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7/16/2019 Gravel Pck Productivity
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PetroPerth Training Centre 39 / 474 Murray Street, PERTH, Western Australia
www.petroperth.com
Well Completion
Master Class
Workshop
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Perth’s Premier Petroleum Consulting Group
PetroPerth 38 / 474 Murray Street, PERTH, Western Australia
+61410477165 www.petroperth.com
2
Joseph Graham
DISCLAIMERS :
The exercises and information contained here has been prepared by PetroPerth Pty Ltd solely for training purposes only and should not be relied upon by any other party. PetroPerth Pty Ltd will not accept responsibility or liability to third parties to whom this may be shown or into whose hands it may come across.
PetroPerth Pty Ltd has made all reasonable efforts to ensure that the
interpretations, conclusions and recommendations presented herein are in
accordance with good oil industry practice. However, PetroPerth Pty Ltd
does not guarantee the correctness of any such interpretations and shall not
be liable or responsible for any loss, costs, damages or expenses incurred or
sustained by anyone resulting from any interpretation or recommendation
made by any of our Directors, Senior Managers, Authorised Representatives
/ Agents / Contractors or employees.
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INDEX
Inflow Performance Relationships ......................................................................... 5
Prob lem 1: IPR – Trans ient (Underst aurated Reserv oir ) ................................ 5
Pro bl em 2: IPR - Stead y S tate, Ski n fac to r.........................................................7
Pro blem 3: IPR - Tw o-ph ase Fl ow , Vogel's Cor relati on .................................. 9 Gravel Pack ........................................................................................................... 11Pro blme 4: Op tim al G rav el Size ..................................................................... 13
Pro du cti vit y o f Gr avel Pac ked Wells - Equatio ns ......................................... 14
Gas Lift .............. .......................................................................................................16
Problem 5: Inj ec ted Gas Pressu re ......................................................................16
Problem 6: Gas Inject ion Pressure, Pomit of inject ion and Well Flow Rate..19
Problem 7: Sustianing Prod uction Rate while Reservoir Pressure Deplets..20
Gas Li ft Perf ormanc e Cu rv e - No tes ..................................................................22
Artificail Lift - Pumps...................................................................................................24
Pro bl em 8: Suck er Rod Pump Speed..................................................................24
Pro bl em 9: Effect iv e Plu ng er Stro ke Len gth ......................................................25
Suck er Rod Pump - Dynam om eter Card Shapes...............................................27 Problem 10: Elect ric al Submersi ble Pump Desig n.............................................28
Hydraulic Fracturing......................................................................................................31
Pro bl em 11: Calc ul atio ns o f Stress es vs . Dept h .................................................31
Pro bl em 12: Ini tial Frac tu re Pres su re...................................................................33
Sand st on e Aci di zin g Treatment ...................................................................................34
Problem 13: Maxim um Inject ion Rate and Press ure...........................................34
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Notes
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Inflow Performance RelationshipsInflow performance relationship for a well presents the well production rate as a function of the
driving force in the reservoir. Usually, it shows how the production rate changes with respect to
changes in the bottom hole pressure.
Problem 1: IPR – Transient
(Under-saturated Reservoir)
Objectives:
Construct Transient IPR curve at different time intervals
Given Data
SolutionThe following equation is used (in this equation, time has to be in hours)
For the giving parameters, the equation takes the following form.
At each time interval required, as set of q vs. Pwf can be calculated and plotted, as follows.
B 1.1bbl/STB
Ø 0.19
r w 0.328
k 8.2md
h 53ft
Pi 5651psi
ct 1.29x10-5 psi-1
µ 1.7cp
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Problem 2: IPR – Steady-state, Skin factor
(Under-saturated Reservoir)
Objectives:
Construct IPR curve for different skin factor values (0, 5, 10, 50)
Given Data
Solution
The following equation is used
For the giving parameters, and for skin factor equals 5, the equation reads,
Similar equations can be found for other skin factor values, just change value for S in the above
equation. The following figure can then be then drawn.
k 8.2md
h 53ft
Pi = Pe 5651psi
ct 1.29x10-5 psi-1
µ 1.7cp
B 1.1bbl/STB
Ø 0.19
r w 0.328
Drainage
radius2980ft
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Problem 3: IPR – Two-Phase Flow – Vogel’s Correlation
Objectives:
Construct IPR curve for two-phase flowing system using Vogel’s Correlation
Given Data
Solution
The following equation is used
For the giving parameters, and for skin factor equals zero, the equation reads,
For different Pwf values, qo can be calculated. The following figure can then be then drawn.
k 13md
h 115ft
Pave 4350psi
µo 1.7cp
Bo 1.1bbl/STB
r w 0.406ft
Drainage
radius1490ft
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Gravel Pack
Formation grain size is usually obtained with sieve analysis by using standard sieve trays. The
following table reads the sieve opening sizes for US standard mesh sizes (From Perry, 1963).
Results from sieve analysis are usually presented as a semi-log plot of cumulative weighof
formation material retained vs. Grain size.
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Optimal Gravel Size Correlations
- Schwartz Correlation
Dg40 = 6Df40
Dg40 is the recommended gravel size
Df40 is the diameter of formation sand at which 40% of grains are of large diameter.
Uniformity coefficient (should be less than or equal 1.5):
Uc = (Dg40)/(Dg90)
Dg, min = 0.615 Dg40
Dg, max = 1.383 Dg40
- Saucier Correlation
Dg50 = 5 Df50 orDg60 = 6 Df50
Dg, min = 0.667 Dg50
Dg, max = 1.5 Dg50
Dg50 is the recommended gravel size
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Problem 4: Optimal Gravel Size
Objectives:
To Determine the optimal gravel size using Schwartz and Saucier correlations
Given DataThe following graph shows grain diameter vs. cumulative Weight % for unconsolidated sand in
California.
Solution1- Schwartz Correlation:
From the graph, Df40 = 0.0135 in
Dg40 = 6 (0.0135) = 0.081in
Dg90 = 0.081/1.5 = 0.054in
Dg, min = (0.615)(0.081) = 0.05in (at 100% weight)
Dg, max = (1.383)(0.081) = 0.11in (at 0% weight)
From the previous table, 0.05in corresponds to mesh size 16 where 0.11 corresponds to
mesh size 7.
2- Saucier Correlation:
From the graph, Df50 = 0.00117 in
Dg50 = 5 (0.00117) = 0.059in or Dg50 = 6 (0.00117) = 0.070inDg90 = 0.081/1.5 = 0.054in
Dg, min = (0.667)(0.059) = 0.039in or (0.667)(0.070) = 0.047in (at 100% weight)
Dg, max = (1.5)(0.059) = 0.088in or (1.5)(0.070)=0.105in (at 0% weight)
From the previous table, those grain sizes correspond to mesh sizes 8 and 16 or 18.
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Productivity of Gravel Packed Wells – EquationsEquations for inside casing-casing gravel packs are presented below (gravel pack skin factor and
non-Darcy follow coefficient for the gravel-filled perforations for gas and oil wells).
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where
Values for “a” and “b” for some common gravel sizes are given below.
Sg Gravel-pack skin factor
Dgg Non-Darcy flow coefficient for gravel-filled perforation in gas wells
Dgo Non-Darcy flow coefficient for gravel-filled perforation in oil wells
kh Formation permeability-thickness produce (md-ft)
lperf Gravel-packed perforation length (in)
kg Permeability of the gravel (md)
Dperf Perforation diameter (in)
µ Fluid Viscosity (cp)
ρ Fluid Denisty (lbm/ft3)
γ g Gas gravity
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Gas Lift
Gas lift is a means of artificial lift. When the bottom hole flowing pressure is less than the tubing
pressure difference caused by the flowing fluid, gas lift may be stand as an ideal artificial lift
method. In a gas lift system, gas can be injected intermittently or continuously. The aim is to lift
the fluid up in the well at a desirable well head pressure while keeping the bottom hole pressure
at certain level that is required to provide good sustainable driving force in the reservoir.
Problem 5: Injected Gas Pressure
Objectives:
To Determine the pressure of injected gas
Given DataDetermine the pressure at the injected point for the following conditions.
γ g 0.7
Depth 8000ft
Psurf 900psi Tsurf 80degF
T inj 160 degF
lperf Gravel-packed perforation length (in)
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SolutionThe Problem is solved by trial and error approach using the following equations.
- Assume value for Pinj, say 1100psi.
- From the following figure, estimate Pc and Tc (668psi, 390 degR)
- Ppr = ((900+1100)/2)/668 = 1.5
- Tpr = (((80+160)/2)+460)/390 = 1.49
- Estimate Z = 0.86
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- Calculate Pinj from the equation above, Pinj = 1110. If the assumed value and calculated
value do not match, repeat steps again.
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Problem 6: Gas injection pressure, Point of Injection and Well
Flow rate
Objectives:
Determination of depth of injecting point
Given DataIPR relationship is given by: qt = 0.39 (Pave – Pwf )
SolutionUsing T, γ , and Z values in this example, the equation Pinj given in the previous example
can be approximated using Taylor series and take the following form.
Re-arranging to calculate Psurf (Pinj at 8000ft is 1100psi)
From the IPR equation, at qt = 500STB/day, Pwf = 1770psi
The Injection point has to be where the pressure between injected gas and production string is
balanced.
Hinj is therefore = 5490ftPinj at point of injection (5490ft) is
As ΔPvalve = 100psi, the pressure inside the tubing would be 940psi
Z 0.9
T 600R
Pinj 1100psi
D 8000ft
GLR 300SCF/STB
Pave 3050psi
ΔPvalve 100psi
γ 0.7
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Problem 7: Sustaining Production rate while reservoir pressure
depletes
Objectives: Determination of depth of injecting point and gas injection rate to sustain
production
Given DataUse data given in Problem 6 to calculate point of injection and gas flow rate after the reservoir
pressure has dropped to 2550psi in order to keep production rate at 500STB/day.
SolutionFrom the IPR equation, qt = 0.39 (Pave – Pwf ), at qt = 500STB/day and Pave = 2550psi,
Pwf is 1268psi.
The injection point is obtained by applying the following equation, leading to Hinj = 7120ft
Pinj can be calculated (as in previous problem). Pinj = 980psi
In order to calculate the gas injection rate required to sustain the oil production from the well,
(a tubing graph similar to the following) needs to be used to estimate the new GLR. From the
graph, GLR is 750SCf/STB.
The new gas rate = (750-300)500 = 2.25x105
SCF/day
N.B.: A tubing graph correlates Depth vs. pressure at different GLR ratios. Each graph is unique
for production rate, tubing size, and average Temperature.
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Gas Lift Performance Curve - Notes
For each production rate (liquid rate) there is a limit GLR where minimum Pwf
will observed. This production rate is the intersection of the IPR curve and gas lift
performance curve.
Other values of GLR (above or below the limit GLR) will result in lower
production rate.
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Limit GLR is not constant with time.
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Artificial Lift – PumpsPumps are usually used to boost the productivity of a well by lowering the bottom-hole
pressure. Unlike gas lift, pumps increase the pressure at the bottom of the tubing to a sufficient
level to lift the liquid up to the surface. Two major types of pumps; Positive Displacement
Pumps (Sucker Rod Pump and Hydraulic Piston Pump) and Dynamic Displacement Pumps(Electrical Submersible Pumps and Jet Pumps).
Problem 8: Sucker Rod Pump Speed
Objectives:
To Determine pump speed required to produce certain production rate
Given DataProduction rate = 250STB/day
Rod pump diameter plunger = 2in
Effective plunger stroke length = 50in
Volumetric efficiency = 0.8
Formation factor = 1.2
SolutionEquation that relates production rate (q) to pump specifications is,
In this equation, q is at reservoir conditions. Hence N is calculated as follows.
N = 16spm. Sucker rod pumps are typically operating at speed ranging from 6 -20spm.
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Problem 9: Effective Plunger Stroke Length
Objectives:
To Determine the effective length for the plunger stroke
Given DataPump set at 3600ft
Well has 3/4in sucker rod and 2.5in tubing
Produced liquid specific gravity = 0.90
Pump speed = 20rpm
Plunger is 2in diameter
Stroke length is 64in.
Liquid is at pump level
SolutionCalculating the effective plunger stroke length requires the usage of the following equation.
At (Tubing area) and Ar (rod area) can be read from the following tables
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At = 1.812 in2 and Ar = 0.442 in2
As fluid level is at the pump level, H = 3600ft
Effective Plunger stroke length = 52.5in.
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Sucker Rod Pumps – Dynamometer card shapes
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Problem 10: Electrical Submersible Pump Design
Objectives:
Choosing ESP parameters (number of stages, horsepower motor load) for certain
well and production requirements
Given DataWell depth = 10000ft
Production rate = 1000STB/day
Average reservoir pressure = 4350psi
Tubing size = 2 7/8in, tubing I.D. = 2.259in
Casing size = 7in
Surface tubing pressure = 100psig
Minimum suction pressure = 200psi3/4in sucker rod and 2.5in tubing
Produced liquid specific gravity = 0.90
Pump speed = 20rpm
Plunger is 2in diameter
Stroke length is 64in.
Liquid is at pump level
K = 13md, h = 115ft, rw = 0.406ft, re = 1490ft
Average reservoir pressure = 4350psi
µ = 1.7cp, Bo = 1.26, γ = 32 API (= 0.865)
Friction factor = 0.0068, mean velocity = 2.94 ft/sec
SolutionUsing Vogel IPR correlation, one can estimate Pwf ,
Pwf = 2300psi (for q = 1000STB/day)
q at reservoir conditions = (1000)(1.26) = 1260bbl/day (this is value is used in the design).
Different ESP pumps have associated design chart. For this case, an ESP pump is required to fit7in casing and produce 1260bbl/day. The following chart can be used.
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Minimum depth for setting the pump,
Now, the minimum depth that the pump can be set at is 4390ft.
Assuming that the pump will be set at 9800ft, a suction pressure needs to be calculated (from
the same equation)
Pressure drop in the tubing ( ΔPPE) and friction pressure ( ΔPF) must be calculated.
ΔPPE = (0.433)(γ )(pump setting depth) = (0.433)(0.865)(9800) = 3670psi
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Total ΔP = 3670 + 71 = 3741psi
Total pump discharge pressure = Surface pressure + ΔP = 100 + 14.7 + 3741 = 3856psi
Pressure increase from the pump = discharge pressure – Suction pressure
= 3856 – 2225 = 1631psi
This pressure difference may be expressed as a head (in ft) = 1631/(0.433)(0.865) = 4350ft
Now, from the previous graph, for a capacity of 1260bbl/day, one can read the head for 100-
stage pump is 2180ft (21.8ft for a single stage)
Number of stages required = 4350/21.8 = 200 stages.
From the same graph, the horsepower required for 100 stage pump at 1260bbl/day is 35hp. In
our case, for a 200 stage pump, the required horsepower = 35 x 2 = 71hp
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Hydraulic Fracturing
Problem 11: Calculation of Stresses vs. Depth
Objectives:
Determination of stresses
Given DataFormation Depth = 10000ft
Formation Density = 165 lb/ft3
Poro-elastic constant = 0.72
Poisson ratio = 0.25
Maximum horizontal stress is 2000psi larger than the minimum horizontal stress
Oil density = 55 lb/ft3
Tensile stress = 1000psiReservoir Pressure = 3800psi
SolutionEquation for vertical stress,
Density is 165 lb/ft3, then
Effective vertical stress,
Effective horizontal stress,
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Minimum horizontal stress,
Maximum horizontal stress,
Using those equation stresses values can be calculated at different depth and plotted, as
follows.
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Problem 12: Initial Fracture Pressure
Objectives:
Calculation of pressure required to initiate a fracture
Given Data(Same date from previous problem)
Formation Depth = 10000ft
Formation Density = 165 lb/ft3
Poro-elastic constant = 0.72
Poisson ratio = 0.25
Maximum horizontal stress is 2000psi larger than the minimum horizontal stress
Oil density = 55 lb/ft3
Tensile stress = 1000psi
Reservoir Pressure = 3800psi
SolutionFrom the previous problem, at 10000ft, minimum horizontal stress is 5700psi, maximum
horizontal stress is 7700psi.
Equation to calculate pressure required to initiate a fracture is,
Pressure required is 6600psi
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Sandstone Acidizing Treatment
Problem 13: Maximum Injection rate and Pressure
Objectives:
To determine maximum injection rate and pressure required for acidizing
treatment in sandstone formations
Given DataReservoir Depth = 9822ft
Acid solution specific gravity = 1.07
Acid solution viscosity = 0.7cp
2-in coiled tubing (roughness = 0.001)
Formation fracture gradient = 0.7psi/ftAverage reservoir pressure = 4500psi
re = 1000ft, rw = 0.328ft
Reservoir Pressure = 4500psi
k = 8.2md
h = 53ft
Oil viscosity = 1.7cp
Skin factor (S) = 10
SolutionThe bottom pressure that will initiate a fracture (the breakdown pressure) is,
The maximum injection flow rate is calculated by using the following equation.
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Solving for qi,
The maximum tubing injection pressure is given by
Where,
And
Where f , the friction factor, is
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And Reynolds Number is
Using these equations, for qi = 250bb/day = 0.17bpm, one can estimate maximum injection
pressure to be less that 2330psi