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HYDRAULICS / WELL CONTROL / PRESSURE ANALYSIS
PART 1 DRILLING HYDRAULICS
1.1 Summary of the Purpose of the Drilling Fluid
1.2 Types of Drilling Fluid
1.3 Types of Fluid - Newtonian or Non-Newtonian
a. Definitions
b. Newtonian Fluids
c. Bingham Fluids
d. Power Law Fluids
e. The Modified Power Law
f. Model Affects on Viscous Flow
1.4 Mud Rheology
1.5 Laminar, Turbulent and Transitional Flow Patterns
a. Laminar Flow
b. Turbulent Flow
c. Determination of Flow Type
d. Derivation of Effective Viscosity
e. Determination of Reynold’s Number
f. Determination of Annular Velocity
g. Use of Reynold’s Number to determine Flow Type
h. Determination of Critical Velocity
1.6 Determination of System Pressure Losses
a. Fanning Friction Factor
b. Drillstring Pressure Losses
c. Annular Pressure Losses
d. Bit Pressure Loss
e. Surface Pressure Losses
http://intro.pdf/
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1.7 Other Hydraulic Calculations
a. Cuttings Slip Velocity
b. Slip Velocity in Turbulent Flow
c. Nozzle Velocity
1.8 Hydraulics Optimization
a. Bit Hydraulic Horsepower
b. Hydraulic Impact Force
c. Optimization
1.9 Equivalent Circulating Density
1.10 Surge and Swab Pressures
Appendix
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1.1 SUMMARY OF THE PURPOSE OF DRILLING MUD
The importance of the drilling mud in the drilling of a well cannot be over emphasised. It
has a critical bearing on all aspects of the operation. Not only does it act as a transporting
medium for cuttings and gas, enabling us to see at surface what is happening downhole, but the properties of the mud will determine how affective the drilling is, how well the
hole and formations are protected, and how well subsurface pressure are controlled.
The principle roles of the mud are:-
• Cuttings removal• Control subsurface pressures• Lubricate and cool the drill bit and the drillstring• Bottom hole cleaning
• Aid in formation evaluation• Protect formation productivity• Aid formation stability
Cuttings removal
This is a very important role of the mud. Not only do the cuttings need to be removed
from the annulus to allow for free movement and rotation of the drillstring, but the
cuttings need to reach the surface in such a condition that they can be used by a geologist
to accurately interprete the downhole geology.
This principle is not only determined by the physical properties of the mud but by the type
of flow pattern present in the annulus. The cuttings need to be removed affectively, but
damage and erosion to the cuttings has to be avoided.
Cuttings density is obviously greater than the mud density, therefore it is normal for a
degree of cuttings slip. Mud properties (viscosity, gel strength) have to be such so as to
minimise this.
Cuttings slip will be affected by the annular velocities:-
If annular velocities are restricted for any reason (eg pump volume, large hole section,
downhole conditions), mud properties would have to be changed to compensate for an
increased degree of slip.
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Control subsurface pressures
Minimum mudweight is optimum for faster drilling rates and to minimise the risk of
damaging formations and losing circulation.
However, in conventional drilling, the mud also has to be of sufficient density to protectthe well against subsurface formation pressures.
The pressure produced at the bottom of the hole, due to the weight of the static vertical
column of mud, is known as the Hydrostatic Pressure.
If the hydrostatic pressure is equal to the formation pressure, the well is said to be at
balance.
If the hydrostatic pressure is greater than the formation pressure, the well is said to be
overbalanced and protected against influxes into the wellbore.
If the hydrostatic pressure is less than the formation pressure, the well is said to be
underbalanced and therefore subject to influxes of formation fluid.
PHYD = ρ x TVD x 0.052 where ρ = mud density (ppg)PHYD = psi
TVD = feet
PHYD = ρ x TVD x 0.433 where ρ = SGPHYD = psi
TVD = feet
PHYD = ρ x TVD x 0.00981 where ρ = kg/m3PHYD = Kpa
TVD = m
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Lubricate and cooling
The drilling action and rotation of the drillstring will produce a lot of heat, at the bit and
throughout the drillstring, due to friction. This heat will be absorbed by the mud and
released, to a degree, at surface.
The mud has to cool the bit and lubricate the teeth to allow for affective drilling and to
minimise damage and wear.
The mud has to affectively remove cuttings from around the bit as rock is newly
penetrated. This is to stop the cuttings building up around the bit and teeth (bit balling)
which would prevent the bit from drilling.
The mud lubricates the drillstring by reducing friction between the string and the borehole
wall - this is achieved by additives such as bentonite, graphite or oil.
Aid in formation evaluation
• To obtain the best possible cuttings for geological analysis (viscosity). The type offlow will determine the degree of erosion and alteration, thus smooth laminar flow is
preferred to chaotic turbulent flow.
• To minimise fluid invasion ( filtrate) - both water and oil invasion would affect theresistivity of the mud making formation evaluations more difficult. Thus, a filter cake
is allowed to build up on the wall of the borehole, restricting fluid movement in both
directions.
NB Filter Cake restricts fluid invasion but may reduce the quality of sidewall cores
• To improve logging characteristics (especially for resistivities).
• To improve formation testing
Formation Stability
• to prevent erosion or collapse of the wellbore;• to prevent swelling and sloughing shales (oil based mud preferred, water based muds
would have to be treated with Ca/K/Asphalt compounds);
• to prevent the ‘dissolving’ of salt sections (use salt saturated or oil based mud to prevent taking the salt into solution.
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1.2 TYPES OF DRILLING MUD (a brief summary)
Water Based including gel and polymer muds
Oil Based
Emulsion
Water Based Muds
1. Clear Water - from freshwater to saturated brine
2. Native Water - water allowed to react with formations containing shales/clays; the
mud will therefore build up a solids content and density naturally.
3. Calcium - reduces swelling and hydration of clays
good for gypsum/anhydrite lithologies because there will be no Ca
contamination
4. Lignosulphate - high density muds (>14ppg)
tolerance to high temperatures
high tolerance for contamination by drilled solids
disadvantages - shales/clays will adsorb water from the mud
permeability will be damaged due to clays dispersing
Not often used
5. KCL/Polymer - inhibits shale sloughing
little permeability damage
provides good bit hydraulics
disadvantages - need good solids control equipment at surface because it has a low
tolerance to solids
unstable at high temperatures > 120oC
6. Salt Saturated - water phase saturated with NaCl
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Oil Based Muds
Emulsion of water in oil (invert emulsion)
Crude oil or diesel is normally the continuous phase, water the dispersed phase (droplets)
Advantages reduces/inhibits any problems caused by shales
reduces torque and drag
stable at high temperatures
preserves natural permeability, not damaging hydrocarbon zones
Disadvantages environmental concerns
flammability
solids removal due to high PV (need good equipment as with polymer
muds)
problems for interpretation of log information
cost
Emulsion Muds
Water is the continuous phase, oil the dispersed phase (normally 5 - 10%)
Oil added to increase ROP, reduce filter loss, improve lubrication, reduce drag and torque
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1.3 TYPES OF FLUID - NEWTONIAN or NON-NEWTONIAN ?
The majority of hydraulic parameters are, first of all, dependent on what type of fluid the
drilling mud is and therefore which model is used for the calculations.
The categories are determined by the fluid behaviour when it is subjected to an applied
force (shear stress). Precisely, in terms of fluid behaviour, we are concerned with:-
• At what point of applied shear stress is movement initiated in the fluid.
• Once movement has been initiated, what is the nature of the fluid movement (ShearRate).
1.3a Definitions:
Shear Rate.....in a simple flow, is the change in fluid velocity divided by the width of the
channel through which the fluid is moving.
v2
Shear Rate = v2 - v1
(γ γ ) h
h
= sec-1
v1
At wellsite, the Shear Rate is determined by the rotational speed of the Fann Viscometer
in which the tests are conducted.
Thus, Shear Stress is recorded at rotational speeds of 600 (shear rate = 1022 sec
-1
), 300(shear rate = 511 sec-1
), 200, 100, 6 and 3 rpm.
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Shear Stress....is the force per unit area required to move a fluid at a given shear rate.
Area F
Shear Stress (ττ) = F/A = lb. ft or lb. ft or dynes
in2 100ft
2 cm
2
The shear stresses recorded for each of the shear rates at the different rotational speeds of
the viscometer can then be plotted to produce an overall behaviour profile across the
‘rheological spectrum’:-
Shear
Stress
lb/100ft2
100 200 300 400 500 600
Shear Rate, rpm
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Fluid Viscosity.....is the fluid’s shear stress divided by the corresponding shear rate.
Fluid Viscosity (µµ) = Shear Stress
Shear Rate
= dynes/cm2 = poise
sec-1
1 poise = 100 centipoise (cP)
1 lb. ft. sec = 47886 cP
ft2
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1.3b Newtonian Fluids
The fluid will begin to move the instant that shear stress is applied. Thereafter, the
degree of movement is proportional to the stress applied...
ie A linear relationship exists between Shear Stress (τ) and Shear Rate (γ ).
τ
gradient = µ
γ
For a Newtonian Fluid...... τ = µ γ where µ = viscosity
Most drilling fluids and cement slurries, however, exhibit non-Newtonian behaviour
where the laminar flow relationship between shear stress and shear rate is non-linear.
These fluids also require a certain amount of shear stress to initiate flow and thereafter,
require additional stress to be applied as the shear rate increases.
The level of shear stress required to initiate fluid flow is known as the fluid’s Yield Point .
Two main models have been used as a standard in the oil industry:-
1. The Bingham Plastic Model
2. The Power Law Model
In recent years, it is generally accepted that both models have merit but that the Power
Law Model is more applicable to the majority of fluids.
A third, widely used, model has been developed, being a combination of both previous
models. This model is known as the Modified Power Law (also known as the Yield
Power Law or Herschel-Bulkley Model).
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1.3c Bingham Fluids
This model predicts that fluid movement will take place only after a minimum value of
Shear Stress has been applied. This minimum value is the Yield Point of the fluid.
Once movement has been initiated, the relationship between τ and γ is linear (ie Newtonian), with the constant being called the Plastic Viscosity (PV).
PV is dependant on both temperature and pressure.
Dial
Reading
θ600 gradient = PV
θ300
YP
γ (rpm)300 600
For Bingham Fluids.... τ = YP + γ .PV
PV = θθ600 - θθ300
YP = θθ300 - PV = ττ0
The Bingham Plastic Model represents, fairly well, the behaviour exhibited by fluids such
as bentonite slurries, class G cements and low gravity oils. A typical Bingham fluid will
have high viscosity but no gel strength.
For more complex fluids, however, the Bingham model is subject to error. Whereas theBingham model simulates fluid behaviour in the high shear rate range (300 to 600 rpm), it
is generally inaccurate in the low shear range. Shear stresses measured at high shear rates
are usually poor indicators of fluid behaviour at low shear rates, the area of interest for
simulating annular flow behaviour. Subject to this error, the calculated Yield Point will
tend to result in calculated pressure losses and equivalent circulating densities that are
larger than those actually observed.
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1.3d Power Law Fluids
The Power Law Model assumes that fluid movement will be initiated immediately that
any shield stress is applied. The model then predicts that fluids will exhibit a non linear
relationship between τ and γ and introduces two ‘index’ values in order to determine therelationship.
Dial
Reading
θ600 θ300
300 600 γ (rpm)
When the log of stress and strain is plotted:-
log τ
100 gradient = n
10
K
1 10 100 1000 log γ
For Power Law Fluids.... ττ = K (γ γ )n
where K = consistency index
n = flow behaviour index
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Determination of ‘n’ and ‘K’:-
n = 3.32 log θθ600
θθ300
K = 1.067 θθ300 (units lb / 100ft2 )
(511)n
K = 5.11 θθ300 (units dynes / cm2 )
(511)n
The Power Law rheological model better fits the behaviour of most fluids, especially polymer based fluids, than the Bingham Plastic Model.
Fluids that follow this model have no shear stress when the shear rate is zero. The draw
back here, is that most fluids have a yield stress but this cannot be accounted for in this
model.
Similar to the Bingham Plastic model, but to a lesser degree, the Power Law model
accurately predicts fluid behaviour at high shear rates but shows a degree of error at the
lower shear rates.
The result of this is that annular pressure losses and ECD’s are ‘under-predicted’ by this
model’s calculations.
In many cases, however, the Power Law Model does closely approximate fluid properties
even when calculated from the high shear rate values.
Different values of ‘n’ are possible, depending on which shear stress/rate pairings are
used in the calculation. Thus, this model can be applied by using data from a range of
annular shear rates, providing a better accuracy in predicting drilling fluid performance.
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Calculation of ‘n and K’ at other shear rates:-
With θθ200 and θθ100 With θθ6 and θθ3
n = 3.32 log θ200/θ100 n = 3.32 log θ6/θ3
K = θ100 / (170.3)n K = θ3 / (5.11)
n
In the extreme case, when n=1, the fluid will become a Newtonian fluid
ie τ = K γ where K will be equal to viscosity µ.
When to use the low shear rate pairing (6 and 3 rpm) ? :-
• to more accurately describe the suspension and hole cleaning potential of a fluid• in large diameter holes• in horizontal drilling applications
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1.3e The Modified Power Law
This model combines the theoretical and practical aspects of the Bingham Plastic and the
Power Law models.
In this model, the ‘n and K’ values are similar to those derived by the Power Law model.The model assumes that fluids will require a certain amount of applied stress before
movement will take place and, for these fluids having a yield stress, the calculated values
of ‘n and K’ will be different.
Shear
Stress
τ0 (yield point or yield stress)
Shear Rate
For Modified Power Law Fluids.... ττ = ττ0 + K (γ γ )n
where K = consistency index
n = flow behaviour index
The value τ0 is the fluid’s yield point at zero shear rate and, in theory, is identicle to theBingham Plastic yield point, though it’s calculated value is different.
When n = 1, the model becomes the Bingham Plastic Model
τ0 = 0, the model becomes the Power Law model
The model works well for both water based and oil based drilling muds because both
exhibit shear thinning behaviour and have a shear stress at zero shear rate.
The problem with the model is that the determination of n, K and τ0 is very complex.
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Rheogram Summary of the Drilling Fluid Models
Shear
Stress Bingham Plastic
Modified P.Law
Power Law
Newtonian
Shear Rate
NOTE, in order for the QLOG system to accurately calculate realtime hydraulics,
the Shear Rate values need to be updated regularly in the Equipment Table.
The data can be entered in any of the 3 standard shear rate pairings
ie θ600 and θ300θ200 and θ100θ6 and θ3
The industry normal is to use the 600/300 pairing but as was seen in this manual, there
are applications when the 6/3 pairing can be more meaningful.
Ideally, if there is a reason for using the 6/3 pairing, it should be discussed and confirmedwith the drilling and mud engineers.
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1.3f Model Affects on Viscous Flow
Newtonian Fluids
Laminar flow through pipes or annulus is characterised by a parabolic velocity profile,with the velocity approaching zero at the walls and being at a maximum in the centre of
the flow.
Non Newtonian Fluids
For these fluids, the flow will not necessarily be parabolic. As the fluid becomes
‘increasingly’ non-Newtonian, the velocity profile will become increasingly flatter
towards the centre. This is known as plugged flow.
Using the Power Law as a basis, when ‘n’ is equal to one, the fluid is Newtonian and thevelocity profile will indeed be parabolic.
As the value of ‘n’ decreases, ie the fluid becomes increasingly non-Newtonian and the
velocity profile will become increasingly flatter. In this flat part of the profile, the shear
rate will be close to zero (ie very little movement between adjacent laminae). Fluids that
exhibit a high viscosity in this near zero shear rate condition offer significant
improvements in hole cleaning efficiency.
Areas of
Low Shear Zone High Shear
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Affect of the value of ‘n’ on velocity profile
n = 1 n = 0.6 n = 0.2
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1.4 MUD RHEOLOGY - principle parameters
Viscosity
Controls the magnitude of shear stress which develops as one layer of fluid slides overanother. It is a measure of the friction between fluid layers, providing a scale for
describing fluid thickness. It will decrease with temperature.
In simple terms, it describes the thickness of the mud when it is in motion.
Normal unit of measurement is the centipoise (CP), where 47886 CP = 1 lb.f.s
ft2
Plastic Viscosity
For a Bingham Fluid, PV is the amount of shear stress, in excess of the yield stress, that
will induce a unit rate of shear. More simply, it is the relationship between shear stress
and shear rate during fluid movement; it is the slope of the straight line that passes
through θ600 and θ300.
Funnel Viscosity
This is a direct measurement from the Funnel (as opposed to Fann) viscometer and ismeasured in secs/qt. Generally used at wellsite for immediate measurements, this is
simply the length of time it takes for one quart of fluid to pass through the funnel.
It is not regarded as being applicable to the analysis of circulating performance.
Apparent Viscosity - simply θ600/2
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1.5 LAMINAR, TURBULENT and TRANSITIONAL FLOW PATTERNS
The type of flow pattern will be governed by the fluid velocity, the annular diameters and
the characteristics of the mud.
In general, the lower the fluid velocity and the greater the annular diameter, the more
likely the flow is to be laminar.
A turbulent flow pattern is more likely when the fluid velocity is high and when there is a
small annular clearance ie around the drill collar section.
1.5a Laminar Flow
A smooth flow pattern will be exhibited with fluid layers travelling in straight lines
parallel to the axis. The velocity will increase towards the centre of the stream. Laminar
flow will develop from low fluid velocities.
There is only one component of fluid velocity - longitudinal.
Shear resistance is caused by sliding action only.
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1.5b Turbulent Flow
The flow pattern is random in time and space, with chaotic and disordered motion of the
fluid particles. This results in two velocity components - longitudinal and transverse.
Shear resistances present are far greater than in Laminar flow.
Turbulent flow will develop at higher fluid velocities with the final velocity profile
tending to be reasonably uniform despite the chaotic components. For this reason,
turbulent flow is actually more affective in cuttings removal, but the disadvantages
outweigh this advantage:-
Disadvantages erosion of cuttings, thereby destroying interpretative propertiesthe possibility of hole erosion
increased pressure losses due to higher frictional forces
removal of mud filter cake
One advantage of turbulent flow is when cementing - helping to dislodge mud cake from
the walls allowing the cement to contact fresh surfaces.
Transitional Flow
In reality, there is not an instantaneous change from laminar to turbulent flow as fluid
velocity increases. There will obviously be a transitional period where the flow changes
from one to the other. This transitional flow will exhibit elements of both laminar and
turbulent flow.
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1.5d Derivation of Effective Viscosity
Bingham Fluid
µe = PV + 300(Dh − D p) YP (imperial) v = av annular velv
= PV + 2874 (Dh − D p) YP (metric)48000 v
Imperial units: µe = cP Metric: µe = cPv = ft/min v = m/sec
D = inches D = mm
YP = lb/100ft2 YP = dynes/cm2 PV = cP PV = cP
Power Law Fluid
µe = [ 2.4 v x 2n + 1 ] n
x 200K (Dh−D p) (imperial)[ Dh−D p 3n ] v
= 1916K (Dh−D p) x [ 4000 v (2n + 1) ]n
4800v [ Dh−D p ( n ) ] (metric)
= [ 200 v x 2n + 1 ] n
x 0.5K (Dh−D p)[ Dh−D p 3n ] v (SI)
Imperial: µe = cP Metric: µe = cP SI: µe = mPa.sv = ft/min v = m/sec v = m/min
D = inches D = mm D = mm
K = lb/100ft2
K = dynes/cm2
K = Poise
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1.5e Determination of the Reynolds Number
Imperial Re = 15.47 Dvρ D = diameter = inchesµe v = average velocity = ft/min
ρ = mud density = ppgµe = effective visc = cP
Metric Re = 1000 DVρ D = mmµe v = m/sec
ρ = kg/litreµe = cP
SI Re = DVρ D = mm60µe v = m/min
ρ = kg/m3
µe = mPa.s
For Reynolds number inside the pipe, D = pipe internal diameter
For Reynolds number in the annulus, D = hole diameter - pipe outside diameter
Note that for fluid velocity, an average velocity is used in the determination of the
Reynolds Number and Effective Viscosity. In reality, as we have seen, the velocity is
least at the walls of the conduit, increasing to a maximum at the centre of the channel.
The average fluid velocity (annular velocity or pipe velocity) is determined using the
following formulae:
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1.5f Determination of Annular Velocity
v (ft/min) = 24.5 Q Q = flowrate (gpm)
Dh2 − D p
2 Dh = hole diameter (in)
D p = pipe outer diam (in)
v (ft/min) = 1030 Q Q = bbls/min
Dh2 − D p
2 Diameters (in)
v (m/min) = 1273000 Q Q = m3/min
Dh2 − D p
2 Diameters (mm)
These formulae can obviously be used to calculate the velocity of the mud within the
drillstring.
In this case, Dh2 would be replaced by Di
2 the inside diameter of the pipe.
‘D p’ would, in this case, be equal to zero.
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1.5g Use of the Reynolds Number in determining Flow Type
The value of the Reynolds number defines the transition between laminar and turbulent
flow.
Bingham
The Critical Reynolds Number is 2100.
If Re < Rec, then the flow is said to be laminar
If Re > Rec, then the flow is said to be turbulent
Power Law
The Critical Reynolds Number is given by ‘3470 - 1370n’
If Re < 3470 - 1370n, the flow is laminar
If Re > 4270 - 1370n, the flow is turbulent
If 3470 - 1370n < Re < 4270 - 1370n, the flow is transitional
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Power Law
1 n
String Vc = 0.6 [ (3470 − 1370n)K ]2−n [ 3n + 1 ]
2−n
[ 1.27 ρ ] [ 1.25 Di n ]
1 n
Annular Vc = 0.6 [ (3470 − 1370n)K ]2−n [ 2n + 1 ]
2−n
[ 2.05 ρ ] [0.64 (Dh−D p)n ]
The units are the same as above.
n and K are the Power Law coefficients.
Further equations to determine the Critical Annular Velocity:-
1 n
Imperial Vc (ft/min) = [ 3.88 x 104K]
2 - n[ ( 2.4 ) (2n + 1) ]
2 - n
[ ρ ] [ (DhD p) ( 3n ) ]
ρ = ppg
D = inchesK = lb / 100ft2
1 n
SI Vc (m/min) = [ 9 x 104K]
2 - n [ ( 200 ) (2n + 1) ]
2 - n
[ ρ ] [ (DhD p) ( 3n ) ]
ρ = kg / m3 D = mm
K = Poise
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1.6 DETERMINATION OF SYSTEM PRESSURE LOSSES
Regarding the well as a whole as a closed system, pressure losses will occur throughout
the system :-
through each drillpipe section
through the bit
through each annular section
through surface lines eg standpipe, kelly hose, pumps and lines
The total of all theses losses ie Total System Pressure Loss should be equal to the actual
pressure measured on the standpipe.
This is a very important part of hydraulic evaluation. Obviously, the maximum pressure
loss possible will be determined by the rating of the pumps and other surface equipment.This maximum is normally far in excess of the pressure loss that will be desired by the
drilling engineer.
The logging engineer’s task is normally to take given parameters from the drilling
engineer, then select, for example, the correct nozzle sizes that will produce the desired
system pressure loss.
The amount of pressure loss will be dependant on flowrate, mud density and rheology, the
length of each section and the diameters of each section.
Whether the flow is laminar or turbulent is also an important influence on the pressureloss - turbulent flow will produce larger pressure losses.
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1.6a Fanning Friction Factor
The frictional forces involved will have a large affect on the actual pressure losses in a
given annular or pipe section.
The frictional forces present will be very different depending on whether the flow is
laminar or turbulent:-
• with laminar flow, the fluid movement is in one direction only - parallel to the conduitwalls, with velocity increasing towards the centre.. Frictional forces will therefore
only be present due to fluid ‘layers’ moving longitudinally against each other.
• with turbulent flow, fluid movement is much more complex and multi-directional, sothat many more frictional forces are present.
For this reason, a coefficient called the Fanning Friction Factor is determined for each
type of flow and whether we are dealing with pipe or annular pressure losses. The friction
factor is determined from the Reynolds Number which has already been calculated for
pipe or annular sections based on annular velocity, diameters, density and effective
viscosity.
Laminar Flow f ann = 24 / Re Re = Annular Reynolds No.
f pipe = 16 / Re Re = Pipe Reynolds No.
Turbulent Flow f turb = a / Re b
where Re = Reynolds number in the pipe or annulus
a = log n + 3.93
50
b = 1.75 - log n
7
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Transitional Flow
f ann = [ Re - c ] x [ ( a ) - (24) ] + 24
[ 800 ] [ (4270 - 1370n) b
( c ) ] c
where Re = Annular Reynolds No.
a = (log n + 3.93) / 50
b = (1.75 - log n) / 7
c = 3470 - 1370n
f pipe = [ Re - c ] x [ ( a ) - (16) ] + 16
[ 800 ] [ (4270 - 1370n) b
( c ) ] c
where Re = Pipe Reynolds No.
a, b, and c are as above
When using the Power Law Model, the values of the Fanning Friction are substituted intoequations in order to calculate pressure losses in the annulus or in the pipe.
When calculating these pressure losses, each individual section has to be calculated
seperately, then totalled to give an overall pipe or annular pressure loss.
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1.6b Drillstring Pressure Losses
Bingham
For laminar flow, Ploss (KPa) = LQ PV + YP L
612.95 Di4 13.26Di
For turbulent flow, Ploss (KPa) = L ρ0.8
Q1.8
PV0.2
901.63 Di4.8
where L = length of section (m)
Q = flowrate (litre/min)
ρ = mud density (kg/litre)PV = plastic viscosity
YP = yield point
Di = pipe inner diameter (inch)
Power Law
Here, there is just one equation to be considered, since whether the flow is laminar or
turbulent has already been accounted for by the Reynolds Number and the Fanning
Friction Factor.
SI Ploss (Kpa) = f p.v2.ρ.L
1800 Di where f p = Friction Factor in the pipe
v = Average velocity in the pipe (m/min)
ρ = Mud density (kg/m3)Di = Pipe inner diameter (mm)
L = Length of section (m)
Imperial Ploss (psi) = f p.v2.ρ.L
92870 Di where v = ft/min
ρ = ppgDi = inches
L = ft
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1.6c Annular Pressure Losses
Bingham
laminar flow, Ploss = L Q PV + YP L
408.63(Dh+D p)(Dh−D p)3 13.26(Dh−D p)
turbulent flow, Ploss = L ρ0.8
Q1.8
PV0.2
706.96 (Dh+D p)1.8
(Dh−D p)3
The units are the same as those used in the drillstring pressure loss formula.
Dh = hole diameter (inch)D p = pipe outer diam (inch)
Power Law
SI Ploss (Kpa) = f a.v2.ρ.L
1800 (Dh - D p) f a = Annular Friction Factor
v = Average annular velocity (m/min)
ρ = Mud density (kg/m3)Dh = Hole diameter (mm)
D p = Pipe outside diameter (mm)
L = Length of section (m)
Imperial Ploss (psi) = f a.v2.ρ.L
92870 (Dh - D p) where v = ft/min
ρ = ppg
Dh = inchesD p = inches
L = ft
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1.6d Bit Pressure Loss
ie the system loses pressure when the mud passes through the nozzles.
Due to the very fast velocities involved and the small area of the nozzles, this will be the
largest singular pressure loss throughout the system.
SI Ploss (Kpa) = ρ. Q2. 277778
(D12 + D2
2+D3
2 +....)
2
where ρ = mud density (kg/m3)Q = flow rate (m
3/min)
Dn = nozzle diameter (mm)
Imperial Ploss (psi) = ρ. Q2. 156
(D12 + D22 +D32 +....)2
where ρ = ppgQ = gpm
Dn = 32nds inch
Unfortunately, these equations (and the QLOG software) will not produce accurate
calculations for diamond bit pressure losses.
Eastman Christensen suggest the following calculations:-
For Radial Flow, Ploss (bar) = 7.3188 ρ0.61 QTFA
For Feeder Collector, Ploss (bar) = 24.738 ρ0.34 Q1.47
TFA1.76
where ρ = mud density (kg/l)Q = flowrate (l/min)
TFA = mm2
1 bar = 100KPa
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1.6e Surface Pressure Losses
The calculation of pressure losses due to surface equipment is not as clear cut as previous
calculated losses and will be dependant on the type of equipment present on the rig.
ie type of pump, length of standpipe and surface lines, length of kelly etc
One method of calculation is based on Binghams formula for turbulent flow pressure
losses, where the main part of the equation, ρ0.8 Q1.8 PV0.2 is multiplied by a constantrepresenting 4 rig types or classifications.
Surface Ploss = E ρ0.8
Q1.8
PV0.2
where Ploss = psi or KPa
ρ = ppg or kg/litreQ = gpm or litre/min
E is the constant representing the 4 rig surface equipment types. The rig type should be
attainable from charts/tables kept on the rig. If not, the usual type and default is Type 4.
Classification E
Imperial Metric
1 2.5 x 10−4 8.8 x 10−
6
2 9.6 x 10−5 3.3 x 10−
6
3 5.3 x 10−5
1.8 x 10−6
4 4.2 x 10−5 1.4 x 10−
6
In practice, this classification is generally not available at wellsite. For this reason,
together with the fact that the method is based on a Bingham fluid, Datalog uses another
method based on mud density and flowrate, together with a constant to represent different
types of rig equipment.
Hence:
Surface Pressure Loss = 0.35 x Factor x Mud Density x Flowrate
Factor represents the value selected in the QLOG equipment table - the surface
connection factor . This value can range from 0.2 to 0.5, with 0.5 being the normal
default value.
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1.7 OTHER HYDRAULIC CALCULATIONS
1.7a Cuttings Slip Velocity
So far, the annular velocities that we have seen calculated only deal with the velocity ofthe fluid. Drilled cuttings are obviously going to be far denser than the mud, so that there
is always going to be a degree of ‘slip’.
How significant the degree of cuttings slip is going to be will be dependent on the relative
densities, viscosity and ‘carrying’ potential of the mud, and particle size.
Net Cuttings Velocity = annular velocity − cuttings slip velocity
SI units Vs = 0.42 D p (ρ p − ρ m)0.667
D p = particle diameter (mm)
ρm0.333
µe0.333
ρ p = particle density (kg/m3)ρm = mud density (kg/m3)µe = effective mud viscosity (mPa.s)Vs = slip velocity (m/min)
Imperial Vs = 175 D p (ρ p − ρ m)0.667
D p = inches
ρm0.333
µe0.333
ρ p = ppgρm = ppgµe = cPVs = ft/min
Cuttings slip when the flow type is turbulent will be clearly different from when the flow
is laminar and constant forces are involved.
With turbulent flow, whether the slip velocity is constant or not is dependant on the
Reynolds Number determined for the cuttings.
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1.7b Slip Velocity in Turbulent Flow
SI units Vs (m/min) = 6.85 [ D p (ρ p - ρm) ]0.5
[ 1.5ρ ]
Imperial Vs (ft/min) = 113.4 [ D p (ρ p - ρm) ]0.5
[ 1.5ρ ]
Note that there are no velocity or viscosity inputs into this equation. It is therefore
dependant on the Particle Reynolds number as to whether the slip velocity will be
constant.
Particle Reynolds Number
SI units Re p = 0.01686. ρ. Vs. D p where ρ = mud density (kg/m3)µe Vs = slip velocity (m/min)
Dp = particle diameter (mm)
µe = effective viscosity (mPa.s)
Imperial Re p = 15.47. ρ. Vs. D p where ρ = ppgµe Vs = ft/min
Dp = inches
µe = cP
If the Particle Reynolds Number > 2000, the particle will fall at the same rate
ie cuttings slip velocity will be constant in turbulent flow
In the determination of slip velocity, a Cuttings Re number is incorporated. So to, because
of the different frictional forces present on the cuttings, is a friction or drag coefficient.
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1.7c Nozzle Velocity
Vn (m/sec) = Q Q = flowrate (litre/min)
38.71A A = total flow area of nozzles (in2)
Vn (ft/sec) = 0.32Q Q = gpm
A A = in2
Nozzle conversion to Total Flow Area
TFA (inch2) =
1/4π (d1
2 + d2
2 + d3
2) =
1/4π Σ d
2
( 322 ) 1024
where d = nozzle size in 32nds of an inch
Alternatively, the nozzle diameters, rather than TFA can be used:
SI units Vn (m/sec) = 21220 Q where Q = m3/min
Σ Dn2 Dn = mm
Imperial Vn (ft/sec) = 418.3 Q where Q = gpm
Σ Dn2 Dn = 32nds inch
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Exercise 1a Use of the Hydraulics Program
Use the following hole and pipe profiles and setups:
13 3/8” casing set at 500m, ID = 12.42” (315.5mm)
12 1/4” (311.2mm) hole drilled to a depth of 1500m
200m x 9 1/2” DC’s, OD 9.5” ID 3.0” (241.3/76.2mm)
100m x 8” DC’s OD 8.0” ID 3.0” (203.2/76.2mm)
300m x HWDP OD 5.0” ID 3.0” (127/76.2mm)
DP OD 5.0” ID 4.28” (127/108.7mm)
Jets 3 x 15 (3 x 11.9mm)MD 9.8ppg
100 SPM at flowrate 2.0 m3/min
θ600 and θ300 60/35 )Surface Conn Factor 0.5 ) set in equipment table
1. What type of flow is present in each annular section ?
What is the Total System and Surface Pressure Loss ?
2. Compare the surface pressure loss using a factor of 0.2
3. What flowrate is required to produce a system pressure of 2500psi ?
4. What new jet sizes are required to reduce the pressure back to 2000psi ?
5. What is the pressure if the mud weight is increased to 10.8ppg ?
6. With a flowrate of 2.0 m3/min, what jet sizes are now required to give a system
pressure of around 2000psi ?
7. Is the flow still laminar in all annular sections ?
8. If transitional flow is acceptable around the 9 1/2” DC’s but not the 8” DC’s, what is
the maximum flowrate ?
9. With this flowrate, how many jets may have washed out if a surface pressure drop to
1650psi was seen ?
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1.8 HYDRAULICS OPTIMIZATION
1.8a Bit Hydraulic Horsepower
This is the power used by the jetting action of the bit, which has to balance maximum
ROP and maximum jetting with effective hole cleaning.
SI units Bit HP (KW) = P b x Q x 0.01667 Q = flowrate (m3/min)
Pb = bit pressure loss (KPa)
Imperial Bit HP (HP) = P b x Q Q = gpm
1714 P b = psi
The Total System Hydraulic Horsepower can be calculated by substituting the Total
System Pressure Loss (in place of Bit Pressure Loss) into the same equation.
1.8b Hydraulic Impact Force
This is the force exerted on the formation due to the fluid exiting the jets. Cleaning is by
direct erosion on the bottom and by cross flow under the bit.
SI units Bit IF (newtons) = ρ Q Vn ρ = mud density (kg/m3)
60 Q = flowrate (m3/min)
Vn = nozzle velocity (m/sec)
Imperial Bit IF (lbs) = ρ Q Vn ρ = ppg1932 Q = gpm
Vn = ft/sec
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1.8c Optimization
Hydraulics can be optimised in two ways:-
1) by maximising the Impact Force of the jets on the bottom of the hole.2) by maximising the hydraulic power expended by the bit.
The power expended (or used up) by the bit is a proportion of the total power available
( HPt). This is determined either by the maximum pressure of the pumps:
where max HP pump = HPt = Pmax Q
1714
or, more typically, it is based on a desired maximum pump pressure together with a
maximum flow rate that will give sufficient annular velocity for cuttings removal.
Once the maximum power available to the system is known, hydraulic performance can
be optimised in the following ways :-
1) Optimise Horsepower by setting Bit HP at 65% of Total Available Power
2) Optimise Impact Force by setting Bit HP at 48% of Total Available Power
Since the hydraulic horsepower at the bit is dependent on jet velocity and therefore on the
pressure loss at the bit, hydraulic performance in practice can simply be optimised by
selecting jet sizes to give:
Bit Pressure Loss = 65% System Pressure Loss
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Use of the QLOG hydraulics programs:
Current Profiles (onhyd)
This is an optimization program that works based on realtime information such as pumpoutput, mud density and pressure losses. These values can be changed should a change in
parameters be the reason for running the optimization program. The minimum and
maximum jet velocities are suggested values.
The program can then be run to give you the parameters required for optimum hydraulics
based on both Hydraulic Impact Force and Hydraulic Horsepower at the bit.
Impact Force relates directly to the erosional force of the drill fluid and is therefore good
optimization for bottom hole cleaning.
Hydraulic Horsepower optimization generally requires lower annular velocities so thatflow type is more likely to be laminar.
New Profiles (offhyd)
This program is offline so that you can input any hole and pipe profiles, mud parameters,
flow rate and jet size and calculate the resulting hydraulic parameters such as pressure
losses, flow types, annular velocities etc.
This program would be used when pre-determining the correct parameters for a new hole
section or bit run. By changing the inputs, you can attempt to optimize the hydraulics.
To optimize for hydraulic horsepower, the %HHP at the bit should be 65% of the Total
HHP.
Since HHP is determined by pressure loss, this equates to Bit Pressure Loss being 65% of
the Total System Pressure Loss.
To optimize for hydraulic impact, the %HHP at the bit should be 48% of the Total HHP.
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Exercise 1b Optimizing Hydraulics
Use the original profiles and set ups used in exercise 1a
1. What is the % Hydraulic Horsepower of the bit ?
2. Using the following ranges and limitations, try to optimise the hydraulics whilst still
retaining laminar flows and good annular velocities for cuttings removals.
Flowrate 1.8 to 2.2 m3/min
Mud density 9.6 to 10.2 ppg
Maximum System Pressure 2800 psi
Minimum Jet sizes 3 x 10mm
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Exercise 1c Optimizing Hydraulics
Use the following hole and pipe profiles and setups:
9 5/8” casing set at 2500m, ID 8.68” (220.4mm)8 1/2” (215.9mm) hole drilled to 4000m
500m x 6 1/2” DC’s OD 6.5”, ID 2.88” (165.1/73.1mm)
400m x HWDP OD 5.0”, ID 3.0” (127/76.2mm)
DP OD 5.0”, ID 4.28” (127/108.7mm)
Flowrate 1.4 m3/min
Mud density 10.5ppg
Surface Conn Factor 0.5
θ600 and θ300 70/42
1. What jets would produce a system pressure of 2500 psi ?
With these setups, what are a) flow types in each annular section
b) annular velocities in each section
c) % HP at the bit
2. With a flowrate of 1.6m3/min, what jets are required to give a system pressure
of 2200psi ?
What now are a) flow types
b) annular velocities
c) % HP at bit
3. Using the following ranges and limits, attempt to optimize the hydraulics whilst
retaining laminar flows in each section and good annular velocities.
Flowrate 1.3 to 1.6 m3/min
Mud density 10.3 to 10.6 ppg
Maximum system pressure 2850 psi
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1.9 Equivalent Circulating Density
The pressure exerted at the bottom of the hole by the static column of mud is known as
the Hydrostatic Pressure.
PHYD = ρ x TVD x 0.052 where ρ = mud density (ppg)PHYD = psi
TVD = feet
PHYD = ρ x TVD x 0.433 where ρ = SG(pressure & depth have the same units)
PHYD = ρ x TVD x 0.00981 where ρ = kg/m3PHYD = Kpa
TVD = m
During circulation, the pressure exerted by the fluid column at the bottom of the hole
increases as a result of frictional forces and annular pressure losses caused by the fluid
movement.
This increased pressure is termed the Dynamic Pressure or Bottom Hole Circulating
Pressure (BHCP).
BHCP = PHYD + ∆ Pa where ∆ Pa is the sum of the annular pressure losses
This in turn means that the ‘acting’ density of the mud will increase when the fluid is
moving. This is termed the Equivalent Circulating Density.
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Determination of Equivalent Circulating Density
a. ECD = ρ + ∆ Pa ECD = ppg EMW (equivalent mudweight)0.052xTVD ∆ Pa = psi
TVD = feetρ = ppg
BHCP can therefore be expressed as ECD x 0.052 x TVD
b. ECD = ρ + ∆ Pa ECD = kg/m3 EMW0.00981xTVD ∆ Pa = KPa
TVD = m
ρ = kg/m3
BHCP can therefore be expressed as ECD x 0.00981 x TVD
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Exercise 1d Equivalent Circulating Density
For each of the following situations, calculate the mud density.
1. TVD 3500 ft
Hydrostatic Pressure 1729psi
2. TVD 3000m
Hydrostatic Pressure 32373 Kpa
For each of the following situations, calculate a) Hydrostatic Pressure
b) Bottom Hole Circulating Pressurec) Equivalent Circulating Density
3. TVD 4000 ft
Mud density 9.5ppg
Annular Pressure Losses 250psi
4. TVD 3000m
Mud density 1150 kg/m3
Annular Pressure Losses 3000 Kpa
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1.10 Surge and Swab Pressures
Surge Pressures result from running pipe into the hole creating a pressure increase. The
size of the pressure increase will be dependent on the pipe running speed, the annular
clearance and whether the pipe is open or closed.
Excessive Bottom Hole Pressure could break down weak or unconsolidated formations.
Swab Pressures result from pulling the pipe out of the hole. The frictional drag combined
with the piston effect will create a reduction in pressure.
This reduction in the hydrostatic could lead to the invasion of formation fluids.
• More than 25% of blowouts result from reduced hydrostatic pressure caused byswabbing.
• Excessive surge pressures can lead to lost circulation. Running casing is a particularlyvulnerable time due to the small annular clearance and the fact that the casing is
closed ended.
• Beside the well safety aspect, invasion of fluids due to swabbing can lead to mudcontamination and necessitate the costly task of replacing the mud.
• Pressure changes due to changing pipe direction, eg during connections, can be particularly damaging to the well by causing sloughing shale, by forming bridges or
ledges, and by causing hole fill requiring reaming.
Calculation of Surge and Swab Pressures.
The same method is used to calculate the differential pressure caused by both surging and
swabbing. To determine the new Hydrostatic Pressure, the differential pressure is either
added or subtracted depending on whether surge or swab respectively.
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Firstly, the Fluid Velocity of the displaced mud caused by the pipe movement has to be
calculated.
For Closed Ended Pipe:
Fluid Vel (ft/min) = [ 0.45 + D p2
] x Vp Vp = pipe speed (ft/min)
[ Dh2 − D p
2 ] Dh = hole diameter (in)
D p = pipe outer diameter (in)
Di = pipe inner diameter (in)
For Open Ended Pipe:
Fluid Vel (ft/min) = [ 0.45 + D p2 − Di
2 ] x Vp
[ Dh2 − D p
2 + Di
2 ]
This fluid velocity then has to be converted to the equivalent flowrate by using the
annular velocity equation, where:-
fluid velocity (ft/min) = 24.5 Q where Q = gpm
Dh2 − D p
2
The change in pressure is then calculated for each annular/pipe section using the Pressure
Loss equations. This is calculated for both laminar and turbulent flow with the largest
value being taken.
The total swab or surge pressure acting on the bottom of the hole is the sum of all of the
pressure losses for each annular/pipe section.
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Use of the QLOG Swab and Surge Program
This program is used to determine the pressures induced by the defined maximum and
minimum running speeds of the pipe. Thus, a safe speed can be deduced in order to avoid
excessive pressures.
Required information:-
Bit depth and hole depth - read from the realtime system, editable if required.
Current surge/swab pressure - read from current recorded pressures, editable if
required.
Current Flow In - read from realtime system, editable if required.
Use Current Profile - ie current hole and pipe profiles, the user should select Y(es).
Maximum and Minimum running speed - limits defined by the user. Negative values
should be used in order to calculate swab pressures. For example, for surge pressure, the
minimum running speed may be 5m/min and the maximum 50m/min. For the same
limits, the swab calculation requires the minimum to be set at -50m/min, and the
maximum at -5m/min.
Current running speed - read from realtime system, editable if required.
Press F7 to calculate the maximum and minimum pressures.
Press F2 to print the data out.
Press F8 to produce a plot. The plot will be pressure against running speed and will show
the pressures against the max/min limits defined together with the current
pressure/running speed situation.
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Exercise 1e Use of the Swab Surge program
This program accesses information from the realtime system.
Therefore:-
Enter the hole and pipe profiles from Exercise 1c into the realtime files.
Enter the following into equipment table a) Mud density override 9.3ppg
b) θ600 and θ300 50/30
Using maximum and minimum running speeds of 20 and 100 m/min, calculate the
swab/surge pressures with the following bit depths:
1000m
2000m
3000m
3500m
3950m
With an increased mudweight of 10.3ppg, calculate, for the same maximum and
minimum running speeds, the swab/surge pressures at 3500 and 3950m.
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APPENDIX - Answers to Training Exercises
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Exercise 1a Use of Hydraulics Program
1. Laminar flow in all sections
System pressure loss 2038 psi
Surface pressure loss 59.6psi
2. 23.8 psi
3. 2.24 m3/min giving a pressure of 2498psi
4. 1 x 13mm, 2 x 14mm giving a pressure of 1994psi
5. 2162psi
6. 1 x 12mm, 2 x 13mm giving a pressure of 1983psi
7. Yes, flow is laminar in each section
8. 2.24 m3/min
9. 1 jet with 12mm jet washout, pressure would be 1658psi
with 13mm jet washout, pressure would be 1671psi
Exercise 1b Optimizing hydraulics
1. 15.0% HP at the bit
2. Two possible situations are:-
a. Mud weight 9.9ppg
Flowrate 2.0 m3/min
Jets 2 x 10, 1 x 11
This gives 60.2% HHP at the bit
2771psi system pressure loss
Laminar flow in all sections with good annular velocities
b. Mud weight 10.15ppg
Flowrate 1.9 m3/min
Jets 3 x 10
This gives 63.9% HHP at the bit
2765psi system pressure loss
Laminar flows, but lower annular velocities
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Exercise 1c Optimizing Hydraulics
1. 3 x 10mm jets, giving system pressure loss of 2523psi
a) laminar in all sections
b) 55 to 92 m/minc) 39.3 %
2. 3 x 14mm jets, giving system pressure loss of 2211psi
a) transitional around collars, laminar in all other sections
b) 63 to 105 m/min
c) 15.3 %
3. Using flowrate 1.3 m3/min
mud weight 10.3 ppg
jets 2 x 9, 1 x 8mm
system pressure loss of 2834psi
% power at bit 52.2 %
Laminar flows in all sections
Annular velocities 51 to 86 m/min
Exercise 1d Equivalent Circulating Densities
1. 9.5 ppg
2. 1100 kg/m3
3. Phyd = 1976 psi
BHCP = 2226 psi
ECD = 10.7 ppg EMW
4. Phyd = 33844 Kpa
BHCP = 36844 Kpa
ECD = 1252 kg/m3 EMW
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Exercise 1e Swab/Surge Program
at 1000m, min/max pressure = 35 + 211 psi
at 2000m, 50 + 280 psi
at 3000m, 69 + 388 psi
at 3500m, 77 + 426 psiat 3950m. 85 + 461 psi
With 10.3ppg mud weight:
at 3500m, min/max pressure = 77 + 460 psi
at 3950m, 85 + 497 psi
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