2 - Ipr

Petroleum Production Engineering • In general, petroleum engineers can specialize in: – – – –

Drilling, Completion and Stimulation Reservoir and Well Testing Production and Artificial Lift Facilities and Fluids Processing

• Production Engineering is that part of petroleum engineering which attempts to maximize production in a cost-effective manner. • In general Production Engineers are involved with the design, analysis, optimization and troubleshooting of production systems. • Production system consists of: – – – –

The reservoir The near wellbore zone The well production and artificial lift completion The production gathering system Mauricio G. Prado – The University of Tulsa

Petroleum Production Engineering • The basic tools used to model the flow of fluids in the production system are: – Conservation Equations • • • •

Mass Momentum Energy Species

– Equations of State and Phase Equilibrium

• Usually the system is modeled from the reservoir up to a certain known boundary condition such as the separator pressure and temperature. • Usually the separator performance, the flow of the fluids streams downstream of the separator gas and liquid measurements is under responsibility of fluid processing and surface facilities engineers.

Mauricio G. Prado – The University of Tulsa

Petroleum Production Engineering • The conservation equations (mass, momentum and energy) describe the flow of fluids across all components of the production system such as the porous media, perforations, gravel pack, casing and tubing, artificial lift equipment, chokes, flowlines, control valves, manifolds, etc… • The specific form of the conservation equations varies according to the component being modeled. • The flow of fluids across the production system is divided in 5 groups. – – – – –

Flow of fluids in the porous media Flow of fluids in the near wellbore region Flow of fluids in horizontal, inclined and vertical pipes Flow of fluids through restrictions Interaction of fluids with artificial lift equipment

• As fluids move through the production system, the pressure, temperature, gas-liquid-solid equilibrium and phase properties will be changing. Those changes are all inter-related but the pressure changes are very important since they determine the production flowrates for the well. Mauricio G. Prado – The University of Tulsa

Equilibrium Flowrate The equilibrium flowrate q is the solution to the following equation

Pr − Ps = ∑ ∆Pc (q) Potential

Pressure Drops in System Components

Mauricio G. Prado – The University of Tulsa

Natural Equilibrium Flowrate 4000

Reservoir

Flowline

Tubing

3500

∆Preservoir (q )

2500

Separator

Pressure (psi)

3000

2000 Flowrate q

1500

∆Ptubing (q )

1000 500

∆Pflowline (q )

0 0

5000

10000 Distance (ft)

Mauricio G. Prado – The University of Tulsa

15000

Equilibrium Flowrate The system components are: reservoir, tubing, flowline, etc... Pr − Ps = ∆Preservoir (q) + ∆Pnear wellbore (q ) + ∆Ptubing (q ) + ∆Pflowline (q ) + ∆Pchoke (q ) + ∆Partificial lift (q )

Press Drop in Porous Media

Press Drop in Pipes and Equipment

Mauricio G. Prado – The University of Tulsa

Equilibrium Flowrate Regrouping the terms: Pr − ∆Preservoir (q) − ∆Pnear wellbore (q) = Ps + ∆Ptubing (q) + ∆Pflowline (q) + ∆Pchoke (q) + ∆Partificial lift (q)

Flow in Porous Media

Flow in Pipes and Equipment

Mauricio G. Prado – The University of Tulsa

Flowrates • The source of the produced fluids is the reservoir. During production, the fluids will flow inside the porous media losing pressure as they flow towards the perforations. • The driving force for the fluids to move inside the porous media is the pressure drop in the reservoir. The fluids will arrive at the mid point of the perforations with a pressure Pwf also called Inflow Bottomhole Pressure.

Pr − ∆Preservoir (q) − ∆Pnear wellbore (q ) = Pwfinflow (q )

Mauricio G. Prado – The University of Tulsa

Equilibrium Flowrate Then: Pr − ∆Preservoir (q) − ∆Pnear wellbore (q) = Ps + ∆Ptubing (q) + ∆Pflowline (q) + ∆Pchoke (q) + ∆Partificial lift (q)

Flow in Porous Media

Pwfinflow (q )

Flow Pipes and Equipment

= Ps + ∆Ptubing (q ) + ∆Pflowline (q ) + ∆Pchoke (q ) + ∆Partificial lift (q )

Mauricio G. Prado – The University of Tulsa

Inflow Performance Relationship • Therefore the rate q of the inflow of fluids from the reservoir into the well is a function of the bottonhole flowing pressure Pwf at the midpoint of the perforations • This function is called: – Inflow Performance Relationship - IPR inflow wf

P

(q)

or

inflow wf

q( P

Mauricio G. Prado – The University of Tulsa

)

Flowrates • After the fluids get into the production system throught the perforations, they will flow through the remaining components of the production system (tubing, flowline, etc…) losing pressure as they flow towards the separator. • The driving force for the fluids to move through those element is the pressure at the bottom of the well. The fluids will require a certain pressure at the mid point of the perforations to flow. This pressure, Pwf is called outflow bottomhole pressure.

Pwfoutflow (q ) = Ps + ∆Ptubing (q) + ∆Pflowline (q ) + ∆Pchoke (q) + ∆Partificial lift (q )

Mauricio G. Prado – The University of Tulsa

Equilibrium Flowrate Then: Pr − ∆Preservoir (q) − ∆Pnear wellbore (q) = Ps + ∆Ptubing (q) + ∆Pflowline (q) + ∆Pchoke (q) + ∆Partificial lift (q)

Flow in Porous Media

Flow Pipes and Equipment

Pr − ∆Preservoir (q ) − ∆Pnear wellbore (q ) = Pwfoutflow (q ) Mauricio G. Prado – The University of Tulsa

Outflow Performance Relationship • Therefore the rate q of the outflow of fluids from the bottom of the well into the separator is a function of the bottonhole flowing pressure Pwf at the midpoint of the perforations • This function is called: – Outflow Performance Relationship – OPR – It is also called System Performance Relationship or Tubing Performance Relationship or Total System Performance Relationship or Tubing Intake Performance.

outflow wf

P

(q)

or

outflow wf

q( P

Mauricio G. Prado – The University of Tulsa

)

Equilibrium Flowrate Then: Pr − ∆Preservoir (q) − ∆Pnear wellbore (q) = Ps + ∆Ptubing (q) + ∆Pflowline (q) + ∆Pchoke (q) + ∆Partificial lift (q)

Flow in Porous Media

inflow wf

P

(q )

Flow Pipes and Equipment

=

outflow wf

P

Mauricio G. Prado – The University of Tulsa

(q )

Equilibrium Flowrate Usually an increase in flowrate causes an increase in the pressure drops inside the tubing and flowline and chokes or restrictions

Pwfoutflow (q ) = Ps + ∆Ptubing (q) + ∆Pflowline (q) + ∆Pchoke (q ) + ∆Partificial lift (q)

An increase in flowrate cause an increse in the pressure drop in the reservoir and near wellbore region

Pwfinflow (q ) = Pr − ∆Preservoir (q ) − ∆Pnear wellbore (q )

Mauricio G. Prado – The University of Tulsa

Equilibrium Flowrate 5000

Pwfo

OPR

4500 4000

Pre ssu re (p si)

3500

Pwf

3000

IPR

2500 2000

qe

1500 1000

i

Pwf

500 0 0

1000

2000

3000

4000

F lo w rate (b p d )

Mauricio G. Prado – The University of Tulsa

5000

6000

Performance Relationships • A well’s Inflow Performance Relationship is a measure of the reservoir ability to produce fluids under an imposed reservoir pressure drop or bottom hole flowing pressure Pwf. • The IPR represents the pressure available at the bottom of the well for a certain flowrate. It represents the pressure available in front of the perforations for the fluids to flow inside the porous media at a certain flowrate • A well’s Outflow Performance Relationship is a measure of the production system requirements to produce fluids under an imposed system pressure drop or bottom hole flowing pressure Pwf • The OPR represents the pressure required at the bottom of the well to flow fluids at a certain flowrate from that location to the surface separator. Mauricio G. Prado – The University of Tulsa

Performance Relationships • The natural equilibrium flowrate is the flowrate at which the Inflow and Outflow Performance Relationships are the same. • The IPR and OPR represent a relationship between the bottonhole flowing pressure and the flowrate. For a certain condition there is only one equilibrium flowrate and only one bottomhole flowing pressure. The equilibrium flowrate is the one that causes the IPR and OPR pressures to be the same. • Proper System design requires – Knowledge of the IPR and OPR at current as well as future reservoir pressure levels and operational conditions.

Mauricio G. Prado – The University of Tulsa

Performance Relationships • The bottomhole flowing pressure is defined as the pressure occurring inside the well at mid perforation depth Pwfoutflow (q ) = Ps + ∆Ptubing (q ) + ∆Pflowline (q ) + ∆Pchoke (q ) + ∆Partificial lift (q )

Pwfinflow (q ) = Pr − ∆Preservoir (q ) − ∆Pnear wellbore (q )

Mauricio G. Prado – The University of Tulsa

Performance Relationships • • • •

The flowrate is usually the flowrate at standard conditions of the fluid of interest For oil wells – Oil flowrate in stock tank conditions For gas wells – Gas flowrate in some standardized condition Sometimes the flowrate can also be the total liquid flowrate (oil and water) but it must be clearly stated which fluid/fluids is being used for the flowrate Flowrate in Standard Conditions for the Fluid of Interest

Bottom hole pressure Mauricio G. Prado – The University of Tulsa

Performance Relationships • The models for solving the balance equations usually require some parameters such as pipe roughness, thermal conductivities, choke discharge coefficient, permeability, etc… • For the production equipment: – Accurate models exist for single phase flow and the characteristic parameters are known or can be easily determined for each component. – For two phase flow, models do exist but experimental closure equations are required (the problem is too complex) and some degree of uncertainty exist in the predictions.

• For the flow in the reservoir and near wellbore region – Accurate models exist for single phase flow but the characteristic parameters such as permeability and porosity distribution are not known and an “averaged effect” must be determined by a well test. – For two phase flow no simple model exist and the predictions rely on correlations or empirical methods

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • We focus now on the prediction of single phase pressure losses inside the reservoir. • In1856 Darcy (1803-1858) performed experiments for the design of sand filters for water purification in France

THE PUBLIC FOUNTAINS OF THE CITY OF DIJON EXPERIENCE AND APPLICATION PRINCIPLES TO FOLLOW AND FORMULAS TO BE USED IN THE QUESTION OF THE DISTRIBUTION OF WATER by HENRY DARCY INSPECTOR GENERAL OF BRIDGES AND ROADS 1856

Mauricio G. Prado – The University of Tulsa

Darcy’s Law •

•

•

•

Henry Darcy's law of fluid flow through porous media forms the basis of hydrogeology. Experiments on water flow through sand led Darcy to formulate the empirical law that he published in1856 as an appendix to his book Les Fontaines publiques de la ville de Dijon. Darcy wrote the book to describe the construction of Dijon's water supply system in 1839–40 and to provide practical guidance to engineers involved in similar projects. Darcy's water supply system transformed Dijon from a pestilential provincial capital to the second-best city in Europe (after Rome) in terms of water supply and quality. As a young Engineer of the Corps of Bridges and Roads assigned to his native city, Darcy gauged nearby springs and selected an abundant spring to divert to Dijon via a 12-km underground aqueduct. In 1839–1840, he built two reservoirs, 13 kilometers of pipes and 115 street fountains in Dijon. These fountains supplied free water for all inhabitants, water for flushing the streets, and water for fire pumps. The book contains 4 parts and an appendix. – Part 1 describes the historical water situation of Dijon and proposals to provide water for its residents. – Part 2 discusses the construction of the aqueduct and the internal distribution system. – Part 3 presents experiments that Darcy conducted on the aqueduct and distribution system. – Part 4 discusses the expropriation of the springs, which belonged to a nearby village. – The appendix contains eight sections on such topics as the water supply systems of London and major French cities, artificial and natural filtration of river water, and pipe making. – Appendix D contains the experiments that Darcy conducted in developing his law of fluid flow through porous media. A separate 28-plate atlas includes Darcy's drawings of the components of the Dijon water supply system, the Pitot tube, and the apparatus Darcy used for the experiments that led to the formulation of Darcy's Law.

Mauricio G. Prado – The University of Tulsa

Darcy Experiment Sand

Viscous Fluid

µ

P + dP

A

q

dx

Mauricio G. Prado – The University of Tulsa

P

Darcy’s Law • After several concluded that:

experiments

Darcy

A dP q is proportional to µ dx k A dP q= µ dx • Where k is defined as the permeability of the porous media. • In 1933 Muskat et al, proposed to measured permeability in a unit called Darcy

Mauricio G. Prado – The University of Tulsa

Darcy’s Law The units of k can be obtained by:

q µ dx k= dP A 3

2

2

L M Lt t L 1 2 k= L =L 2 2 2 t t L MLL • The permeability k has the units of area Mauricio G. Prado – The University of Tulsa

Darcy’s Law • The definition of 1 Darcy is the permeability of a porous media that will allow the flow of 1 cm3/s of a fluid with 1 cp viscosity when the pressure gradient is 1 atm/cm and the flow area is 1 cm2

q µ dx k= dP A cm3 cp cm −13 2 1 Darcy = 1 = 9 . 869 10 m s atm cm 2

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • The value of 1 Darcy by definition is then:

1 D = 9.869 10 −13 m 2 • Curiosity: Human hair thickness ~ 60 microns = 6 x 10-5 m • Human hair cross section area

Hair Cross Section Area =

πd 4

2

=

π (6 10

)

−5 2

4

= 28.27 10 −10 m 2

Hair Cross Section Area = 2865 Darcy

Mauricio G. Prado – The University of Tulsa

Darcy’s Law

k A dP q= µ dx q A

µ

- Liquid flow rate - Cross Section Area - Liquid Viscosity

dP - Pressure Gradient dx Mauricio G. Prado – The University of Tulsa

Darcy’s Law • Darcy’s law is in reality an expression for the steady state momentum conservation equations (or pressure losses) through the porous media channels. • This can be verified by comparing Darcy’s equation with a simplified model for laminar flow in the porous media based on momentum balance equations for pipes.

Mauricio G. Prado – The University of Tulsa

Darcy’s Law and Laminar Flow in Pipes • Darcys Law: Sand

Viscous Fluid

µ

P + dP

A

q

P

dx ft 2

mD bpd

k A dP q= µ dx

psi/ft

0.001127 k A dP q= µ dx cp Mauricio G. Prado – The University of Tulsa

Darcy’s Law and Laminar Flow in Pipes • Laminar Flow in Pipes:

P + dP q bbl/d

P

d

dl in psi/ft

1 d 4 dP q= 7.9628 10 −6 µ dl cp bbl/d

ft psi/ft

12 4 d 4 dP q= 7.9628 10 −6 µ dl cp Mauricio G. Prado – The University of Tulsa

Darcy’s Law and Laminar Flow in Pipes • Darcy’s Law and the Laminar flow in pipes are described by similar equations ft 2

mD bpd

psi/ft

0.001127 k A dP q= µ dx

Darcy’s Law

cp

ft

bbl/d Laminar Flow in Pipes

psi/ft

12 4 d 4 dP q= 7.9628 10 −6 µ dx cp Mauricio G. Prado – The University of Tulsa

Darcy’s Law • Lets obtain a version of Darcy’s in cylindrical coordinates system.

Mauricio G. Prado – The University of Tulsa

P

rw

dr

r

r

q A h

Mauricio G. Prado – The University of Tulsa

Darcy’s Law

k A dP q= µ dr q=

A = 2π r h

2π k h

µ

dP r dr

Mauricio G. Prado – The University of Tulsa

Darcy’s Law

q=

2π k h

µ

dP r dr

• This ordinary differential equation (ODE) represents the steady stage momentum balance equation for the pressure losses in the porous media and must be solved in conjunction with a mass balance equation Mauricio G. Prado – The University of Tulsa

Darcy’s Law • In field units mD

ft

bpd psi/ft

kh

dP q = 0.00708 r µ dr cp

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • In Darcys equation, the flowrate is the flowrate at in-situ conditions. • Even for low compressibility fluids, the difference between the fluid volumes at reservoir condition and the fluid volumes at the surface separator conditions (where fluids are measured and sold) needs to be considered. • This is done taking into account the fluid compressibility or the fluid formation volume factor B to convert from in-situ conditions to surface or standard conditions • This may be important for two reasons: – The standard conditions or surface conditions flowrate is a measure of mass rate and may help in solving the mass balance equation – We are interested in obtaining the IPR which will relate the bottomhole pressure available for producing a certain amount of fluids on the surface that can be sold. Mauricio G. Prado – The University of Tulsa

Darcy’s Law • The fluid compressibility is defined as: 1 ∂V c=− V ∂P

c=

1 ∂ρ ρ ∂P

• The formation volume factor is defined as: Vin − situ B= Vstd

• As a consequence: c=−

1 ∂B B ∂P

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • Two important cases exist: • Liquids – Constant compressibility 1 ∂B c=− B ∂P

B=e

− c ( P − Pstd )

c=

1 ∂ρ ρ ∂P

ρ = ρ std e c ( P − P

std

• Ideal gases 1 dB c=− B dP

P B = std P

Pstd = PB

Pstd ρ std = Pρ

ρ=

Pstd ρ std P

Mauricio G. Prado – The University of Tulsa

c=

1 P

)

Darcy’s Law • Then Darcy’s law can be modified to: mD

ft

std bpd psi/ft

qstd

k h dP = 0.00708 r µ B dr cp

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • The single phase reservoir model consists of a transient mass balance equation and the steady state momentum balance equation represented by Darcy’s Law:

∂ qρ 1 ⎧ ∂ρ ⎪ ∂t = 2 π φ r h ∂r ⎪ ⎨ ⎪ q = 2 π k h r ∂P ⎪⎩ ∂r µ • This system must be solved with the proper initial condition as well as boundary conditions.

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • Reservoir Engineers are familiar with the Diffusivity equation which can be obtained by substituting Darcy law in the mass balance equation:

1 ∂ qρ ∂ρ = ∂t 2 π φ r h ∂r

2 π k h r ∂P q= µ ∂r

∂ρ ∂ ⎛ 2 π k h rρ ∂P ⎞ 1 ⎜⎜ ⎟⎟ = ∂t 2 π φ r h ∂r ⎝ µ ∂r ⎠ ∂ρ 1 ∂ ⎛ k rρ ∂P ⎞ ⎜⎜ ⎟⎟ φ = ∂t r ∂r ⎝ µ ∂r ⎠ Mauricio G. Prado – The University of Tulsa

Darcy’s Law • But from the compressibility we have: 1 ∂ρ c= ρ ∂P

1 ∂ρ ∂t c= ρ ∂t ∂P

∂ρ ∂P = cρ ∂t ∂t

∂ρ 1 ∂ ⎛ k rρ ∂P ⎞ ⎜⎜ ⎟⎟ φ = ∂t r ∂r ⎝ µ ∂r ⎠

∂P 1 ∂ ⎛ k rρ ∂P ⎞ ⎜⎜ ⎟⎟ φ cρ = ∂t r ∂r ⎝ µ ∂r ⎠

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • The diffusivity equation becomes then:

∂P 1 ∂ ⎛ k rρ ∂P ⎞ ⎜⎜ ⎟⎟ φ cρ = ∂t r ∂r ⎝ µ ∂r ⎠

psi-1

mD

psi

∂P 1 ∂ ⎛ k rρ ∂P ⎞ ⎜⎜ ⎟⎟ = 162.6 φ cρ ∂t r ∂r ⎝ µ ∂r ⎠ lb/ft3 day

ft cp

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • Once the diffusivity equation is solved yielding a solution for the pressure as a function of time and radial position, Darcy’s law can be used to find a solution for the flowrate as a function of time and radial position. Finally the solution at the wellbore yields the bottomhole pressure and sandface production as a function of time

φ cρ

∂P 1 ∂ ⎛ k rρ ∂P ⎞ ⎜⎜ ⎟⎟ = ∂t r ∂r ⎝ µ ∂r ⎠

P(r , t )

q=

2 π k h r ∂P µ ∂r

q(r , t ) Mauricio G. Prado – The University of Tulsa

Pwf (t )

qsf (t )

Liquid Diffusivity Equation for Homogeneous Reservoirs • For liquid the compressibility can be considered constant and for homogeneous reservoir we have:

162.6 φ cρ

162.6

162.6

∂P 1 ∂ ⎛ k rρ ∂P ⎞ ⎜⎜ ⎟⎟ = ∂t r ∂r ⎝ µ ∂r ⎠

φ cρ µ ∂P k

∂t

=

1 ∂ ⎛ ∂P ⎞ ⎜ rρ ⎟ r ∂r ⎝ ∂r ⎠

φ cρ µ ∂P

162.6

1 ⎛ ∂ ⎛ ∂P ⎞ ∂P ⎛ ∂ρ ⎞⎞ = ⎜⎜ rρ ⎜ ⎟ + r ρ + ⎜ ⎟ ⎟⎟ ∂t r ⎝ ∂r ⎝ ∂r ⎠ ∂r ⎝ ∂r ⎠⎠

k

φ cρ µ ∂P k

1 ⎛ ∂ ⎛ ∂P ⎞ ∂P ⎛ ∂P ⎞⎞ = ⎜⎜ rρ ⎜ ⎟ + rc ρ ρ + ⎜ ⎟ ⎟⎟ ∂t r ⎝ ∂r ⎝ ∂r ⎠ ∂r ⎝ ∂r ⎠⎠ Mauricio G. Prado – The University of Tulsa

c=

1 ∂ρ ρ ∂P

Liquid Diffusivity Equation for Homogeneous Reservoirs • For liquid the compressibility can be considered constant and for homogeneous reservoir we have:

162.6

φ cρ µ ∂P k

162.6

162.6

1 ⎛ ∂ ⎛ ∂P ⎞ ∂P ⎛ ∂P ⎞⎞ = ⎜⎜ rρ ⎜ ⎟ + + ρ ⎟ ⎟⎟ ⎜ rcρ ∂t r ⎝ ∂r ⎝ ∂r ⎠ ∂r ⎝ ∂r ⎠⎠

φ c µ ∂P k

1 ⎛ ∂ ⎛ ∂P ⎞ ∂P ⎛ ∂P ⎞⎞ = ⎜⎜ r ⎜ ⎟ + rc 1 + ⎜ ⎟ ⎟⎟ ∂t r ⎝ ∂r ⎝ ∂r ⎠ ∂r ⎝ ∂r ⎠⎠

1 ⎛⎜ ∂ ⎛ ∂P ⎞ ∂P ⎛ ∂P ⎞ + rc⎜ ⎟ = r ⎜ ⎟+ ∂t r ⎜⎝ ∂r ⎝ ∂r ⎠ ∂r ⎝ ∂r ⎠

φ c µ ∂P k

162.6

φ c µ ∂P k

1 ⎛ ∂ ⎛ ∂P ⎞ ∂P ⎞ = ⎜⎜ r ⎜ ⎟ + ⎟⎟ ∂t r ⎝ ∂r ⎝ ∂r ⎠ ∂r ⎠

Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟ ⎟ ⎠

Liquid Diffusivity Equation for Homogeneous Reservoirs • Finally the diffusivity equation becomes:

162.6

φ c µ ∂P k

1 ⎛ ∂ ⎛ ∂P ⎞ ∂P ⎞ = ⎜⎜ r ⎜ ⎟ + ⎟⎟ ∂t r ⎝ ∂r ⎝ ∂r ⎠ ∂r ⎠

cp

psi-1

162.6

mD

psi

φ c µ ∂P k

1 ⎛ ∂ ⎛ ∂P ⎞ ⎞ = ⎜⎜ ⎜ r ⎟ ⎟⎟ ∂t r ⎝ ∂r ⎝ ∂r ⎠ ⎠ day

ft

Mauricio G. Prado – The University of Tulsa

Liquid Diffusivity Equation for Homogeneous Reservoirs • In this case the mass balance equation can also be modified: ∂ρ ∂ qρ 1 = ∂t 2 π φ r h ∂r 1 ∂ρ ∂q⎞ ⎛ ∂ρ = +ρ ⎟ ⎜q ∂t 2 π φ r h ⎝ ∂r ∂r ⎠ 1 ∂ρ ∂P ∂q⎞ ⎛ ∂ρ ∂P ρ = q + ⎟ ⎜ ∂P ∂t 2 π φ r h ⎝ ∂P ∂r ∂r ⎠ cρ

c

1 ∂P ∂P ∂q⎞ ⎛ ρ ρ = c q + ⎟ ⎜ ∂t 2 π φ r h ⎝ ∂r ∂r ⎠ 1 ∂P ⎛ ∂P ∂q⎞ = + ⎟ ⎜c q ∂t 2 π φ r h ⎝ ∂r ∂r ⎠ Mauricio G. Prado – The University of Tulsa

c=

1 ∂ρ ρ ∂P

Liquid Diffusivity Equation for Homogeneous Reservoirs • In this case the mass balance equation can also be modified:

c

1 ∂P ⎛ ∂P ∂q⎞ = + ⎟ ⎜c q ∂r ∂r ⎠ ∂t 2 π φ r h ⎝

psi

psi-1

c

bpd

∂P 0.8935 ⎛ ∂ P ∂ q ⎞ = + ⎟ ⎜c q φrh ⎝ ∂t ∂r ∂r ⎠

day

ft

Mauricio G. Prado – The University of Tulsa

Darcy’s Law • In principle, an infinite number of solutions to the diffusivity equations can be obtained depending on initial and boundary conditions imposed. – Initial conditions refer to the pressure and flowrate distribution inside the porous media at a certain moment. – Boundary conditions refer to the conditions both at the wellbore as well as at the edge of the reservoir. •

The BC at the reservoir edger could be: – Infinite reservoir – Sealed reservoir (no flow at the edge) – Constant pressure (pressure maintenance at the edge)

• The BC at the wellbore could be: – Constant bottomhole pressure – Constant flowrate – Relationship between the flowrate and bottomhole pressure (OPR)

Mauricio G. Prado – The University of Tulsa

Darcy’s Law •

Lets imagine a reservoir at a certain initial state where a disturbance is created. Examples of this case could be: – Opening a well to produce at a constant flowrate or constant bottomhole pressure. – Changing the well production conditions (modifying the flowrate or modifying the bottomhole flowing pressure) – Shutting in a well.

• • •

In all those cases there is a period of time, called transient, where the pressure response in the reservoir is not affected by the presence of the outer boundaries, since the perturbation created at the wellbore is still traveling across the well. In this condition, the well appears to be infinite in extent. This condition applies for a small period of time and mainly occur when wellbore conditions are deliberately changed. – Opening a well – Changing operational conditions (chokes, artificial lift) – Well testing disturbances

Pr

Pwf rw Mauricio G. Prado – The University of Tulsa

rr

Darcy’s Law • •

At a certain point in time after the disturbance is generated, the pressure response will start to feel the effects of the outer boundary. The outer boundary can be: – Closed boundary – representing a sealed reservoir – Constant pressure – representing a water influx or water injection.

•

• •

In reality the conditions at the wellbore (flowrate or pressure) may be constantly slowly changing with time even for a stable well. But those changes do not compare with disturbances created by well testing operations or changes in choke opening or artificial lift parameters to produce more or less fluids. For practical purposes we can assume that during normal stable operation the conditions at the wellbore are reasonably constant. When operational conditions (choke or artificial lift) are changed we need to estimate the time it will take for the transitent period to die out and for the well to stabilize again. For those conditions two other solutions for the diffusivity equation are possible – Semi Steady State condition for a sealed reservoir – Steady State condition for a constant pressure at the edge.

•

In summary the solutions to the diffusivity equation can be goruped in: – Transient condition – Stabilized Flow conditions • Semi Steady State • Steady State Mauricio G. Prado – The University of Tulsa

Transient Condition • This conditions is applicable to a relatively short period after some disturbance has been created. • The condition is mainly applied to the analysis of well test in which the wellbore condition is changed and the pressure and flworate response is measured and analyzed. Comparison of the measured values with a model can help in determining key reservoir parameters. • This condition can also be used to determine the transient IPR • The mathematical solution to the diffusivity equation is very complex.

Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition • Recall that this is an approximation for the stabilized flow condition and we can assume that the well flowrate is reasonably constant. Also we have no flow at the edge of the reservoir. Then ⎧ φ c µ ∂P 1 ⎛ ∂ ⎛ ∂P ⎞ ⎞ = ⎜⎜ ⎜ r ⎟ ⎟⎟ ⎪162.6 ∂ ∂ ∂ k t r r r ⎠⎠ ⎝ ⎝ ⎪ ⎪ ⎨q (rw , t ) = cte ⎪ ∂P ⎪ =0 r ∂ ⎪⎩ rr

Pr

P

Pwf rw

rr Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

Since the well is producing at a constant flowrate, the following mass balance can be written for changes in the reservoir average pressure

c=−

1 ∂V V ∂P

c

1 ∂V dP =− dt V ∂t

bpd

psi

dP q =− = cte dt cπ re2 − rw2 hφ

(

dP q = −1.787 2 dt c re hφ

)

day

ft

Pr P

Pwf rw

rr Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

It can be shown that for this condition: ∂P ∂P = cte = ∂t ∂t

•

Which means that the pressure profiles are parallel.

Pr

P

Pwf

rr

rw Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

Then the diffusivity equation becomes:

⎧ φ c µ ∂P 1 ⎛ ∂ ⎛ ∂P ⎞ ⎞ = ⎜⎜ ⎜ r ⎟ ⎟⎟ ⎪ ⎪ k ∂t r ⎝ ∂r ⎝ ∂r ⎠ ⎠ ⎪ ⎨q (rw , t ) = cte ⎪ ∂P ⎪ =0 ⎪⎩ ∂r rr

⎧ 1 ⎛ ∂ ⎛ ∂P ⎞ ⎞ qµ r ⎜ ⎟ = − ⎟⎟ ⎪ ⎜ ⎜ 2 r r r r hk ∂ ∂ π ⎝ ⎠ ⎝ ⎠ r ⎪ ⎪ ⎨q (rw , t ) = cte ⎪ ∂P ⎪ =0 ⎪⎩ ∂r rr

Mauricio G. Prado – The University of Tulsa

dP q =− dt cπ rr2 hφ

Semi Steady State Condition •

The solution then becomes

qµ 1 ⎛ ∂ ⎛ ∂P ⎞ ⎞ ⎜⎜ ⎜ r ⎟ ⎟⎟ = − 2 r ⎝ ∂r ⎝ ∂r ⎠ ⎠ π rr h k qµ ∂ ⎛ ∂P ⎞ ⎟=− 2 r ⎜r π rr h k ∂r ⎝ ∂r ⎠ r •

qµ ∂P 2 =− r +A 2 2π rr h k ∂r

Using the outer boundary condition

0=−

qµ 2 r r + A 2π rr2 h k

A=

qµ 2π h k

Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

The solution then becomes

∂P q µ ⎛ r2 ⎞ ⎜⎜1 − 2 ⎟⎟ = r ∂r 2π h k ⎝ rr ⎠ ∂P qµ ⎛1 r ⎞ ⎜⎜ − 2 ⎟⎟ = ∂r 2π h k ⎝ r rr ⎠ qµ ⎛ r2 ⎞ ⎜⎜ ln(r ) − 2 ⎟⎟ + B P= 2π h k ⎝ 2rr ⎠ •

Using the inner boundary condition

rw2 ⎞ qµ ⎛ ⎜⎜ ln(rw ) − 2 ⎟⎟ + B Pwf = 2π h k ⎝ 2rr ⎠

rw2 ⎞ qµ ⎛ ⎜⎜ ln(rw ) − 2 ⎟⎟ B = Pwf − 2π h k ⎝ 2rr ⎠

Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

Then finally:

P=

r ⎞ qµ ⎛ ⎜⎜ ln(r ) − 2 ⎟⎟ + B 2π h k ⎝ 2rr ⎠ 2

rw2 ⎞ qµ ⎛ ⎜⎜ ln(rw ) − 2 ⎟⎟ B = Pwf − 2π h k ⎝ 2rr ⎠

rw2 ⎞⎟ q µ ⎛⎜ ⎛ r ⎞ r 2 P = Pwf + ln⎜⎜ ⎟⎟ − 2 + 2 ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 2rr 2rr ⎠ •

Then pressure at the edge of the reservoir can be calculated as:

q µ ⎛⎜ ⎛ rr ⎞ 1 rw2 ⎞⎟ Pr = Pwf + ln⎜⎜ ⎟⎟ − + 2 ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 2 2rr ⎠ q µ ⎛⎜ ⎛ rr ⎞ 1 ⎞⎟ Pr = Pwf + ln⎜⎜ ⎟⎟ − ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 2 ⎠ Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

The reservoir average pressure is given by: rr

P=

•

∫ P(r ) 2π hφ rdr

rw

π (rr2 − rw2 )hφ

r

r 2 P ≈ 2 2 ∫ P (r ) rdr rr − rw rw

Then:

2 P≈ 2 2 rr − rw

2 P≈ 2 2 rr − rw

2 ⎛ ⎛ ⎛ r ⎞ r2 ⎞⎞ r q µ w ⎜ Pwf + ⎜ ln⎜ ⎟ − 2 + 2 ⎟ ⎟ rdr ∫r ⎜ ⎜ ⎜ r ⎟ 2r h k 2 π 2rr ⎟⎠ ⎟⎠ r ⎝ ⎝ w⎠ w⎝ rr

2 ⎛ ⎛ ⎛ r ⎞ r2 ⎞⎞ r q µ w ⎜ Pwf + ⎜ ln⎜ ⎟ − 2 + 2 ⎟ ⎟ rdr ∫r ⎜ ⎜ ⎜ r ⎟ 2r h k 2 π 2rr ⎟⎠ ⎟⎠ r ⎝ ⎝ w⎠ w⎝ rr

Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

The reservoir average pressure is given by:

2 P≈ 2 2 rr − rw

2 ⎛ ⎛ ⎛ r ⎞ r2 ⎞⎞ r q µ w ⎜ Pwf + ⎜ ln⎜ ⎟ − 2 + 2 ⎟ ⎟ rdr ∫r ⎜ ⎜ ⎜ r ⎟ 2r h k 2 π 2rr ⎟⎠ ⎟⎠ r ⎝ ⎝ w⎠ w⎝ rr

⎛ ⎛ 2 2 ⎞⎞ 2 ⎛ rr ⎞ ⎜ ⎜ rw − rr + 2rr ln⎜ ⎟ ⎟⎟ ⎟ ⎜ 2 2 4 4 2 2 2 2 ⎜ ⎛ r − rw ⎞ qµ ⎜ ⎝ rw ⎠ − rr − rw + rr − rw rw ⎟ ⎟ ⎟⎟ + P ≈ 2 2 ⎜ Pwf ⎜⎜ r ⎟⎟ 4 8rr2 4rr2 rr − rw ⎜ ⎝ 2 ⎠ 2π h k ⎜⎜ ⎟⎟ ⎜ ⎟⎟ ⎜ ⎝ ⎠⎠ ⎝

(

⎛ ⎛ ⎞⎞ 2 ⎛ rr ⎞ ⎜ ⎜ ⎟⎟ 2rr ln⎜⎜ ⎟⎟ 2 2 2 ⎜ rw ⎠ rr + rw rw ⎟ ⎟ qµ ⎜ 1 ⎝ P ≈ ⎜ Pwf + ⎜ − 4 + 4 r 2 − r 2 − 8r 2 + 4 r 2 ⎟ ⎟ π h k r w r r ⎟⎟ ⎜ ⎜ ⎜ ⎟⎟ ⎜ ⎝ ⎠⎠ ⎝

(

)

Mauricio G. Prado – The University of Tulsa

)

Semi Steady State Condition •

And Finally:

⎞ ⎛ 2 ⎛ rr ⎞ ⎟ ⎜ 2rr ln⎜⎜ ⎟⎟ 2 2 2 qµ ⎜ 1 ⎝ rw ⎠ − rr + rw + rw ⎟ P ≈ Pwf + − + π h k ⎜⎜ 4 4 rr2 − rw2 8rr2 4rr2 ⎟⎟ ⎟ ⎜ ⎠ ⎝

(

P = Pwf +

)

q µ ⎛⎜ ⎛ rr ⎞ 3 ⎞⎟ ln⎜⎜ ⎟⎟ − ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 4 ⎠

Mauricio G. Prado – The University of Tulsa

Semi Steady State Condition •

In summary for the semi steady state condition we have:

q µ ⎛⎜ ⎛ rr ⎞ 1 ⎞⎟ Pr = Pwf + ln⎜⎜ ⎟⎟ − ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 2 ⎠

⎛ ⎛ rr ⎞ 1 ⎞ qµ ⎜ ln⎜ ⎟ − ⎟ Pwf = Pr − 0.00708 h k ⎜⎝ ⎜⎝ rw ⎟⎠ 2 ⎟⎠

q µ ⎛⎜ ⎛ rr ⎞ 3 ⎞⎟ P = Pwf + ln⎜⎜ ⎟⎟ − ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 4 ⎠

Pwf = P −

Mauricio G. Prado – The University of Tulsa

⎛ ⎛ rr ⎞ 3 ⎞ qµ ⎜ ln⎜ ⎟ − ⎟ 0.00708 h k ⎜⎝ ⎜⎝ rw ⎟⎠ 4 ⎟⎠

Steady State IPR • For the steady state solution which is the solution for a bounded reservoir with constant pressure at the edge.

Pr

Pr

Pwf rw

rr

Mauricio G. Prado – The University of Tulsa

Steady State IPR •

Then we have the diffusivity equation as:

2 π k h r ∂P ⎧ ⎪q= ∂r µ ⎨ ⎪P = P @ r r r ⎩

∂P qµ 1 = ∂r 2 π k h r

qµ P= ln(r ) + A 2π k h

Mauricio G. Prado – The University of Tulsa

Steady State IPR •

Using the boundary condition

Pr =

qµ ln(rr ) + A 2π k h

A = Pr −

qµ ln (rr ) 2π k h

qµ P= ln(r ) + A 2π k h P=

⎛r⎞ qµ ln⎜⎜ ⎟⎟ + Pr 2 π k h ⎝ rr ⎠

P = Pr −

qµ ⎛r ⎞ ln⎜ r ⎟ 2π k h ⎝ r ⎠

Mauricio G. Prado – The University of Tulsa

Steady State IPR •

The IPR can be written as: Pwf = Pr −

•

⎛r ⎞ qµ ln⎜⎜ r ⎟⎟ 2 π k h ⎝ rw ⎠

And the pressure inside the reservoir can also be written as: P = Pr −

qµ ⎛r ⎞ ln⎜ r ⎟ 2π k h ⎝ r ⎠

P = Pwf −

qµ ⎛r ⎞ ln⎜ w ⎟ 2π k h ⎝ r ⎠

Mauricio G. Prado – The University of Tulsa

Steady State IPR •

The reservoir average pressure is given by: rr

P=

•

∫ P(r ) 2π hφ rdr

rw

r

r 2 P ≈ 2 2 ∫ P (r ) rdr rr − rw rw

π (rr2 − rw2 )hφ

Then:

2 P≈ 2 2 rr − rw

⎛ ⎛ ⎞⎞ ⎜ Pwf + q µ ln⎜ r ⎟ ⎟ rdr ∫⎜ 2π h k ⎜⎝ rw ⎟⎠ ⎟⎠ rw ⎝ rr

Mauricio G. Prado – The University of Tulsa

Steady State IPR •

The reservoir average pressure is given by:

2 P≈ 2 2 rr − rw

⎛ ⎛ ⎞⎞ ⎜ Pwf + q µ ln⎜ r ⎟ ⎟ rdr ∫⎜ 2π h k ⎜⎝ rw ⎟⎠ ⎟⎠ rw ⎝ rr

⎛ ⎛ 2 2 ⎛ ⎞ ⎞⎞ ⎜ ⎜ rw − rr + 2rr2ln⎜ rr ⎟ ⎟ ⎟ ⎜ r ⎟ ⎟⎟ qµ ⎜ 2 ⎜ ⎛ rr2 − rw2 ⎞ ⎝ w⎠ ⎟⎟ + P ≈ 2 2 ⎜ Pwf ⎜⎜ ⎟⎟ ⎜ rr − rw ⎜ ⎝ 2 ⎠ 2π h k ⎜ 4 ⎟⎟ ⎟⎟ ⎜ ⎜ ⎠⎠ ⎝ ⎝ ⎛ ⎞⎞ ⎛ 2 ⎛ rr ⎞ ⎟ ⎜ ⎜ 2rr ln⎜⎜ ⎟⎟ ⎟ ⎜ qµ ⎜ 1 ⎝ rw ⎠ ⎟ ⎟ P ≈ ⎜ Pwf + − + ⎜ 4 4 r2 − r2 ⎟⎟ π h k r w ⎟⎟ ⎜ ⎜ ⎟⎟ ⎜ ⎜ ⎠⎠ ⎝ ⎝

(

Mauricio G. Prado – The University of Tulsa

)

Steady State IPR •

And Finally:

⎞ ⎛ 2 ⎛ rr ⎞ ⎜ 2rr ln⎜⎜ ⎟⎟ ⎟ qµ ⎜ 1 ⎝ rw ⎠ ⎟ P ≈ Pwf + − + π h k ⎜⎜ 4 4 rr2 − rw2 ⎟⎟ ⎟ ⎜ ⎠ ⎝

(

P = Pwf +

)

q µ ⎛⎜ ⎛ rr ⎞ 1 ⎞⎟ ln⎜⎜ ⎟⎟ − ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 2 ⎠

Mauricio G. Prado – The University of Tulsa

Steady State IPR •

In summary for the steady state condition we have:

⎛ rr ⎞ qµ ln⎜⎜ ⎟⎟ Pr = Pwf + 2 π k h ⎝ rw ⎠

Pwf = Pr −

⎛r ⎞ qµ ln⎜⎜ r ⎟⎟ 0.00708 h k ⎝ rw ⎠

q µ ⎛⎜ ⎛ rr ⎞ 1 ⎞⎟ P = Pwf + ln⎜⎜ ⎟⎟ − ⎟ ⎜ 2π h k ⎝ ⎝ rw ⎠ 2 ⎠

Pwf = P −

Mauricio G. Prado – The University of Tulsa

⎛ ⎛ rr ⎞ 1 ⎞ qµ ⎜ ln⎜ ⎟ − ⎟ 0.00708 h k ⎜⎝ ⎜⎝ rw ⎟⎠ 2 ⎟⎠

Steady State IPR •

Finally in field units:

psi

bpd

cp

µq

⎛ rr ⎞ P (r ) = Pr − ln⎜ ⎟ 0.00708 k h ⎝ r ⎠ ft mD

Mauricio G. Prado – The University of Tulsa

Steady State IPR •

Finally in field units:

bpd

cp

psi/ft

dP qµ 1 = dr 0.00708 k h r mD

Mauricio G. Prado – The University of Tulsa

ft

Darcy’s Law µq

⎛ rr ⎞ P (r ) = Pr − ln⎜ ⎟ 0.00708 k h ⎝ r ⎠

6000

Pressure (psi)

5000 4000 3000 2000 1000

Reservoir Pressure

5000

psi

Viscosity

8

cp

Flowrate

2000

bpd

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

0 0

200

400

600

800

Radial Position (ft)

Mauricio G. Prado – The University of Tulsa

1000

1200

Darcy’s Law

6000

Pressure (psi)

5000 4000

Reservoir Pressure

5000

psi

Viscosity

8

cp

Flowrate

2000

bpd

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

3000 2000

µq

⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 0.00708 k h ⎝ r ⎠

1000 0 0

20

40

60

Radial Position (ft) Mauricio G. Prado – The University of Tulsa

80

100

Darcy’s Law dP qµ 1 = dr 0.00708 k h r

Radial Pressure Gradient (psi/ft)

1400 1200 1000 800 600 400

Reservoir Pressure

5000

psi

Viscosity

8

cp

Flowrate

2000

bpd

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

200 0 0.1

1

10 Radial Position (ft)

Mauricio G. Prado – The University of Tulsa

100

1000

Darcy’s Law µq

Pr

⎛ rr ⎞ P (r ) = Pr − ln⎜ ⎟ 0.00708 k h ⎝ r ⎠

6000

Pressure (psi)

5000 4000 3000

i wf

2000

P

1000

Reservoir Pressure

5000

psi

Viscosity

8

cp

Flowrate

2000

bpd

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

0 0

200

400

600

800

Radial Position (ft)

Mauricio G. Prado – The University of Tulsa

1000

1200

Darcy’s Law Well Centerline Sand Face

3000

µq

⎛ rr ⎞ P (r ) = Pr − ln⎜ ⎟ 0.00708 k h ⎝ r ⎠

Pressure (psi)

2600

2200

1800

Pwfi

1400

Reservoir Pressure

5000

psi

Viscosity

8

cp

Flowrate

2000

bpd

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

rw 1000 0

2

4

6

Radial Position (ft) Mauricio G. Prado – The University of Tulsa

8

10

Darcy’s Law µq

⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 0.00708 k h ⎝ r ⎠

Flowrate (bpd)

6000

1000 1500

Pressure (psi)

5000 4000

2000

3000 2500

2000 1000

Reservoir Pressure

5000

psi

Viscosity

8

cp

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

0 0

200

400

600

800

Radial Position (ft)

Mauricio G. Prado – The University of Tulsa

1000

1200

Darcy’s Law µq

7000

⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 0.00708 k h ⎝ r ⎠

Reservoir Pressure (psi) 6000

6000

5000

5500

4500

Pressure (psi)

5000 4000 3000 2000 1000

Flowrate

2000

bpd

Viscosity

8

cp

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

0 0

200

400

600

800

Radial Position (ft)

Mauricio G. Prado – The University of Tulsa

1000

1200

Darcy’s Law µq

⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 0.00708 k h ⎝ r ⎠

Fluid Viscosity (cp)

6000 8

Pressure (psi)

5000

6

4

10

4000 3000 2000 1000

Reservoir Pressure

5000

psi

Flowrate

2000

bpd

Permeability

500

mD

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

0 0

200

400

600

800

Radial Position (ft)

Mauricio G. Prado – The University of Tulsa

1000

1200

Darcy’s Law µq

⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 0.00708 k h ⎝ r ⎠

Permeability (mD)

6000 500

800

400

5000

Pressure (psi)

600

4000 3000 2000 1000

Reservoir Pressure

5000

psi

Flowrate

2000

bpd

Fluid Viscosity

8

cp

Thickness

10

ft

Well Diameter

9

in

Reservoir Radius

1000

ft

0 0

200

400

600

800

Radial Position (ft) Mauricio G. Prado – The University of Tulsa

1000

1200

Darcy’s Law µq

⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 0.00708 k h ⎝ r ⎠

Thickness (ft)

6000 20

50

30

10

Pressure (psi)

5000 4000 3000 2000 1000

Reservoir Pressure

5000

psi

Flowrate

2000

bpd

Fluid Viscosity

8

cp

Permeability

500

mD

Well Diameter

9

in

Reservoir Radius

1000

ft

0 0

200

400

600

800

Radial Position (ft) Mauricio G. Prado – The University of Tulsa

1000

1200

Darcy’s Law µq

Reservoir Radius (ft)

6000

1000

500

2000

3000

⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 0.00708 k h ⎝ r ⎠

Pressure (psi)

5000 4000 3000 2000 1000

Reservoir Pressure

5000

psi

Flowrate

2000

bpd

Fluid Viscosity

8

cp

Permeability

500

mD

Well Diameter

9

in

Thickness

10

ft

0 0

500

1000

1500

2000

Radial Position (ft) Mauricio G. Prado – The University of Tulsa

2500

3000

3500

Steady State IPR •

And the IPR can be written as:

psi

bpd

cp

⎡ ⎛ rr ⎞ 1 ⎤ Pwf = P − ⎢ln⎜⎜ ⎟⎟ − ⎥ 0.00708 k h ⎣ ⎝ rw ⎠ 2 ⎦

µ qsc B

mD

Mauricio G. Prado – The University of Tulsa

ft

Summary of Stabilized Flow Equations •

Circular Steady State Flow ⎡ ⎛ rr ⎞ 1 ⎤ Pwf = P − ⎢ln⎜⎜ ⎟⎟ − ⎥ 0.00708 k h ⎣ ⎝ rw ⎠ 2 ⎦

µ qsc B

•

Circular Pseudo Steady State Flow: ⎛ ⎛ rr ⎞ 3 ⎞ ⎜ ln⎜ ⎟ − ⎟ Pwf = P − 0.00708 k h ⎜⎝ ⎜⎝ rw ⎟⎠ 4 ⎟⎠

µ qsc B

Mauricio G. Prado – The University of Tulsa

Pseudo Steady State Flow in Irregular Drainage Areas • •

•

Rarely wells will drain from a regularly shaped area. The drainage area is usually distorted by the presence of natural boundaries or because of lopsided production rates in adjoining wells. The drainage area is then shaped by the production duty of a particular well in a field.

Mauricio G. Prado – The University of Tulsa

Pseudo Steady State Flow in Irregular Drainage Areas •

Circular Pseudo Steady State Flow ⎛ ⎛ rr ⎞ 3 ⎞ ⎜ ln⎜ ⎟ − ⎟ Pwf = P − 0.00708 k h ⎜⎝ ⎜⎝ rw ⎟⎠ 4 ⎟⎠

µ qsc B

•

But

⎛ ⎞ ⎛ ⎞ ⎛ 34 ⎞ 1 ⎛ rr2 ⎞ 1 ⎛ 32 ⎞ 1 ⎜ rr2 ⎟ 1 ⎜ 4π rr2 ⎟ ⎛ rr ⎞ 3 1 ⎛ rr2 ⎞ 3 1 ⎛ rr2 ⎞ ln⎜⎜ ⎟⎟ − = ln⎜⎜ 2 ⎟⎟ − = ln⎜⎜ 2 ⎟⎟ − ln⎜⎜ e ⎟⎟ = ln⎜⎜ 2 ⎟⎟ − ln⎜⎜ e ⎟⎟ = ln⎜ 3 ⎟ = ln⎜ 3 ⎟ 4 2 4 2 2 r 2 2 2 r r r ⎝ w⎠ ⎝ w⎠ ⎝ w⎠ ⎝ w⎠ ⎜ e2r2 ⎟ ⎜ e 2 4π r 2 ⎟ ⎝ ⎠ ⎝ ⎠ w ⎠ ⎝ w⎠ ⎝ ⎛ ⎞ ⎛ rr ⎞ 3 1 ⎜ 4 Adrainage ⎟ ln⎜⎜ ⎟⎟ − = ln⎜ 3 ⎟ r 4 2 ⎝ w⎠ ⎜ e 2 4π r 2 ⎟ w ⎠ ⎝

Mauricio G. Prado – The University of Tulsa

Pseudo Steady State Flow in Irregular Drainage Areas •

Introducing a Dietz shape factor and the Euler constant 3 2

γ = 1.78

e 4π = 56.32 = γ C A

•

Then we have: 1 ⎛ 4 Adrainage ⎞ ⎟ Pwf = P − ln⎜⎜ 2 ⎟ 0.00708 k h 2 ⎝ γ C A rw ⎠

µ qsc B

•

And the shape factor for a circular area is C ACircle =

56.32

γ

=

56.32 = 31.62 1.78

Mauricio G. Prado – The University of Tulsa

Steady State Flow in Irregular Drainage Areas •

Circular Pseudo Steady State Flow ⎛ ⎛ rr ⎞ 1 ⎞ ⎜ ln⎜ ⎟ − ⎟ Pwf = P − 0.00708 k h ⎜⎝ ⎜⎝ rw ⎟⎠ 2 ⎟⎠

µ qsc B

•

But ⎛ rr ⎞ 1 1 ⎛ rr2 ⎞ 1 1 ⎛ rr2 ⎞ 1 1 ⎛ rr2 ⎞ 1 ⎛ 4π rr2 ⎞ ⎟ ln⎜⎜ ⎟⎟ − = ln⎜⎜ 2 ⎟⎟ − = ln⎜⎜ 2 ⎟⎟ − ln(e ) = ln⎜⎜ 2 ⎟⎟ = ln⎜⎜ 2 ⎟ 2 ⎝ e rw ⎠ 2 ⎝ e 4π rw ⎠ ⎝ rw ⎠ 2 2 ⎝ rw ⎠ 2 2 ⎝ rw ⎠ 2

⎛ r ⎞ 1 1 ⎛ 4 Adrainage ⎞ ⎟ ln⎜⎜ r ⎟⎟ − = ln⎜⎜ 2 ⎟ r 2 2 e 4 π r w ⎠ ⎝ w⎠ ⎝

Mauricio G. Prado – The University of Tulsa

Steady State Flow in Irregular Drainage Areas •

Introducing a Dietz shape factor and the Euler constant γ = 1.78

e 4π = 34.159 = γ C A

•

Then we have: 1 ⎛ 4 Adrainage ⎞ ⎟ Pwf = P − ln⎜⎜ 2 ⎟ 0.00708 k h 2 ⎝ γ C A rw ⎠

µ qsc B

•

And the shape factor for a circular area under steady state is: C ASteady − State =

34.159

γ

=

34.159 − drive = 19.19 = C Water A 1.78

Mauricio G. Prado – The University of Tulsa

Summary of Stabilized Equations •

For Stabilized Flow we have: Pwf = P −

•

1 ⎛ 4 Adrainage ⎞ ⎟ ln⎜⎜ 2 ⎟ 0.00708 k h 2 ⎝ γ C A rw ⎠

µ qsc B

And the IPR can be written as: mD

ft psi

bpd

qsc =

0.00708 k h Bµ

2 ( P − Pwf ) ⎛ 4A ⎞ ⎟ ln⎜⎜ drainage 2 ⎟ ⎝ γ C A rw ⎠ ft2

cp ft

Mauricio G. Prado – The University of Tulsa

Summary of Stabilized Equations •

For Stabilized Flow we have in terms of the drainage area: mD

ft psi

bpd

qsc =

0.00708 k h Bµ

2 (P − Pwf ) A 97888 ⎛ drainage ⎞ ⎟⎟ ln⎜⎜ 2 C A rw ⎝ ⎠ acres

cp ft mD

ft

bpd

psi

qsc =

0.00708 k h Bµ

2 ( P − Pwf ) ⎛ 14095820 Adrainage ⎞ ⎟⎟ ln⎜⎜ 2 γ C A rw ⎝ ⎠ acres

cp in Mauricio G. Prado – The University of Tulsa

Summary of Stabilized Equations •

Or in terms of the drainage radius: Adrainage = π re2

qsc =

qsc =

0.00708 k h Bµ

0.00708 k h Bµ

2 ⎛ 4π r ln⎜⎜ ⎝γ C r

2 e 2 A w

⎞ ⎟⎟ ⎠

(P − P ) wf

P − Pwf ⎛ 2.657 ⎛ re ⎞ ln⎜⎜ ⎟⎟ + ln⎜ ⎜ C ⎝ rw ⎠ A ⎝

Mauricio G. Prado – The University of Tulsa

⎞ ⎟ ⎟ ⎠

Summary of Stabilized Equations •

Or in terms of the drainage radius: mD

ft

psi

bpd

qsc =

0.00708 k h Bµ cp

acres

13866 Adrainage = re2

ft2

P − Pwf ⎛ 2.657 ⎛ re ⎞ ln⎜⎜ ⎟⎟ + ln⎜ ⎜ C ⎝ rw ⎠ A ⎝

⎞ ⎟ ⎟ ⎠

acres

1996516 Adrainage = re2

Mauricio G. Prado – The University of Tulsa

in2

Summary of Stabilized Equations • •

Another important question to be answered is how long the well should be producing under reasonably stable flow conditions so that Pseudo Steady State conditions are valid. Earlougher has shown that the time when pseudo steady state conditions are valid can be expressed in terms of a dimensionless time

t pss =

φµcA

psi-1 cp

ft

hours

k

t da acres

psi-1

2

cp hours

t pss =

φµcA 0.000264 k

t da

t pss = mD

φµcA −9

6.054 10 k

t da

psi-1 cp

ft2

hours

t pss = 11900

φ µ c re2 k

t da mD

Mauricio G. Prado – The University of Tulsa

mD

Summary of Stabilized Equations •

The following table is a summary of Dietz shape factors and the dimensionless time to pseudo steady state

Mauricio G. Prado – The University of Tulsa

Single Phase IPR Darcy’s Law can also be written as: Pressure “drawdown” represents the pressure drop in the porous rock

Flowrate in standard conditions (stb)

(

q = J P − Pwfi

)

Productivity Index has the units of stb/psi

J=

0.00708 k h Bµ

1 ⎛ 2.657 ⎛ re ⎞ ln⎜⎜ ⎟⎟ + ln⎜ ⎜ C ⎝ rw ⎠ A ⎝

Mauricio G. Prado – The University of Tulsa

⎞ ⎟ ⎟ ⎠

Single Phase IPR Darcy’s Law can also be written as: Bottom hole flowing pressure

q qmax

= 1−

Pwfi P

Maximum flowrate that would occur if the bottom hole flowing pressure could be zero. It is also known as “Absolute Open Flow” or “AOF”

Reservoir Average Pressure

qmax = J P

J=

0.00708 k h Bµ

1 ⎛ 2.657 ⎛ re ⎞ ln⎜⎜ ⎟⎟ + ln⎜ ⎜ C ⎝ rw ⎠ A ⎝

Mauricio G. Prado – The University of Tulsa

⎞ ⎟ ⎟ ⎠

Single Phase IPR Pwf

Pe

q = J (P − Pwf )

dq J =− dPwf

qmax = J P q qmax Mauricio G. Prado – The University of Tulsa

Single Phase IPR • The incompressible single phase or straight line IPR is valid when the fluids flowing inside the reservoir are in single phase incompressible conditions. – Whenever Pwf is above the bubble point Pressure – Very High Water Cuts – Very low GOR – Dead Oil Reservoirs

Mauricio G. Prado – The University of Tulsa

Single Phase IPR • The single phase productivity index J can be calculated from:

– Reservoir and Fluid properties or – Calculated from well test data

Mauricio G. Prado – The University of Tulsa

Single Phase IPR Example 1:

k - 20 mD h - 60 ft

µ - 10 cP rr - 600 ft rw - 3.5” Pr - 1250 psi

Mauricio G. Prado – The University of Tulsa

Single Phase IPR Example 1:

k - 20 mD h - 60 ft

µ - 10 cP rr - 600 ft rw - 3.5”

0.00708 20 60 J= ⎛ 600 ⎞ 10 ln⎜ ⎟ ⎝ 3.5 / 12 ⎠ J = 0.1114 stb/d/psi

Pr - 1250 psi

Mauricio G. Prado – The University of Tulsa

Single Phase IPR Example 1:

k - 20 mD h - 60 ft

(

q = 0.1114 P − P

µ - 10 cP rr - 600 ft

qmax = 0.1114 1250

rw - 3.5” Pr - 1250 psi

qmax = 139.25 stb/d

Mauricio G. Prado – The University of Tulsa

i wf

)

Single Phase IPR Example 1:

Pwf

Pr = 1250 psig

(

q = 0.1114 P − P

i wf

)

1250

qmax = 139.25 stb/d

q 139.25 Mauricio G. Prado – The University of Tulsa

Single Phase IPR Example 2:

Pr - 1250 psi Well Test 600 stb/d @ Pwf = 900 psi

Mauricio G. Prado – The University of Tulsa

Single Phase IPR Example 2:

Pr - 1250 psi Well Test 600 stb/d @ Pwf = 900 psi

(

q= J P −P 600 = J (1250 − 900 )

i wf

)

J = 1.71 stb/d/psi

Mauricio G. Prado – The University of Tulsa

Single Phase IPR Example 2:

Pr - 1250 psi Well Test 600 stb/d @ Pwf – 900 psi

(

q= J P −P qmax = 1.71 (1250 - 0)

i wf

)

qmax = 2137.5 stb/d

Mauricio G. Prado – The University of Tulsa

Single Phase IPR Example 2: Pr = 1250 psig

Pwf 1250

(

q = 1.71 P − P

i wf

)

qmax = 2137.5 stb/d

q 2137.5 Mauricio G. Prado – The University of Tulsa

Homework 1 Well Tests 30 stb/d @ Pwf = 1000 psi 60 stb/d @ Pwf = 800 psi

Calculate: Pr , J and Qmax

Mauricio G. Prado – The University of Tulsa

Homework 1 Well Tests 30 stb/d @ Pwf = 1000 psi 60 stb/d @ Pwf = 800 psi Well Tests 30 = J ( Pr – 1000) 60 = J ( Pr – 800)

(

q= J P −P

( P − 800) 2= (P − 1000)

30 = J ( Pr – 1000)

i wf

Pr = 1200 psi

J = 3/20 stb/d/psi

30 = J ( 1200 – 1000) Qmax = J Pr = 3/20 (1200) = 180 stb/d Mauricio G. Prado – The University of Tulsa

)

Pwf

Pr = 1200 psi

Homework 1

30 stb/d @ Pwf = 1000 psi 60 stb/d @ Pwf = 800 psi

qmax = 180 stb/d

Q Mauricio G. Prado – The University of Tulsa

As we produce fluids from the reservoir, a depletion in reservoir pressure will take place. What is the effect of depletion on the Inflow Performance Relationship ? Can we estimate Future IPR ?

Mauricio G. Prado – The University of Tulsa

Future Linear IPR Pwf

Pr

As time t incresases, reservoir pressure Pr decreases and cumulative production Np increases.

(

q= J P −P

i wf

) q

qmax Mauricio G. Prado – The University of Tulsa

Other topics • Transient single phase IPR • Include B and viscosity – integration with pressure for single phase • Gas IPR • Other models – partial penetration • Skin • Etc….

Mauricio G. Prado – The University of Tulsa

IPR • The IPR equation is valid for a well without damage or stimulation effects • In actual wells, sometimes a damage area around the wellbore exists. • This area usually is of very small radius and the permeability is also reduced. This causes an different pressured drop behavior that must be taken into account. • In some wells, stimulation practices can create a small radius area around the wellbore where the permeability is higher than the original reservoir permeability and this effect must also be considered. Mauricio G. Prado – The University of Tulsa

Skin Effect • The pressure profile changes in this region. • The region has a very small radius, and because of this, the reduction or increase in pressure is called skin effect. • We can use Steady State Darcys Law for single phase flow to illustrate the problem

µ qsc B ⎛ rr ⎞ ln⎜ ⎟ P(r ) = Pr − 2π k h ⎝ r ⎠

Mauricio G. Prado – The University of Tulsa

Skin Effect rw

ri

rr

k skin

Pwf

k

Pr

Pi

P(r ) = Pi −

µ qsc B ⎛ ri ⎞ ln⎜ ⎟ 2 π k skin h ⎝ r ⎠ P(r ) = Pr −

µ qsc B ⎛ rr ⎞ ln⎜ ⎟ 2π k h ⎝ r ⎠

Mauricio G. Prado – The University of Tulsa

Skin Effect • The pressure at the edge of the affected zone can be obtained by: P(r ) = Pr −

µ qsc B ⎛ rr ⎞ ln⎜ ⎟ 2π k h ⎝ r ⎠

µ qsc B ⎛ rr ⎞ Pi = Pr − ln⎜⎜ ⎟⎟ 2 π k h ⎝ ri ⎠

• The bottonhole flowing pressure can be obtained by: P(r ) = Pi −

µ qsc B ⎛r ⎞ ln⎜ i ⎟ 2 π k skin h ⎝ r ⎠

Pwf = Pi −

⎛r ⎞ µ qsc B ln⎜⎜ i ⎟⎟ 2 π k skin h ⎝ rw ⎠

Mauricio G. Prado – The University of Tulsa

Skin Effect • Then, the bottonhole flowing pressure can be written as: µ qsc B ⎛ rr ⎞ Pi = Pr − ln⎜⎜ ⎟⎟ 2 π k h ⎝ ri ⎠ Pwf = Pi −

Pwf = Pr −

⎛r ⎞ µ qsc B ln⎜⎜ i ⎟⎟ 2 π k skin h ⎝ rw ⎠

⎛r ⎞ µ qsc B ⎛ rr ⎞ µ qsc B ln⎜⎜ ⎟⎟ − ln⎜⎜ i ⎟⎟ 2 π k h ⎝ ri ⎠ 2 π k skin h ⎝ rw ⎠

µ qsc B ⎛⎜ 1 ⎛ rr ⎞ 1 ⎛ ri ⎞ ⎞⎟ Pwf = Pr − ln⎜⎜ ⎟⎟ + ln⎜⎜ ⎟⎟ ⎟ ⎜ 2 π h ⎝ k ⎝ ri ⎠ k skin ⎝ rw ⎠ ⎠ Mauricio G. Prado – The University of Tulsa

Skin Effect •

The bottonhole flowing pressure that would exist if the well had no skin would be: no − skin wf

P

•

µ qsc B ⎛ rr ⎞ = Pr − ln⎜⎜ ⎟⎟ 2 π k h ⎝ rw ⎠

Defining the Skin pressure drop as the difference between the real bottonhole flowing pressure and the bottonhole flowing pressure without skin: no − skin wf

P

µ qsc B ⎛ rr ⎞ = Pr − ln⎜⎜ ⎟⎟ 2 π k h ⎝ rw ⎠

no − skin wf

∆Pskin = P

− Pwf

∆Pskin

µ qsc B ⎛⎜ 1 ⎛ rr ⎞ 1 ⎛ ri ⎞ ⎞⎟ Pwf = Pr − ln⎜⎜ ⎟⎟ + ln⎜⎜ ⎟⎟ ⎟ ⎜ 2 π h ⎝ k ⎝ ri ⎠ k skin ⎝ rw ⎠ ⎠

µ qsc B ⎛ rr ⎞ µ qsc B ⎛⎜ 1 ⎛ rr ⎞ 1 ⎛ ri ⎞ ⎞⎟ = Pr − ln⎜ ⎟ − Pr + ln⎜ ⎟ + ln⎜ ⎟ 2 π k h ⎜⎝ rw ⎟⎠ 2 π h ⎜⎝ k ⎜⎝ ri ⎟⎠ k skin ⎜⎝ rw ⎟⎠ ⎟⎠

Mauricio G. Prado – The University of Tulsa

Skin Effect •

The skin pressure drop becomes: ∆Pskin

µ qsc B ⎛ rr ⎞ µ qsc B ⎛⎜ 1 ⎛ rr ⎞ 1 ⎛ ri ⎞ ⎞⎟ = Pr − ln⎜ ⎟ − Pr + ln⎜ ⎟ + ln⎜ ⎟ 2 π k h ⎜⎝ rw ⎟⎠ 2 π h ⎜⎝ k ⎜⎝ ri ⎟⎠ k skin ⎜⎝ rw ⎟⎠ ⎟⎠ ∆Pskin

∆Pskin

µ qsc B ⎛⎜ 1 ⎛ rr ⎞ 1 ⎛ ri ⎞ 1 ⎛ rr ⎞ ⎞⎟ = ln⎜ ⎟ + ln⎜ ⎟ − ln⎜ ⎟ 2 π h ⎜⎝ k ⎜⎝ ri ⎟⎠ k skin ⎜⎝ rw ⎟⎠ k ⎜⎝ rw ⎟⎠ ⎟⎠

µ qsc B ⎛⎜ 1 ⎛ ri ⎞ 1 ⎛ ri ⎞ 1 ⎛ rr ⎞ ⎞⎟ = − ln⎜ ⎟ + ln⎜ ⎟ − ln⎜ ⎟ 2 π h ⎜⎝ k ⎜⎝ rr ⎟⎠ k skin ⎜⎝ rw ⎟⎠ k ⎜⎝ rw ⎟⎠ ⎟⎠ ∆Pskin =

µ qsc B ⎛⎜ 1 ⎛ ri ⎞ 1 ⎛ ri ⎞ ⎞⎟ − ln⎜⎜ ⎟⎟ + ln⎜⎜ ⎟⎟ ⎟ ⎜ 2 π h ⎝ k ⎝ rw ⎠ k skin ⎝ rw ⎠ ⎠

∆Pskin =

µ qsc B ⎛ ri ⎞⎛ 1 1⎞ − ⎟⎟ ln⎜⎜ ⎟⎟⎜⎜ 2 π h ⎝ rw ⎠⎝ k skin k ⎠

Mauricio G. Prado – The University of Tulsa

Skin Effect •

The skin pressure drop becomes: ∆Pskin

µ qsc B ⎛ ri ⎞⎛ 1 1⎞ ⎜ ⎟ ⎜ = − ⎟⎟ ln⎜ ⎟⎜ 2 π h ⎝ rw ⎠⎝ k skin k ⎠

∆Pskin =

∆Pskin

•

µ qsc B ⎛ ri ⎞ k − k skin ln⎜⎜ ⎟⎟ 2 π h ⎝ rw ⎠ k skin k

µ qsc B ⎛ ri ⎞ k − k skin = ln⎜⎜ ⎟⎟ 2 π k h ⎝ rw ⎠ k skin

The skin effect is usually determined with a pressure build up test. Usually it is related to a skin factor S which is a parameter determined by the build up test analysis: ∆Pskin

µ qsc B ⎛ ri ⎞ k − k skin µ qsc B = ln⎜⎜ ⎟⎟ =S 2 π k h ⎝ rw ⎠ k skin 2π k h Mauricio G. Prado – The University of Tulsa

⎛ ri ⎞ k − k skin S = ln⎜⎜ ⎟⎟ ⎝ rw ⎠ k skin

Skin Effect • •

Summary S – Positive S indicates a damage – Zero indicates no skin – Negative indicates stimulation

•

S is obtained by a pressure build up analysis ⎛ r ⎞ k − k skin S = ln⎜⎜ i ⎟⎟ ⎝ rw ⎠ k skin

•

The skin pressure drop is psi

∆Pskin = S

µ qsc B 2π k h

∆Pskin = S

bpd

cp

µ qsc B 0.00708 k h ft mD

Mauricio G. Prado – The University of Tulsa

Skin Effect •

The pressure in the unnafected region is given by: psi

cp

P (r ) = Pr −

bpd

µ qsc B

⎛r ⎞ ln⎜ r ⎟ 0.00708 k h ⎝ r ⎠ mD

•

ft

The pressure in the affected region is

P (r ) = Pi −

µ qsc B

⎛r ⎞ ln⎜ i ⎟ 0.00708 k h ⎝ r ⎠

Mauricio G. Prado – The University of Tulsa

Skin Effect •

The bottonhole flowing pressure is given by Pwfno − skin = Pwf + ∆Pskin

⎛ rr ⎞ µ qsc B Pwf = Pr − ln⎜⎜ ⎟⎟ + S 0.00708 k h ⎝ rw ⎠ 0.00708 k h

µ qsc B

psi

cp

bpd

⎛ ⎛ rr ⎞ ⎞ ⎜ ⎜⎜ ⎟⎟ + S ⎟ ln Pwf = Pr − ⎟ 0.00708 k h ⎜⎝ ⎝ rw ⎠ ⎠

µ qsc B

mD

ft

Mauricio G. Prado – The University of Tulsa

Skin Effect P(r ) = Pr − 3000

µ qsc B

⎛r ⎞ ln⎜ r ⎟ 0.00708 k h ⎝ r ⎠

Pressure (psi)

2500

P(r ) = Pi −

2000

µ qsc B

⎛r ⎞ ln⎜ i ⎟ 0.00708 k h ⎝ r ⎠

1500

∆Pskin = S

1000

0.00708 k h

⎛ r ⎞ k − k skin S = ln⎜⎜ i ⎟⎟ ⎝ rw ⎠ k skin

ri

500

µ qsc B

0 0

2

4

6

Radial Position (ft)

Pwf = Pr −

⎛ ⎛ rr ⎞ ⎞ ⎜ ln⎜ ⎟ + S ⎟ ⎟ 0.00708 k h ⎜⎝ ⎜⎝ rw ⎟⎠ ⎠

µ qsc B

Mauricio G. Prado – The University of Tulsa

8

10

Skin Effect •

The Steady State Linear IPR for a reservoir that presents some Skin effect is then: bpd

mD

qsc =

ft

psi

(

0.00708 k h Pr − Pwfi ⎛ ⎛r ⎞ ⎞ µ B⎜⎜ ln⎜⎜ r ⎟⎟ + S ⎟⎟ ⎝ ⎝ rw ⎠ ⎠

)

cp

psi

bpd

cp

∆Pskin = S

µ qsc B 0.00708 k h ft mD

Mauricio G. Prado – The University of Tulsa

Skin Effect qsc =

Bottomhole Flowing Pressure (psi)

6000

(

0.00708 k h Pr − Pwfi ⎛ ⎛ rr ⎞ ⎞ µ B ⎜⎜ ln⎜⎜ ⎟⎟ ⎟⎟ ⎝ ⎝ rw ⎠ ⎠

5000

4000

) Reservoir Pressure

5000

psi

Viscosity

10

cp

Permeability

500

mD

Thickness

10

ft

Reservoir Radius

1000

ft

Well Radius

6

in

3000

2000

1000

0 0

500

1000

1500

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

2000

2500

Skin Effect qsc =

Bottomhole Flowing Pressure (psi)

6000

5000

(

)

0.00708 k h Reservoir Pressure Pr − Pwfi ⎛ ⎛ r ⎞⎞ Viscosity µ B ⎜⎜ ln⎜⎜ r ⎟⎟ ⎟⎟ Permeability ⎝ ⎝ rw ⎠ ⎠ 0.00708 k h Thickness ( Pr − Pwfi ) qsc = ⎛ ⎛r ⎞ ⎞ Reservoir Radius µ B ⎜⎜ ln⎜⎜ r ⎟⎟ + S ⎟⎟ Well Radius ⎝ ⎝ rw ⎠ ⎠

5000

psi

10

cp

500

mD

10

ft

1000

ft

6

in

4000

3000 ∆Pskin = S

2000

µ qsc B

Skin Factor

0.00708 k h

0

1000

3

0 0

500

1000

1500

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

2000

2500

Skin Effect •

A similar analysis yield similar results for the semi-steady state IPR as well in terms of average pressure

•

In summary we have: psi

bpd

cp

∆Pskin = S

µ qsc B 0.00708 k h ft mD

Circular Steady State

Circular Pseudo Steady State mD

mD

ft

psi

ft

psi

bpd

bpd

(

0.00708 k h qsc = P − Pwfi ⎛ ⎛r ⎞ 1 ⎞ µ B ⎜⎜ ln⎜⎜ r ⎟⎟ − + S ⎟⎟ ⎝ ⎝ rw ⎠ 2 ⎠

)

qsc =

(

0.00708 k h P − Pwfi ⎛ ⎛r ⎞ 3 ⎞ µ B ⎜⎜ ln⎜⎜ r ⎟⎟ − + S ⎟⎟ ⎝ ⎝ rw ⎠ 4 ⎠

cp

cp Mauricio G. Prado – The University of Tulsa

)

Skin Effect • In general we have: psi

bpd

cp

∆Pskin = S

µ qsc B 0.00708 k h ft mD

mD

ft psi

bpd

qsc =

0.00708 k h Bµ

⎛ ln⎜⎜ ⎝

1 ⎛ 2.657 re ⎞ ⎟⎟ + ln⎜ ⎜ C rw ⎠ A ⎝

⎞ ⎟+S ⎟ ⎠

cp

Mauricio G. Prado – The University of Tulsa

(P − P ) i wf

Skin Effect • Or in terms of the productivity index and AOF: mD

ft

bpd/psi

(

q= J P −P

i wf

)

J=

0.00708 k h Bµ

⎛ ln⎜⎜ ⎝

1 ⎛ 2.657 re ⎞ ⎟⎟ + ln⎜ ⎜ C rw ⎠ A ⎝

⎞ ⎟+S ⎟ ⎠

cp

mD

ft

bpd

q qmax

= 1−

Pwfi P

qmax =

psi

0.00708 k h Bµ

⎛ ln⎜⎜ ⎝

cp

Mauricio G. Prado – The University of Tulsa

1 ⎛ 2.657 re ⎞ ⎟⎟ + ln⎜ ⎜ C rw ⎠ A ⎝

⎞ ⎟+S ⎟ ⎠

P

Skin Effect • A comparison with the skinless variables yields:

(

q= J P −P

q qmax

= 1−

i wf

)

q=J

no − skin

Pwfi

q

P

no skin qmax

(P − P

= 1−

no − skin wf

)

J J no skin

Pwfno − skin

=

J = FE J

no skin

P − Pwfi

P − Pwfno − skin qmax = no skin qmax P − Pwfi

P

• This led to the definition of the Flow Efficiency: FE =

P − Pwfno − skin

P − Pwfno − Skin P − Pwfi

no skin qmax = FE qmax

Mauricio G. Prado – The University of Tulsa

Skin Effect •

How to relate the Flow Efficiency with the Skin S:

FE =

J=

P − Pwfno − Skin P − Pwfi

0.00708 k h Bµ

⎛ ln⎜⎜ ⎝

no skin qmax = FE qmax

J = FE J no skin

1 ⎛ 2.657 re ⎞ ⎟⎟ + ln⎜ ⎜ C rw ⎠ A ⎝

⎞ ⎟+S ⎟ ⎠

J no skin =

0.00708 k h Bµ

⎛ 2.657 ⎞ ⎛ r ⎞ ⎟ ln⎜⎜ e ⎟⎟ + ln⎜ ⎜ C A ⎟⎠ ⎝ rw ⎠ ⎝ FE = ⎛ 2.657 ⎞ ⎛ r ⎞ ⎟+S ln⎜⎜ e ⎟⎟ + ln⎜ ⎜ ⎟ ⎝ rw ⎠ ⎝ CA ⎠

Mauricio G. Prado – The University of Tulsa

1 ⎛ 2.657 ⎛ r ⎞ ln⎜⎜ e ⎟⎟ + ln⎜ ⎜ C ⎝ rw ⎠ A ⎝

⎞ ⎟ ⎟ ⎠

Skin Effect •

How to calculate the skin pressure drop from the flow efficiency: FE =

(

P − Pwfno − Skin P − Pwfi

)

FE P − Pwfi = P − Pwfno − Skin FE P − FE Pwfi = P − Pwfno − Skin FE P − P − FE Pwfi + Pwfi = Pwfi − Pwfno − Skin

(FE − 1)P − (FE − 1) Pwfi (FE −1)(P − Pwfi ) = Pwfi

= Pwfi − Pwfno − Skin

− Pwfno − Skin

∆Pskin = Pwfno − skin − Pwf

(

∆Pskin = (1 − FE ) P − Pwfi

)

Mauricio G. Prado – The University of Tulsa

Skin Effect •

Finally how to relate FE with well test data

(

∆Pskin = (1 − FE ) P − Pwfi

)

∆Pskin FE = 1 − P − Pwfi

∆Pskin = S

bpd

cp

FE = 1 −

µ qsc B S P − Pwfi 0.00708 k h psi

mD

Mauricio G. Prado – The University of Tulsa

ft

µ qsc B 0.00708 k h

Skin Effect • •

Example The following data is available for a well without damage

•

•

Reservoir Pressure

5000

psi

Viscosity

10

cp

Permeability

500

mD

Thickness

10

ft

Reservoir Radius

1000

ft

Well Radius

6

in

After a workover job, the following production test data is available. Flowing Pressure

3000

psi

Flowrate

750

bpd

Estimate the skin factor for this well and the production for the same bottonhole flowing pressure if the skin is removed

Mauricio G. Prado – The University of Tulsa

Skin Effect •

We first calculate the IPR for the well with damage q=

(

0.00708 k h Pr − Pwfi ⎛ ⎛ re ⎞ ⎞ ⎜ µ ⎜ ln⎜⎜ ⎟⎟ + S ⎟⎟ ⎝ ⎝ rw ⎠ ⎠

)

⎛ rr ⎞ 0.00708 k h i S= Pr − Pwf − ln⎜⎜ ⎟⎟ µq ⎝ rw ⎠

(

S=

)

0.00708 500 10 (5000 − 3000) − ln⎛⎜ 1000 ⎞⎟ 10 750 ⎝ 0.5 ⎠

S = 1.84 Mauricio G. Prado – The University of Tulsa

Skin Effect •

The production without skin for the same bottonhole flowing pressure is q=

q=

(

0.00708 k h Pr − Pwfi ⎛ ⎛ r ⎞⎞ µ ⎜⎜ ln⎜⎜ e ⎟⎟ ⎟⎟ ⎝ ⎝ rw ⎠ ⎠

)

0.00708 500 10 (5000 − 3000) 1000 ⎛ ⎞ 10 ln⎜ ⎟ 0 . 5 ⎝ ⎠

q = 931 bpd

Mauricio G. Prado – The University of Tulsa

Apparent Well Radius •

Another common way of refering to the skin is through the apparent well radius: qsc =

qsc =

•

0.00708 k h Bµ

0.00708 k h Bµ

⎛ ln⎜⎜ ⎝

1 ⎛ 2.657 re ⎞ ⎟⎟ + ln⎜ ⎜ C rw ⎠ A ⎝

i wf

(P − P ) ⎞

1

i wf

⎛ 2.657 ⎛ e S rwa ⎞ ⎛ re ⎞ ⎟ ⎟⎟ + ln⎜ ⎟⎟ + ln⎜⎜ ln⎜⎜ ⎜ ⎟ ⎝ rw ⎠ ⎝ rwa ⎠ ⎝ CA ⎠

If:

rwa = •

⎞ ⎟+S ⎟ ⎠

(P − P )

rw eS

Then mD

ft

psi

bpd

qsc =

0.00708 k h Bµ

1

(P − P ) ⎞

⎛ 2.657 ⎛ r ⎞ ⎟ ln⎜⎜ e ⎟⎟ + ln⎜ ⎜ ⎟ ⎝ rwa ⎠ ⎝ CA ⎠

cp Mauricio G. Prado – The University of Tulsa

i wf

Multi Phase IPR • Not all of the producing wells are in conditions where the linear IPR is valid. • Usually inside the reservoir we have a mixture of fluids flowing that will go to multiphase conditions when the pressure is lower than the bubble point. • The fluids are not incompressible, their properties change with pressure. • The relative permeability of the reservoir to a specific fluid is function of the fluid saturation. • The saturation of fluids changes with position and time • If n-phases are present, the problem can theoretically be solved using n mass balance equations and n momentum balance equations. • For single phase flow, Darcy’s law is a representation of the steady state momentum balance equation. • Can Darcy’s law be written for two-phase flow conditions ?

Mauricio G. Prado – The University of Tulsa

Darcy’s Law Viscous Fluid

Sand 100% Saturated with Fluid

µ

P + dP

A

q

dx

k A dP q= µ dx

Mauricio G. Prado – The University of Tulsa

P

Darcy’s Law Sand Partially Saturated with Fluid

Viscous Fluid

µ

P + dP

A

q

dx k f A dP q= µ dx k f = k f (Rock , Fluids, Saturation )

Mauricio G. Prado – The University of Tulsa

P

Darcy’s Law 0.6

Absolute Permeability - k

Permeability (mD)

0.5 0.4 0.3 0.2

Critical Saturation

0.1 0.0 0.0

0.2

0.4

0.6

Fluid Saturation (Fraction)

Mauricio G. Prado – The University of Tulsa

0.8

1.0

Darcy’s Law • The relative permeability is defined as the ratio between the permeability to a fluid and the rock absolute permeability:

k rf =

kf k

k f A dP k rf dP =kA q= µ dx µ dx

Mauricio G. Prado – The University of Tulsa

Darcy’s Law 1.0

Relative Permeability

0.8

0.6

0.4

0.2

0.0 0.0

0.2

0.4

0.6

Fluid Saturation (Fraction) Mauricio G. Prado – The University of Tulsa

0.8

1.0

Multi Phase IPR •

• •

As said before, the problem of flow of n phases inside the porous media is described by a system of n mass balance equations and n momenutm balance equations (Darcy’s Law). The variables are the pressure, the saturation for n-1 phases and n phase velocities. The permeability can be a function of position. With the proper boundary and initial conditions this system can be solved for the pressure, phase fractions and phase velocities fields

k = k (r ,θ , z ) P = P(r ,θ , z , t )

k f = k f (S f , k )

S f = S f (r ,θ , z , t )

k f = k f (r , θ , z , t ) Mauricio G. Prado – The University of Tulsa

k rf = k rf (r ,θ , z , t )

Multi Phase IPR • Also for a real fluid, the fluid properties are function of pressure. • Specifically, the fluid compressibility and viscosity are function of the pressure. • Darcy law is valid for the actual fluid flowrate or velocity occuring at the pressure and temperature conditions inside the porous media. • Since the fluid is compressible, this flowrate is not the same that is measured at surface conditions. • For that reason it would be interesting to relate the IPR with the liquid flowrate at specific surface or standard conditions. • This is done with the use of the fluid formation volume factor

q = B f qsc Mauricio G. Prado – The University of Tulsa

Darcy’s Law k rf dP q=kA µ f dx

q = B f qsc kf

dP qsc = A B f µ f dx Mauricio G. Prado – The University of Tulsa

Darcy’s Law For radial flow we have : qsc = A

kf

dP B f µ f dr

A = 2π r h

qsc =

2 π r k f h dP µ f B f dr

Mauricio G. Prado – The University of Tulsa

Darcys Law in Cylindrical Coordinates Fluid properties are function of pressure and saturation or relative permeability is function of position, then :

dr ∫ kf r =

2π h q scf

q = 2π h sc f

∫B

∫B

1 f

1 f

µf

µf

dP

dP

dr ∫ kf r

Mauricio G. Prado – The University of Tulsa

Multi Phase IPR • As the pressure inside the reservoir goes below the bubble point value, gas goes out of solution reducing the oil saturation and relative permeability, and increasing oil viscosity. The oil productivity is reduced, since now the driving force for fluid movement is spent moving the liquid and the gas phases. • The constant PI concept is no longer valid.

Mauricio G. Prado – The University of Tulsa

Single Phase and Multi Phase IPR Pwf

Pr

2π k h dP q= ∫ ⎛ re ⎞ Pwf µ ln⎜⎜ ⎟⎟ ⎝ rw ⎠ Pr

q = 2π h sc f

∫B

1 f

dr ∫ kf r

q Mauricio G. Prado – The University of Tulsa

µf

dP

Multi Phase IPR • IPR under multiphase flow conditions can not be easily calculated. • The most accurate method is by solving the equations governing the flow in the porous media through a reservoir simulator. • The IPR is so important to Production Engineers that simplified or empirical methods to estimate it are necessary. • The most common correlations are Vogel and Fetkovich

Mauricio G. Prado – The University of Tulsa

Vogel IPR • Vogel used a numerical reservoir simulator to generate the IPR. He studied several cases for a specific condition: – Mechanism of production – Solution Gas Drive – No water production – Reservoir pressure below bubble point – Saturated conditions

• He changed several other conditions such as fluid and rock properties • He then plotted the results of the simulation for the several cases

Mauricio G. Prado – The University of Tulsa

Vogel IPR 3500 3000

Pressure (psi)

2500 2000 1500 1000 500 0 0

500

1000 Flowrate (sbpd)

Mauricio G. Prado – The University of Tulsa

1500

2000

Vogel IPR • Like expected, all IPR’s showed a curved shape. • He then tried to find a “common” shape to describe all the IPRs. • He tried to “normalize” the curves. For each case, he divided the pressure by the reservoir pressure and the flowrate by the maximum flowrate. • The result was not a perfect correlation, but the points were clustered along a curved line.

Mauricio G. Prado – The University of Tulsa

Vogel IPR Pwf

1

P Pressure (dimensionless)

0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6

Flowrate (dimensionless)

0.8

q qmax

Mauricio G. Prado – The University of Tulsa

1

Vogel IPR • A linear relationship is clearly applicable. • Vogel tried then a quadratic form

q qmax

⎛ Pwf = a + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ + c ⎜ ⎜ ⎠

Pwf ⎝ P

Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟⎟ ⎠

not

Vogel IPR Pwf

1

q

P Pressure (dimensionless)

0.8

qmax

⎛ Pwf = a + b ⎜⎜ ⎝ P

2

Pwf ⎞ ⎟⎟ ⎝ P ⎠

⎛ ⎞ ⎟⎟ + c ⎜ ⎜ ⎠

0.6

0.4

0.2

0 0

0.2

0.4

0.6

Flowrate (dimensionless)

0.8

1

q qmax

Mauricio G. Prado – The University of Tulsa

Vogel IPR q qmax

⎛ Pwf = a + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ + c ⎜ ⎜ ⎠

Pwf ⎝ P

2

⎞ ⎟⎟ ⎠

The following conditions must be met by this expression q = qmax

for

Pwf = 0

and q=0

for

Pwf = P

Mauricio G. Prado – The University of Tulsa

Vogel IPR The results of those conditions are: a=1 and c=-(1+b) So the proposed expression becomes:

q qmax

⎛ Pwf = 1 + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

2

Pwf ⎞ ⎟⎟ ⎝ P ⎠

Mauricio G. Prado – The University of Tulsa

1

q qmax

D imensionless Pressure

0.8

⎛ Pwf = 1 + b ⎜⎜ ⎝ P

-1

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎝ P

0.2 0.4

- 0.8

0.6

0.6

- 0.6 - 0.4

0.4 - 0.2

b

0 0.2

0 0

0.2

0.4

0.6

D imensionless Flow rate Mauricio G. Prado – The University of Tulsa

0.8

1

2

⎞ ⎟⎟ ⎠

Vogel IPR

q qmax

⎛ Pwf = 1 + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

2

Pwf ⎞ ⎟⎟ ⎝ P ⎠

-1 < b < 0

Mauricio G. Prado – The University of Tulsa

Vogel IPR Vogel then used his numerical results from the simulations to get the best value of b that would fit his data.

Mauricio G. Prado – The University of Tulsa

Vogel IPR 1

Pressure (dimensionless)

0.8

b = - 0.2 0.6

0.4

0.2

0 0

0.2

0.4

0.6

Flowrate (dimensionless)

Mauricio G. Prado – The University of Tulsa

0.8

1

q

1

qmax

⎛ Pwf = 1 − 0.2 ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − 0.8 ⎜ ⎜ ⎠

2

Pwf ⎞ ⎟⎟ ⎝ P ⎠

Pressure (dimensionless)

0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6

Flowrate (dimensionless)

Mauricio G. Prado – The University of Tulsa

0.8

1

Vogel IPR • Vogel IPR can be obtained only from well tests as opposed to the linear IPR that can be determined from well tests of rock and fluid properties. • Although the method was develloped for solution gas drive reservois, the equation is generally accepted and used for other drive mechanisms as well. • It is found to give excellent results for any well with a reservoir pressure below the oil bubble point, i.e., saturated reservoirs. WHY ?

Mauricio G. Prado – The University of Tulsa

Vogel Type IPR • The best value for b according to Vogel’s numerical results is -0.2 • Fetkovich following a more analytical approach also proposed an IPR with a b value of 0 • Several other investigators obtained different values for b as well.

Mauricio G. Prado – The University of Tulsa

Vogel Type IPR IPR

b

Linear

-1

Wiggins (Water - Multiphase)

-0.72

Wiggins (Oil)

-0.52

Vogel

-0.2

Klins (Quadratic)

-0.1225

Fetkovich

0

Caution: When using a quadratic type IPR the equation must follows: q qmax

⎛ Pwf = 1 + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎝ P

Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟⎟ ⎠

Vogel Type IPR

q qmax

⎛ Pwf = 1 + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎝ P

1 Dimensionless Pressure

0.9 Vogel

0.8

Klins

0.7 0.6

Linear

Fetkovich

0.5 0.4 Wiggins - Water

0.3 0.2

Wiggins - Oil

0.1 0 0

0.2

0.4

0.6

Dimensionless Flowrate

Mauricio G. Prado – The University of Tulsa

0.8

1

2

⎞ ⎟⎟ ⎠

Vogel IPR • Example: Saturated Reservoir Pr = 1500 psi

Test qo = 200 bpd @ Pwf = 1400 psi Determine Vogel IPR (b = - 0.2)

Mauricio G. Prado – The University of Tulsa

Vogel IPR • Example:

q qmax

⎛ Pwf = 1 − 0.2 ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − 0.8 ⎜ ⎜ ⎠

Pwf ⎝ P

2

1400 ⎞ ⎛ 200 ⎛ 1400 ⎞ = 1 − 0.2 ⎜ ⎟ − 0.8 ⎜ ⎟ qmax ⎝ 1500 ⎠ 1500 ⎝

qmax = 1717.56 stb/d

Mauricio G. Prado – The University of Tulsa

⎠

2

⎞ ⎟⎟ ⎠

1600 1400

Pressure (psi)

1200 1000 800 600 400 200 0 0

500

1000 Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

1500

2000

Homework 1 • Well Test – 100 bpd @ Pwf = 1400 psi – Pr = 2000 psi

• • • •

We know the reservoir is saturated Calculate and Plot the Vogel IPR Calculate and Plot a linear IPR What can you say of the use of linear IPR for saturated reservoirs ?

Mauricio G. Prado – The University of Tulsa

Homework 1 - Vogel q qmax

⎛ Pwf = 1 − 0.2 ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − 0.8 ⎜ ⎜ ⎠

2

Pwf ⎞ ⎟⎟ ⎝ P ⎠

2

1400 ⎞ ⎛ 100 ⎛ 1400 ⎞ = 1 − 0.2 ⎜ ⎟ − 0.8 ⎜ ⎟ qmax 2000 ⎝ ⎠ 2000 ⎝

qmax = 213.7 stb/d

Mauricio G. Prado – The University of Tulsa

⎠

Homework 1 - Vogel q = J ( P − Pwf ) 100 = J (2000 − 1400) J = 1/6 stb/d/psi Qmax = J Pr = 1/6 2000 = 333.33 stb/d

Mauricio G. Prado – The University of Tulsa

Homework 1 B ottomhole flow ing pressure (psi)

2000 1800 1600 ove r pre dicting re gion for line a r IP R

1400 1200 unde r pre dicting re gion for line a r IP R

1000 800 600 400 200 0 0

30

60

90

120

150

180

210

Oil Flow rate (bpd) Mauricio G. Prado – The University of Tulsa

240

270

300

330

Homework 1a • Well Test – 100 bpd @ Pwf = 1400 psi – Pr = 2000 psi

• We know the reservoir is saturated • Calculate and Plot the Vogel, Fetkovich, Klins, Wiggins and the linear IPR • Compare the results

Mauricio G. Prado – The University of Tulsa

Homework 1a - Vogel q qmax

⎛ Pwf = 1 + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎝ P

2

⎞ ⎟⎟ ⎠

2

1400 ⎞ ⎛ 100 ⎛ 1400 ⎞ = 1 + b⎜ ⎟ − (1 + b) ⎜ ⎟ qmax ⎝ 2000 ⎠ 2000 ⎝

⎠

IPR

b

qmax

Linear

-1

333

Wiggins

-0.52

250

Vogel

-0.2

214

Klins

-0.1225

206

Fetkovich

0

196

Mauricio G. Prado – The University of Tulsa

Homework 1a 2000 1800

Pressure (psi)

1600 1400 Linear

1200 1000

Wiggins

800

Vogel

600 Klins

400 200

Fetkovich

0 0

50

100

150

200

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

250

300

350

Productivity Index • The value of the productivity index J needs to be redefined for the case of the saturated IPR. The value of J is given by: ⎛ dq J = −⎜ ⎜ dP ⎝ wf

⎞ ⎟ ⎟ ⎠

• For saturated reservoirs we have ⎛ Pwf q = 1 + b⎜⎜ qmax ⎝ P

q J = max P

⎞ ⎛ Pwf ⎟⎟ − (1 + b)⎜⎜ ⎠ ⎝ P

Pwf ⎛ ⎞ ⎜⎜ 2(1 + b) − b ⎟⎟ P ⎝ ⎠

Mauricio G. Prado – The University of Tulsa

⎞ ⎟⎟ ⎠

2

Productivity Index • In particular, the values of J* and J0 are defined as: ⎛ dq J = −⎜ ⎜ dP ⎝ wf *

⎞ ⎟ ⎟ ⎠ Pwf = Pr

⎛ dq J = −⎜ ⎜ dP ⎝ wf 0

⎞ ⎟ ⎟ ⎠ Pwf =0

• For saturated reservoirs we have

J

*

( 2 + b ) qmax = P

− b qmax J = P 0

Mauricio G. Prado – The University of Tulsa

Productivity Index for Saturated IPR 2000

J* =

(2 + b ) qmax P

Pressure (psi)

1500

1000

− b qmax J = P 0

500

0 0

400

800 1200 Oil Flowrate (stb/d)

1600

Mauricio G. Prado – The University of Tulsa

2000

IPR Errors - Example • Assume we have a very reliable value of 1500 psi for Pr for a saturated reservoir • The well test gives us the following data – q = 300 bpd @ Pwf = 800 psi

• Assume the test flowrate error can be off by 5% • Assume the test pressure error can be off by 1% • So the test pressure and flowrate ranges are: – 285 bpd < q < 315 bpd – 792 psi < Pwf < 808 psi

• Estimate the error in the Vogel IPR

Mauricio G. Prado – The University of Tulsa

IPR Errors - Example (q

1600

(q, P )

1400

Pressure (psi)

, Pwf+

)

High Flowrate and Pressure

wf

1200

+

Test Data

1000 800

(q

600 400

−

, Pwf−

)

Low Flowrate and Pressure

200

q

− max

qmax

+ qmax

0 0

100

200

300

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

400

500

IPR Errors - Example • Lets calculate the value of qmax for the test data:

qt

qmax = ⎛ Pwf 1 − 0.2 ⎜⎜ ⎝ P

2

Pwf ⎞ ⎟ ⎟ P ⎠ ⎝

⎛ ⎞ ⎟ − 0.8 ⎜ ⎟ ⎜ ⎠

Mauricio G. Prado – The University of Tulsa

IPR Errors - Example • Lets calculate the value of qmax for the two extreme cases: qt+

+ = qmax

⎛P 1 − 0.2 ⎜⎜ ⎝ P

+ wf

+ ⎛ Pwf ⎞ ⎟ − 0.8 ⎜ ⎟ ⎜ P ⎠ ⎝

2

⎞ ⎟ ⎟ ⎠

qt−

− = qmax

⎛P 1 − 0.2 ⎜⎜ ⎝ P

− wf

− ⎛ P ⎞ ⎟ − 0.8 ⎜ wf ⎟ ⎜ P ⎠ ⎝

Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟ ⎟ ⎠

IPR Errors - Example • The values for qmax are:

qmax = 450 q

+ max

= 477

q

− max

= 424

424 bpd < qmax < 477 bpd

qmax ≈ 450 ± 6% Mauricio G. Prado – The University of Tulsa

IPR Errors - Example 1600 1400

Pressure (psi)

1200 1000 800 600 400

q

200

− max

qmax

+ qmax

0 0

100

200

300

400

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

500

600

IPR Errors - Homework • Assume we have a very reliable value of 1500 psi for Pr • The well test gives us the following data – q = 300 bpd @ Pwf = 1450 psi • Assume the test flowrate can be off by 5% • Assume the test pressure can be off by 1% • So the test pressure and flowrate ranges are: – 285 bpd < q < 315 bpd – 1435 psi < Pwf < 1465 psi • Estimate the IPR from this test • Estimate the errors in qmax. • Compare the errors in qmax with the errors from the previous example. Why do the errors increase so much ? • What are the consequences ?

Mauricio G. Prado – The University of Tulsa

IPR Errors - Homework • The values for qm are:

qmax = 5075

q

+ max

= 7473

q

− max

= 3754

5075 – 26% bpd < qmax < 5075 + 47% bpd

Mauricio G. Prado – The University of Tulsa

IPR Errors - Example 1600 1400

Pressure (psi)

1200 1000 800 600 400

q

200

qmax

− max

+ qmax

0 0

2000

4000 Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

6000

8000

Can we estimate the behavior of multiphase IPR in the future ?

Mauricio G. Prado – The University of Tulsa

Future IPR • The prediction of the future IPR is very important to forecast future well production. It is necessary to have simplified methods for estimating future IPR. • One of those methods (Eickemeier) states that the productivity index J* is proportional to square of reservoir pressure. • In this way, if we know the evolution of the reservoir pressure with depletion, we can also estimate the evolution of the IPR.

Mauricio G. Prado – The University of Tulsa

Future IPR

q = sc f

∫ 2π k h B µ sc

1 f

µ rf

dr ∫ krf r

Mauricio G. Prado – The University of Tulsa

dP

Future IPR Vogel’s Type Productivity Index is defined as

(2 + b) qmax J = P *

Eickemeir assumption is that J* is proportional to (Pr)2

⎛ P1 ⎞ J = ⎜⎜ ⎟⎟ J ⎝ P2 ⎠ * 1 * 2

2

Then

qmax1 ⎛ P1 ⎞ = ⎜⎜ ⎟⎟ qmax2 ⎝ P2 ⎠

Mauricio G. Prado – The University of Tulsa

3

Future IPR - Homework Data for a Saturated Reservoir: Conditions Today Pr

= 1500 psi

qmax = 1200 bpd

Estimate the Vogel IPR’s for: Pr = 1500, 1350, 1200, 1050, 900 and 750 psig

Mauricio G. Prado – The University of Tulsa

Future IPR Pr

qmax

1500

1200

1350 1200 1050 900 750 Mauricio G. Prado – The University of Tulsa

J*

Future IPR Pr

qmax

J*

1500

1200

1.44

1350

874.8

1.1664

1200

614.4

0.9216

1050

411.6

0.7056

900

259.2

0.5184

750

150

0.36

Mauricio G. Prado – The University of Tulsa

Future IPR 1600 1400

Pressure (psi)

1200 1000 800 600 400 200 0 0

200

400

600

800

1000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

1200

1400

Future IPR • This is not the only available method to estimate future Vogel type IPR for saturated reservoirs. • Usually all those methods try to use a Vogel type IPR (quadratic equation) and use some form of relationship between the value of the productivity index J* or the maximum flowrate qmax and the reservoir pressure Pr. Mauricio G. Prado – The University of Tulsa

Future IPR Vogel’s type Productivity Index is defined as

(2 + b) qmax J = P *

Assuming that changes in J* are a function F of changes in (Pr)

⎛ P1 ⎞ J 2* = F ⎜⎜ ⎟⎟ * J1 ⎝ P2 ⎠ We can obtain changes in qmax as a function of changes in (Pr)

qmax 2 P1 ⎛ P1 ⎞ = F ⎜⎜ ⎟⎟ qmax1 P2 ⎝ P2 ⎠ Mauricio G. Prado – The University of Tulsa

Future IPR Method

⎛ P1 ⎞ J 2* = F ⎜⎜ ⎟⎟ * J1 ⎝ P2 ⎠

Eickmeier

⎛ P1 ⎞ ⎜⎜ ⎟⎟ ⎝ P2 ⎠

Fetkovich

⎛ P1 ⎞ ⎜⎜ ⎟⎟ ⎝ P2 ⎠

Wiggins – Oil

⎛ P1 ⎞ 0.16 + 0.84 ⎜⎜ ⎟⎟ ⎝ P2 ⎠

Wiggins - Water

⎛ P1 ⎞ 0.6 + 0.4 ⎜⎜ ⎟⎟ ⎝ P2 ⎠ Mauricio G. Prado – The University of Tulsa

2

Future IPR 1 J* Productivity Index Ratio

Wiggins - Water

0.8

0.6 Wiggins - Oil

0.4 Eickmeier

Fetkovich

0.2

0 0

0.2

0.4

0.6

Reservoir Pressure Ratio Mauricio G. Prado – The University of Tulsa

0.8

1

Future IPR - Homework Data for a Saturated Reservoir: Conditions Today Pr

= 1500 psi

qmax = 1200 bpd Using Fetkovich, Eickemeier and Wiggins methods for evolution of the value of J*, estimate the Vogel IPR’s (with b = -0.2) for the following reservoir pressures Pr = 1500, 1350, 1200, 1050, 900 and 750 psig Mauricio G. Prado – The University of Tulsa

Future IPR Future Vogel IPR (b = - 0.2) Fetkovich

Eickmeier

Wiggins

Pr qmax 1500

1200

J*

qmax

J*

1200

1350 1200 1050 900 750

Mauricio G. Prado – The University of Tulsa

qmax 1200

J*

Future IPR Future Vogel IPR (b = - 0.2) Fetkovich

Eickmeier

Wiggins

Pr qmax

J*

qmax

J*

qmax

J*

1500

1200

1.44

1200

1.44

1200

1.44

1350

972

1.30

875

1.17

989

1.32

1200

768

1.15

614

0.92

799

1.20

1050

588

1.01

412

0.71

628

1.08

900

432

0.86

259

0.52

478

0.96

750

300

0.72

150

0.36

348

0.84

Mauricio G. Prado – The University of Tulsa

Future IPR 1600 1400 Pressure (psi)

1200 1000 800

Wiggins Fetkovich

600

Eickmeier

400 200 0 0

200

400

600

800

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

1000

1200

1400

IPR • The linear IPR is valid for single phase flow of fluids in the reservoir. It is not valid for compressible flow. • For saturated reservoirs, the linear IPR is no longer valid and correlations should be used. Vogel type correlations although develloped for solution gas drive reservoirs have been applied successfully in fields producing with other mechanisms. They are only valid for saturated reservoirs. Mauricio G. Prado – The University of Tulsa

The linear IPR concept is valid when the flow of fluids in the reservoir is single phase – Single Phase Reservoir The Vogel IPR concept is valid when the flow of fluids in the reservoir is always in two phase flow – Saturated Reservoir Mauricio G. Prado – The University of Tulsa

Can we estimate the IPR for undersaturated reservoirs ?

Mauricio G. Prado – The University of Tulsa

Single Phase Reservoir – Linear IPR

P

Pr

Pwf q

rr

rw Mauricio G. Prado – The University of Tulsa

Saturated Reservoir – Vogel Approach P Pb Pr

Pwf q

rr

rw Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir - ??? P

Pr Pb

Pwf q

rr

rw Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir P

Pr Pwf Pb For undersaturated reservoirs, if the bottomhole flowing pressure is above the bubble point, the linear IPR approach can be used

rr

rw Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir P q Pr Pwf = Pb For undersaturated reservoirs, the linear IPR approach can be used for low flowrates. There is a upper limit that the linear IPR can be used in this case. The limit is given by the flowrate that causes the bottomhole flowing pressure to be equal to the bubble point.

rr

rw Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir P q Pr Pb Beyond this limit, we will have two regions in the reservoir. One close to the wellbore where free gas is present and the pressure is below the bubble point. The other region far away from the wellbore the fluid is single phase since the pressure is above the bubble point.

Pwf

rr

rw Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir P

P

Pr

Pr

Pb

Pb

q

q

In an undersaturated reservoir, for low flowrates, the IPR is linear and for higher flowrates, the IPR will not follow the linear behavior any longer, since there is a region close to the wellbore that contains free gas. Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir q = J (P − Pwf )

P

For the linear part, we can use the linear IPR equation, but for higher flowrates we can not use Vogel approach directly since in his work it was assumed that the whole reservoir was below the bubble point. The important question then is how to adapt Vogel approach for this case.

Pb

? q qb

qmax Mauricio G. Prado – The University of Tulsa

Single Phase Reservoir Pr

P

q = 2π h sc f

∫

Pwf

1

dP

Bf µ f rr

∫

rw

=

dr kf r

Pr

2π hk Bf µ f

∫ dP dr ∫r

P P Pb

Pwf qf q rw

re Mauricio G. Prado – The University of Tulsa

Saturated Reservoir Pr

P

q = 2π h sc f

∫

Pwf

1 Bf µ f rr

∫

Pb

rw

dP

dr kf r

Pr P P Pb

Pwf qf q rw

re Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir Pr

q = 2π h sc f

∫

Pwf

1 Bf µ f rr

Pb

dP = 2π h

∫

Pwf

1 Bf µ f

dP + ∫

Pb

1

dP

Bf µ f

rr

dr ∫r k f r w

dr ∫r k f r w

P

Pr

Pr Pb

Pb

q scf = 2 π h

∫

Pwf

1 Bf µ f rb

dr ∫r k f r wf

Pwf qf

rw

re Mauricio G. Prado – The University of Tulsa

Pr

dP = 2π h

∫B

Pb

rr

1 f

µf

dr ∫r k f r b

dP

Undersaturated Reservoir Pr

q scf = 2 π h

P Pr

∫

Pwf

1 Bf µ f

dP

rr

dr ∫k r rw f

qb

Pb

Pb

q scf = 2 π h

q

Pwf

rw

re Mauricio G. Prado – The University of Tulsa

∫

Pwf

1 Bf µ f

Pr

dP + ∫

1

Bf µ f Pb

rr

dr ∫k r rw f

dP

Undersaturated Reservoir Pr

Pwf > Pb

⇒ q = 2π h sc f

∫

Pwf

1 Bf µ f rr

dr ∫r k f r w Pr

Pwf = Pb

dP

⇒ q = qb = 2 π h sc f

Mauricio G. Prado – The University of Tulsa

∫B

Pb

1 f

rr

µf

dr ∫r k f r w

dP

Undersaturated Reservoir Pb

Pwf < Pb

⇒ q scf = 2 π h

∫

Pwf

Pr

1

dP + ∫

Bf µ f

Pb

⇒ q scf = 2 π h

Bf µ f

rr

dr ∫r k f r w Pb

Pwf < Pb

1

∫

Pwf

1 Bf µ f rr

dr ∫r k f r w

Mauricio G. Prado – The University of Tulsa

dP + qb

dP

P

Undersaturated Reservoir q = J (P − Pwf )

P

Pb

Pb

q scf − qb = 2 π h

∫

Pwf

1

dP

Bf µ f rr

dr ∫r k f r w

q qb

qmax Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir Pb

P

q − qb = 2 π h sc f

Pb

∫

Pwf

1 Bf µ f

P ' = Pb

dP

rr

dr ∫r k f r w

q’ = q - qb q qb

qmax Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir Pr'

P Pr

qsc = 2 π h '

∫

Pwf

P ' = Pb

q’ = q - qb qmax - qb

Mauricio G. Prado – The University of Tulsa

1 Bf µ f rr

dr ∫r k f r w

dP

Undersaturated Reservoir P Pr

⎛ Pwf = 1 + b ⎜⎜ ' qmax ' ⎝P q'

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

P ' = Pb

q’ = q - qb qmax’ = qmax - qb Mauricio G. Prado – The University of Tulsa

2

Pwf ⎞ ⎟ ' ⎟ ⎝ P ⎠

P

Undersaturated Reservoir q = J (P − Pwf )

P

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

Pb

2

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎞ ⎟⎟ ⎝ Pb ⎠

q qb

qmax Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir J* = J

P

J

b

( 2 + b )(qmax − qb ) = Pb

Pb

qmax

J b Pb = + qb (2 + b )

q qb

qmax Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir Parameters : J and P '

P Pb Parameters : Pb , qb and Jb

q qb

qmax Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir Parameters : J and P

P

q Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir Parameters : J , P and Pb

P Pb

q Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir Parameters : J , P and Pb

P

Pb Parameters : qb and Jb

q Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir Parameters : J , P and Pb

P

qb = J (Pr − Pb )

Pb Parameters : Jb

q Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir Parameters : J , P and Pb

P

qb = J (P − Pb )

Pb

Jb = J

q Mauricio G. Prado – The University of Tulsa

P

Undersaturated Reservoir q = J (P − Pwf )

P

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

Pb

2

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎞ ⎟⎟ ⎝ Pb ⎠

qb = J (P − Pb ) qmax

J Pb = + qb (2 + b ) q

qb

qmax Mauricio G. Prado – The University of Tulsa

Homework - Undersaturated Reservoir Pwf

q

J = 2 bpd/psi

2000

0

Pb = 1500 psi

1750

qb

=

1500

qmax =

1000

Pr = 2000 psi

500 0

Mauricio G. Prado – The University of Tulsa

Homework - Undersaturated Reservoir Pwf

q

J = 2 bpd/psi

2000

0

Pb = 1500 psi

1750

500

Pr = 2000 psi

1500 1000 qb = 2 (2000 – 1500) = 1000 qmax = 2 1500 / 1.8 + 1000 = 2666

1000 1851 500

2407

0

2666

Mauricio G. Prado – The University of Tulsa

Homework - Undersaturated Reservoir 2500

Pressure (psi)

2000

1500

1000

500

0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Can we summarize now the behavior of future IPR ?

Mauricio G. Prado – The University of Tulsa

Evolution of IPR 2000

J = Constant

1800

J = J b*

1600

Pb

Pressure (psi)

1400 1200

⎛P⎞ J* = F ⎜⎜ ⎟⎟ * Jb ⎝ Pb ⎠

1000 800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Evolution of IPR - Homework

Initial Reservoir Pressure Pr = 2000 psi Bubble Point Pressure Pb = 1400 psi Initial Productivity Index J* = 2 stb/d/psi Assume Eickmeier model for future IPR

Mauricio G. Prado – The University of Tulsa

Evolution of IPR - Homework Pr

2000 1800 1600 1400 1200 1000

Pb

1400 1400 1400 1400 1400 1400 1400

J* qb

2

2

2

800

2 0

qm

Mauricio G. Prado – The University of Tulsa

0

0

0

Evolution of IPR - Homework Pr

2000

1800

1600

1400

1200

1000

800

Pb

1400

1400

1400

1400

1400

1400

1400

J*

2

2

2

2

1.469

1.020

0.653

qb

1200

800

400

0

0

0

0

qm

2755

2355

1955

1555

979

566

290

Mauricio G. Prado – The University of Tulsa

Evolution of IPR 2000

J = Constant

1800 1600

Pb

Pressure (psi)

1400 1200

⎛P⎞ J = ⎜⎜ ⎟⎟ * J b ⎝ Pb ⎠ *

1000 800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

2

Skin Effect • •

•

•

• •

The problem now is how to obtain expressions for the IPR for saturated and undersaturated reservoirs that present skin. The linear portion is a trivial exercise, but for the saturated region, you must remember that the methods of Vogel, Fetkovich, etc... Are valid only for reservoirs without damage. The problem is that the real data obtained from production tests for reservoir that have a damage reflects a bottomhole flowing pressure with the skin effect. For those cases to be properly analysed, we must have information on the skin effect (skin factor and reservoir properties or flow efficiency) so that we can recover the tests bottonhole flowing pressures without skin. Once the test data is corrected by the skin, we can estimate the IPR parameters After the parameters for the IPR without skin have been obtained, we can estimate the real IPR by re-incorporating the skin effect on the bottomhole flowing pressures Mauricio G. Prado – The University of Tulsa

Skin Effect • The first method was proposed by Standing as an extension to Vogel method. • The follwoing procedure illustrates Standing Method

Mauricio G. Prado – The University of Tulsa

Skin Effect • •

•

Example The following data is available for a well test Reservoir Pressure

2400

psi

Bubble Point Pressure

3000

psi

Bottomhole Test Pressure

1800

psi

Test Flowrate

70

bpd

Flow Efficiency

0.7

Determine the IPR for this well.

Mauricio G. Prado – The University of Tulsa

Skin Effect Bottomhole Flowing Pressure (psi)

3000 2500 2000 1500 1000 500 0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

Skin Effect • •

The data seems to be from a saturated reservoir. The well presents a damage, so we can not use Vogel equation direclty.

• The first step is to correct the test data for the skin

Pwftest = 1800

(

∆Pskin = (1 − FE ) Pr − Pwf

)

∆Pskin = (1 − 0.7 )(2400 − 1800) ∆Pskin = 180

Pwftestno − skin = Pwftest + ∆Pskin Pwftestno − skin = 1800 + 180 = 1980 Mauricio G. Prado – The University of Tulsa

Skin Effect Bottomhole Flowing Pressure (psi)

3000 2500 2000 1500 1000 500 0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

Skin Effect •

There is no need to calculate the reservoir pressure, since it is known.

Pr = 2400 •

The maximum flowrate parameter for Vogel Equation (b=-0.2) for the undamaged reservoir:

qmax =

qmax =

q ⎛P⎞ ⎛P⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b )⎜⎜ ⎟⎟ ⎝ Pr ⎠ ⎝ Pr ⎠

2

70 ⎛ 1980 ⎞ ⎛ 1980 ⎞ 1 − 0.2 ⎜ ⎟ − 0.8 ⎜ ⎟ 2400 2400 ⎝ ⎠ ⎝ ⎠

qmax = 241bpd

Mauricio G. Prado – The University of Tulsa

2

Skin Effect •

The IPR for the undamaged reservoir is then:

⎛ Pwf no − skin ⎞ ⎛ Pwf no − skin ⎞ q ⎜ ⎟ ⎟⎟ = 1+ b ⎜ − (1 + b) ⎜⎜ ⎟ qmax ⎝ Pr ⎠ ⎝ Pr ⎠

⎛ Pwf no − skin ⎞ ⎛ Pwf no − skin ⎞ q ⎟⎟ ⎟⎟ − 0.8⎜⎜ = 1 − 0.2 ⎜⎜ 241 ⎝ 2400 ⎠ ⎝ 2400 ⎠

Mauricio G. Prado – The University of Tulsa

2

2

Skin Effect Bottomhole Flowing Pressure (psi)

3000

q qmax

2500

⎛ Pwf no − skin ⎞ ⎛P ⎞ ⎟⎟ − (1 + b) ⎜⎜ wf no − skin ⎟⎟ = 1 + b ⎜⎜ ⎝ Pr ⎠ ⎝ Pr ⎠

2

2000 1500 1000 500 0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

Skin Effect •

We can now use the undamage IPR to calculate the damaged IPR:

⎛ Pwf no − skin ⎞ ⎛ Pwf no − skin ⎞ q ⎟⎟ ⎟⎟ − 0.8⎜⎜ = 1 − 0.2 ⎜⎜ 241 ⎝ 2400 ⎠ ⎝ 2400 ⎠

2

(

∆Pskin = (1 − FE ) Pr − Pwf

Pwf no − skin = Pwf + ∆Pskin

(

Pwf no − skin = Pwf + (1 − FE ) Pr − Pwf

)

Pwf no − skin = (1 − FE )Pr + FE Pwf

Mauricio G. Prado – The University of Tulsa

)

Skin Effect •

The IPR for the damage well becomes: ⎛ Pwf no − skin ⎞ ⎛P ⎞ q ⎟⎟ − 0.8⎜⎜ wf no − skin ⎟⎟ = 1 − 0.2 ⎜⎜ 241 ⎝ 2400 ⎠ ⎝ 2400 ⎠

2

Pwf no − skin = (1 − FE )Pr + FE Pwf ⎛ (1 − FE )Pr + FE Pwf q = 1 − 0.2 ⎜ ⎜ 241 2400 ⎝

•

⎞ ⎛ (1 − FE )Pr + FE Pwf ⎟ − 0.8⎜ ⎟ ⎜ 2400 ⎠ ⎝

For a flow efficiency of 0.7 we have ⎛ 0.3Pr + 0.7 Pwf q = 1 − 0.2 ⎜ ⎜ 241 2400 ⎝ ⎛ 720 + 0.7 Pwf q = 1 − 0.2 ⎜ ⎜ 241 2400 ⎝

⎞ ⎛ 0.3Pr + 0.7 Pwf ⎟ − 0.8⎜ ⎟ ⎜ 2400 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

⎞ ⎛ 720 + 0.7 Pwf ⎟ − 0.8⎜ ⎟ ⎜ 2400 ⎠ ⎝

2

Mauricio G. Prado – The University of Tulsa

⎞ ⎟ ⎟ ⎠

2

⎞ ⎟ ⎟ ⎠

2

Skin Effect ⎛ Pwf no − skin ⎞ ⎛ Pwf no − skin ⎞ q ⎜ ⎟ ⎟⎟ = 1+ b ⎜ − (1 + b) ⎜⎜ ⎟ qmax ⎝ Pr ⎠ ⎝ Pr ⎠

Bottomhole Flowing Pressure (psi)

3000 2500

q qmax

2000

2

P ⎞ Pwf ⎞ ⎛ ⎛ ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE wf ⎟⎟ = 1 + b ⎜⎜ (1 − FE ) + FE Pr ⎠ Pr ⎠ ⎝ ⎝ Sandface IPR

1500 1000 500

IPR

0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

2

Skin Effect •

The maximum flowrate for the IPR with skin is: ⎛ 720 + 0.7 Pwf q = 1 − 0.2 ⎜ ⎜ 241 2400 ⎝

⎞ ⎛ 720 + 0.7 Pwf ⎟ − 0.8⎜ ⎟ ⎜ 2400 ⎠ ⎝

q ⎛ 720 ⎞ ⎛ 720 ⎞ = 1 − 0.2 ⎜ ⎟ ⎟ − 0.8⎜ 241 ⎝ 2400 ⎠ ⎝ 2400 ⎠

2

q = 209 bpd

Mauricio G. Prado – The University of Tulsa

⎞ ⎟ ⎟ ⎠

2

Skin Effect q

Bottomhole Flowing Pressure (psi)

3000

qmax

⎛ Pwf no − skin ⎞ ⎛P ⎞ ⎟⎟ − (1 + b) ⎜⎜ wf no − skin ⎟⎟ = 1 + b ⎜⎜ ⎝ Pr ⎠ ⎝ Pr ⎠

q

2500

qmax

2

P ⎞ Pwf ⎞ ⎛ ⎛ ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE wf ⎟⎟ = 1 + b ⎜⎜ (1 − FE ) + FE Pr ⎠ Pr ⎠ ⎝ ⎝

2000

2

The IPR with skin is less curved than the skinless IPR.

1500

Why ???

1000

Sandface IPR

500

IPR

0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

Skin Effect Bottomhole Flowing Pressure (psi)

3000 Similar

µq ⎛ re ⎞ P r P ( ) ln = − ⎜ ⎟ r Saturation 0.00708 k h ⎝ r ⎠

2500

Profiles

2000 1500

∆Pskin = Pwfno − skin − Pwf

1000 500 0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

Skin Effect •

For a saturated IPR we have:

⎛ Pwf no − skin ⎞ ⎛ Pwf no − skin ⎞ q ⎟⎟ − (1 + b) ⎜⎜ ⎟⎟ = 1 + b ⎜⎜ qmax ⎝ Pr ⎠ ⎝ Pr ⎠

2

Pwf no − skin = (1 − FE )Pr + FE Pwf •

Or:

q qmax

Pwf ⎞ Pwf ⎞ ⎛ ⎛ ⎜ ⎟ ⎜ ⎟⎟ = 1 + b ⎜ (1 − FE ) + FE − (1 + b) ⎜ (1 − FE ) + FE ⎟ Pr ⎠ Pr ⎠ ⎝ ⎝

Mauricio G. Prado – The University of Tulsa

2

Skin Effect For a saturated IPR using Vogel we obtain: Bottomhole Flowing Pressure / Reservoir Pressure

•

q

1.0

qmax

P ⎞ P ⎞ ⎛ ⎛ = 1 + b ⎜⎜ (1 − FE ) + FE wf ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE wf ⎟⎟ Pr ⎠ Pr ⎠ ⎝ ⎝

0.9 0.8

1.0

0.7

0.9 0.8

0.6 0.5 0.4

0.7

0.3

Flow Efficiency

0.6

0.2

0.5

0.1 0.0 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Flowrate / Skinless Maximum Flowrate Mauricio G. Prado – The University of Tulsa

0.8

0.9

1.0

2

Skin Effect •

The value of J is

d Pwf noskin dq dq J =− =− d Pwf d Pwf noskin d Pwf •

The saturated IPR is

⎛ Pwf no − skin ⎞ ⎛ Pwf no − skin ⎞ q ⎟⎟ − (1 + b) ⎜⎜ ⎟⎟ = 1 + b ⎜⎜ qmax ⎝ Pr ⎠ ⎝ Pr ⎠ •

2

Pwf no − skin = (1 − FE )Pr + FE Pwf

Then:

⎡ b 2 (1 + b) ⎤ dq = qmax ⎢ P − wf no − skin ⎥ dPwf no − skin 2 Pr ⎣ Pr ⎦

dPwf no − skin = FE dPwf Mauricio G. Prado – The University of Tulsa

Skin Effect •

The value of J is

d Pwf noskin dq dq J =− =− d Pwf d Pwf noskin d Pwf ⎡ b 2 (1 + b) ⎤ − J = −qmax ⎢ Pwf no − skin ⎥ FE 2 Pr ⎦ ⎣ Pr Pwf no − skin = (1 − FE )Pr + FE Pwf •

Then:

⎡ b (1 − FE )Pr + FE Pwf J = −qmax ⎢ − 2 (1 + b) 2 P P r ⎣⎢ r

Mauricio G. Prado – The University of Tulsa

⎤ ⎥ FE ⎦⎥

Skin Effect •

The value of J is

⎡ b (1 − FE )Pr + FE Pwf J = −qmax ⎢ − 2 (1 + b) 2 Pr ⎢⎣ Pr •

⎤ ⎥ FE ⎥⎦

The value of Jo is

⎡ b ( 1 − FE )Pr ⎤ J = −qmax ⎢ − 2 (1 + b) ⎥ FE 2 Pr ⎦ ⎣ Pr o

J o = − FE •

qmax [ b − 2 (1 + b) (1 − FE )]FE Pr

For Jo to be zero we have:

0 = b − 2 (1 + b) (1 − FE )

FE = 1 −

b 2 (1 + b)

Mauricio G. Prado – The University of Tulsa

FE =

2+b 2 + 2b

Skin Effect •

The maximum value of FE is:

FE =

2+b 2 + 2b

b

FEmax

IPR Type

0

1

Fetkovick

-0.2

1.125

Vogel

-0.4

1.333

-0.8

3

-0.9

5.5

-1

Infinite Mauricio G. Prado – The University of Tulsa

Linear

Skin Effect For Vogel the maximum value of FE is 1.125:

Bottomhole Flowing Pressure / Reservoir Pressure

•

1.0

Flow Efficiency

0.9

1.1

0.8

1.125

0.7

1.5

0.6 0.5

0.8

0.4

0.9 0.3 0.2

1.0 0.1 0.0 0.0

0.2

0.4

0.6

0.8

Flowrate / Skinless Maximum Flowrate Mauricio G. Prado – The University of Tulsa

1.0

1.2

Skin Effect – Harrison Extension •

The value of J is

⎡ b (1 − FE )Pr + FE Pwf J = −qmax ⎢ − 2 (1 + b) 2 Pr ⎢⎣ Pr •

The value of J is negative when:

(1 − FE )Pr + FE Pwf b − 2 (1 + b) ≥0 2 Pr Pr b Pr − 2 (1 + b) (1 − FE )Pr − 2 (1 + b) FE Pwf ≥ 0

Pwf Pr

≤

b − 2 (1 + b) (1 − FE ) 2 (1 + b) FE

Mauricio G. Prado – The University of Tulsa

⎤ ⎥ FE ⎥⎦

Skin Effect Vogel IPR for J Positive: 1 Bottomhole Flowing Pressure / Reservoir Pressure

•

Flow Efficiency 3.00

0.9

2.50 0.8

2.25 2.00

0.7

1.75

0.6

0.6

0.5

1.50 0.7

0.4

0.8 0.3

1.25 0.9

b − 2 (1 + b) (1 − FE ) ≤ Pr 2 (1 + b) FE

0.2

Pwf

0.1

1.0 1.125

0 0

0.2

0.4

0.6

0.8

Flowrate / Skinless Maximum Flowrate

Mauricio G. Prado – The University of Tulsa

1

1.2

Skin Effect – Harrison Extension

•

Harrison proposed a second way of avoiding negative values for J. The method is based on a modification of the IPR equation for saturated reservoirs. The equation proposed by Harrison is as follows: q qmax

Bottomhole Flowing Pressure / Reservoir Pressure

• •

1.792

Pwf

= 1.2 − 0.2 e

noskin

q

Pr

qmax

Pwf

= 1.2 − 0.2 6

noskin

Pr

1 0.9 Pwf

0.8

q

0.7

qmax

= 1.2 − 0.2 6

0.6 0.5 0.4 0.3

q

0.2

qmax

⎛ Pwf ⎞ ⎟⎟ = 1 − 0.2 − 0.8 ⎜⎜ Pr P ⎝ r ⎠ Pwf

2

0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Flowrate / Skinless Maximum Flowrate

Mauricio G. Prado – The University of Tulsa

0.9

1

noskin

Pr

Skin Effect – Harrison Extension •

The IPR with skin then becomes: q qmax

Pwf

= 1.2 − 0.2 6

q qmax

•

noskin

Pwf no − skin = (1 − FE )Pr + FE Pwf

Pr

= 1.2 − 0.2 6

(1− FE )+ FE

Pwf Pr

The value of J IS ALWAYS POSITIVE !!!!! J =−

dq dPwf noskin

dPwf noskin dq dq =− dPwf dPwf noskin dPwf

q = −0.2 ln(6) max 6 Pr

Pwf

noskin

Pr

dPwf no − skin = FE dPwf

q J = 0.2 ln(6) FE max 6 Pr

Pwf

noskin

Pr

Mauricio G. Prado – The University of Tulsa

Skin Effect But ..... This second Harrison method is not good since it limits the maximum flowrate even for very high flow efficiencies: Bottomhole Flowing Pressure / Reservoir Pressure

•

1 0.9

q

0.8

qmax

0.7

= 1.2 − 0.2 6

(1− FE )+ FE

Pwf Pr

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

Flowrate / Skinless Maximum Flowrate

Mauricio G. Prado – The University of Tulsa

1.6

1.8

2

Skin Effect Harrison observed that: 1 Bottomhole Flowing Pressure / Reservoir Pressure

•

Flow Efficiency 3.00

0.9

2.50 0.8

2.25 2.00

0.7

1.75

0.6

0.6

0.5

1.50 0.7

0.4

0.8 0.3

1.25 0.9

b − 2 (1 + b) (1 − FE ) ≤ Pr 2 (1 + b) FE

0.2

Pwf

0.1

1.0 1.125

0 0

0.2

0.4

0.6

0.8

Flowrate / Skinless Maximum Flowrate

Mauricio G. Prado – The University of Tulsa

1

1.2

Skin Effect •

Harrison observed that: 10

⎛ Pwf 1− ⎜ ⎜ P ⎝ r

⎞ ⎟ ⎟ ⎠

2

1

0.1

0.01 0.01

0.1

1

Flowrate / Skinless Maximum Flowrate

Mauricio G. Prado – The University of Tulsa

10

Skin Effect •

Harrison proposed to approximate the IPRs when the FE is greater than the critical value by: 10

⎛ Pwf 1− ⎜ ⎜ P ⎝ r

⎞ ⎟ ⎟ ⎠

2

⎛ ⎛P q wf = C ⎜1 − ⎜ ⎜ ⎜ P qmax ⎝ ⎝ r

⎞ ⎟ ⎟ ⎠

2

⎞ ⎟ ⎟ ⎠

n

1

0.1

0.01 0.01

0.1

1

Flowrate / Skinless Maximum Flowrate

Mauricio G. Prado – The University of Tulsa

10

Skin Effect – Harrison Extension • •

The equation proposed by Harrison still needs two conditions so that the expoent n and the coefficient C can be determined. The method proposed by Harrison is the following: – Determine the qmax for the skinless IPR as before – Determine the FE critical value: FEcrit =

2+b 2 + 2b

– IF FE is greater thant the critical value, calculate the critical value for the bottomhole flowing pressure : Pwfcrit = Pr

b − 2 (1 + b) (1 − FE ) 2 (1 + b) FE

– Determine two extra points on the IPR where the regular behavior is observed: PA = Pwfcrit +

(

4 Pr − Pwfcrit 6

)

PB = Pwfcrit +

(

5 Pr − Pwfcrit 6

)

– Calculate the flowrates at those two points by using the regular equation for the quadratic IPR with flow efficiency. ⎛ ⎛ P ⎞ P ⎞ qA = 1 + b ⎜⎜ (1 − FE ) + FE A ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE A ⎟⎟ Pr ⎠ Pr ⎠ qmax ⎝ ⎝

2

⎛ ⎛ P ⎞ P ⎞ qB = 1 + b ⎜⎜ (1 − FE ) + FE B ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE B ⎟⎟ Pr ⎠ Pr ⎠ qmax ⎝ ⎝

Mauricio G. Prado – The University of Tulsa

2

Skin Effect – Harrison Extension • •

The equation proposed by Harrison still needs two conditions so that the expoent n and the coefficient C can be determined. The method proposed by Harrison is the following: – Determine the values of C and n as: ⎛ qA ⎞ ln⎜⎜ B ⎟⎟ ⎝q ⎠ n= ⎛ Pr2 − PA2 ⎞ ⎟ ln⎜⎜ 2 2 ⎟ ⎝ Pr − PB ⎠

C=

qA qmax ⎛ ⎛P ⎜1 − ⎜ A ⎜ ⎜⎝ Pr ⎝

⎞ ⎟⎟ ⎠

2

⎞ ⎟ ⎟ ⎠

n

– If Flow Efficiency is smaller than the critical value, the IPR is: q qmax

P ⎞ P ⎞ ⎛ ⎛ = 1 + b ⎜⎜ (1 − FE ) + FE wf ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE wf ⎟⎟ Pr ⎠ Pr ⎠ ⎝ ⎝

– If Flow Efficiency is greater than the critical value, the IPR is: ⎛ ⎛P q wf = C ⎜1 − ⎜ ⎜ ⎜ P qmax ⎝ ⎝ r

⎞ ⎟ ⎟ ⎠

2

⎞ ⎟ ⎟ ⎠

n

Mauricio G. Prado – The University of Tulsa

2

Bottomhole Flowing Pressure / Reservoir Pressure

Skin Effect q

1

qmax

0.9

P ⎞ Pwf ⎞ ⎛ ⎛ ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE wf ⎟⎟ = 1 + b ⎜⎜ (1 − FE ) + FE Pr ⎠ Pr ⎠ ⎝ ⎝

2

⎛ ⎛P q wf = C ⎜1 − ⎜ ⎜ ⎜ P qmax ⎝ ⎝ r

0.8 0.7

⎞ ⎟ ⎟ ⎠

2

⎞ ⎟ ⎟ ⎠

n

0.6 0.5 0.4 0.3

FEcrit =

0.2

2+b 2 + 2b

0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

Flowrate / Skinless Maximum Flowrate

Mauricio G. Prado – The University of Tulsa

1.6

1.8

2

Skin Effect – Undersaturated Case • •

Lets examine the case of an undersaturated reservoir First we calculate the new linear part. q = J FE ( Pr − Pwf )

•

Now we can calculate the new flowrate that would occur at the bubble point for this FE q A = J FE ( Pr − Pb )

•

The maximum flowrate for the saturated IPR with this FE is: qC =

•

J FE Pb + qA 2+b

The saturated IPR equation is: ⎛ Pwf q − qA = 1 + b ⎜⎜ qC − q A ⎝ Pb

⎛P ⎞ ⎟⎟ − (1 + b)⎜⎜ wf ⎝ Pb ⎠

Mauricio G. Prado – The University of Tulsa

⎞ ⎟⎟ ⎠

2

Skin Effect – Undersaturated Case •

The undersaturated IPR is given by:

⎧ Pwf ≥ Pb ⇒ q = J FE ( Pr − Pwf ) ⎪⎪ Pr ≥ Pb ⇒ ⎨ J FE Pb ( ) < ⇒ = − + P P q J FE P P wf b r b ⎪ 2+b ⎪⎩

⎡ ⎛P ⎢1 + b ⎜⎜ wf ⎢⎣ ⎝ Pb

Mauricio G. Prado – The University of Tulsa

⎞ ⎛P ⎟⎟ − (1 + b)⎜⎜ wf ⎠ ⎝ Pb

⎞ ⎟⎟ ⎠

2

⎤ ⎥ ⎥⎦

Skin Effect ⎧ Pwf ≥ Pb ⇒ q = J FE ( Pr − Pwf ) ⎪⎪ Pr ≥ Pb ⇒ ⎨ J FE Pb ⎪ Pwf < Pb ⇒ q = J FE ( Pr − Pb ) + 2 + b ⎪⎩

4000

Bottomhole Flowing Pressure (psi)

3500

⎡ ⎛P ⎢1 + b ⎜⎜ wf ⎢⎣ ⎝ Pb

⎛P ⎞ ⎟⎟ − (1 + b)⎜⎜ wf ⎝ Pb ⎠

⎞ ⎟⎟ ⎠

2

⎤ ⎥ ⎥⎦

3000 1.0

Flow Efficiency

1.2

2500

1.4

0.5

2000

1.6 1.8

0.6

2.0 0.7

1500

0.8 0.9

1000 500 0 0

1000

2000

3000

4000

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

5000

6000

Skin Effect – Summary •

The saturated IPR is given by: ⎧ ⎪ FE < ⎪ ⎪ Pr < Pb ⇒ ⎨ ⎪ ⎪ FE ≥ ⎪⎩ Pwfcrit = Pr

•

Pwf ⎞ P ⎞ ⎛ ⎛ 2+b q ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE wf ⎟⎟ ⇒ = 1 + b ⎜⎜ (1 − FE ) + FE 2 + 2b qmax Pr ⎠ Pr ⎠ ⎝ ⎝ 2 n ⎧ ⎞ ⎛ ⎞ ⎛ P 2+b ⎪ q wf ⎟ ⎜ ⎟ ⎜ ⇒⎨ = C 1− ⎟ ⎜ ⎟ ⎜ 2 + 2b ⎪ qmax ⎝ Pr ⎠ ⎠ ⎝ ⎩ ⎛ qA ⎞ ln⎜⎜ B ⎟⎟ ⎝q ⎠ n= ⎛ P 2 − PA2 ⎞ ⎟ ln⎜⎜ r2 2 ⎟ ⎝ Pr − PB ⎠

b − 2 (1 + b) (1 − FE ) 2 (1 + b) FE

PA = Pwfcrit +

(

)

PB = Pwfcrit

(

)

4 Pr − Pwfcrit 6 5 + Pr − Pwfcrit 6

⎛ ⎛ qA P ⎞ P ⎞ = 1 + b ⎜⎜ (1 − FE ) + FE A ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE A ⎟⎟ qmax P Pr ⎠ r ⎠ ⎝ ⎝

2

⎛ ⎛ qB P ⎞ P ⎞ = 1 + b ⎜⎜ (1 − FE ) + FE B ⎟⎟ − (1 + b) ⎜⎜ (1 − FE ) + FE B ⎟⎟ qmax Pr ⎠ Pr ⎠ ⎝ ⎝

2

2

C=

qA qmax ⎛ ⎛P ⎜1 − ⎜ A ⎜ ⎜⎝ Pr ⎝

The undersaturated IPR is given by: ⎧ Pwf ≥ Pb ⇒ q = J FE ( Pr − Pwf ) ⎪⎪ Pr ≥ Pb ⇒ ⎨ J FE Pb ( ) < ⇒ = − + P P q J FE P P wf b r b ⎪ 2+b ⎪⎩

⎡ ⎛P ⎢1 + b ⎜⎜ wf ⎢⎣ ⎝ Pb

Mauricio G. Prado – The University of Tulsa

⎞ ⎛P ⎟⎟ − (1 + b)⎜⎜ wf ⎠ ⎝ Pb

⎞ ⎟⎟ ⎠

2

⎤ ⎥ ⎥⎦

⎞ ⎟⎟ ⎠

2

⎞ ⎟ ⎟ ⎠

n

Skin Effect - Summary •

•

• • • • •

Get the test data (usually 2 points of bottomhole flowing pressure and flowrates). The data does not need to be collected at the “same time”. But you need to make sure that nothing major has happened between the tests (workover, high depletion, long time between tests, etc...). You need to be confident that the tests are for the same IPR !!! Check if there is any information on damage or stimulation valid for that IPR such as a build up test. If not, check if you have any information that may allow you to estimate the damage such as a previous production test. This could be a production test prior to a workover operation (to check if there was damage or stimulation in the well) or it could be a old production test. If it is an old production test (at a higher reservoir pressure) you may need to estimate the undamaged IPR for the current reservoir pressure level. If there is damage or stimulation, get the S value and the Flow Efficiency FE. You can now use the FE to calculate the values of the no-skin bottomhole flowing pressure for the tests you want to analyse With the no-skin test data, you can determine the skinless IPR parameters. With the skinless IPR parameters, you can use the equations to calculate the IPR with skin. If you need to estimate Future IPRs, remeber that this should be done based on the skinless IPR parameters and that the skin effect should be added later. You should get future skinless IPRs and then calculate the future IPRs with skin. Mauricio G. Prado – The University of Tulsa

Skin Effect - Summary •

You will need then 3 programs (spreadsheets) to help you analysing the IPR. – First Program – Skinless IPR • Based on two production tests (without skin or corrected to remove skin effect) this program calculates the skinless IPR parameters (Reservoir Pressure, Bubble Point Flowrate, Productivity Index, Productivity Index at Bubble Point and Maximum Flowrate). The program should graph the skinless IPR as well as the test data.

– Second Program – Skinless Future IPR • Based on the skinless IPR parameters (Reservoir Pressure, Bubble Point Flowrate, Productivity Index, Productivity Index at Bubble Point and Maximum Flowrate) and the Bubble Point Pressure, this program calculate the parameters of future skinless IPRs (Productivity Index,Productivity Index at Bubble Point and Maximum Flowrate as a function of Reservoir Pressure. The program should graph the future IPR for several user input reservoir pressure levels

– Third Program – Skin Analysis • Based on a test data (possibly with skin) and the skinless IPR parameters (Reservoir Pressure, Bubble Point Flowrate, Productivity Index, Productivity Index at Bubble Point and Maximum Flowrate), this program will determine the skin pressure loss, and the flow efficiency. The program should graph the skinless IPR, the test data and the IPR with skin.

– Fourth Program – IPR with Skin • Based on IPR parameters and flow efficiency, this program will calculate and graph the skinless IPR and the IPR with skin.

Mauricio G. Prado – The University of Tulsa

Skin Effect • This method has several disadvantages and given the nature of the approximation given by the IPR expressions another simpler method is proposed. • In this method we just recover from the test data and skin data the undamaged bottom hole flowing pressure. • Then the undamage bottom hole flowing pressure and the real bottom hole flowing pressure are used to calculate the IPR parameters for the undamaged and real IPRs Mauricio G. Prado – The University of Tulsa

Skin Effect • •

•

Example The following data is available for a well test Reservoir Pressure

2400

psi

Bubble Point Pressure

3000

psi

Bottomhole Test Pressure

1800

psi

Test Flowrate

70

bpd

Flow Efficiency

0.7

Determine the IPR for this well.

Mauricio G. Prado – The University of Tulsa

Skin Effect Bottomhole Flowing Pressure (psi)

3000 2500 2000 1500 1000 500 0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

Skin Effect • The first step is to correct the test data for the skin Pwftest = 1800

(

∆Pskin = (1 − FE ) Pr − Pwf

)

∆Pskin = (1 − 0.7 )(2400 − 1800) ∆Pskin = 180

Pwftestno − skin = Pwftest + ∆Pskin Pwftestno − skin = 1800 + 180 = 1980 Mauricio G. Prado – The University of Tulsa

Skin Effect Bottomhole Flowing Pressure (psi)

3000 2500 2000 1500 1000 500 0 0

50

100

150

200

Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

250

300

Skin Effect •

There is no need to calculate the reservoir pressure, since it is known.

Pr = 2400 •

The maximum flowrate parameter for Vogel Equation (b=-0.2) for the undamaged reservoir:

no − skin = qmax

q ⎛P⎞ ⎛P⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b )⎜⎜ ⎟⎟ ⎝ Pr ⎠ ⎝ Pr ⎠

2

no − skin = qmax

70 ⎛ 1980 ⎞ ⎛ 1980 ⎞ 1 − 0.2 ⎜ ⎟ − 0.8 ⎜ ⎟ 2400 2400 ⎝ ⎠ ⎝ ⎠

no − skin qmax = 241bpd

⎛ Pwf ⎞ ⎛ Pwf ⎞ q ⎟⎟ ⎟⎟ − 0.8⎜⎜ = 1 − 0.2 ⎜⎜ 241 ⎝ 2400 ⎠ ⎝ 2400 ⎠ Mauricio G. Prado – The University of Tulsa

2

2

Skin Effect •

For the damaged reservoir we have:

qmax =

q ⎛P⎞ ⎛P⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b )⎜⎜ ⎟⎟ ⎝ Pr ⎠ ⎝ Pr ⎠

2

qmax =

70 ⎛ 1800 ⎞ ⎛ 1800 ⎞ − 1 − 0.2 ⎜ 0 . 8 ⎟ ⎜ ⎟ ⎝ 2400 ⎠ ⎝ 2400 ⎠

qmax = 175 bpd

⎛ Pwf ⎞ ⎛ Pwf ⎞ q ⎟⎟ ⎟⎟ − 0.8⎜⎜ = 1 − 0.2 ⎜⎜ 175 ⎝ 2400 ⎠ ⎝ 2400 ⎠

Mauricio G. Prado – The University of Tulsa

2

2

Skin Effect Bottomhole Flowing Pressure (psi)

3000 2500 Undamaged IPR

2000 1500

Standing - Vogel IPR

1000 Vogel IPR

500 0 0

50

100

150

200

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

250

300

How do we calculate the IPR in real cases ?

Mauricio G. Prado – The University of Tulsa

Vogel Undersaturated Reservoir - Summary P

q = J (P − Pwf )

P

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

Pb

2

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎞ ⎟⎟ ⎝ Pb ⎠

q qb

qmax Mauricio G. Prado – The University of Tulsa

IPR from Well Tests • The main advantage of this simplified IPR procedure is to have a simple set of analytical expressions to represent the reservoir performance. • The set of equations defining the IPR for the undersaturated case depend on 5 parameters: – – – – –

qmax qb Pb J Pr

• The bubble pressure can be obtained from a fluid sample. • The remaining 4 parameters are not independent, since the IPR should be continuous and smooth at the bubble point. Mauricio G. Prado – The University of Tulsa

Continuity at the Bubble Point P

q = J (P − Pwf )

P

qb = J (P − Pb )

Pb ⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

2

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎞ ⎟⎟ ⎝ Pb ⎠

q qb

qmax Mauricio G. Prado – The University of Tulsa

Smoothness at the Bubble Point P

q = J (P − Pwf )

P dq J = Jb = − dPwf

Pb

Jb = − Pwf = Pb

dq dPwf

= Pwf = Pb

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

(2 + b )(qmax − qb ) Pb

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎞ ⎟⎟ ⎝ Pb ⎠

q qb

qmax Mauricio G. Prado – The University of Tulsa

2

Smoothness at the Bubble Point P

q = J (P − Pwf )

P

qmax Pb

J Pb = + qb 2+b ⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pwf ⎞ ⎟⎟ ⎝ Pb ⎠

q qb

qmax Mauricio G. Prado – The University of Tulsa

2

IPR from Well Tests • As a consequence, if the bubble point is known, the undersaturated IPR depends only on 2 of the 4 parameters below – – – –

qmax qb J Pr

• The remaining 2 parameters can be calculated by the auxiliary equations for continuity and smoothness

Mauricio G. Prado – The University of Tulsa

Undersaturated Reservoir - Summary P

q = J (P − Pwf )

P

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

Pb

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

2

Pwf ⎞ ⎟⎟ ⎝ Pb ⎠

qb = J (P − Pb ) qmax

J Pb = + qb 2+b q

qb

qmax Mauricio G. Prado – The University of Tulsa

IPR from Well Tests • We can then obtain the undersaturated IPR if 2 production tests are known. • Several cases are possible depending on the location of the well test data on the Pwf x Q plot • 4 Cases are possible – Both tests are above the bubble point – One test above and one test below the bubble point – Two tests below the bubble point and the reservoir is undersaturated. – Two tests below the bubble point and the reservoir is saturated

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests 2000 1800 1600

Pressure (psi)

1400

Pb

1200 1000 800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests 2000 1800 1600

Pressure (psi)

1400

Pb

1200 1000 800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests 2000 1800 1600

Pressure (psi)

1400

Pb

1200 1000 800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests • The main objective is to use the test data in order to obtain the values of: – – – –

qmax qb J Pr

• The mathematical procedure is very simple. • The test data must satisfy the appropriate equation for the IPR • The IPR must be continuous and smooth at the bubble point

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests • Case A – Both tests are above the bubble point

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case A q = J (P − P )

2000

wf

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

1800 1600

Pwf ⎝ Pb

Pb

1400 Pressure (psi)

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

2

⎞ ⎟⎟ ⎠

1200

qb = J (P − Pb )

1000 800 600

qmax

400

J Pb = + qb (2 + b )

200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests – Case A • Since both tests are in the linear region, we have:

q1 = J (P − P1 )

q2 = J (P − P2 ) • For the IPR to be continuous and smoth at the bubble point we have: qb = J (P − Pb ) qmax

J Pb = + qb (2 + b )

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case A • Solving the first two equations for the reservoir pressure and the productivity index:

q1 = J (P − P1 )

q2 = J (P − P2 )

q1 J (P − P1 ) (P − P1 ) = = q2 J (P − P2 ) (P − P2 ) q1 − q2 = J (P − P1 ) − J (P − P2 ) = J (P2 − P1 )

Mauricio G. Prado – The University of Tulsa

q2 P1 − q1 P2 P= q2 − q1

q2 − q1 J= P1 − P2

IPR from Well Tests – Case A • Finally: q2 P1 − q1 P2 P= q2 − q1

q2 − q1 J= P1 − P2 qb = J (P − Pb ) qmax

J Pb = + qb (2 + b )

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case A q2 P1 − q1 P2 P= q2 − q1

2000 1800

q2 − q1 J= P1 − P2

1600

Pb

Pressure (psi)

1400

qb = J (P − Pb )

1200 1000

qmax

800 600

J Pb = + qb (2 + b )

400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests - Homework • Determine the equations to calculate the IPR parameters (qmax, J, Pr, and qb) when we have the information on 2 production tests and the bubble point pressure. Use the equations to calculate the Vogel IPR for the following case: • Pb= 1400 – Tests - Pwf = 1500 q = 600 and Pwf = 1700 q = 200

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case A q2 P1 − q1 P2 P= q2 − q1

2000 1800

q2 − q1 J= P1 − P2

1600

Pb

Pressure (psi)

1400

qb = J (P − Pb )

1200 1000

qmax

800 600

J Pb = + qb (2 + b )

400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Case A Example Test 1

Test 2

Bubble Point

Pwf = 1700 q = 200

Pwf = 1500 q = 600

Pb= 1400

q2 P1 − q1 P2 P= = q2 − q1

qb = J (P − Pb ) =

q2 − q1 J= = P1 − P2

qmax

J Pb = + qb = 1.8

Mauricio G. Prado – The University of Tulsa

Case A Example Test 1

Test 2

Bubble Point

Pwf = 1700 q = 200

Pwf = 1500 q = 600

Pb= 1400

600 1700 − 200 1500 P= = 1800 600 − 200

600 − 200 J= =2 1700 − 1500

qb = 2(1800 − 1400) = 800

qmax

2 1400 = + 800 = 2355 1.8 Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case A 2000 1800 1600

Pb

Pressure (psi)

1400 1200 1000 800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests • Case B – One test above and one test below the bubble point

Mauricio G. Prado – The University of Tulsa

2000

IPR from Well Tests – Case B q = J (P − P ) wf

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

1800 1600

1

Pressure (psi)

1400

Pb

1200 1000

Pwf ⎝ Pb

600

qmax

400

J Pb = + qb (2 + b )

200 0 0

500

1000

⎞ ⎟⎟ ⎠

qb = J (P − Pb )

2

800

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

2

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests – Case B • Since the tests are in different regions, we have:

q1 = J (P − P1 )

⎛ P2 ⎛ P2 ⎞ q2 − qb = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P qmax − qb ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

• For the IPR to be continuous and smoth at the bubble point we have: qb = J (P − Pb ) qmax

J Pb = + qb (2 + b )

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case B • We can start calculating J. For example:

⎛ P2 ⎛ P2 ⎞ q2 − qb = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P qmax − qb ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

qmax

2

(q2 − qb )(2 + b ) = 1 + b ⎛⎜ P2 ⎞⎟ − (1 + b) ⎛ P2 ⎞ ⎜P ⎟ ⎝ b⎠

J Pb

(q

2

⎜⎜ ⎟⎟ ⎝ Pb ⎠

J Pb = + qb (2 + b )

qb = J (P − Pb )

⎛ P2 ⎛ P2 ⎞ − J (P − Pb ))(2 + b ) = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P J Pb ⎝ Pb ⎠ ⎝ b Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟⎟ ⎠

IPR from Well Tests – Case B • We can start calculating J. For example:

(q

2

⎛ P2 ⎛ P2 ⎞ − J (P − Pb ))(2 + b ) = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P J Pb ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

q1 = J (P − P1 )

⎛ ⎞⎞ ⎛q ⎜ q2 − J ⎜ 1 + P1 − Pb ⎟ ⎟(2 + b ) ⎛ P2 ⎛ P2 ⎞ ⎠⎠ ⎝J ⎝ = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P J Pb ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

⎛ ⎜ ⎛ P2 ⎛ P2 ⎞ q2 (2 + b ) − (2 + b )q1 − J (2 + b )P1 + J (2 + b )Pb = J Pb ⎜ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P ⎜⎜ ⎝ Pb ⎠ ⎝ b ⎝ Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟⎟ ⎠

⎞ ⎟ ⎟ ⎟⎟ ⎠

IPR from Well Tests – Case B • We can start calculating J. For example: 2 ⎛ ⎞ ⎜ ⎛ P2 ⎞ ⎟ ⎛ P2 ⎞ ⎟ q2 (2 + b ) − (2 + b )q1 − J (2 + b )P1 + J (2 + b )Pb = J Pb ⎜ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎟ ⎜ P ⎟ ⎟⎟ ⎜⎜ ⎝ Pb ⎠ ⎝ b⎠⎠ ⎝ 2 ⎛ ⎞ ⎜ ⎛ P2 ⎞ ⎟ ⎛ P2 ⎞ ⎟ + J (2 + b )P1 − J (2 + b )Pb = (2 + b )(q2 − q1 ) J Pb ⎜ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎟ ⎜ P ⎟ ⎟⎟ ⎜⎜ ⎝ Pb ⎠ ⎝ b⎠⎠ ⎝ 2 ⎛ ⎞ P2 ⎜ J ⎜ Pb + b P2 − (1 + b) + (2 + b )P1 − (2 + b )Pb ⎟⎟ = (2 + b )(q2 − q1 ) Pb ⎝ ⎠

J=

(2 + b )(q2 − q1 ) ⎛ P22 ⎞ (2 + b ) P1 + b P2 − (1 + b ) ⎜⎜ Pb + ⎟⎟ Pb ⎠ ⎝ Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case B • Now that we have J, we can calculate Pr:

q1 = J (P − P1 ) q1 Pr = + P1 J • Once we know J and Pr, we have: qb = J (P − Pb ) qmax

J Pb = + qb (2 + b )

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case B J=

2000 1800 1600

1

Pressure (psi)

1400

(2 + b )(q2 − q1 ) ⎛ P22 ⎞ (2 + b ) P1 + b P2 − (1 + b ) ⎜⎜ Pb + ⎟⎟ Pb ⎠ ⎝

Pb

1200 1000

P=

2

800

qmax

J Pb = + qb (2 + b )

500

1000

600 400

q1 + J P1 J

qb = J (P − Pb )

200 0 0

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests - Homework • Determine the equations to calculate the IPR parameters (qmax, J, Pr, and qb) when we have the information on 2 production tests and the bubble point pressure. Use the equations to calculate the Vogel IPR for the following case: • Pb= 1400 – Tests - Pwf = 1190 q = 1192 and Pwf = 1700 q = 200

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case B J=

2000 1800 1600

1

Pressure (psi)

1400

(2 + b )(q2 − q1 ) ⎛ P22 ⎞ (2 + b ) P1 + b P2 − (1 + b ) ⎜⎜ Pb + ⎟⎟ Pb ⎠ ⎝

Pb

1200 1000

P=

2

800

qmax

J Pb = + qb (2 + b )

500

1000

600 400

q1 + J P1 J

qb = J (P − Pb )

200 0 0

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Case B Example Test 1

Test 2

Bubble Point

Pwf = 1700 q = 200

Pwf = 1190 q = 1192

Pb= 1400

J=

1.8(q2 − q1 ) 2 2

P 1.8 P1 − 0.2 P2 − 0.8 Pb − 0.8 Pb

P=

=

q1 + J P1 = J

qb = J (P − Pb ) =

qmax

J Pb = + qb = 1.8

Mauricio G. Prado – The University of Tulsa

Case B Example Test 1

Test 2

Bubble Point

Pwf = 1700 q = 200

Pwf = 1190 q = 1192

Pb= 1400

J=

P=

1.8(1192 − 200 ) 2

1190 1.8 1700 − 0.2 1190 − 0.8 1400 − 0.8 1400

200 + 2 1700 = 1800 2

qmax

=2

qb = 2(1800 − 1400) = 800

2 1400 = + 800 = 2355 1.8 Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case B 2000 1800 1600

1

Pressure (psi)

1400

Pb

1200 1000

2

800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests • Cases C1 and C2 – Both tests are below the bubble point – The reservoir pressure can be above or below the bubble point. – We start first by assuming the reservoir pressure is above the bubble point. – We must test this assumption – The test involves the determination of qb.

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case C1 q = J (P − Pwf )

2000 1800

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

1600

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

Pressure (psi)

1400

Pb

1200

⎞ ⎟⎟ ⎠

qb = J (P − Pb )

1

1000

Pwf ⎝ Pb

2

800

2

600

qmax

400

J Pb = + qb (2 + b )

200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests – Case C1 • Since both tests are in the saturated region, we have:

⎛ P1 ⎛ P1 ⎞ q1 − qb = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P qmax − qb ⎝ Pb ⎠ ⎝ b ⎛ P2 ⎛ P2 ⎞ q2 − qb = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P qmax − qb ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

2

⎞ ⎟⎟ ⎠

• For the IPR to be continuous and smoth at the bubble point we have: qb = J (P − Pb ) J Pb qmax = + qb (2 + b ) Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case C1 • If the bubble point flowrate is greater than zero, then the reservoir is subsaturated. The bubble point flowrate is calculated by:

⎛P ⎛P⎞ q1 − qb = 1 + b ⎜⎜ 1 ⎟⎟ − (1 + b) ⎜ 1 ⎜P qmax − qb ⎝ Pb ⎠ ⎝ b q1 − qb

⎛ P1 ⎛ P1 ⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

⎛ P2 ⎛ P2 ⎞ q2 − qb ⎜ ⎟ = 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎜P qmax − qb ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

q2 − qb

= qmax − qb

q2 − qb

⎛ P2 ⎛ P2 ⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

⎛ P2 ⎛ P2 ⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P ⎝ Pb ⎠ ⎝ b =

q1 − qb

⎛ P1 ⎛ P1 ⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P ⎝ Pb ⎠ ⎝ b

Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟⎟ ⎠

2

⎞ ⎟⎟ ⎠

2

⎞ ⎟⎟ ⎠

= qmax − qb

IPR from Well Tests – Case C1 • If the bubble point flowrate is greater than zero, then the reservoir is subsaturated. The bubble point flowrate is calculated by: q2 − qb

⎛ P2 ⎛ P2 ⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

=

q1 − qb

⎛ P1 ⎛ P1 ⎞ 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜1 + b P2 − (1 + b )⎜ P2 ⎟ ⎟ q − ⎜1 + b P1 − (1 + b )⎜ P1 ⎟ ⎟ q ⎜P ⎟ ⎟ 1 ⎜ ⎜P ⎟ ⎟ 2 ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ qb = ⎝ 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ P P P P 2 2 1 1 ⎜1 + b − (1 + b )⎜ ⎟ ⎟ − ⎜1 + b − (1 + b )⎜ ⎟ ⎟ ⎜P ⎟ ⎟ ⎜ ⎜P ⎟ ⎟ ⎜ Pb Pb b ⎠ ⎝ ⎝ b⎠ ⎠ ⎝ ⎠ ⎝

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case C1 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜1 + b P2 − (1 + b )⎜ P2 ⎟ ⎟ q − ⎜1 + b P1 − (1 + b )⎜ P1 ⎟ ⎟ q ⎜P ⎟ ⎟ 1 ⎜ ⎜P ⎟ ⎟ 2 ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ qb = ⎝ 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ P P P P 2 2 1 1 ⎜1 + b − (1 + b )⎜ ⎟ ⎟ − ⎜1 + b − (1 + b )⎜ ⎟ ⎟ ⎜P ⎟ ⎟ ⎜ ⎜P ⎟ ⎟ ⎜ Pb Pb b ⎠ ⎝ ⎝ b⎠ ⎠ ⎝ ⎠ ⎝

2000 1800 1600

Pressure (psi)

1400

Pb

1200

qb ≥ 0

1

1000 800

2

600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests – Case C1 • Once we confirm that the reservoir is undersaturated and we have obtained the value of qb, we proceed to calculate the reservoir pressure:

⎛ P2 ⎛ P2 ⎞ q2 − qb = 1 + b ⎜⎜ ⎟⎟ − (1 + b) ⎜ ⎜P qmax − qb ⎝ Pb ⎠ ⎝ b

2

⎞ ⎟⎟ ⎠

qmax

J Pb = + qb (2 + b ) 2

(q2 − qb )(2 + b ) = 1 + b ⎛⎜ P2 ⎞⎟ − (1 + b) ⎛ P2 ⎞ J Pb

J=

⎜P ⎟ ⎝ b⎠

⎜⎜ ⎟⎟ ⎝ Pb ⎠

(2 + b ) (q2 − qb ) 2 ⎛ ⎛ ⎞ P2 P2 ⎞⎟ ⎜ Pb 1 + b − (1 + b) ⎜⎜ ⎟⎟ ⎜ Pb Pb ⎠ ⎟ ⎝ ⎝ ⎠ Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case C1 • Now we calculate the remaining parameters: qb = J (P − Pb )

P = Pb +

qmax

J Pb = + qb (2 + b )

Mauricio G. Prado – The University of Tulsa

qb J

2000

IPR from Well Tests – Case C1 ( 2 + b )(q2 − qb ) J= 2 ⎛ ⎞ ⎛ ⎞ P P Pb ⎜1 + b 2 − (1 + b) ⎜⎜ 2 ⎟⎟ ⎟ ⎜ Pb Pb ⎠ ⎟ ⎝ ⎝ ⎠

1800 1600

Pressure (psi)

1400

Pb

1200

P = Pb +

1

1000

qb J

800

qmax =

2

600

J Pb + qb (2 + b )

400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests - Homework • Determine the equations to calculate the IPR parameters (qmax, J, Pr, and qb) when we have the information on 2 production tests and the bubble point pressure. Use the equations to calculate the Vogel IPR for the following case: • Pb= 1400 – Tests - Pwf = 770 q = 1808 and Pwf = 980 q = 1528

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case C1 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜1 + b P2 − (1 + b )⎜ P2 ⎟ ⎟ q − ⎜1 + b P1 − (1 + b )⎜ P1 ⎟ ⎟ q ⎜P ⎟ ⎟ 1 ⎜ ⎜P ⎟ ⎟ 2 ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ qb = ⎝ 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ P P P P 2 2 1 1 ⎜1 + b − (1 + b )⎜ ⎟ ⎟ − ⎜1 + b − (1 + b )⎜ ⎟ ⎟ ⎜P ⎟ ⎟ ⎜ ⎜P ⎟ ⎟ ⎜ Pb Pb b ⎠ ⎝ ⎝ b⎠ ⎠ ⎝ ⎠ ⎝

2000 1800 1600

Pressure (psi)

1400

Pb

1200

qb ≥ 0

1

1000 800

2

600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Case C1 Example Test 1

Test 2

Bubble Point

Pwf = 980 q = 1528

Pwf = 770 q = 1808

Pb= 1400

2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ P P P P ⎜1 − 0.2 2 − 0.8⎜ 2 ⎟ ⎟ q − ⎜1 − 0.2 1 − 0.8⎜ 1 ⎟ ⎟ q ⎜P ⎟ ⎟ 1 ⎜ ⎜P ⎟ ⎟ 2 ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ ⎝ qb = = 2 2 ⎞ ⎞ ⎛ ⎛ ⎛ ⎞ ⎛ ⎞ P P P P ⎜1 − 0.2 2 − 0.8⎜ 2 ⎟ ⎟ − ⎜1 − 0.2 1 − 0.8⎜ 1 ⎟ ⎟ ⎜P ⎟ ⎟ ⎜P ⎟ ⎟ ⎜ ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ ⎝

Mauricio G. Prado – The University of Tulsa

Case C1 Example Test 1

Test 2

Bubble Point

Pwf = 980 q = 1528

Pwf = 770 q = 1808

Pb= 1400

2

2

2

2

⎛ P2 ⎞ 770 P2 ⎛ 770 ⎞ ⎜ ⎟ 1 − 0.2 − 0.8⎜ ⎟ = 1 − 0.2 − 0.8⎜ ⎟ = 0.648 1400 Pb ⎝ 1400 ⎠ ⎝ Pb ⎠ ⎛ P1 ⎞ 980 P1 ⎛ 980 ⎞ 1 − 0.2 − 0.8⎜⎜ ⎟⎟ = 1 − 0.2 − 0.8⎜ ⎟ = 0.468 1400 Pb ⎝ 1400 ⎠ ⎝ Pb ⎠

0.648 1528 − 0.468 1808 qb = = 800 > 0 0.648 − 0.468 Mauricio G. Prado – The University of Tulsa

2000

IPR from Well Tests – Case C1 ( 2 + b )(q2 − qb ) J= 2 ⎛ ⎞ ⎛ ⎞ P P Pb ⎜1 + b 2 − (1 + b) ⎜⎜ 2 ⎟⎟ ⎟ ⎜ Pb Pb ⎠ ⎟ ⎝ ⎝ ⎠

1800 1600

Pressure (psi)

1400

Pb

1200

P = Pb +

1

1000

qb J

800

qmax =

2

600

J Pb + qb (2 + b )

400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Case C1 Example Test 1

Test 2

Bubble Point

Pwf = 980 q = 1528

Pwf = 770 q = 1808

Pb= 1400

J=

1.8 (q2 − qb ) ⎛ ⎛ P2 ⎞ P2 ⎜ Pb 1 − 0.2 − 0.8⎜⎜ ⎟⎟ ⎜ Pb Pb ⎠ ⎝ ⎝

qb P = Pb + = J

qmax

2

⎞ ⎟ ⎟ ⎠

=

J Pb = + qb = 1.8

Mauricio G. Prado – The University of Tulsa

Case C1 Example Test 1

Test 2

Bubble Point

Pwf = 980 q = 1528

Pwf = 770 q = 1808

Pb= 1400

J=

1.8 (1808 − 800 ) ⎛ 770 ⎛ 770 ⎞ ⎜ − 0.8⎜ 1400 1 − 0.2 ⎟ ⎜ 1400 ⎝ 1400 ⎠ ⎝

800 P = 1400 + = 1800 2 qmax

2 1400 = + 800 = 2355 1.8 Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟ ⎟ ⎠

=2

IPR from Well Tests – Case C1 2000 1800 1600

Pressure (psi)

1400

Pb

1200

1

1000 800

2

600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests • Cases C1 and C2 – When both tests are below the bubble point but qb is less than zero, the reservoir is saturated and we should re-calculate the parameters. – The test for the bubble point flowrate is stilll the same as before.

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case C2 2000 1800 1600

q

Pb

qmax

Pressure (psi)

1400

⎛ Pwf = 1 + b ⎜⎜ ⎝ P

⎛ ⎞ ⎟⎟ − (1 + b) ⎜ ⎜ ⎠

1200 1000

Pwf ⎝ P

2

⎞ ⎟⎟ ⎠

1

800 600

2

400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests – Case C2 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜1 + b P2 − (1 + b )⎜ P2 ⎟ ⎟ q − ⎜1 + b P1 − (1 + b )⎜ P1 ⎟ ⎟ q ⎜P ⎟ ⎟ 1 ⎜ ⎜P ⎟ ⎟ 2 ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ qb = ⎝ 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ P P P P 2 2 1 1 ⎜1 + b − (1 + b )⎜ ⎟ ⎟ − ⎜1 + b − (1 + b )⎜ ⎟ ⎟ ⎜P ⎟ ⎟ ⎜ ⎜P ⎟ ⎟ ⎜ Pb Pb b ⎠ ⎝ ⎝ b⎠ ⎠ ⎝ ⎠ ⎝

2000 1800 1600

Pb

Pressure (psi)

1400 1200 1000

1

800

qb < 0

600

2

400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

IPR from Well Tests – Case C2 • Once we confirm that the reservoir is saturated, we proceed to calculate the reservoir pressure: 2

P1 ⎞ ⎛ q1 ⎛ P1 ⎞ = 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎟ qmax P ⎝ ⎠ P ⎠

⎝

q1 2

P1 ⎞ ⎛ ⎛ P1 ⎞ 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎟ ⎝P⎠ P ⎝

⎠

= qmax

P2 ⎛ q2 ⎛ P2 ⎞ = 1 + b ⎜ ⎟ − (1 + b) ⎜ qmax ⎝P⎠ P ⎝

q2

P2 ⎛ ⎛ P2 ⎞ 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎝P⎠ P

Mauricio G. Prado – The University of Tulsa

⎝

2

⎞ ⎟ ⎠

2

⎞ ⎟ ⎠

= qmax

IPR from Well Tests – Case C2 • Once we confirm that the reservoir is saturated, we proceed to calculate the reservoir pressure: q1 2

P1 ⎞ ⎛ ⎛ P1 ⎞ 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎟ ⎝P⎠ P ⎝

q2

= qmax

P2 ⎛ ⎛ P2 ⎞ 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎝P⎠ P

⎠

⎛ P2 ⎜ ⎛ ⎛ P2 ⎞ q1 ⎜1 + b ⎜ ⎟ − (1 + b) ⎜ ⎝P⎠ ⎜ ⎝P ⎝

⎝

2

⎞ ⎟ ⎠

2

⎞ ⎟ ⎠

= qmax

2 ⎞ ⎛ ⎞ P1 ⎞ ⎟ ⎟ ⎜ ⎛ ⎛ P1 ⎞ ⎟ = q2 ⎜1 + b ⎜ ⎟ − (1 + b) ⎜ ⎟⎟ P ⎝ ⎠ ⎟ ⎜ ⎟ P ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎜ ⎛ P2 ⎛ ⎛ P2 ⎞ ⎛ P1 ⎞ ⎞ (q1 − q2 ) + b ⎜⎜ q1 ⎜ ⎟ − q2 ⎜ ⎟ ⎟⎟ − (1 + b)⎜ q1 ⎜ ⎝ P ⎠⎠ ⎝ ⎝P⎠ ⎜ ⎝ P ⎝ Mauricio G. Prado – The University of Tulsa

2

⎞ ⎟ ⎠

⎞ P ⎛ 1⎞ ⎟=0 ⎜ ⎟⎟ ⎝ P ⎠ ⎟⎠ 2

− q2

IPR from Well Tests – Case C2 • Once we confirm that the reservoir is saturated, we proceed to calculate the reservoir pressure: ⎛ ⎜ ⎛ P2 ⎛ ⎛ P2 ⎞ ⎛ P1 ⎞ ⎞ (q1 − q2 ) + b ⎜⎜ q1 ⎜ ⎟ − q2 ⎜ ⎟ ⎟⎟ − (1 + b)⎜ q1 ⎜ ⎝ P ⎠⎠ ⎝ ⎝P⎠ ⎜ ⎝ P ⎝

2

⎞ ⎟ ⎠

⎞ P ⎛ 1⎞ ⎟=0 ⎜ ⎟⎟ ⎝ P ⎠ ⎟⎠ 2

− q2

− b (q1 P2 − q2 P1 ) − b 2 (q1 P2 − q2 P1 ) 2 + 4 (1 + b ) (q1 − q2 ) (q1 P2 − q2 P1 ) P= 2 (q1 − q2 ) 2

Mauricio G. Prado – The University of Tulsa

2

IPR from Well Tests – Case C2 • And the remaining parameters are: qmax =

q1 2

P1 ⎞ ⎛ ⎛ P1 ⎞ 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎟ ⎝P⎠ P ⎝

⎠

=

q2

P2 ⎛ ⎛ P2 ⎞ 1 + b ⎜ ⎟ − (1 + b) ⎜ ⎝P⎠ P

( 2 + b ) qmax J= P

Mauricio G. Prado – The University of Tulsa

⎝

2

⎞ ⎟ ⎠

IPR from Well Tests – Case C2 − b (q1 P2 − q2 P1 ) − b 2 (q1 P2 − q2 P1 ) 2 + 4 (1 + b ) (q1 − q2 ) (q1 P2 − q2 P1 ) P= 2 (q1 − q2 ) 2

2000 1800 1600

P >0

Pb

Pressure (psi)

1400 1200

1

1000

qmax =

800

q ⎛P⎞ ⎛P⎞ 1 + b ⎜ ⎟ − (1 + b )⎜ ⎟ ⎝P⎠ ⎝P⎠

600

( 2 + b ) qmax J=

2

400

2

P

200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

2

IPR from Well Tests - Homework • Determine the equations to calculate the IPR parameters (qmax, J, Pr, and qb) when we have the information on 2 production tests and the bubble point pressure. Use the equations to calculate the Vogel IPR for the following case: • Pb= 1400 – Tests - Pwf = 560 q = 1232 and Pwf = 770 q = 1008

Mauricio G. Prado – The University of Tulsa

IPR from Well Tests – Case C2 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜1 + b P2 − (1 + b )⎜ P2 ⎟ ⎟ q − ⎜1 + b P1 − (1 + b )⎜ P1 ⎟ ⎟ q ⎜P ⎟ ⎟ 1 ⎜ ⎜P ⎟ ⎟ 2 ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ qb = ⎝ 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ P P P P 2 2 1 1 ⎜1 + b − (1 + b )⎜ ⎟ ⎟ − ⎜1 + b − (1 + b )⎜ ⎟ ⎟ ⎜P ⎟ ⎟ ⎜ ⎜P ⎟ ⎟ ⎜ Pb Pb b ⎠ ⎝ ⎝ b⎠ ⎠ ⎝ ⎠ ⎝

2000 1800 1600

Pb

Pressure (psi)

1400 1200 1000

1

800

qb < 0

600

2

400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Case C2 Example Test 1

Test 2

Bubble Point

Pwf = 770 q = 1008

Pwf = 560 q = 1232

Pb= 1400

2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ P P P P ⎜1 − 0.2 2 − 0.8⎜ 2 ⎟ ⎟ q − ⎜1 − 0.2 1 − 0.8⎜ 1 ⎟ ⎟ q ⎜P ⎟ ⎟ 1 ⎜ ⎜P ⎟ ⎟ 2 ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ ⎝ qb = = 2 2 ⎞ ⎞ ⎛ ⎛ ⎛ ⎞ ⎛ ⎞ P P P P ⎜1 − 0.2 2 − 0.8⎜ 2 ⎟ ⎟ − ⎜1 − 0.2 1 − 0.8⎜ 1 ⎟ ⎟ ⎜P ⎟ ⎟ ⎜P ⎟ ⎟ ⎜ ⎜ Pb P b ⎝ b⎠ ⎠ ⎝ b⎠ ⎠ ⎝ ⎝

Mauricio G. Prado – The University of Tulsa

Case C2 Example Test 1

Test 2

Bubble Point

Pwf = 770 q = 1008

Pwf = 560 q = 1232

Pb= 1400

2

⎛ P2 ⎞ 560 P2 ⎛ 560 ⎞ 1 − 0.2 − 0.8⎜⎜ ⎟⎟ = 1 − 0.2 − 0.8⎜ ⎟ = 0.792 1400 Pb ⎝ 1400 ⎠ ⎝ Pb ⎠ 2

2

⎛ P1 ⎞ 770 P1 ⎛ 770 ⎞ ⎜ ⎟ 1 − 0.2 − 0.8⎜ ⎟ = 1 − 0.2 − 0.8⎜ ⎟ = 0.648 1400 Pb ⎝ 1400 ⎠ ⎝ Pb ⎠

0.792 1008 − 0.648 1232 qb = =0 0.792 − 0.648 Mauricio G. Prado – The University of Tulsa

2

IPR from Well Tests – Case C2 − b (q1 P2 − q2 P1 ) − b 2 (q1 P2 − q2 P1 ) 2 + 4 (1 + b ) (q1 − q2 ) (q1 P2 − q2 P1 ) P= 2 (q1 − q2 ) 2

2000 1800 1600

P >0

Pb

Pressure (psi)

1400 1200

1

1000

qmax =

800

q ⎛P⎞ ⎛P⎞ 1 + b ⎜ ⎟ − (1 + b )⎜ ⎟ ⎝P⎠ ⎝P⎠

600

( 2 + b ) qmax J=

2

400

2

P

200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

2

Case C2 Example Test 1

Test 2

Bubble Point

Pwf = 770 q = 1008

Pwf = 560 q = 1232

Pb= 1400

(q1 P2 − q2 P1 ) − (q1 P2 − q2 P1 ) 2 + 80 (q1 − q2 ) (q1 P2 − q2 P1 ) P= 10 (q1 − q2 ) 2

qmax =

q ⎛P⎞ ⎛P⎞ 1 − 0.2 ⎜ ⎟ − 0.8 ⎜ ⎟ ⎝P⎠ ⎝P⎠

2

1.8 qmax J= P

Mauricio G. Prado – The University of Tulsa

2

Case C2 Example Test 1

Test 2

Bubble Point

Pwf = 770 q = 1008

Pwf = 560 q = 1232

Pb= 1400

(1008 560 − 1232 770) − (1008 560 − 1232 770) 2 + 80 (1008 − 1232) (1008 560 2 − 1232 770 2 ) P= 10 (1008 − 1232)

P = 1400 qmax =

1008 ⎛ 770 ⎞ ⎛ 770 ⎞ 1 − 0.2 ⎜ ⎟ − 0.8 ⎜ ⎟ ⎝ 1400 ⎠ ⎝ 1400 ⎠

2

= 1555

Mauricio G. Prado – The University of Tulsa

1.8 1555 J= =2 1400

IPR from Well Tests – Case C2 2000 1800

Pb

1600

Pressure (psi)

1400 1200

1

1000

2

800 600 400 200 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

IPR from Well Tests • With this procedure we are able to use only two tests to determine the four parameters: – – – –

Reservoir Pressure Productivity Index Bubble Point Flowrate Absolute Open Flow

• With those parameters determined we can draw or calculate the IPR for each case. • Usually we will either have: – Saturated Reservoir – Under saturated Reservoir

Mauricio G. Prado – The University of Tulsa

Summary q = J (P − Pwf )

2000 1800

⎛ Pwf q − qb = 1 + b ⎜⎜ qmax − qb ⎝ Pb

Bottom Hole Flowing Pressure (psi)

1600

⎞ ⎛ Pwf ⎟⎟ − (1 + b) ⎜⎜ ⎠ ⎝ Pb

1400 1200 1000

⎛ Pwf q = 1 + b ⎜⎜ qmax ⎝ P

800

⎛ Pwf ⎞ ⎟⎟ − (1 + b) ⎜⎜ ⎝ P ⎠

⎞ ⎟⎟ ⎠

2

600 400 200 0 0

500

1000

1500 Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

2000

2500

⎞ ⎟⎟ ⎠

2

Summary q = J (P − Pwf )

2000

⎛ Pwf q − qb = 1 + b ⎜⎜ ' qmax − qb ⎝P

1800

Bottom Hole Flowing Pressure (psi)

1600

⎞ ⎛P ⎟⎟ − (1 + b) ⎜⎜ wf' ⎠ ⎝P

⎞ ⎟⎟ ⎠

2

Pb

1400 1200 1000

qmax

800

J P' = + qb 2+b

600

qb = J (P − Pb )

qb = 0

400

P ' = Pb

P' = P

200 0 0

500

1000

1500 Flowrate (bpd)

Mauricio G. Prado – The University of Tulsa

2000

2500

Examples 3000

Test 1 Pressure 2500 psi Flowrate 250 bpd

Pb = 2000 psi Test 1

2500

Bottom Hole Flowing Pressure (psi)

Test 2 Pressure 2200 psi Flowrate 750 bpd

Test 2 2000

1500

b = -0.4 qmax = 3166 bpd Pr = 2650 psi

b=0 qmax = 2750 bpd Pr = 2650 psi

1000

500

b = -0.2 qmax = 2935 bpd Pr = 2650 psi

0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

3500

Examples 3000

b = 0 (Fetkovick)

Test 2 Pressure 2200 psi Flowrate 750 bpd

Test 1

2500

Bottom Hole Flowing Pressure (psi)

Test 1 Pressure 2500 psi Flowrate 250 bpd

Test 2 2000

Pb = 1600 psi qmax = 3083 bpd Pr = 2650 psi

1500

Pb = 2000 psi qmax = 2750 bpd Pr = 2650 psi

1000

Pb = 1800 psi qmax = 2916 bpd Pr = 2650 psi

500

0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

3500

Examples 3500

Test 1 Pressure 2500 psi Flowrate 250 bpd

Pb = 2000 psi

Bottom Hole Flowing Pressure (psi)

3000

Test 2 Pressure 1600 psi Flowrate 750 bpd

Test 1

2500

2000 Test 2

b = -0.4 qmax = 1255 bpd Pr = 2935 psi

1500

b=0 qmax = 1122 bpd Pr = 2930 psi

1000

b = -0.2 qmax = 1181 bpd Pr = 2932 psi

500

0 0

200

400

600

800

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

1000

1200

1400

Examples 3500

b = 0 (Fetkovick)

Bottom Hole Flowing Pressure (psi)

3000

Test 1 Pressure 2500 psi Flowrate 250 bpd

Test 2 Pressure 1600 psi Flowrate 750 bpd

Test 1

2500

2000

Pb = 1800 psi qmax = 1150 bpd Pr = 2944 psi

Test 2 1500

Pb = 2000 psi qmax = 1122 bpd Pr = 2930 psi

1000

Pb = 1900 psi qmax = 1134 bpd Pr = 2938 psi

500

0 0

200

400

600

800

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

1000

1200

1400

Examples 2500

Test 1 Pressure 1700 psi Flowrate 250 bpd

Pb = 2000 psi

Test 2 Pressure 1400 psi Flowrate 400 bpd

Bottom Hole Flowing Pressure (psi)

2000

Test 1 1500

Test 2

b = -0.4 qmax = 831 bpd Pr = 2132 psi

1000

b=0 qmax = 716 bpd Pr = 2110 psi 500

b = -0.2 qmax = 769 bpd Pr = 2120 psi 0 0

100

200

300

400

500

600

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

700

800

900

Examples 2500

b = 0 (Fetkovick)

Test 1 Pressure 1700 psi Flowrate 250 bpd

Test 2 Pressure 1400 psi Flowrate 400 bpd

Bottom Hole Flowing Pressure (psi)

2000

Test 1 1500

Test 2

Pb = 1800 psi qmax = 716 bpd Pr = 2133 psi

1000

Pb = 2000 psi qmax = 716 bpd Pr = 2110 psi 500

Pb = 1900 psi qmax = 716 bpd Pr = 2118 psi 0 0

100

200

300

400

500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

600

700

800

Examples 2500

Test 1 Pressure 1600 psi Flowrate 300 bpd

Pb = 2350 psi

Test 2 Pressure 1400 psi Flowrate 400 bpd

Bottom Hole Flowing Pressure (psi)

2000

Test 1 1500

Test 2

b = -0.4 qmax = 846 bpd Pr = 2116 psi

1000

b=0 qmax = 726 bpd Pr = 2088 psi 500

b = -0.2 qmax = 782 bpd Pr = 2100 psi 0 0

100

200

300

400

500

600

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

700

800

900

What About Comingling Production ?

Mauricio G. Prado – The University of Tulsa

Multiple Zones IPR q

q1 = q1 ( Pwf ) Pr1 Pwf

h Pr2 q2 = q2 ( Pwf ) Mauricio G. Prado – The University of Tulsa

Multiple Zones IPR q

q1 Pr1 Pwf

q ( Pwf ) = q1 ( Pwf ) + q2 ( Pwf ) Pr2 q2 Mauricio G. Prado – The University of Tulsa

Multiple Zones IPR 1600

Zone 1

1400

Zone 2 Pressure (psi)

1200 1000 800 600

q1

q1 q2

400 200

q 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Multiple Zones IPR 1600

Zone 1

1400

Zone 2

Zones 1 + 2

Pressure (psi)

1200 1000 800 600

q1

q1 q2

400 200

q 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Multiple Zones IPR 1600 1400

Zones 1 + 2 Pressure (psi)

1200 1000 800 600 400 200 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Multiple Zones IPR with Interflow 1600

Zone 1 Zone 2

1400

Zones 1 + 2 Pressure (psi)

1200 1000 800 600

q1

400

q2

q1

200

q

0 -500

0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Multiple Zones IPR with Interflow 1600 1400

Zones 1 + 2

Pressure (psi)

1200 1000 800 600 400 200 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Can we estimate the IPR when the well is producing oil and water from different zones ?

Mauricio G. Prado – The University of Tulsa

Composite IPR q

qoil Pro Pwf

q ( Pwf ) = qo ( Pwf ) + qw ( Pwf ) Prw qwater Mauricio G. Prado – The University of Tulsa

Composite IPR 1600 1400

Oil Zone

Pressure (psi)

1200

Water Zone

1000 800 600 400 200 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Composite IPR 1600 1400

Oil Zone

Pressure (psi)

1200

Water Zone 1000 800

Oil + Water Production 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Composite IPR 1

Water Cut (Fraction)

0.8

0.6

0.4

0.2

0 0

500

1000

1500

2000

Total Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Composite IPR 1600

Oil Zone 1400

Pressure (psi)

1200

Water Zone 1000 800 600 400 200 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Composite IPR 1600

Oil Zone 1400

Pressure (psi)

1200

Water Zone

1000

Oil + Water Production

800 600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Composite IPR 1 0.9

Water Cut (Fraction)

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

500

1000

1500

2000

Total Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

Composite IPR q

qwater qoil Pr Pwf

q ( Pwf ) = qo ( Pwf ) + qw ( Pwf )

Mauricio G. Prado – The University of Tulsa

Composite IPR 1600

Oil Production

1400

Water Production

Pressure (psi)

1200 1000

Pb

800 600 400 200 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Composite IPR 1600

Oil Production

1400

Water Production

Pressure (psi)

1200 1000

Pb

800

Oil + Water Production

600 400 200 0 0

500

1000

1500

2000

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2500

3000

3500

Composite IPR qw WC = q w + qo

qw WOR = qo

WC = WC P

* w

*

wf

= Pr

WOR = WOR P

J = * * Jw + Jo

*

wf

Mauricio G. Prado – The University of Tulsa

= Pr

J = J

* w * o

Composite IPR 0.7 0.68

Water Cut (Fraction)

0.66 0.64 0.62 0.6 0.58 0.56 0.54 0.52 0.5 0

500

1000

1500

2000

2500

Total Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

3000

3500

Homework - Composite IPR Reservoir Presure Pr = 1800 psi Bubble Point Pressure Pb = 1300 psi 1 Zone producing Oil and Water Well Test qt = 600 bpd @ 1400 psi WC = 30 %

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR qo = 600 * 0.7 = 420 bpd @ 1400 psi qw = 600 * 0.3 = 180 bpd @ 1400 psi OIL – Combined IPR – Undersaturated Reservoir

Jo = qo / ( Pr - Pt ) = 420 / ( 1800 – 1400 ) = 1.05 bpd/psi qb = Jo ( Pr – Pb) = 1.05 ( 1800 – 1300 ) = 525 bpd qmax = Jo Pb /1.8 + qb = 1.05 1300 / 1.8 + 525 = 1283.33 bpd

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR qo = 600 * 0.7 = 420 bpd @ 1400 psi qw = 600 * 0.3 = 180 bpd @ 1400 psi Water – Linear IPR

Jw = qw / ( Pr - Pt ) = 180 / ( 1800 – 1400 ) = 0.45 bpd/psi

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR Water

qw = 0.45 (1800 − Pwf )

Oil

qo = 1.05 (1800 − Pwf )

Pwf > 1300

qo − 525 = 1 − 0.2 − 0.8 758.33 1300 1300 2 Pwf

Pwf2

Mauricio G. Prado – The University of Tulsa

Pwf < 1300

Homework - Composite IPR Pwf

Qo

Qw

Qt

WC

1800 1550 1300 1200 1000 800 600 400 200 0 Mauricio G. Prado – The University of Tulsa

WOR

Homework - Composite IPR Pwf

Qo

Qw

Qt

WC

WOR

1800

0

0

0

0.30

0.43

1550

262.5

112.5

375

0.30

0.43

1300

525

225

750

0.30

0.43

1200

626

270

896

0.30

0.43

1000

807

360

1168

0.31

0.45

800

960

450

1410

0.32

0.47

600

1084

540

1624

0.33

0.50

400

1179

630

1809

0.35

0.53

200

1245

720

1966

0.37

0.58

0

1283

810

2093

0.39

0.63

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR qo = 1.05 (1800 − Pwf )

2000 1800

2

⎛ Pwf ⎞ ⎛ Pwf ⎞ qo − 525 ⎟⎟ − 0.8 ⎜ = 1 − 0.2 ⎜⎜ ⎟⎟ ⎜ 758.333 1300 ⎝ ⎠ 1300

1600 Pressure (psi)

1400

⎝

1200

qw = 0.45 (1800 − Pwf )

Pb

1000

⎠

800 600

qo+qw

400 200 0 0

500

1000

1500

Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Homework - Composite IPR 0.39

Water Cut (Fraction)

0.37 0.35 0.33 0.31 0.29 0.27 0.25 0

500

1000

1500

Total Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

2000

2500

Homework - Composite IPR - WOR 0.65

Water Oil Ratio (Fraction)

0.6 0.55 0.5 0.45 0.4 0.35 0.3 0.25 0

200

400

600

800

Oil Flowrate (bpd) Mauricio G. Prado – The University of Tulsa

1000

1200

1400

Economics When producing oil and water not always it is desirable to maximize oil production

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR - Economics • Assuming that the unitary costs of production for each phase are: – Co = 7 US$/stb

– Cw = 6 US$/stb

• Calculate the profit for each total flowrate when the oil sale price is So = 11, 12 and 13 US$/stb

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR - Economics Oil

Price

US$/stb

11

12

13

Co

Cw

P

P

P

Pwf

Qo

Qw

Qt

1800

0

0

0

1550

262.5

112.5

375

1300

525

225

750

1200

626

270

896

1000

807

360

1168

800

960

450

1410

600

1084

540

1624

400

1179

630

1809

200

1245

720

1966

0

1283

810

2093

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR - Economics Oil

Price

US$/stb

11

12

13

Pwf

Qo

Qw

Qt

Co

Cw

P

P

P

1800

0

0

0

0

0

0

0

0

1550

262.5

112.5

375

1837.5

675

375

637.5

900

1300

525

225

750

3675

1350

750

1275

1800

1200

626

270

896

4382

1620

884

1510

2136

1000

807

360

1168

5649

2160

1068

1875

2682

800

960

450

1410

6720

2700

1140

2100

3060

600

1084

540

1624

7588

3240

1096

2180

3264

400

1179

630

1809

8253

3780

936

2115

3294

200

1245

720

1966

8715

4320

660

1905

3150

0

1283

810

2093

8981

4860

272

1555

2838

Mauricio G. Prado – The University of Tulsa

Homework - Composite IPR - Economics 3500

S o (US$/stb)

3000

Profit (U S $/d)

2500

13 2000 1500

12

1000

11

500 0 0

200

400

600

800

Oil Flow rate (stb/d) Mauricio G. Prado – The University of Tulsa

1000

1200

1400