Drilling Engineering Ahmed-3
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Transcript of 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.
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
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.
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Drill Bit Hydraulics
bb QPHP
Thus, for maximum bit hydraulic horsepower,
Since
Drill Bit Hydraulics
0
Q
QP
Q
HP bb
b
m
p PcQP m
pb cQPP
0
Q
cQPQ
Q
HP m
bb
Differentiation gives
Since
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Drill Bit Hydraulics
.
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
Drill Bit Hydraulics
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.
Drill Bit Hydraulics
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.
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Drill Bit Hydraulics
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
Frictional pressure loss at 300 gpm = 2334.4 psi
Frictional pressure loss at 400 gpm = 3760.2 psi
Determine the optimum flow parameters for operating window-1 using
Maximum jet impact force criteria
Drill Bit Hydraulics