Gravel Pack PI

PetroPerth Training Centre 39 / 474 Murray Street, PERTH, Western Australia

www.petroperth.com

Well Completion

Master Class

Workshop

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

Problem 1: IPR – Transient (Understaurated Reservoir) ................................ 5 Problem 2: IPR - Steady State, Skin factor.........................................................7 Problem 3: IPR - Two-phase Flow, Vogel's Correlation .................................. 9

Gravel Pack ........................................................................................................... 11 Problme 4: Optimal Gravel Size ..................................................................... 13 Productivity of Gravel Packed Wells - Equations ......................................... 14

Gas Lift .............. .......................................................................................................16 Problem 5: Injected Gas Pressure......................................................................16 Problem 6: Gas Injection Pressure, Pomit of injection and Well Flow Rate..19 Problem 7: Sustianing Production Rate while Reservoir Pressure Deplets..20 Gas Lift Performance Curve - Notes..................................................................22

Artificail Lift - Pumps...................................................................................................24 Problem 8: Sucker Rod Pump Speed..................................................................24 Problem 9: Effective Plunger Stroke Length......................................................25 Sucker Rod Pump - Dynamometer Card Shapes...............................................27 Problem 10: Electrical Submersible Pump Design.............................................28 Hydraulic Fracturing......................................................................................................31 Problem 11: Calculations of Stresses vs. Depth.................................................31 Problem 12: Initial Fracture Pressure...................................................................33 Sandstone Acidizing Treatment...................................................................................34 Problem 13: Maximum Injection Rate and Pressure...........................................34




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Notes




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Inflow Performance Relationships Inflow 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

Solution The 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

rw 0.328

k 8.2md

h 53ft

Pi 5651psi

ct 1.29x10-5 psi-1

µ 1.7cp




6




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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

rw 0.328

Drainage radius

2980ft




8




9

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

rw 0.406ft

Drainage radius

1490ft




10




11

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 or Dg60 = 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 Data The following graph shows grain diameter vs. cumulative Weight % for unconsolidated sand in California.

Solution 1- 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) = 0.070in Dg90 = 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 – Equations Equations 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)

γg 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 Data Determine the pressure at the injected point for the following conditions.

γg 0.7

Depth 8000ft

Psurf 900psi Tsurf 80degF

Tinj 160 degF

lperf Gravel-packed perforation length (in)




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Solution The 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 Data IPR relationship is given by: qt = 0.39 (Pave – Pwf)

Solution Using 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 = 5490ft Pinj 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 Data Use 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.

Solution From 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.




21




22

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.




23

Limit GLR is not constant with time.




24

Artificial Lift – Pumps

Pumps 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 Data Production rate = 250STB/day Rod pump diameter plunger = 2in Effective plunger stroke length = 50in Volumetric efficiency = 0.8 Formation factor = 1.2

Solution Equation 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 Data Pump 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

Solution Calculating 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.




27

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 Data Well 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 = 200psi 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 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

Solution Using 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 fit 7in casing and produce 1260bbl/day. The following chart can be used.




29

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 Data 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

Solution Equation 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

Solution From 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 Data Reservoir 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/ft Average reservoir pressure = 4500psi re = 1000ft, rw = 0.328ft Reservoir Pressure = 4500psi k = 8.2md h = 53ft Oil viscosity = 1.7cp Skin factor (S) = 10

Solution The 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

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Gravel Pack PI

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