is the rate of change of velocity at which one layer of fluid passes over an adjacent layer.

Mathematically:

dv

dy

F

A

It has been shown experimentally that the force per unit area (shear stress) applied to a fluid is proportional to the velocity change of two fluid layers (shear rate) a unit distance apart:

where n (power index) depends on the type of fluid.

DRILLING FLUIDS

Time dependent Fluid

Newtonian Fluids

Non Newtonian Fluids

Pseudoplastic

Power law Herschel-Bulkley

Bingham plastic

Dilatant

DRILLING FLUIDS

Newtonian fluid Newtonian fluids are those whose flow behavior call be fully described by a single term called the Newtonian viscosity, .

For these fluids. examples of which include water and light oil.

The shear stress is related to shear rate linearly with the proportionality constant being the Newtonian constant viscosity, .

In engineering units,

is in dyne/cm2 = 4.79 lbs/100 ft2

is in s-1

is in poise= dyne x s/cm2

The field unit of viscosity is the centipoise (1 poise = 100 centipoise). The field unit of shear stress-is lbs/100 ft2.

non-Newtonian fluid

•A fluid whose viscosity is not constant at all shear rates and does not behave like a Newtonian fluid.

•Most successful drilling fluids are non-Newtonian.

•Within that group are several general types and rheological mathematical models to describe them.

•Pseudoplastic is a general type of shear-thinning, non-Newtonian behavior that is desirable for drilling fluids.

Pseudoplastic

Pseudoplastic is a general type of shear-thinning, (i.e., the apparent viscosity decreases as the shear rate

increases) non-Newtonian behavior that is desirable for drilling fluids.

Pseudoplastic rheology: low viscosity at high shear rates and high viscosity at low shear rates, benefits several aspects of drilling-higher drilling rate and improved cuttings lifting.

Bingham plastic and power-law models describe a psuedoplastic behavior using only two measurements (two parameters).

The Herschel-Bulkley model is a three-parameter rheological model

n

K

y p

n

y K

Source / Illustrations: glossary.oilfield.slb.com

Bingham plastic model. Fluids that conform to the Bingham plastic model do not have a constant viscosity and require a certain minimum stress to initiate flow. The yield point, or threshold stress, is the y intercept. Bingham Plastics include thickened hydrocarbon greases, certain asphalts and bitumen, some emulsions

PV should be as low as possible for fast drilling and is best achieved by minimizing colloidal solids.

YP must be high enough to carry cuttings out of the hole, but not so large as to create excessive pump pressure when starting mud flow.

DRILLING FLUIDS

y p

Power-law fluid

A fluid described by the two-parameter rheological model of a pseudo plastic fluid, or a fluid whose viscosity decreases as shear rate increases

In this equation, K is the consistency index and n is the flow behavior index. The flow behavior index is readily determined as the slope of a plot of vs on logarithmic coordinates. The value of n is less than unity for Power Law .

Example: Water-base polymer muds, especially those made with XC polymer

n

K

DRILLING FLUIDS

Herschel-Bulkley fluid (Yield Power Law)

A fluid described by a three-parameter rheological model. A Herschel-Bulkley fluid can be described mathematically as follows:

Many clay-water behave as Herschel-Bulkley fluid

n

y K

DRILLING FLUIDS

Some Important Definitions

Viscosity : Viscosity is the internal resistance of a fluid to flow. It is equal to the ratio of shearing stress to the rate of shearing strain.

Plastic Viscosity : The plastic viscosity is the part of the flow resistance of the fluid caused by mechanical friction within the fluid.

This mechanical friction is due to

1. the interaction of individual solid particles,

2. solid and liquid particles

3. the deformation (shearing) of the liquid particles under shear stress.

The amount of solid particles, their size, distribution and their shape have direct effect on the plastic viscosity.

Yield Stress : The yield stress is the part of the flow resistance of the fluid caused by electrochemical forces within the fluid.

These electrochemical forces are due to

1. the electrical charges on the surface of reactive particles,

2. the electrical charges on the sub-micron particles

3. the presence of the electrolytes in the case of water-base muds.

Effective Circulating Viscosity

Most drilling muds commonly used in the field exhibit viscous properties which are shear rate dependent. During normal drilling, the mud being circulated experiences different velocities in the various sections of the circulating system ranging from practically 0 ft/s in the pits to over 300 ft/s across the jets of the drill bit. These wide ranging velocities give rise to mud shear rates of less than 5 sec-1 in mud pits to over 100,000 sec-1 across the bit jets.

The effective circulating viscosity: therefore, can be defined as the viscosity of the mud at a given shear rate in a particular section of the circulating system under given conditions of pressure and temperature.

where e is in cp, K is in lbf/ft2(sec-1)n, annular velocity, va, is in ft/s and wellbore diameter, Dw, and pipe diameter, Dp, are in inches.

1

47913.6

n

ae

w p

vK

D D

1. Place a freshly stirred sample in a container and immerse the rotor-sleeve exactly to the scribed line.

2. Start the motor by placing the switch in the high-speed position with the gear shift all the way down. Wait for a steady indicator dial value, and record the 600 RPM reading. Change gears only when motor is running.

3. Change switch to the 300-RPM speed. Wait for a steady value and record 300-RPM reading.

4. Plastic viscosity in centipoise = 600 reading minus 300 reading (see Figure).

5. Yield Point in lb/100 ft2 = 300 reading minus plastic viscosity in centipoise.

6. Apparent viscosity in centipoise = 600 reading divided by 2.

Field measurements of viscosity:

Newtonian Fluid :

where is the Newtonian viscosity in cp, N is the viscometer dial reading at rotational speed, N.

Bingham Plastic Fluid :

For N=300 and 600 rpm:

where p is the plastic viscosity in cp, y is the yield point in lbf/100 ft2

300 N

N

y p 300yN

pN N

600 300p 300 6002y

DRILLING FLUIDS

Power Law Fluid and Herschel-Bulkley fluid (Yield Power Law):

In filed units, ( Power Law )

where n is the Power Law index, K is the consistency index in eq.cp and o is the zero gel in lbf/100 ft2. The shear rate, (sec-1), can be expressed in terms of N as

n

K

23.32log N o

N o

n

N o

n

N

K

510 N

n

N

K

DRILLING FLUIDS

1.704 N

Flow Regimes: A range of stream flows having similar bed forms, flow resistance, and means of transporting .

Laminar flow is a streamline flow where all fluid particles move along lines parallel to the axis of the conduit, and adjacent fluid layers slip past each other with no mixing of particles. In steady state conditions, inside circular conduits, laminar flow can be visualized as a series of concentric cylinders slipping pass one another as shown in figure. As the fluid velocity becomes higher, The particle travel in irregular paths with no observable pattern and no definite layer. Turbulent flow is characterized by the irregular movement of particles of the fluid.

http://en.wikipedia.org/wiki/File:Airplane_vortex_edit.jpg

Turbulent Flow is the type of flow regime found inside the drill string during drilling operations. Since high mud velocities are required to achieve turbulent flow, this results in high pressure drops. This type of flow is generally not desired in the annulus due to its tendency to cause excessive hole erosion and high “equivalent circulating densities”. However, turbulent flow can move the mud like a plug, causing the mud to move at approximately the same rate. This provides for better hole cleaning and is sometimes required on high angle holes.

The Reynolds number Re is a dimensionless number that gives a measure of the ratio of inertial forces to viscous forces. It is generally used to determine whether a flow regime is laminar or turbulent. For Newtonian straight pipe flow, it has been established experimentally that the critical Reynolds number, i.e., the number above which flow is no more laminar, has a value approximately equal to 2100. Mathematically, Reynolds number is given by

where is the fluid density, v is the average fluid velocity, D is the pipe diameter and is the fluid viscosity. In field units, NRe can be expressed as

where is in ppg, v is in ft/s, D is in in and is in cp.

Re

v DN

Re

928 v DN

Filtrate/Water Loss and Filter Cake Thickness

These two properties shall be dealt with together, as it is the filtration of mud that causes the build up of filter cake.

Filtration is defined as the loss of the liquid phase of a drilling fluid into permeable formations.

What happen during the filtration?

when the permeability is such that it allows fluid to pass through the pore spaces, Loss of fluid (usually water and soluble chemicals) from the mud to the formation occurs. As fluid is lost, a build up of mud solids occurs on the face of the wellbore. This is the filter cake.

Permeability: The ability, or measurement of a rock's ability, to transmit fluids

DRILLING FLUIDS

DRILLING FLUIDS

Two types of filtration occur; dynamic, while circulating and static, while the mud is at rest. Dynamic filtration reaches a constant rate when the rate of erosion of the filter cake due to circulating matches the rate of deposition of the filter cake. Static filtration will cause the cake to grow thicker with time, which results in a decrease in loss of fluids with time. The problems associated with high fluid loss are: Formation damage Formation evaluation difficulties Wellbore instability Cementing difficulties

Mud measurements are confined to the static filtration. Filtration characteristics of a mud are determined by means of a filter press.

The standard test is conducted at surface temperature at 100 psi and is recorded as the number of ml's of fluid lost in 30 minutes.

An API high pressure/high temperature (Hp/Ht) test is conducted at 300° F and 500 psi.;

DRILLING FLUIDS

The residue deposited on a permeable medium when a slurry, such as a drilling fluid, is forced against the medium under a pressure.

Filtrate is the liquid that passes through the medium, leaving the cake on the medium.

Most of these problems are caused by the filter cake and not the amount of filtration because the aim is to deposit a thin, impermeable filter cake.

The cake quality depend on

low water loss

Solids size and it’s distribution into mud system

Differential Pressure also affects filtration by compressing the filter cake, reducing its permeability

Increased temperature has the effect of reducing the viscosity of the liquid phase and hence increasing filtration.

DRILLING FLUIDS

Filter cake

http://www.glossary.oilfield.slb.com/Display.cfm?Term=permeable

http://www.glossary.oilfield.slb.com/Display.cfm?Term=slurry

http://www.glossary.oilfield.slb.com/Display.cfm?Term=pressure

http://www.glossary.oilfield.slb.com/Display.cfm?Term=filtrate

Fluid loss agents

Clays(such az:Bentonite)

Starch

Water soluble polymers

Sodium Carboxyl-Methyl Cellulose (CMC)

polyanoinic cellulose

The following are the high point summary in discussion of filtration:

Aim is Deposition of thin, compressible and impermeable filter cake

Minimization of fluid loss is not equivalent to cake thickness minimization

Amount and type of solids are very important to the type of filter cake that will be deposited

Bentonite is an excellent filtration additive

Formation characteristics must be considered and controlled

The mud in all cases must be designed to have a minimum amount of total solids

DRILLING FLUIDS

Hydrostatic head is determined by: Example:

After penetrating a hydrocarbon bearing formation, the

fluid depth configuration is developed as shown in

figure. What is the pressure at the bottom of the well ?

Note that the circulation is stopped.

True vertical depth : 3700 ft. Casing inner diameter : 8 in. Bit diameter : 77/8 in. Pipe outer diameter : 41/2 in. Drill collar outer diameter : 63/4 in. Mud density : 10.2 ppg Oil density : 5 ppg Gas density : 1 ppg

DRILLING FLUIDS

fP g h

P = 400 psi

Casing shoe

@ 2600 ft

750 ft

420 ft

400 ft

GOC

MGC

Mud- weighting mathematical relations

Material balance equation:

are the final mixture weight, the original liquid weight. and the added material weight respectively.

Volume balance equation:

are the final mixture volume, the original liquid volume. and the added material volume respectively.

DRILLING FLUIDS

Since By using the equation above, the weight of material that has to be added can be derived as: With a similar approach, the amount of liquid volume, VL, with density, L, required to prepare a mud of total volume, Vf, having density, f, can be determined as:

W

V

final added initial added initial

final final added initial

W W W W W

1

final initial

added initial

final

initial

added

W W

1

1

f

addedL f

L

added

V V

For practical purposes, the following equations are commonly used based on the

similar approach presented above. In order to calculate the amount of weight material

required to increase the mud weight (or density) of 100 bbl of mud and the resulting

increase in the volume of the mud is given, respectively, by

and

where X is the number of 100 lb-sacks of weighting material required, Y is the

increase in the volume due to the weighting material in bbl, sp.gr. is the specific

gravity of the weighting material, i.e., for barite, 4.3, 1 is the original mud weight in

ppg, and 2 is the final mud weight in ppg.

2 1

2

350 . .

8.33 . .

sp grX

sp gr

2 1

2

100

8.33 . .Y

sp gr

Also, the number of sacks of weighting material required to prepare exactly 100 bbl

of mud of weight 2 in ppg, is determined by

The volume of original mud of weight 1 ppg which will be required is given by

2 1

1

350 . .

8.33 . .

sp grX

sp gr

2

1

100 8.33 . .

8.33 . .

sp grY

sp gr

Example #1: Calculate how many sacks of barite are required to increase the density of an 800 barrel mud system from 12.7 lb/gal to 14.5 lb/gal.

Example Problem #2: Calculate how much water and barite are required to make 800 barrels of a 10.5 lb/gal water-based drilling mud.

Example: The geometric configuration of a drilling well is as follows:

True vertical depth : 4000 ft, Casing shoe @ 3000 ft, Casing ID : 8 in, Pipe OD : 41/2 in, Pipe ID : 4 in, Drill collar OD : 61/2 in, Drill collar ID : 4 in, Length of collars : 600 ft, Bit size : 71/2 in, Mud pit : 10 ft x 20 ft x 7 ft, Surface mud lines : 5 bbl, Mud weight : 9.6 ppg, Specific gravity of barite : 4.3.

How much weighting material is required to increase the mud weight to 11 ppg ?

DRILLING FLUIDS SELECTION: how to plan a mud program

- obtain pore pressures and casing program; set mud weights

- look for geological hazards, check hole geometry, ECDs and hydraulics;

set optimum viscosities

- establish maxi fluid loss by intervals; set other critical properties (sand

content, pH, …)

- select mud type by interval; check mud program against other phases of

well plan for possible conflict

- write mud systems composition; determine material requirements

- develop contingency plan for kicks, hole trouble

Note: Drilling fluids specialists, usually associated with Service Companies

can provide valuable expertise/assistance to the well planner.

Time dependent

Fluid

Newtonian Fluids

Non Newtonian

Fluids

Power law Fluids

Yield Power law Fluids

Dilatant Fluids

Bingham Fluids

DRILLING FLUIDS

)511(

300

nK

300

600log32.3

n

23.32log N o

N o

n

N o

n

N

K

DRILLING FLUIDS

1.704 N n

K

510 N

n

N

K

Power Law Fluid and Herschel-Bulkley fluid (Yield Power Law):

There are different models proposed to express this relation between the shear stress and shear rate as shown in the figure.

n

K y p

DRILLING FLUIDS

n

y K

DRILLING FLUIDS

Newtonian Fluids : Newtonian fluids are so-named because their rheological behavior follows the postulation of Newton that fluids might be expected to respond to an applied shearing stress by flowing in a manner such that the velocity gradient is strictly proportional to the applied stress.

where is the constant of proportionality in the relationship characteristic of the Newtonian fluid, which is known as the viscosity of the fluid and this single property defines the rheological behavior of fluids which follow the relation presented above. Viscosity can be determined from the slope of the rheogram ( vs ) of a Newtonian fluid. Water, air and glycerin are the examples of Newtonian fluids.

DRILLING FLUIDS

Pseudoplastic Fluids : Pseudoplastic fluids are those fluids whose behavior is time-independent, for which an infinitesimal shear stress will initiate motion, and for which the rate of increase in shear stress with velocity gradient decreases with increasing velocity gradient. This type of behavior is widely encountered in solutions or suspensions in which large molecules or fine particles form loosely bounded aggregates or alignment groupings that are stable and reproducible at any given shear rate, but which rapidly and reversibly break down or reform with increase or decrease in shear rate. A number of drilling fluids fall in this category. There is no single or simple form of constitutive equation that accurately describes the rheological behavior of pseudoplastics, although several empirical equations are useful over limited ranges of velocity gradient. Even the simplest bust most limited of these involves two constants to characterize the fluid as compared with the single property for Newtonians. Some of these equations incorporate three or more such constants.

DRILLING FLUIDS

Power Law is the most widely used model in engineering calculations. It is presented as

In this equation, K is the consistency index and n is the flow behavior index. The flow behavior index is readily determined as the slope of a plot of vs on logarithmic coordinates. The value of n is less than unity for pseudoplastics.

Other well-known pseudoplastic models available in the literature are:

Prandtl-Eyring

Ellis

Reiner-Philippoff

Sisko

Cross

Meter

n

K

DRILLING FLUIDS

suspensions of clay, fly ash, finely divided minerals, quartz, and paint systems. Some of the drilling muds fall in this category. The constitutive equation for a Bingham fluid is

where y is the yield stress and p is the coefficient of rigidity, so-called plastic viscosity. The yield stress is obtained by extrapolation to zero velocity gradient and the coefficient of rigidity corresponds with the slope of the line of the plot of a vs graph.

Yield-Pseudoplastics : A number of materials exhibit a yield stress, as in the case of Bingham plastics, but the relationship between the shearing stress in excess of that initiating flow and the resulting velocity gradient is not linear. Commonly, the relationship exhibits convexity to the shear stress axis, and fluids showing this behavior may be called yield-pseudoplastics.

y p

DRILLING FLUIDS

Many clay-water and similar suspensions like drilling muds behave as yield-pseudoplastics. As in the case of pseudoplastics, there is no single theoretically based constitutive equation for yield-pseudoplastics, but any of the empirical equations for pseudoplastics may be modified for yield-pseudoplastics. Illustrating this for the Power Law equation;

This three-parameter equation is also known as Hershel-Bulkley equation, describes the behavior of yield-pseudoplastics reasonably well except at high shear rates.

n

y K

DRILLING FLUIDS

• The optimum utilization of mud pump horsepower is of significant importance in rotary drilling operations. Drill bit hydraulics is generally associated with the use of jet bits

The purpose of the jet nozzles is to improve the cleaning action of the drilling fluid at the bottom of the hole.

Drilled cuttings of rocks that are not efficiently removed from underneath the bit, will be re-grinded and, hence, will cause wasteful bit wear and drilling time, and therefore, higher drilling cost.

Current field practice in bit hydraulic design involves the selection of an optimum flow rate and corresponding optimum bit nozzle sizes that maximize one of the following parameters:

a) Bit hydraulics b) Jet impact force c)Jet nozzle velocity

We are now facing with wells which are more difficult to drill. Also, we are required to make the wells more cost effective.

The result of these requirements is to put more emphasis on the well design process. Easy for implementation. The design should also provide flexibility Cost effectiveness.

Significant importance in rotary drilling operations:

Trajectory design Select optimal mud weight Determinate the casing setting depths Hydraulic program design Bit hydraulic design Design the casing Design the cement program Design the drill string

Current field practice in bit hydraulic design involves the selection of an optimum flow rate and corresponding optimum bit nozzle sizes that maximize one of the following parameters:

a) Bit hydraulics

b) Jet impact force

c)Jet nozzle velocity

The purpose of the jet nozzles is to improve the cleaning action of the drilling fluid at the bottom of the hole.

Insufficient hole cleaning

re-grinded

wasteful bit wear

high drilling time

higher drilling cost

Pump Pressure Requirement in Rotary Drilling

The available pump surface

pressure

p s dp dc adc adp bP P P P P P P

bfp PPP

Hydraulic Power Requirements Hydraulic power is defined as the product of a flow rate, Q, and a

corresponding pressure, P, such that

where P is in psi, Q is in gpm and HP is in horsepower. Using Pp definition,

1714p

PQHP

bfp PPP bfp QPQPQP

p f bHP HP HP

Example

A 15000 ft well is to be drilled. It has been determined that, for a flow rate of 350 gpm, the pressure losses in the well will be as follows; inside the drillpipes, 0.1 psi/ft, inside the drill collars, 0.3 psi/ft, between the wellbore and drill collars, 0.07 psi/ft, between the wellbore and drillpipes, 0.01 psi/ft, pressure loss at the bit is 2200 psi, and at the surface connections, equivalent to 600 ft drillpipe. Drill collar length is 1000 ft. Determine the minimum required pump pressure and hydraulic horsepower if the pump efficiency is 80%.

psi

Since the pump efficiency is 80%, HPp will be

HP

adcadpdcdpsf PPPPPP

0.1 14000 0.3 1000 0.07 1000 0.01 14000 0.1 600 2200 4170pP

4170 350 11064.4

1470 0.8PHP

Simple Equation To Perform Pressure Drop Calculations

For a Newtonian fluid in Laminar region,

For a Newtonian fluid in Turbulent region,

2

32P v

L D

QPf

22 ff vP

L D

2fQPf

In a well, we have a mixture of flow regimes.

c is a constant and a function of mud flow properties, hole geometry, pipe geometry, depth, etc,

m is a flow exponent that varies between 1 and 2, and can be measured in the field.

If no information is available, a value of 1.86 is assumed.

Therefore,

b

m

p PcQP

m

f cQP

Determination of m m

f cQP

Tak

ing

lo

garith

m

QmccQP m

f lnlnlnln

If arbitrary two flow-rate vs frictional pressure loss data is known, m can be determined by

2

1

2

1

ln

ln

Q

Q

P

P

mf

f

Determination of m

In the field, during a drilling operation, data is always recorded regarding with the flow rate, pump pressure, nozzle information, etc.

22

2

2

5

2

22

2

1

5

1

103.8

103.8

dt

pf

dt

pf

CA

QPP

CA

QPP

Therefore, for arbitrary points, Pf can be expressed as

bfp PPP

Example

A well is being drilled at 12000 ft while using a 15.5 ppg mud

and a drill bit having 14-14-14 nozzles. At flow rates of 400

gpm and 300 gpm, the circulating stand pipe pressure were

recorded to be 4622 psi and 2880 psi, respectively.

Determine the flow exponent, m&c

Example From an exploration well, the following system losses

or parasitic losses have been measured at different well depths (Table). show the continuous pressure drop-flow rate behavior

depth(ft) Pressure Loss (psi) Q(gpm)

4000 1400 600 4000 2225 800 7200 1500 500 7200 3200 800

10500 1800 500 10500 3845 800

Typical optimization criteria are The maximization of the hydraulic energy delivered

through the bit nozzles,

The maximization of the jet impact force. (Drilling Engineering, J.J.Azar & G. Robello Samuel Publisher:

PennWell , 2007 page 141-170)

Although these criteria seem reasonable at a first glance,

a closer look at the total hydraulic system reveals that they may have limitations.

hydraulic optimization problem in a nontraditional way (Modern Well Design Bernt S. Aadnoy (1999), 2nd Edition, A.A Balkema page 25-31)

Operating Windows for Pumps

Pumps used at the field have basically two operating windows as shown in the figure.

Drill Bit Hydraulics

Flow Rate

HP

, P HP

Pressure

Region I

Region II

1714p

PQHP

bfp PPP

Operating Windows for Pumps

•For a constant liner size, in the first operating region, pump pressure is constant and horsepower in use increases. This increment continues until maximum horsepower of the pump is reached. At this point, second operating region is active, where horsepower is constant, i.e., maximum horsepower of the pump, and pump pressure decreases.

Maximizing Hpb involves minimizing Hpf, or in other words, the lowest flow rate and the highest pump pressure will result in the highest Hpb.

However, the “lowest flow rate” will usually result in inadequate bottom hole cleaning. To compensate for this, bottom hole pressure can be increased by using smaller jet nozzles.


Maximum Bit Hydraulic Horsepower Criterion

Hydraulic horsepower is based on the theory that cuttings are best removed from beneath the bit by delivering the most power to the bottom of the hole.

The amount of pressure lost at the bit, or bit pressure drop, is essential in determining the hydraulic horsepower.

Maximum bit hydraulic horsepower criterion assumes that optimum bottom hole cleaning achieved if hydraulic horsepower across the bit is maximized with respect to the flow rate, Q.

Thus, expressing this hydraulic horsepower explicitly in terms of Q and then using differential calculus will lead to the desired flow rate that is believed to ensure optimum bottom hole cleaning. Mathematically, the maximum bit hydraulic horsepower is achieved when


0bHP

Q

Limitations There are two limitation to the hydraulic bit horsepower that need to be considered;

i) limitations due to the maximum allowed surface operating pressure, Pp, and

ii) ii) limitations due to the maximum available pump hydraulic horsepower, HPp.

Case-i From hydraulic horsepower definition, bit hydraulic horsepower can be

written as


bb QPHP

Thus, for maximum bit hydraulic horsepower,

Since


0

Q

QP

Q

HP bb

b

m

p PcQP m

pb cQPP

0

Q

cQPQ

Q

HP m

bb

Differentiation gives

Since


1 0m

pP c m Q

m

f cQP fp PmP 1

pf Pm

P1

1

Hence, on the basis of the maximum

bit hydraulic horsepower criteria: the optimum bit hydraulics will be achieved if friction pressure loss in the circulating system is maintained at an optimum value of

max1

1Pfopt P

mP

Optimum Values for Flow Rate and Nozzle Area Optimum flow rate can be determined using the frictional pressure loss information obtained using either maximum hydraulic horsepower, or maximum jet impact force. It has been shown that


m

f

f

optP

PQQ

opt

1

1

1

.

2 optQ Q

2

1

2

1

ln

ln

Q

Q

P

P

mf

f

Optimum Values for Flow Rate and Nozzle Area Therefore, optimum bit pressure loss can be determined as

Once optimum flow rate and optimum bit pressure is determined, nozzle diameters can be optimized by using these variables. In field units, total nozzle area can be determined as


optbd

opt

totalnPC

QA

2

25103.8

.

foptpoptb PPP

Optimum Values for Flow Rate and Nozzle Area

where is in ppg, Q is in gpm, P is in psi and A is in in2. Cd is called

discharge factor, and is usually assumed as 0.95. If nozzle sizes are assumed

to be constant, nozzle size (in) can be determined as

n= number of nozzles


.

n

Ad

nopt

optn

4

Example Given the following data:

Tricone bits used.

Mud density = 15.5 ppg

Flow exponent = 1.657

Maximum and minimum flow rates = 404 gpm and 268 gpm

Maximum allowable pump pressure = 5440 psi

Maximum hydraulic horsepower = 1600 HP

Frictional pressure loss at 300 gpm = 2334.4 psi

Determine the optimum flow parameters for operating region-1 using Maximum hydraulic horsepower criteria

Example You are given the following well data:

• Drill pipe:

OD = 4.5 in.

ID =3.64in.

Weight = 20 lbs/ft

• Drill collar:

OD=7in.

ID =2 in.

Weight = 120.lbs/ft

Length= 1.000 ft

• Mud:

Bingham

θ600= 21: θ300 = 29; 10s/10min; gel = 15/25 lbs/l00 ft2

Weight= 15.5 lbs/gal

• Pump: National duplex (double acting) Maximum allowed operating pressure=, 5,440 psi Hy1iraulic horsepower= 1,600 hp Volumetric efficiency= 80% Minimum required annular fluid velocity: 85 ft/min average • Well geometry: Shape as shown in figure 2 Last intermediate casing is 9 7/8 in. at 12,000 ft TVD Drill bit: 12 7/8 in. tri-cone with 3-14 nozzles to 12,000 ft; next bit to be used is 87/8 in. Tri-cone (assume that the hole is washed out to 97/8 in.)

Field data: At 12,000 ft. while using the 8 7/8 in. tri-cone 3-14 nozzles.

Q1= 300 gpm Ppl = 2,966 psi

Q2 = 400 gpm P p2 = 4,883 psi

Using the bit hydraulic horsepower criterion, determine the optimum nozzle sizes to be used to drill the next depth interval. Assume current depth is 12,000 ft.

7000 ft

7000 ft

30deg

12000 ft

3000 ft

Case-ii Bit hydraulic horsepower can be expressed as the difference between the pump hydraulic horsepower and the hydraulic horsepower lost due to frictional losses:

fpb HPHPHP

1 m

ff cQQPHP

1 m

pb cQHPHP


For maximizing bit hydraulic horsepower

Thus, only solution for this equality is that, Q=0, which is a trivial solution. Mathematically, this means that bit hydraulic horsepower cannot be maximized for operating region-2.


0bHP

Q

1

0

m

pHP cQ

Q

Maximum Jet Impact Force Criterion •Hydraulic (Jet) Impact Force is based on the theory that cuttings are best removed from beneath the bit when the force of the fluid leaving the jet nozzles and striking the bottom of the hole is the greatest.

•Jet impact force can be derived using Newton’s second law and Bernoulli’s energy conservation equation.


02

2

n

b

vP

2

2

vP g h W F

Maximum Jet Impact Force Criterion • vn is the average velocity at the nozzle downstream. Thus, solving for nozzle velocity gives

•As mentioned in the previous chapters, Newton’s second law states that

•Thus, inserting the nozzle velocity into this equation, jet impact force, Fj, can be derived as


b

n

Pv

2

.j n

mF m v v Qv

t t

bj PQF

Maximum Jet Impact Force Criterion • where ζis a constant and equal to 0.01823Cd.

•The maximum jet impact force assumes that optimum bottom hole cleaning is achieved by maximizing the jet impact force with respect to the flow rate.

•Thus, expressing this impact force, Fj, in terms of flow rate, Q, and using differential calculus will yield the desired value of Q that makes Fj maximum. Mathematically, maximum jet impact force is achieved when

•Since


m

pj cQPQF

0jF

Q

m

pb cQPP

Maximum Jet Impact Force Criterion •Prior to carrying out the differentiation, there are two limitation to the hydraulic impact force that need to be considered; i) limitations due to the maximum allowed surface operating pressure, Pp, and ii) limitations due to the maximum available pump hydraulic horsepower, HPp.

•Case-i


max

m

j pF Q P cQ

max

2 2m

j pF P Q cQ

Maximum Jet Impact Force Criterion

Differentiating the above equation with respect to Q yields


max2

2pf P

mP

max

12 2 0m

pP Q c m Q

max2pb P

m

mP

.

Maximum Jet Impact Force Criterion

Case-ii

Since


pf Pm

P2

1

pb P

m

mP

2

1

.

p pHP P Q

2m

j pF HP Q cQ

12 0m

pP Q c m Q

Differentiating

Typical optimization criteria Maximum Bit Hydraulic Horsepower Criterion (region I)

Maximum Bit Hydraulic Horsepower Criterion (region II) (cannot be

maximized for operating region-2.)

Maximum Jet Impact Force Criterion (region I)

Maximum Jet Impact Force Criterion (region II)

max1

1Pfopt P

mP

max2

2pfopt P

mP

pfopt Pm

P2

1

Example Given the following data:

Tricone bits used.

Mud density = 15.5 ppg

Flow exponent = 1.657

Maximum and minimum flow rates = 404 gpm and 268 gpm

Maximum allowable pump pressure = 5440 psi

Maximum hydraulic horsepower = 1600 HP



Determine the optimum flow parameters for operating window-1 using

Maximum jet impact force criteria


drilling engineering ahmed-3

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