Reservoir Engineering Notes

PTRL6001

RESERVOIR ENGINEERING I

by

Val Pinczewski

School of Petroleum EngineeringUniversity of New South Wales

Sydney NSW 2052.AUSTRALIA

March, 2002

Prepared for

PETROLEUM ENGINEERING DISTANCELEARNING PROGRAM

UNSW

IMPORTANT NOTICE

2002 University of New South Wales. All rights are reserved.

This copy of the manual and accompanying software was prepared in accordancewith copyright laws for the sole use of students enrolled in a course at the Uni-versity of New South Wales. It is illegal to reproduce any of this material or touse it for any other purpose.

2

Contents

1 INTRODUCTION 7

1.1 BASIC CONCEPTS . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Accumulation of Sediments . . . . . . . . . . . . . . . . . 7

1.1.2 Origin of Petroleum . . . . . . . . . . . . . . . . . . . . . . 9

1.1.3 Hydrocarbon Traps . . . . . . . . . . . . . . . . . . . . . . 9

1.1.4 Classification of Traps . . . . . . . . . . . . . . . . . . . . 11

1.2 OIL RECOVERY PROCESSES . . . . . . . . . . . . . . . . . . . 14

1.2.1 Residual Oil Resource (Target for EOR ) . . . . . . . . . . 15

1.2.2 Residual Oil is Trapped or By-passed . . . . . . . . . . . . 15

1.2.3 Recovery Processes . . . . . . . . . . . . . . . . . . . . . . 15

1.2.4 Primary Recovery Mechanisms . . . . . . . . . . . . . . . 17

1.2.5 Secondary Recovery . . . . . . . . . . . . . . . . . . . . . . 18

1.2.6 Tertiary Recovery — EOR Processes . . . . . . . . . . . . . 19

1.3 WHAT IS RESERVOIR ENGINEERING? . . . . . . . . . . . . . 20

2 RESERVOIR DESCRIPTION 23

2.1 RESERVOIR DESCRIPTION PROGRAM . . . . . . . . . . . . . 31

2.2 SOURCES OF DATA . . . . . . . . . . . . . . . . . . . . . . . . 35

2.2.1 Coring And Core Analysis . . . . . . . . . . . . . . . . . . 36

2.2.2 Wireline Logging . . . . . . . . . . . . . . . . . . . . . . . 37

1

2.2.3 Pressure and Production Testing . . . . . . . . . . . . . . 38

2.2.4 Fluid Sampling . . . . . . . . . . . . . . . . . . . . . . . . 39

2.3 INTEGRATED FORMATION EVALUATION PROGRAM . . . 42

2.4 AQUIFER DESCRIPTION . . . . . . . . . . . . . . . . . . . . . 46

3 VOLUMETRICS AND INITIAL HYDROCARBON VOLUME 48

3.1 STRUCTURE MAPS . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.2 ISOPACH MAPS . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.3 VOLUMETRIC METHOD FOR DETERMINING ORIGINALOIL-IN-PLACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.3.1 Reservoir Volume . . . . . . . . . . . . . . . . . . . . . . 54

3.3.2 Average Porosity . . . . . . . . . . . . . . . . . . . . . . . 55

3.3.3 Average Initial Water Saturation . . . . . . . . . . . . . . 55

3.3.4 Average Oil Formation Volume Factor . . . . . . . . . . . 56

3.3.5 Determining Initial Oil-In-Place . . . . . . . . . . . . . . . 56

3.4 DETERMINATION OF OIL-IN-PLACE — MATERIAL BAL-ANCE METHOD . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.5 ESTIMATING RESERVES . . . . . . . . . . . . . . . . . . . . . 68

3.6 ESTIMATING DECLINE IN OIL PRODUCTION RATES . . . . 69

4 HYDROSTATIC PRESSURE DISTRIBUTION IN RESER-VOIRS 70

4.1 SUBSURFACE PRESSURES . . . . . . . . . . . . . . . . . . . . 71

4.1.1 Water zone pressures . . . . . . . . . . . . . . . . . . . . . 71

4.1.2 Oil zone pressures . . . . . . . . . . . . . . . . . . . . . . . 72

4.1.3 Gas cap pressures . . . . . . . . . . . . . . . . . . . . . . . 73

4.2 HYDROSTATIC PRESSURE DISTRIBUTION IN A RESER-VOIR CONTAINING OIL, WATER AND GAS . . . . . . . . . . 74

2

4.3 GEOTHERMAL GRADIENT . . . . . . . . . . . . . . . . . . . . 84

5 FLUID PROPERTIES 86

5.1 PHASE BEHAVIOR . . . . . . . . . . . . . . . . . . . . . . . . . 87

5.1.1 Pure Hydrocarbons . . . . . . . . . . . . . . . . . . . . . . 87

5.1.2 Hydrocarbons Mixtures . . . . . . . . . . . . . . . . . . . . 89

5.1.3 Classification of Hydrocarbon Reservoirs . . . . . . . . . . 92

5.2 PVT PROPERTIES . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.2.1 Pressure Dependence of PVT Properties . . . . . . . . . . 98

5.3 CALCULATION OF GAS PROPERTIES . . . . . . . . . . . . . 103

5.3.1 Single Gas Component . . . . . . . . . . . . . . . . . . . . 103

5.3.2 Multi-Component Gas Mixtures . . . . . . . . . . . . . . . 105

5.4 DETERMINATION OF OIL PVT DATA FROM LABORATORYEXPERIMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

5.4.1 Flash Expansion Test . . . . . . . . . . . . . . . . . . . . 113

5.4.2 Differential Liberation Test . . . . . . . . . . . . . . . . . 113

5.4.3 Separator Flash Expansion Test . . . . . . . . . . . . . . . 116

5.4.4 Procedure for calculating PVT parameters from laboratorydata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

5.5 FLUID SAMPLING . . . . . . . . . . . . . . . . . . . . . . . . . 119

5.6 PVT TESTS FOR GAS CONDENSATE FIELDS . . . . . . . . . 119

5.7 GAS HYDRATES . . . . . . . . . . . . . . . . . . . . . . . . . . 121

5.8 SURFACE TENSION . . . . . . . . . . . . . . . . . . . . . . . . 124

5.8.1 Estimating Surface Tension . . . . . . . . . . . . . . . . . 124

5.9 CORRELATIONS FOR PROPERTIES OF RESERVOIR FLUIDS 126

6 MATERIAL BALANCE EQUATIONS 133

6.1 ORIGINAL OIL VOLUME BALANCE . . . . . . . . . . . . . . . 134

3

6.1.1 Gas Cap Expansion . . . . . . . . . . . . . . . . . . . . . . 135

6.1.2 Released Gas Volume . . . . . . . . . . . . . . . . . . . . . 138

6.1.3 Remaining Oil Volume . . . . . . . . . . . . . . . . . . . . 140

6.1.4 Rock and Connate Water Expansion . . . . . . . . . . . . 142

6.1.5 Water Influx . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.1.6 General Material Balance Equation . . . . . . . . . . . . . 146

6.2 PRIMARY RECOVERY MECHANISMS . . . . . . . . . . . . . . 148

6.2.1 Typical Performance Characteristics for the Different DriveMechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . 149

6.3 USING MATERIAL BALANCE EQUATIONS . . . . . . . . . . 153

6.3.1 Average Reservoir Pressure . . . . . . . . . . . . . . . . . 153

6.3.2 Knowns and Unknowns . . . . . . . . . . . . . . . . . . . . 153

6.4 MATERIAL BALANCE FOR A CLOSED OIL RESERVOIR . . 157

6.5 MATERIAL BALANCE FOR A CLOSED GAS RESERVOIR . . 159

6.5.1 Water Drive Gas Reservoirs . . . . . . . . . . . . . . . . . 162

7 RESERVOIR ROCK PROPERTIES AND CORE ANALYSISPROCEDURES 165

7.1 POROSITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

7.1.1 Effective and Total Porosity . . . . . . . . . . . . . . . . . 167

7.1.2 Laboratory Measurement of Porosity . . . . . . . . . . . . 168

7.2 PERMEABILITY . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

7.2.1 Measurement of Permeability . . . . . . . . . . . . . . . . 175

7.2.2 Laboratory Measurement of Permeability . . . . . . . . . . 179

7.2.3 The Klinkenberg Effect . . . . . . . . . . . . . . . . . . . . 180

7.3 POROSITY-PERMEABILITY RELATIONSHIPS . . . . . . . . . 182

7.3.1 Capillary Tube Model . . . . . . . . . . . . . . . . . . . . 182

4

7.3.2 Fractured medium model . . . . . . . . . . . . . . . . . . . 185

7.4 ROCK COMPRESSIBILITY . . . . . . . . . . . . . . . . . . . . 186

7.4.1 Pore Volume Compressibility . . . . . . . . . . . . . . . . 189

7.4.2 Measurement of Formation Compressibility . . . . . . . . . 190

7.4.3 Use of Rock Compressibilities . . . . . . . . . . . . . . . . 192

8 FLUID FLOW 194

8.1 DARCY’S LAW . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

8.1.1 Pressure Potential . . . . . . . . . . . . . . . . . . . . . . 195

8.2 STEADY-STATE FLOW . . . . . . . . . . . . . . . . . . . . . . . 197

8.2.1 Horizontal Linear Flow of an Incompressible Fluid . . . . . 197

8.2.2 Radial Flow of an Incompressible Fluid . . . . . . . . . . . 201

8.2.3 Wellbore Damage . . . . . . . . . . . . . . . . . . . . . . . 206

8.2.4 Relationship between s and the size of the altered zone . . 210

8.2.5 Effective Wellbore Radius . . . . . . . . . . . . . . . . . . 212

8.2.6 Flow Efficiency . . . . . . . . . . . . . . . . . . . . . . . . 213

8.3 UNSTEADY STATE FLOW . . . . . . . . . . . . . . . . . . . . . 215

8.3.1 Radial Diffusivity Equation . . . . . . . . . . . . . . . . . 216

8.3.2 Liquids Having Small and Constant Compressibility . . . . 219

8.3.3 Pseudo-Steady-State Radial Flow . . . . . . . . . . . . . . 221

8.3.4 Flow Equations in terms of Average Reservoir Pressure . . 223

8.3.5 Dietz Shape Factors for Vertical Wells . . . . . . . . . . . 225

8.3.6 Approximating Complex Geometries . . . . . . . . . . . . 228

8.4 WELL PRODUCTIVITY . . . . . . . . . . . . . . . . . . . . . . 229

8.4.1 Productivity Index . . . . . . . . . . . . . . . . . . . . . . 229

8.4.2 Partial Penetration . . . . . . . . . . . . . . . . . . . . . . 230

5

8.5 GAS FLOW . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

8.5.1 Low pressure approximation — p < 3, 000psi and ∆p small 236

8.5.2 High pressure approximation — p > 3, 000psi and ∆p small 237

8.5.3 ∆p is not small . . . . . . . . . . . . . . . . . . . . . . . . 238

8.5.4 Steady and Pseudo-Steady State Radial Gas Flow . . . . . 239

8.5.5 Non-Darcy flow . . . . . . . . . . . . . . . . . . . . . . . . 240

8.5.6 Pressure-squared approximation . . . . . . . . . . . . . . . 241

8.5.7 High Pressure approximation . . . . . . . . . . . . . . . . 241

8.5.8 Gas Well Back Pressure Equation . . . . . . . . . . . . . . 245

8.6 HORIZONTAL WELLS . . . . . . . . . . . . . . . . . . . . . . . 249

8.6.1 Drainage Area . . . . . . . . . . . . . . . . . . . . . . . . . 251

8.6.2 Productivity of Horizontal Wells . . . . . . . . . . . . . . . 261

6

Chapter 1

INTRODUCTION

For those unfamiliar with reservoir engineering or reservoir geology the followingsection gives a brief summary of some basic concepts. If you are familiar withthis material skip this section.

1.1 BASIC CONCEPTS

1.1.1 Accumulation of Sediments

The accumulation of sediments in a basin depends on the balance between theenergy of the depositional environment (water velocity) and the sedimentationvelocity of the particles. The sedimentation velocity depends on the size anddensity of the particles. Sediments carried by high velocity streams may bedeposited in the delta region of the river where flow velocities are much slower.Smaller particle sediments may be further moved by waves and currents to otherlocations where the environmental energy is insufficient to carry the particlesfurther. This process leads to sorting of sediments with the accumulation of sandgrains in one area and clay and silt particles in another area.

The depositional energy at a particular location varies with time. This results in asequential deposition of sand (large particles) and shale (fine particles) producingsequences of sand and shale. (layering). This layering or vertical heterogeneity isthe single most important characteristic determining reservoir performance andrecovery.

7

Figure 1.1: Schematic of a typical depositional sequence

8

1.1.2 Origin of Petroleum

The precursors of petroleum, organic matter from dead plants and animals, aredeposited together with fine-grained sediments in shallow marine environmentsduring low energy periods of basin formation. Quite water is deficient in oxygenleading to the creation of anaerobic conditions and preservation of the organicmatter. Anaerobic bacteria decomposes the organic matter to produce com-pounds of carbon, hydrogen (hydrocarbons) and oxygen. The conversion of theorganic matter occurs over geological time as sediments become progressivelyburied and temperature and pressure increase.

The minimum temperature for oil and gas formation is about 150oF and themaximum is about 320oF. Since temperature increases with depth, this results ina burial depth window of between 7,000 ft to 23,000 ft for fine grained sedimentscontaining organic matter (source rocks) to produce oil and gas.

Increasing burial causes the fine-grained sediments to undergo compaction andthe source rocks eventually become effectively impermeable. As the particlescompact the generated hydrocarbon particles (oil or gas) are squeezed from thesource rock. This process is called primary migration. The expelled hydrocar-bon particles, in the form of colloidal solutions (micelles) or individual drops orbubbles, enter the overlying and underlying water saturated permeable sand lay-ers which have retained their porosity and permeability because the coarse sandgrains are stronger than the fine-grained silt and clay particles and thereforebetter withstand the increasing compaction forces with increasing burial depth.

1.1.3 Hydrocarbon Traps

The expelled hydrocarbon is lighter than the water in the interconnected porespace of the permeable sand layers and moves upwards as a result of buoyancy.This is called secondary migration. The upward migration of generated hydrocar-bon continues until it is halted by an impermeable barrier or trap. As hydrocarbonaccumulates under the trap a reservoir is formed.

The characteristics of hydrocarbon traps are illustrated by considering a porouspermeable formation between two impermeable layers which has been folded bytectonic action into an anticline (see the attached figure). This is called an an-ticlinal trap. The hydrocarbon contained in the reservoir may be oil or gas orboth.

Here are some commonly used terms to describe petroleum reservoirs. You willprobably be familiar with most of these terms.

9

Figure 1.2: Cross-section for an anticlinal reservoir

Formation: the rocks, These may be either clastic rocks (sandstones) or lime-stone and dolomite (carbonates).

Spill point: the lowest point of the trap that can hold hydrocarbon.

Trap closure: the distance between the crest and the spill point.

Reservoir: the part of the formation which contains hydrocarbon (oil and/orgas) and connate water.

Connate water: water in the pore space occupied by hydrocarbons.

Gas cap: gas filled zone or gas reservoir.

Oil zone: oil reservoir.

Water zone or aquifer: the body of water bearing rock in hydraulic commu-nication with the reservoir.

Gas-oil contact: lowest depth at which gas can be produced.

Oil-water contact: lowest depth at which oil can be produced.

Bottom water: water below the oil-water contact.

Edge water: water laterally adjacent the oil-water contact.

10

Figure 1.3: Anticline structural trap

Figure 1.4: Faulted structural trap

1.1.4 Classification of Traps

Hydrocarbon traps are classified as either structural or stratigraphic.

Structural traps

Structural traps are formed by tectonic processes acting on sedimentary layersafter deposition. They may be classified as,

Fold traps - formed by compressional or compactional anticlines.

Fault traps - formed by displacement of blocks of rock as a result of unequaltectonic forces.

Diapiric traps - formed by intrusion of salt or mud diapirs.

11

Figure 1.5: Salt dome structural trap

Stratigraphic traps

Stratigraphic traps are produced by facies (rock type) changes in the formationsuch as pinchouts and lenticular sand bodies surrounded by impermeable shales.The processes involved in the formation of stratigraphic traps are complex becausethey involve changes in the depositional environment.

Stratigraphic traps may be associated with unconformities. An uncomformityforms when a site of sediments is uplifted, eroded and buried again under newlayers of sediments that may form the trap. Unconformities generally separateformations formed under very different depositional conditions.

Study the attached figures in the context of the above discussion. This willimprove your understanding of the basic concepts. Note the following importantpoints in the last of the figures:

(i) The oil and gas zones in the central block are not in direct pressure com-munication with the oil and gas zones in the left and right blocks. The oiland gas zones have different contacts.

(ii) However, it cannot be concluded from the section alone that the centralblock is completely isolated from the rest of the formation. The blocks maybe in pressure communication through the aquifer.

12

Figure 1.6: Stratigraphic trap

Figure 1.7: Schematic of an oil and gas reservoir with a faulted central block anddifferent fluid contacts

13

Figure 1.8: USA oil resource

1.2 OIL RECOVERY PROCESSES

Oil recovery operations are generally classified into three groups.

Primary Recovery - production using only natural reservoir energy (naturalwater drive, gas cap expansion, solution gas drive and pressure depletiondrive).

Secondary Recovery - water or gas injection to maintain reservoir pressure(waterflooding and immiscible gas injection to supplement natural reservoirenergy).

Tertiary Recovery - enhanced oil recovery processes (EOR).

An EOR process is any process which does a better job of recovering oil thanconventional technology (primary and secondary recovery processes). In an EORprocess conventional water or gas is replaced by a more effective (more expensive)recovery agent.

14

1.2.1 Residual Oil Resource (Target for EOR )

Conventional primary recovery methods are usually very inefficient and, on theworld average, recover approximately 1/3 of the OOIP.

- 2/3 OOIP cannot be produced by conventional recovery technology.

- current EOR technology can produce about 20% —30% of the residual oilresource.

1.2.2 Residual Oil is Trapped or By-passed

Oil which cannot be recovered using existing facilities or infrastructure (existinginvestment) is trapped (microscale) or by-passed (macroscale) - trapping of oiloccurs on all reservoir length scales.

- Trapping on the micro or pore scale is by capillary forces.

- Trapping on the macro or field scale is caused by areal and vertical by-passing.

Trapping on all scales is strongly influenced by heterogeneity.

1.2.3 Recovery Processes

Unrecovered oil may be classified into two categories: Unrecovered mobile oil andimmobile or residual oil.

- Unrecovered mobile oil can be recovered by conventional processes by im-proved access to the reservoir. Reservoir access may be improved by,

- infill drilling.

- horizontal and multilateral wells.

- removal of formation damage caused by completion operations.

- fracking.

- perforating unperforated layers.

- selective shut-off of water and gas producing zones.

15

Figure 1.9: Residual is trapped or bypassed on all reservoir length scales

16

Figure 1.10: Recovery processes

- Immobile oil cannot be recovered by primary and secondary recoveryprocesses such as waterflooding and immiscible gas injection. This oil canonly be recovered by tertiary EOR processes.

All EOR techniques attempt to recover residual oil by:

(i) Improving reservoir sweep efficiency .

(ii) Mobilizing immobile or residual oil.

1.2.4 Primary Recovery Mechanisms

An estimate of the likely primary production mechanism for a reservoir may bemade on the basis of geological data even before the reservoir has been produced.

17

For example:

(i) Thin, flat reservoirs with low permeability will most likely produce by asolution-gas drive mechanism.

(ii) Reservoirs having high closure and high permeability may experience thebeneficial effects of gravity segregation of oil and gas. This may supplementthe effectiveness of the original gas cap, or form a secondary gas cap.

(iii) High permeability reservoirs in contact with an extensive aquifer will benefitto some degree from natural water influx (water drive).

Material balance calculations may be used to determine the relative importance ofthe various natural drive mechanisms for a particular reservoir. Material balancecalculations require:

(i) Accurate and comprehensive production history.

(ii) Representative fluid samples for PVT analysis.

(iii) Periodic pressure surveys.

The importance of instituting a program to gather this data as early as possiblein the development and production life of a reservoir cannot be over emphasized.A good data gathering program is a key component of an effective reservoirmanagement strategy.

1.2.5 Secondary Recovery

Predictions of future reservoir performance are required, amongst other things,to identify the need and timing for secondary gas or water injection. Usefulestimates of future production trends can be obtained from a knowledge of:

(i) The size of the gas cap relative to the size of the oil reservoir.

(ii) The size of the aquifer relative to the size of the oil zone.

(iii) Formation permeability can be estimated from core data and flow tests onwells.

Drill-stem tests on dry holes and structurally low wells (wells penetrating theaquifer) may yield important indications of aquifer permeability and continuity.Sometimes it is useful to complete dry holes as aquifer observation wells. Pressuredata from such wells can be important in characterizing reservoir-aquifer systems.The aquifer can have a profound effect on reservoir performance.

18

Figure 1.11: Steam flooding process

1.2.6 Tertiary Recovery — EOR Processes

The same techniques used to determine the need for secondary recovery are usedto assess the need for tertiary developments. Decisions regarding the actual im-plementation of EOR processes are not made until sufficient field data is availableto accurately assess primary and secondary performance.

EOR processes may be classified into three major categories:

1. Chemical

— Micellar Polymer Flooding

— Polymer Flooding

— Caustic or Alkaline Flooding

2. Thermal

— Steam Flooding

— Fire Flooding

19

Figure 1.12: Carbon dioxide flooding process

3. Miscible

— Enriched Hydrocarbon Gas Flooding

— CO2 Flooding

— Nitrogen and Flue Gas Flooding

The complexity of these processes requires the gathering of large quantities of ad-ditional laboratory fluid and core data and the use of considerably more complexanalysis techniques.

1.3 WHAT IS RESERVOIR ENGINEERING?

Reservoir Engineering is the science of understanding the productioncharacteristics of oil and gas fields under primary (natural pressuredepletion), secondary (water flooding, immiscible gas flooding) andtertiary (EOR) drive mechanisms.

A basic understanding of reservoir engineering concepts is necessary for the plan-ning, development and production of oil and gas fields.

20

Figure 1.13: Chemical flooding process

Geologists and Petrophysicists provide a description of the structure of the forma-tion or reservoir and vital physical parameters describing the reservoir internalstructure and initial distribution of fluids. The reservoir engineer utilizes thisdata together with well production rates and measured reservoir pressures toanalyze and interpret the data in order to optimize oil and gas recovery.

Reservoir Engineers are required to answer the following three vital questions:

1. How much hydrocarbon (oil and/or gas) does a reservoir initiallycontain(initial hydrocarbon in-place)?

2. How much of the hydrocarbon initially in-place can ultimately berecovered?

3. How will the production rates of wells depend on the physical pa-rameters of the reservoir, reservoir geometry (shape), well num-ber and development pattern and how will the rate decline withtime?

The first question may be answered if the shape and size of the pay zone is knownand the distribution of porosity and water saturation in the zone are also known.

21

This data is provided by the geologist and the petrophysicist. The reservoirengineer may assist by estimating the positions of gas-oil and water-oil contactsif the positions of these contacts is unknown.

The answer to the second question is very complex. It requires the use of sophisti-cated mathematical models (usually computer based reservoir simulators). Theseare required to model the many alternative developments which result in differentultimate recoveries and profitability. The process of producing production fore-casts for the different recovery mechanisms and development plans will always beassociated with some degree of uncertainty because geologic formations are highlyheterogeneous and the geological description, no matter how sophisticated, cannever capture all the geologic variability.

The answer to the third question relies on detailed analysis of pressure and indi-vidual well test data.

22

Chapter 2

RESERVOIR DESCRIPTION

The basis for a sound understanding of reservoir performance is a good descriptionof reservoir geology (geological model) and the distribution of fluids (oil, waterand gas) it contains. This is the first step in a reservoir engineering study.

Rock and Reservoir Description

Reservoir engineering studies require that the physical makeup of a reservoir berepresented in a usable manner. This should include;

(i) a description of reservoir stratification (reservoir lithology),

(ii) a description of reservoir geometry, both areally and vertically,

(iii) information on porosity, permeability, and water saturation throughout thereservoir, and

(iv) a description of the size and permeability of the adjoining aquifer.

This information is represented by various types of maps and cross-sections.These are usually prepared by production geologists.

Reservoir Heterogeneity

Most reservoirs are layered because of variations that existed in the depositionalenvironment. Depositional conditions at any instant also vary from one locationto another which results in lateral as well as vertical changes within the reservoirand within individual rock units. As we will see later, these changes result in

23

Figure 2.1: Highly detailes geostatistical models for reservoir characterization

24

variations in porosity, fluid distribution, and permeability. Permeability is ameasure of the relative ease with which fluids flow through the rock.

Wireline logs

Electric and other types of logs can be very important tools for determiningreservoir layering. Reservoir cross-sections based on logs can show if permeablezones are continuous throughout the reservoir or if they are lenses having limitedareal extent.

Logs also show the net sand at each well. Net sand is the rock that contains recov-erable hydrocarbons and possesses permeability. A geologist should be consultedwhen analyzing logs for reservoir zonation and net sand thickness. Generally, logsare available for all wells drilled and represent the most complete set of reservoirdescriptive data. Methods of analyzing logs and techniques of using logs for de-termining net sand thickness and evaluating sand continuity are beyond the scopeof this course.

Core analysis

Core analysis data provide an additional basis for determining net sand thicknessand reservoir zonation. Core analysis data is more quantitative than logs indescribing the reservoir, but generally, only a fraction of the wells are cored.However, in most cases logs are more effective when supported by core data. Thelogs can be calibrated by core data to make them more useful from a quantitativestandpoint. Under some conditions, the calibrated logs can be used to obtainfairly good estimates of the porosity and permeability profiles at all wells in areservoir.

Stratification

Stratification or heterogeneity occurs on all length scales in a reservoir. Althoughreservoir stratification is usually considered only in terms of net sand layers andimpermeable streaks of sand or shale, stratification also exists within the net sandlayers themselves and the degree of porosity and permeability can vary greatlywithin each strata.

Porosity and Permeability

The two most important properties of net sand are porosity and permeability.Porosity of net sand usually is in the range from 10 to 30 %. Some limestone anddolomite reservoirs locally may have porosities as high as 60 %. The porosityin reservoir rocks usually occurs in the spaces between rock particles and this iscalled primary porosity or intergranular porosity. Primary porosity can also existin fractures and vugs. Primary porosity may be modified by post depositionalevents. For example, small mineral particles may be deposited in the space be-

25

Figure 2.2: Vertical heterogeneity or layering

26

Figure 2.3: Examples of structural and geological complexity

27

Figure 2.4: Schematic of wireline logging operation

tween sand grains to produce microporosity. These digenetic changes may greatlyreduce the original porosity and permeability of a rock.

Typically, the permeabilities of the net sand portion of a reservoir will vary from10 md or less to 500 md or more. The ability of a rock to allow fluid flow isproportional to permeability, so fluid flow profiles in the reservoir may be veryuneven.

The process of reservoir description is sometimes referred to as formation evalu-ation and reservoir characterization. It involves the following steps:

(i) Gathering of data on the physical characteristics of formation rocks.

(ii) Gathering of data on the characteristics, occurrences and distribution offluids within these rocks.

(iii) Interpretation of the above data for accuracy and reliability. These dataare used to evaluate initial volumes of oil and gas in the reservoir

(iv) Evaluation of potential sources of reservoir production energy. e.g. extentof aquifer and size of gas cap.

28

Figure 2.5: Formation microscanner and resistivity logs

29

Figure 2.6: Schematic of a rotory core barrel

30

Figure 2.7: Diamond coring bits

Data collection is expensive (wireline logging, sampling, pressure testing etc.) andthe collection of poor quality or inaccurate data is contradictory and wasteful. Itis therefore necessary to place considerable effort into preparing an effective datagathering program.

2.1 RESERVOIR DESCRIPTION PROGRAM

The primary objectives of a data gathering program are to answer the followingquestions:

(i) Does the formation contain commercial quantities of oil and gas?

(ii) How should the reservoir be produces to maximize economic return?

or, more specifically,

(i) How much stock tank oil and/or free gas is initially in place?

(ii) What is the likely primary recovery mechanism and will it be necessary tosupplement this energy by water or gas injection?

(iii) What are the oil and gas reserves (production volumes for a particular fielddevelopment)?

(iv) How will individual well rates decline, and how can the decline in productionrate be arrested?

31

Figure 2.8: Schematic of a sidewall coring gun

32

Figure 2.9: Full diameter core

33

Figure 2.10: Length scales for laboratory core analysis

Figure 2.11: Sampling anisotropic core

34

The above questions cannot all be answered immediately after drilling of a discov-ery well. However, provided that the data gathering program is well thought-out,the engineer can arrive at reasonable planning estimates.

As the field is developed and produced, additional data becomes available, andthis is integrated with the existing data to reduce the level of uncertainty associ-ated with the initial estimates.

The overall data gathering program is thus a continuing process over the life of thefield. It is important to recognize this when developing the initial data gatheringprogram.

2.2 SOURCES OF DATA

The following is a guide to the type of data which may be collected with thedrilling of the first well in a reservoir:

- Original reservoir pressure and temperature.

- Gross reservoir thickness at the well.

- Lithology of the reservoir rock.

- Stratigraphic sequence of rock at the well.

- Reservoir porosity.

- Initial fluid saturations.

- Well productivity (permeability).

- Characteristics of reservoir fluid.

The above information is obtained from:

(i) Core samples.

(ii) Wireline logs.

(iii) Fluid samples.

(iv) Pressure and production testing.

35

Additional delineation or development wells drilled after the first discovery wellshould provide the following data:

- Reservoir thickness variations to allow mapping of the field.

- Areal variations in permeability, porosity and water saturation.

- Continuity of stratigraphic units between wells.

- Vertical permeability variations (vertical barriers to flow) in the reservoir.

- Variations of sub-sea depth of reservoir top and base for structure maps.

- Depth Of gas-oil and oil-water contacts.

- Variations in fluid compositions within the reservoir.

These factors are determined in the same manner as for the first well. Thepresence of multiple wells allows interference testing between wells to determineaverage permeabilities between wells and to test the continuity of individual sandunits.

Not all of the above data will be collected from each well drilled. The engineermust continually assess the data gathering process and only collect that datawhich materially reduces the level of uncertainty in estimating the importantreservoir parameters.

When areal variations in rock properties are large, it may be necessary to collectall the data from all the wells drilled and to drill additional data wells.

2.2.1 Coring And Core Analysis

Coring is the most basic formation evaluation tool. It provides the engineer withthe only opportunity to physically inspect a piece of the reservoir. It providesthe only means of determining,

(i) Reservoir wettability.

(ii) Capillary pressure.

(iii) Relative permeability.

(iv) Residual oil saturation.

36

These are important in determining formation production characteristics andreservoir recovery factors.

Measurements on reservoir core samples are also required for:

(i) Calibration of wireline logs, essential for quantitative log interpretation.

(ii) Identifying potential causes of formation damage.

It is difficult to base an entire reservoir description entirely on core analysisdata. This is because even in a heavily cored reservoir, the total cored volumeconstitutes only a very small fraction of the entire reservoir volume. As a result,it is very difficult to assess the statistical significance of the data. At least onewell in the reservoir should be cored over the entire producing interval. Thedata obtained provides valuable information for describing vertical variations inreservoir rock properties.

2.2.2 Wireline Logging

Wireline logging provides information on:

(i) lithology,

(ii) porosity,

(iii) water saturation,

of the formation penetrated by a well. Whereas only a few wells are fully cored,it is common practice to log all wells in the field.

Logs provide the basis for determining,

(i) gross and net formation thickness,

(ii) correlations identifying individual reservoir sand units,

(iii) continuity of reservoir sand units,

37

There are three basic types of log:

1. Electric — fundamental log run in all wells drilled. Most electric logs mustbe run in an uncased hole containing conducting fluid.

2. Sonic — Usually run in open hole.

3. Radioactivity — can be run successfully under nearly all wellbore conditions.

Although logs provide valuable data for reservoir description, the main purposeof logging is to identify a suitable completion interval for the well.

A logging program should be designed to designed to provide the data requiredfor reservoir description at a minimum cost. Lithology and borehole conditionsmust be considered in the selection of a suitable suite of logs. This will usuallyrequire a trial-and-error process of selecting the most effective combination oflogs for a particular reservoir. Prior experience is a major fracture in the designof an effective logging program.

2.2.3 Pressure and Production Testing

Pressure and production tests are designed to determine,

(i) fluid content,

(ii) formation productivity.

Production tests repeated on a yearly basis over the life of a well provide valuableinformation on,

(i) wellbore plugging,

(ii) reservoir pressure decline,

(iii) invasion of the wellbore by water or gas.

Pressure build-up and draw-down tests involve flowing a single well at aconstant rate for a predetermined period of time and then shutting-in the welland observing the rate at which the wellbore pressure rises or falls with time.An analysis of the pressure response allows an estimate of formation productivityand permeability.

38

Figure 2.12: Wireline formation interval tester

Interference tests provide checks on formation permeability and continuity.These tests involve flowing a production well and measuring the resulting pressuredecline at nearby wells.

Drill-stem tests are usually run at the time the well is drilled to determine ifthe well should be completed over the interval identified by the well log. The DSTtool allows us to make a temporary completion to conduct pressure build-up anddraw-down tests. DST’s are useful for:

(i) testing the potential production intervals,

(ii) locating the positions of gas-oil and oil-water contacts,

(iii) Determining average reservoir pressure.

2.2.4 Fluid Sampling

Reservoir oil usually contains a considerable amount of dissolved gas. When theoil is produced to the surface the gas comes out of solution and the oil volumeconsequently shrinks. The oil that fills a barrel at surface conditions will haveoccupied between 10%—50% at reservoir conditions.

39

Figure 2.13: Initial and production formation pressure data with flowmeter survey

40

Figure 2.14: Single and multi-well pressure tests

Figure 2.15: Rate and pressure history for an interference test

41

To determine oil-in-place it is necessary to know the number of stock tank barrelsoccupied by a reservoir barrel of oil.

Other important properties of oil, such as density and viscosity, also change whensolution gas is released. These properties are very important in reservoir fluidflow calculations and must be accurately known.

In large reservoirs having high closure, fluid properties may vary significantly bothareally and vertically. Enough samples should be taken to adequately describethese variations.

Reservoir fluid may be sampled in two ways:

1. Subsurface — wireline samplers in the wellbore at the perforations or aspart of the initial open-hole formation evaluation program.

2. Recombined surface samples — oil and gas samples from the test sepa-rator recombined in the ratio of the produced gas-oil ratio.

2.3 INTEGRATED FORMATION EVALUA-

TION PROGRAM

No single wireline tool or procedure is capable of providing all the data requiredto characterize a reservoir and its contents. It is therefore necessary to develop anintegrated formation evaluation program which consists of a carefully consideredmix of:

(i) Logs - these provide most of the data for reservoir characterization.

(ii) Cores - used to calibrate logs for quantitative interpretation.

(iii) Well testing - provide estimates of permeability and productivity.

(iv) Fluid sampling - only means of determining reservoir fluid properties.

The best approach is to adopt a Key Well Program. The following steps aretaken in the development of such a program:

1. A number of Key wells are selected to provide a representative coverageover the reservoir. A rule of thumb is that at least one well is needed foreach 640 acres. For heterogeneous reservoirs this is reduced to 320 acres orless.

42

Figure 2.16: Bottom hole sampling tool

43

Figure 2.17: Surface sampling for laboratory recombination

2. A data gathering program is designed for each key well. This will includelogging, coring and well testing to provide complete reservoir coverage overthe reservoir.

3. Determining which logs provide the best quantitative data by comparinglog interpretations with core data. Logs run in non-key wells are theninterpreted according to the correlations developed for the key wells.

4. Any available production data from key wells is also used to aide log inter-pretation.

5. The above data gathering procedures are continuously monitored with theobjective of eliminating any unnecessary or ineffective procedures.

The above steps ensure a data gathering procedure which is both simple and costeffective.

A reservoir description program is needed during all phases of the producing lifeof a field.

1. Early in the development of the field this information is required toachieve proper well spacing and productive completions.

2. After field development the information is used for reservoir energycontrol in order to achieve high ultimate recovery.

44

Figure 2.18: Key well program

45

Figure 2.19: Aquifer models - limited data and interference

3. Secondary and Tertiary recovery. To effectively engineer theseprocesses it is necessary to have an accurate reservoir description and anestimate of hydrocarbon recovery.

In addition to production wells it may be necessary to drill observation wells.These wells can provide data which includes:

(i) Reservoir pressure.

(ii) Pressure gradients in the reservoir.

(iii) Position and movement of contacts.Contacts can be detected by production tests, cased hole logging and 4Dhigh-resolution interwell seismic.

2.4 AQUIFER DESCRIPTION

The aquifer is the total volume of porous water-bearing rock in pressure commu-nication with a hydrocarbon reservoir. The size and permeability of an aquiferwill control how much water drive energy is available to the reservoir. The aquifer

46

energy will determine if the reservoir will produce primarily by water drive or byliquid or gas expansion within the reservoir. Reservoir drive mechanisms will bediscussed in detail later in the course. At the present time, we are concerned onlywith methods of determining the size and permeability of an aquifer.

If an aquifer is large, most of the information on it must come from wells drilledoutside the reservoir area. Logs and drill-stem tests on dry holes are about theonly source of aquifer data. Logs will show the thickness of water-bearing rockthat is in communication with the reservoir. If enough dry holes are available,the areal extent of the aquifer can be estimated. An estimate of the aquiferpermeability can be determined from the results of drill-stem tests on dry holes.Equations for calculating permeability from flow test data will be studies in theportion of the course entitled fluid flow.

The aquifer can have a major effect on the production-pressure per-formance of a reservoir. This topic will be covered in detail in the section onwater drive reservoirs.

47

Chapter 3

VOLUMETRICS AND INITIALHYDROCARBON VOLUME

The first step in a reservoir study is to accurately determine the initial hydro-carbon volume. This requires data which will allow us to calculate the size andgeometry of the reservoir and the fluid volumes which the reservoir contains. Inthis section we will briefly look at how maps and cross-sections are used to de-scribe the geometry of a reservoir and calculate the hydrocarbon volume in-place.

Contour maps are commonly used to show reservoir geometry and the distributionof important reservoir parameters.

3.1 STRUCTURE MAPS

Structure maps show the geometric shape of a reservoir or formation. The mapsmay show the top or the bottom of a structure or reservoir unit. Examples oftop of structure maps are attached. These maps also show the positions of fluidcontacts in the reservoir.

Structure maps are prepared by geologists. The data on which the maps aredrawn usually come from;

(i) well control,

(ii) geophysical data usually in the form of time maps, and

(iii) geological models of depositional and post-depositional events.

Gross thickness isopach maps show the total interval between the top andthe base of the reservoir rock for each well. Structure maps on the top and on

48

Figure 3.1: Isometric schematic of the Ekofisk structure

49

Figure 3.2: Top-of-structure map of a hydrocarbon reservoir

the base of the reservoir can provide data for the isopach map. Frequently, mostof the wells are not drilled to the base of the reservoir so a base structure mapcannot be drawn. In this case, the gross sand interval must be estimated fromreservoir cross-sections based on logs from the well which penetrated the entireinterval.

The gross pay isopach map for an oil reservoir is more descriptive of the hydro-carbon reservoir geometry than the gross thickness isopach.

3.2 ISOPACH MAPS

Isopach maps show the distribution and thickness of reservoir properties of in-terest. The contour lines connect points of equal vertical interval. Examples ofcommon isopach maps are:

Gross oil thickness isopach map: contours gross pay - the depth of the topof the oil column minus the top of the bottom of the oil column.

Net oil thickness isopach map: contours net pay - gross pay minus non-reservoir intervals such as shales.

Net oil isopach maps are commonly used to calculate volumes of hydrocarbonsin-place.

50

Figure 3.3: Net sand isopach map for a hydrocarbon reservoir

Other useful maps include:

Net-to-gross ratio maps: fraction of the total hydrocarbon interval which con-tributes to recovery.

Iso-porosity map: contours average porosity over net-pay portions of the de-sired formation.

Iso-water saturation map: contours average water saturation over net-payportions of the desired formation.

51

Figure 3.4: Isoporosity map for a hydrocarbon reservoir

Figure 3.5: Iso-water saturation map for a hydrocarbon reservoir

52

3.3 VOLUMETRIC METHOD FOR DETER-

MINING ORIGINAL OIL-IN-PLACE

The original oil-in-place (OOIP) contained in a reservoir expressed in Stock TankBarrels (STB) is designated by the symbol N and is given by the equation (fieldunits):

N =7758Vbφ(1− Swi)

Boiwhere,

Vb = reservoir bulk volume (acre-feet)φ = average porosity (fraction)Swi = average initial water saturation (fraction)Boi = average initial oil formation volume factor (RB/STB)

or, in any set of self consistent units:

N =Vbφ(1− Swi)

Boi

In this and the following reservoir engineering courses it is assumed that you arefamiliar with units and unit conversions i.e., starting with the above equation forany set of self consistent units, you should be able to calculate the constant 7758in the preceeding equation. If you wish to review the topic of units go to thesection Units - unit conversions on page 108 of Dake, Fundamentals of reservoirengineering.

The original gas-in-place (OGIP) contained in a reservoir, expressed in StandardCubic Feet (SCF), is designated by the symbol G and is given by the equation(in field units),

G =7758Vbφ(1− Swi)

Bgi

where,

Bgi = average initial gas formation volume factor (RB/SCF)

or, in any set of self consistent units:

G =Vbφ(1− Swi)

Bgi

53

Figure 3.6: Schematic of the microstructure of a sandstone showing sand grainsand interconnected pore space which allows fluid flow

3.3.1 Reservoir Volume

The following information is required to calculate net reservoir volume:

1. Bulk reservoir Volume.

2. Reservoir Stratification (and net-to-gross ratio).

The calculation procedure involves the following steps:

1. Prepare a map of gross reservoir thickness.

The map is based on log data and requires the construction of structuremaps for;

(i) the top of the reservoir.

(ii) the bottom of the reservoir.

The gross thickness is the difference in depth between the top and thebottom of the reservoir zone.

2. Construct a gross sand isopach map by removing intervals containingonly gas or water. These intervals will generally lie above the gas-oil contact(GOC) and below the oil-water contact (OWC).

54

3. Construct a net sand isopach map by eliminating all non-reservoir rockintervals (shale, siltstone, coal seam etc.).

This procedure will normally involve estimates of minimum porosity andpermeability cut-offs. The actual intervals to be excluded are picked fromporosity logs which have been calibrated using core data.

The calculation is usually performed numerically using digitized maps.

3.3.2 Average Porosity

Average reservoir porosity is determined by mapping individual well porosityvalues. These are determined from core data and from sonic and radioactivitylogs calibrated with core data.

3.3.3 Average Initial Water Saturation

The methods used to determine reservoir water saturation include:

1. Logging — induction and focused resistivity logs.

These logs measure formation resistivity which may be related to water sat-uration. This requires a knowledge of the wettability state of the reservoirand special core test data.

2. Coring with oil-based muds.

Provided that invasion of drilling fluids has not changed reservoir wettabil-ity, coring with oil-based mud can result in core which gives a good indi-cation of the irreducible water saturation. This corresponds to the watersaturation in the reservoir above the transition zone.

In the transition zone the core will indicate low water saturation becausethe oil from the oil-based mud will have displaced some of the mobile waterfrom the core during the coring operation.

3. Restored state tests.

These tests attempt to restore the wettability of core in the laboratory tothat in the actual reservoir. The restored state core is then used in thelaboratory to duplicate the displacement processes by which the reservoirwater saturation was initially established.

55

The data gathering program should include water saturation data determined byall of the above methods. Although cutting core with oil-based mud is expensive,at least one well should be cored in this manner to provide data for calibrationof log resistivity.

Laboratory measurements on restored state core also provide capillary pressuredata which can be used to calculate water saturations in the transition zone. Thisdata is also used to calibrate logs. The overall objective is to collect sufficientdata to develop meaningful correlations between log measured resistivity andwater saturations in restored state laboratory tests.

3.3.4 Average Oil Formation Volume Factor

The average oil formation volume factor is required to convert the reservoir oil orgas filled hydrocarbon volume to the equivalent volumes at surface or stock tankconditions.

Oil and gas formation volume factors are determined in laboratory PVT testsconducted with representative samples of reservoir oil and gas.

Estimates of oil formation volume factors may be obtained from empirical corre-lations if the following are known:

(i) Initial gas-oil ratio (solution GOR).

(ii) Reservoir temperature and pressure.

These are usually estimated from DST tests conducted on exploration wells.

3.3.5 Determining Initial Oil-In-Place

In order to illustrate the calculation of initial oil-in-place and introduce Mathcad- the spreadsheet program which we will be using throughout the course, weconsider the very simple case of a homogeneous reservoir with unity net-to-grossratio where the bottom of the reservoir is the horizontal surface formed by thewater-oil contact (WOC). For these conditions a top-of-structure map and averageporosity and water saturation are all that is needed to calculate the initial ororiginal oil-in-place.

56

Figure 3.7: Schematic of interstitual, irreducible or connate water in a waterwetporous medium

57

Figure 3.8: Reservoir pressure survey map used to determine average reservoirpressure

Example 3.1 - Calculation of OOIP from a reservoir structure map[IHIP.mcd]

Calculate the initial volume of oil-in-place for the Apache Pool reservoir (top-of-structure map attached). The average porosity, average connate water saturationand average oil formation volume factor have been determined and these valuesare given below.

Data:

φ = 0.27Swi = 0.38Boi = 1.2715 (RB/STB)

We use this exercise to introduce you to Mathcad. It is assumed that you haveinstalled Mathcad on your PC and that you have worked through the basics inthe on-line tutorial. In order to see the solution to this exercise you will needto change some of the inputs to the spreadsheets. The areas of the spreadsheetwhere these changes may need to be made are highlighted in yellow.

[Answer: 340 MMSTB]

Solution outline

1. The first step is to determine the area inside each depth contour. Thecontour areas must be determined from the map. This is usually done

58

Figure 3.9: Structure map for the Apache field

59

using specialized computer software and digitized maps. For this exampleI have digitized each of the contours by hand and used Simpson’s rule tonumerically calculate the area. The use of Simpson’s rule for numericalintegration is shown in the attached figures. Study the three figures to besure that you understand what is being done.

To determine the area of a depth contour I first printed an enlarged copyof the Apache structure map and overlayed this with a 1 cm by 1 cm grid.I then divided each contour into an upper and lower curve as shown in theattached figure and read-off the (x, y) points defining each curve. Thesepoints are entered into a file for each contour eg., [C2000.prn] in the folder[Contours]. If you open one of these files you will see the how the pointsare entered for the lower and upper contours. Note that the end-points forthe lower and upper contours are the same. [Contour.mcd] , also locatedin the [Contours] folder, reads the contour files and calculates the areas,and plots the contour. This is done in the manner outlined in the coursenotes and you should have no trouble following the flow of the calculation.When Mathcad reads a file like [C2000.prn] all it ”sees” is a matrix of(x, y) numbers - it ignors blanks or anything which it does nor recognise asa numerical input. This makes it easy to annotate or ”pretty-up” the inputfile so that it is understandable to the casual reader.

You can use as many data points as you like to define a contour - the morepoints you use, the smoother the contour and the more accurate the areacalculation. The only limitation is that the total number of points definingthe lower and upper contours must be the same. This makes it a little easierto program the calculations.

When you open [Contour.mcd] you will see that I have left it doing onecontour over and over. You will need to change the other input contour filenames to see the areas and plots for each of the contours.

The areas within each contour, A0, A1, A2, etc. are multiplied by the mapscale-factor squared to convert the map areas to actual field values.

60

2. The bulk volume between two successive contours is given by the pyramidalformula:

∆Vi =h

3

µAi +Ai+1 +

qAiAi+1

¶where,

∆Vi = bulk volume between contours i and i+ 1.Ai = the area enclosed by the lower contour.Ai+1 = the area enclosed by the upper contour.h = the vertical height between the contours, or

the contour interval.

3. The initial volume of oil-in-place in volume element ∆Vi,in any consistentset of units (as used by Mathcad), is given by,

∆Ni =∆Viφi(1− Swc,i)

Boi,i

4. The initial oil-in-place is the sum of the values for each volume element.

N =Xi

∆Vi

See the attached figures which show how Simpson’s rule is used to calculatecontour areas from digitized map data.

61

1. Contour is divided into upper and lower curves

Upper curve

Lower curve

Figure 3.10: Computer calculation of contour areas

Simpson's rule to calculate the area under the upper and lower curves

A =X yi − yi+ 1

2∆x

yiyi+1

∆x

AU

∆x

yiyi+1

AL

Figure 3.11: Computer calculation of contour areas

62

Contour Area

AU

AL

AC = AU − AL

Figure 3.12: Computer calculation of contour areas

63

Problem 3.1 - Calculation of OGIP from a reservoir structure map

How much gas would the Apache Pool contain if the initial reservoir pressure was1500 psia, the initial reservoir temperature was 200oF, and the gas compressibilityfactor was 0.877 ? All other data as for Exercise 3.1.

[Answer: 223 BSCF]

Solution hint

The gas formation volume factor is given by,

Bg = 0.00504zT

P[RB/SCF]

where T - the average reservoir temperature - is in oR (oR=oF +460) and P -the average reservoir pressure - is in psia. Use this equation to calculate the gasformation factor.

You can use the areas, volumes etc., calculated in the previous exercise to helpyou solve this problem by hand calculation or using your favourite spread-sheet.If you do this be sure to be careful with units. If you want to master Mathcadyou can modify the files used in the previous example to do the calculation. Ifyou choose to do this make sure that you have looked at the Units section of theon-line Mathcad tutorial before making the necessary changes.

Problem 3.2 - Calculation of OGIP from a reservoir structure map

Estimate the initial volume of gas-in-place for the Marlin Field (top-of-structuremap attached). The average reservoir and fluid properties are given below.

Data:

φ = 0.21Swi = 0.19T = 210oFP = 2300 psiaz = 0.89

[Answer: 6.48 TSCF]

64

Figure 3.13: Marlin field top-of-structure map

Solution hint

You are free to solve this problem any way you want including using any com-mercial or in-house software to which you may have access. If you have beenfollowing the Mathcad route - I recommend this strongly - this problem will bringyou right up the learning curve.

Print a full-size (or larger) copy of the Marlin structure map. Create a full copyof the folder for Example 3.1 and in order to modify the files to solbe the presentproblem.

Follow the outline for Example 3.1 and construct the data files for contours (thereare 9 of these). Pay special attention to the particular manner in which thesefiles are constructed! Modify the [Contour.mcd] by copy-and-paste to calculatethe areas for all nine contours. Modify the final graph which plots the contoursto plot all nine contours. You will probably need to revise the ’graphs’ section ofthe on-line manual to do this.

You will need to make the same changes to [Contour.mcd] made for Problem3.1 to convert it to a gas in-place calculation. Note that for the Marlin contoursthe contour depth intervals are not all the same size. The best way to handlethis is to read-in individual values of h in the same way as we already read-inindividual values of porosity, initial water saturation, areas etc. You can do this

65

by adding a column of numbers to the input data file and making the appropriatechanges in the spreadsheet. These changes will involve reading-in the new values,introducing a new array for h, and modifying the equations to calculate with thearray elements hi. Describing these changes is a lot more difficult than makingthem!

If you succede in making the above changes you will have learnt all you need toknow to reproduce any of the Mathcad spreadsheets in this and other courses.

66

3.4 DETERMINATION OF OIL-IN-PLACE —

MATERIAL BALANCE METHOD

Material balance methods are another way of determining the initial hydrocarboncontent of a reservoir. Material balance calculations:

1. Require some production history (reservoir pressure and fluid productionrates with time). The more production history available, the greater theaccuracy of the calculation.

2. The method is most precise for reservoirs where water influx is not sig-nificant.

The method is most accurate for gas-cap-drive reservoirs if the pressuredrop is at least 100 psi.

The material balance method is the preferred method for determiningthe original gas-in-place (OGIP) for gas reservoirs with no significantwater influx.

For reservoirs where water influx is significant the analysis procedure is asfollows:

1. OOIP is determined using the Volumetric Method.

2. The Material Balance Equation is used to calculate the water influx.

3. The computed water influx and measured reservoir pressures are used tocharacterize the size and strength of the aquifer.

4. The Material Balance Equation is then used to predict future water influxand hence future reservoir performance.

For Material Balance calculations to produce realistic results it is necessary to:

1. Accurately determine initial reservoir pressure.This should be measured as part of the data gathering program for thediscovery well.

2. Initiate a program for periodic reservoir pressure measurementimmediately after commencement of production. This program (onan annual basis) should be continued throughout the producing life of thefield.

67

3. Accurately measure and diligently report the production volumesfor all the fluids produced. Although produced oil volumes are accu-rately measured as a matter of routine, material balance calculations requireequally accurate water and gas volumes.

4. Accurately determine fluid PVT data. This requires representativeoil and gas samples at reservoir conditions. The best oil samples are ob-tained before the reservoir is produced because the pressure drop whichaccompanies production may result in the release of solution gas and thismay preclude the possibility of obtaining a representative fluid sample. Gassampling does not have this problem.

3.5 ESTIMATING RESERVES

Reserves for a particular field are estimated from an analysis of the availableengineering data. The implication of this is that the more reliable the data(production history data), the more accurate will be the estimate ofreserves.

The procedure for estimating reserves involves the following steps:

1. Establish the primary production mechanism.

2. Apply the analysis procedure appropriate to the drive mechanism to es-timate reserves. This will usually involve the use of numerical reservoirsimulators.

Rock property data (porosity and permeability), fluid property data, reser-voir description and production performance data (pressure decline andfluid production rates) are the basis for the analysis.

Collecting all production data from early in the life of a field is ,again, ofcritical importance.

68

3.6 ESTIMATING DECLINE IN OIL PRO-

DUCTION RATES

The decline in field and individual well production rates is estimated on the basisof:

1. Decline curve analysis.

Decline curve extrapolation is satisfactory only for wells which are producedcontinuously at maximum or near maximum rates. For wells producing atlimited draw-down (below their maximum rate) extrapolated decline ratesare not meaningful.

Decline curve analysis are studied in detail in course PTRL6007 - ReservoirEngineering II.

2. Productivity index method.

The productivity index for a well is the well production rate per psi ofpressure draw-down. This is determined from well flow tests.

The productivity index is usually a function of flow rate and reservoir pres-sure (oil relative permeability, oil viscosity). Well flow tests must thereforebe conducted at different rates and the productivity index must be adjustedfor falling reservoir pressure.

For wells not subject to severe water and/or gas coning this method providesreliable estimates of production decline rates.

3. Reservoir simulation.

For wells subject to severe water and/or gas coning reservoir simulationoffers the most reliable method for predicting well decline rates and overallreservoir performance.

Reservoir engineering is studied in detail in course PTRL6004 - ReservoirSimulation.

69

Chapter 4

HYDROSTATIC PRESSUREDISTRIBUTION INRESERVOIRS

This section deals with the hydrostatic or initial pressure distribution in a reser-voir - the pressure which exists prior to significant production from the reservoir.We will consider the following specific topics:

1. Introduction to hydrostatic pressures in reservoirs.

2. Determination of gas-oil and water-oil contacts from pressure data.

3. Determination of oil water and gas densities and gradients from pressuredata.

4. Calculation of pressure kicks on penetrating a hydrocarbon zone.

5. Locating exploration wells searching for oil in a water- or gas-bearing for-mation.

70

Symbols and Units

g gravity constant m/s2

D depth to subsurface mp pressure N/m2

ρ density kg/m3

Dowc depth of initial oil-water contact mDgoc depth of initial gas-oil contact mDgwc depth of initial gas-water contact m

g = 9.81 m/s2

1 Pascal (Pa) = 1 N/m2

1 bar = 105 N/m2

1 Atmosphere = 1.01325 bar or 14.695 psia

Conversion factors

lb/ft3 × 16.02 = kg/m3

psi × 6.9 = kPapsi × 0.069 = barft × 3.28 = m

All of the above units may be embedded into working Mathcad files and can beused interchangeably or you can even mix units if you want. Normally we wouldnot elect to mix units!

4.1 SUBSURFACE PRESSURES

4.1.1 Water zone pressures

In normally pressured sedimentary basins water contained in the pore space ofa reservoir (connate water) is in pressure communication with the atmospherethrough the oceans or outcrops. Since the pore spaces which comprise the effectiveporosity of the reservoir are all interconnected, and the reservoir and associatedaquifer are initially in static equilibrium, pressure varies only with depth.

The difference in hydrostatic pressure between any two depth points in a forma-tion containing water is,

p2 − p1 = ρwg(D2 −D1)

71

where, ρw is the average density of water over the interval (D2 − D1) and g isthe gravitational acceleration. ρw varies with depth (pressure) and salinity. Sincewater is only slightly compressible, it is customary to neglect the overbar symboland simply write,

p2 − p1 = ρwg(D2 −D1)where it is understood that ρw is the average pressure over the interval of interest.

If we take p0 to be the pressure at the surface (atmospheric) where D1 = 0, theequation may be written as,

pwD − p0 = ρwgD

or,pwD = p0 + ρwgD

The term ρwg(D2 −D1) is the hydrostatic pressure difference which has units ofpressure. If we divide this pressure difference by (D2 −D1), we obtain the watergradient, γw, which is equal to ρwg and has units of pressure per unit depth. Ingeneral we define a phase pressure gradient as

γi = ρig

where the subscript i may be o-oil, w-water or g-gas.

Using the definition of the fluid gradient, we can write that the pressure differencebetween any two depths, ∆pi, is

∆pi = γi∆D

where ∆D is the difference in depth.

For the water zone the pressure difference is

∆pw = γw∆D

The gradient for fresh water is 0.433 psi/ft.

4.1.2 Oil zone pressures

The initial pressure, poD , at depth D in the oil zone may be expressed in terms ofthe oil pressure at the initial OWC (a commonly used datum point for reservoirengineering calculations) as,

poD − poDOWC= ρog(D −DOWC)

72

Figure 4.1: Hydrostatic pressure distribution in a reservoir

which may be rearranged to give,

poD = poDOWC− ρog(DOWC −D)

or,poD = poDOWC

− γo(DOWC −D)

where ρo is the average density of oil at reservoir conditions.

4.1.3 Gas cap pressures

The initial pressure, pgD , at depth D in the gas cap may be expressed in termsof the gas pressure at the initial GOC as,

pgD − pgDGOC = ρgg(D −DGOC)

or,pgD = pgDGOC − ρgg(DGOC −D)

or,pgD = pgDGOC − γg(DGOC −D)

where ρg is the average density of gas at reservoir conditions.

73

4.2 HYDROSTATIC PRESSURE DISTRIBU-

TION IN A RESERVOIR CONTAINING

OIL, WATER AND GAS

In the absence of capillary pressure, the WOC and GOC are sharp and the fluidpressures at the contacts are equal. The resulting hydrostatic pressure distribu-tion in the reservoir is shown in the attached figure.

Note that the contacts correspond to the intersections of the fluid gradient lines.This can be used estimate the positions of the contacts from measured pressure-depth data.

74

Figure 4.2: Hydrostatic pressure distribution in a reservoir assuming negligiblecapillary pressure

75

Example 4.1 - Calculation of regional hydrostatic pressures[REGPRES.mcd]

Five widely separated exploration wells were drilled in a large sedimentary basin.While penetrating the aquifer, the following pressure measurements were madein each of the wells:

Well A B C D ERotary table elevation (ft above MSL) 2133 1312 2707 3937 197Measured depth (ft) 9744 13993 8235 17388 9383Gauge Pressure (psia) 4340 5656 2480 6005 4090

Determine if all the wells belong to the same pressure system.

In answering this question you can use the Mathcad spreadsheet [REGPRES.mcd]which reads the above data and calculates the required fluid gradients.

[Answer: Well-A is clearly in an overpressured separate hydraulic system]

Solution hints

The relationship between true vertical depth sub-sea DTV D, measured depthDMD, and depth above mean sea level DMSL is given by

DTV D = DMD −DMSL

The average water gradient (sea-level to depth of measurement), γw, is given by

γw =∆p

DTV D

76

Example 4.2 - Calculation of fluid contacts and fluid gradients frompressure data [CONTACTS.mcd]

The following formation tester pressure measurements were made in an explo-ration well. Estimate:

(i) the depths of the contacts

(ii) the density and the nature of the fluids present in the formation

Depth Pressure(ft) (psia)8120 29128202 29208284 29278366 29508448 29788530 30058612 30388694 30758776 3114

Solution hints

Plot the data on a depth-pressure plot. Identify lines of constant slope, these showzones of different fluid saturation. Determine the gradients of depth-pressure linesand from these calculate the fluid densities. The densities identify the fluids.

[CONTACTS.mcd] is a general spreadsheet which I have found very useful inanalysing a wide range of formation pressure-depth data. If you have access tocommercial or in-house software which does similar things, feel free to use it.

[Answer: DGOC= 8302 ft ss., DOWC= 8573 ft ss., ρg= 13.2 lb/cubic ft., ρo= 48.3lb/cubic ft., ρw= 66.8 lb/cubic ft. ]

77

Figure 4.3: Hydrostatic pressure-depth data plot

78

Problem 4.1 - Optimal Location of an Exploration Well [EXPLOR.mcd]

Exploration well EX-1 intersected a hydrocarbon bearing sand with a verticalthickness of 200 ft between 6560 and 6760 ft-subsea. A wireline formation testerrecovered some gas and mud and recorded a pressure of 3850 psia at 6660 ft-ss.

A second exploration well EX-2 encountered the same sand between 7780 ft-ssand 7980 ft-ss but found it to be only water-bearing. Mechanical problems withthe downhole gauge prevented a pressure measurement from being taken. Thetwo wells are 12,000 ft apart.

On the basis of previous experience and sampling of reservoir fluids we have deter-mined that the regional water gradient is 0.524 psi/ft. The formation is normallypressured. The pressure gradient for the gas phase at reservoir conditions is cal-culated to be 0.084 psi/ft. The oil gradient, at reservoir conditions, is estimatedto be 0.296 psi/ft.

Questions

(i) Is the above data consistent with the assumption that the formation couldbe oil-bearing between 6660 and 7880 ft-ss ?

(ii) Where would you locate an additional exploration well which would defi-nitely find oil if any oil is indeed present?

(iii) What is the maximum possible thickness of the oil zone?

[Answers: (i) yes, (ii) new well should intersect the sand at a depth of 7445 ft ss.,(iii) 910 ft.]

Solution hints

Each of the wells gives us a fluid and a corresponding fluid pressure. Where amechanical problem prevented the measurement of a water phase pressure, wecan estimate the pressure by assuming that the formation is normally pressured(i.e., water gradient times depth).

The first well gives us the depth of lowest known gas. The second well gives thedepth of highest known water. If an oil zone exists, it must be between thesedepths.

Knowing the gas and water gradients, and a fluid pressure and correspondingdepth in the gas and water zones, we can draw the gas and water gradient lines

79

Figure 4.4: Maximum thickness of an oil zone between gas and water contacts

on a pressure-depth plot.

If there is no oil present (only gas and water), the depth of the GWC can befound from the intersection of the gas and water gradient lines. If a drop of oilis present it must gravitate to this depth splitting the GWC into a WOC and aGOC. The third well must intersect this depth to prove if any oil is present inthe reservoir.

The maximum possible thickness of the oil zone occurs when either the lowestknown gas or the highest known water corresponds to an oil contact. How do weselect which of the two is a possible contact? Simple, draw the oil gradient linethrough each of the points and select the one which produces a set of contacts(OWC and GOC) which are consistent with lowest known gas and highest knownwater.

If you understand the principles involved and the steps required to solve the prob-lem you should have little problem in entering the correct data into [EXPLOR.mcd]and manipulating the graph by setting appropriate contact and oil gradient linesto produce the correct answer.

80

Problem 4.2 - Calculation of a Pressure Kick when Penetrating a Hy-drocarbon Zone [PKICK.mcd]

Calculate the pressure kick to be expected when a well first penetrates a reservoirat 5000 ft subsea if the reservoir is known to have an OWC at 5500 ft and anGOC at 5200 ft.

The formation is known to be normally pressured with the following phase gra-dients;

water gradient 0.45 psi/ftoil gradient 0.35 psi/ftgas gradient 0.08 psi/ft

What would the pressure kick be if the entire hydrocarbon zone contained gas?

[Answer: 104 psi]

Solution hints

When drilling through the cap rock which forms the reservoir seal, the drill bitexperiences a water hydrostatic pressure gradient because the seal is saturatedwith water and in a normally pressured system this water is in hydraulic com-munication with the aquifer. Just before the bit penetrates the reservoir the bitpressure is the of the water gradient and depth.

On penetrating the hydrocarbon column the reservoir pressure is equal to thepressure at the contact minus the gravity head of hydrocarbon column from thecontact to the top of the reservoir. Since a hydrocarbon column is lighter thana water column of equivalent height, the reservoir pressure at the top of thereservoir is higher than the pressure at the base of the seal. This difference givesrise to a pressure kick.

You do not really need a spreadsheet to solve this problem. If you know what youare doing you can solve the problem faster with a hand calculator! I have preparedthe spreadsheet [PKICK.mcd] as an example of how a simple calculation can belayed out in Mathcad. Enter the correct numbers and you will get the solution.

81

Figure 4.5: Calculation of a pressure kick

Problem 4.3 - Interpretation of Repeat Formation Tester (RFT) data

The RFT allows an unlimited number of spot pressure measurements to be madein a single trip in an open hole. This allows the engineer to determine pressure-depth profiles across reservoir sections of interest. Comparing pressure-depthprofiles in different wells can provide valuable information on field fluid contactsand the degree of areal and vertical communication - sealing or non-sealing faults.

Two wells were drilled in a remote region. Well-A was drilled down-dip of thestructure and encountered only the aquifer. The operator had the foresight tomeasure water pressures over a 160 foot interval to determine the water pressureregime in this new area. The second Well-B, was drilled up-structure severalkilometers from Well-A and discovered a 50 foot oil column with good porosityand permeability. Six RFT pressures were measured over this interval.

82

The pressures recorded for the two wells are summarized below. Note that inorder to meaningfully compare the pressures in the wells it is necessary to convertmeasured depths to a common datum (depth sub-sea) and to express pressuresin absolute units (psia).

Well-A Well-BDepth (ft.ss) Pressure (psia) Depth (ft.ss) Pressure (psia)6075 2662 5771 26026091 2669 5778 26046108 2677 5790 26086220 2725 5800 26106232 2731 5806 2612

5816 2615

What conclusions can you draw from this data with regard to the fluids, thereservoir and the aquifer?

[Answer: you tell me]

Solution hints

Try using CONTACTS.mcd.

83

Figure 4.6: Lithostatic gradient

4.3 GEOTHERMAL GRADIENT

The geothermal gradient, GT , varies considerably from location to location. Onthe average it is about 1oF/100 ft. We can use this to estimate subsurface for-mation (reservoir) temperatures, Tf , as,

Tf = Ts +GTDf

where Ts is the surface temperature and Df is the formation depth.

84

Figure 4.7: Geothermal gradient

85

Chapter 5

FLUID PROPERTIES

Production from a reservoir is usually accompanied by a decline in reservoirpressure. When sufficient production has taken place for the pressure to fallbelow the oil bubble point pressure, gas initially dissolved in the oil is releasedfrom solution.

The volume of solution gas released and the associated changes in oil propertieshave a significant influence on the production characteristics of the reservoir. Theattached figure shows a schematic of solution gas release in an oil reservoir.

The fluid properties of interest in the analysis of reservoirs include:

- Amount of gas in solution.

- Oil shrinkage.

- Density, compressibility and viscosity of all the fluids present.

These properties are determined in special laboratory tests which require rep-resentative reservoir fluid samples. The properties are called PVT (pressure-volume-temperature) properties of the reservoir fluids. Before considering thespecific PVT parameters of interest to reservoir engineering it is necessary toreview some general but basic concepts in hydrocarbon phase behavior.

86

Figure 5.1: Oil reservoir above and below the bubble point

5.1 PHASE BEHAVIOR

All substances can exist as gases, liquids or solids. Whether a particular substanceexists as a gas, a liquid or a solid (or a mixture of liquid and gas or a mixture ofall three) depends on:

- Pressure.

- Temperature.

- Composition

5.1.1 Pure Hydrocarbons

For a pure hydrocarbon such as ethane, propane, butane, methane etc., temper-ature and pressure fully define the state of the hydrocarbon. The state of a puresubstance can be represented on a P-T phase diagram.

The P-T phase diagram can be constructed by experimentally observing the be-havior of the pure substance in a windowed high pressure test cell as shown inthe attached figure.

Over the pressure and temperature range shown in the attached figure, the vaporpressure curve, which terminates in a critical point (C), divides the P-T spaceinto liquid and gas regions.

To the right of the critical point lies a single phase super critical region. A gasat a temperature above it’s critical temperature, may be compressed indefinitelywithout passing through a phase transition.

87

Figure 5.2: Surface production system for a gas reservoir

Figure 5.3: Surface production system for an oil reservoir

88

To the left of the critical point the fluid is;

- liquid for points above the vapor pressure curve.

- vapor for points below the vapor pressure curve.

- mixture of liquid and vapor for points on the vapor pressure curve.

The P-T phase diagram shows that:

- For a liquid state at high pressure, reducing the pressure at constant tem-perature can result in a phase change from liquid to vapor when the pressurefalls to the vapor pressure. Point F represents a liquid state at high pressure.If the pressure is reduced to the fluid vapour pressure (the intersection ofthe line AF and the vapour pressure curve) both liquid and vapour phasescan coexist. When the pressure falls below the vapour pressure to point A,the fluid is in a single vapour state.

- For a gas state at high temperature, increasing the pressure at constanttemperature can result in a phase change from vapor to liquid when thepressure increases to the vapor pressure.

- For pure substances at temperatures above the critical temperature changingpressure (states E to D) at constant temperature does not result in a phasechange.

- For a gas state at temperatures below the critical temperature, state A,increasing temperature at constant pressure (states A to B) does not resultin a phase change.

5.1.2 Hydrocarbons Mixtures

The behavior of hydrocarbon mixtures is more complex than that of a pure fluid.For purposes of illustration we consider a two component mixture. The behaviorof more complex mixtures is similar.

The attached figure shows the P -T phase diagram for a 50:50 mixture of propaneand heptane. Also plotted are the pure component vapor pressure curves.

The major differences between the P -T phase diagrams for a single componentand a two-component mixture are:

89

Figure 5.4: Pressure-temperature diagram for ethane

90

Figure 5.5: Pressure-temperature diagram for ethane-heptane mixtures

1. For the mixture there now exists a two-phase region or a range of pressuresover which two phases can coexist at constant temperature.

2. The two-phase region is bound by a bubble point line and a dew point linewhich meet at the mixture critical point.

The bubble point pressure is the pressure at which the first bubble of gasappears as the pressure of a liquid is reduced at constant temperature.

The dew point pressure is the pressure at which the first drop of liquidappears as the pressure of a gas is increased at constant temperature.

3. Within the two-phase region are lines of constant liquid (or gas) content.

4. The critical point is the temperature and pressure at which the propertiesof the liquid and vapor phases are identical.

We note that for mixtures, liquid and gas phases mixtures may exist atconditions above the critical point.

5. The size and shape of the two-phase region and the location of the criticalpoint vary with mixture composition.

P − T phase diagrams are used to classify reservoir types.

91

Figure 5.6: Pressure-temperature diagram for a multi-component mixture

5.1.3 Classification of Hydrocarbon Reservoirs

Oil Reservoir

The phase diagram for a typical oil reservoir having a temperature T is shown inthe attached figure.

The initial condition of the reservoir (temperature and pressure) is at point A’.

With production the reservoir pressure will fall at constant temperature T alongthe line A’ABC.

The reservoir is said to be undersaturated because the initial reservoir pressure ishigher than the bubble point pressure which is at point A.

As the pressure continues to fall below the bubble point pressure along the lineABC, free gas appears in the reservoir as a result of solution gas release.

As the pressure continues to fall in the two-phase region, the percentage liquidvolume decreases.

92

Figure 5.7: Pressure-temperature diagram for a reservoir oil

For an oil reservoir T , the reservoir temperature, lies to the left of the criticalpoint.

Retrograde Condensate Gas Reservoir

The phase diagram for a typical retrograde condensate gas reservoir having atemperature T is shown in the attached figure.

When pressure falls this reservoir enters the two-phase region through a dewpoint producing a condensate phase in the reservoir. This behavior is termedretrograde.

For a retrograde condensate gas reservoir T , the reservoir temperature, lies to theright of the critical point.

93

Figure 5.8: Pressure-temperature diagram for a gas condensate reservoir

Figure 5.9: Pressure-temperature diagram for a gas reservoir

94

Gas Reservoir

The phase diagram for a typical gas reservoir having a temperature T is shownin the attached figures.

When pressure falls these reservoirs do not enter the two-phase region.

The reservoirs may be further classified as wet gas if the separator condition fallsin the two-phase region, and dry gas if the separator condition is single-phase.

For a gas reservoir T lies to the right of the two-phase region.

5.2 PVT PROPERTIES

The following definitions refer to the PVT properties used in reservoir engineeringto characterize oil and gas reservoirs.

Parameter Symbol Definition

Oil Formation Bo The reservoir volume in barrels (RB)Volume Factor that is occupied by one stock tank

barrel (STB) of oil and its dissolved gas.

Gas Formation Bg The reservoir volume in barrels (RB)Volume Factor that is occupied by one standard

cubic foot (SCF) of gas.

Solution Gas-Oil Rs The volume of gas in SCF dissolved in oneRatio STB of oil at a specific reservoir pressure.

Oil Compressibility co The fractional reduction in oil volume thatresults from a pressure increase of one psi.

Water Compressibility cw The fractional reduction in water volume thatresults from a pressure increase of one psi.

Bubble Point pb The pressure at which the first bubble of gasPressure evolves from solution.

95

Figure 5.10: Production of an oil reservoir above and below the bubble point

The above properties are used to relate surface volumes (mass) to reservoir vol-umes.

Calculation of underground withdrawal rates from surface measure-ments

Oil and gas production rates, Qo (STB/D)and Qg (SCF/D), are measured at thesurface at a time when the reservoir pressure is P . This pressure is below thebubble point pressure Pb. What are the corresponding underground withdrawalrates in (RB/D)?

The producing gas-oil ratio is

R = Qg/Qo (SCF/STB)

This means that for every STB of oil which is produced at this pressure, weproduce R SCF of gas at the surface.

Since every STB of oil produced to the surface produces Rs SCF of solution gas,the volume of free gas produced from the reservoir is

(R−RS) (SCF/STB)

96

Figure 5.11: Use of PVT properties to relate surface measured production ratesto reservoir withdrawal volumes

Each STB of oil produced to the surface requires the withdrawal of Bo RB of oiland is accompanied by (R−Rs)Bg RB of free gas.If we produce Qo STB of oil, the total underground withdrawal is

Qo (Bo + (R−Rs)Bg)

Example 5.1 - Relating surface rates to subsurface withdrawals[SUBSURV.mcd]

The oil production rate is 2500 STB/D and the gas production rate is 2.125MMSCF/D. If the reservoir pressure at this time is 2400 psia, what are theunderground withdrawal rates?

If the density of stock-tank oil is 52.8 lb/ft3 and the gas gravity is 0.67 (densityof air at standard conditions is 0.0763 lb/ft3) calculate the oil density and gasdensity at reservoir conditions when the reservoir pressure is 2400 psia.

The PVT data at 2400 psia are;

Bo=1.1822 RB/STB Rs=352 SCF/STB Bg=0.00119 RB/SCF

97

[Answer: qo= 2956 RB/day, qg= 1482 RB/day, ρo=47.4 lb/cubic ft., ρg=7.65lb/cubic ft.]

Solution hint

To calculate reservoir fluid densities use 1 STB oil and 1 SCF gas as a basis.This gives Bo and Bg as the reservoir volumes of oil and gas, respectively. Usethe PVT properties and the surface densities to calculate the corresponding fluidmasses at reservoir conditions.

Enter the correct numbers into the spreadsheet [SUBSURV.mcd] and see the solu-tion. Study the spreadsheet carefully in order to fully understand the calculationof densities at reservoir conditions.

5.2.1 Pressure Dependence of PVT Properties

Typical relationships between PVT properties and pressure are shown schemati-cally in the attached figure.

Solution GOR

The volume of dissolved gas decreases from Rsi, the initial solution GOR at thebubble point (Pb), to zero at atmospheric pressure.

At pressures above Pb the gas in solution remains constant at Rsi.

Oil Formation Volume Factor

Bo is generally greater than 1.0 at 0 psig because reservoir temperature is greaterthan the standard reference temperature of 60oF (oil expands with increasingtemperature at constant pressure).

Bo increases with pressure to Pb as a result of the increasing solubility of gas (themass of the oil phase increases).

98

Figure 5.12: Dependence of oil PVT properties on reservoir pressure

99

For pressures greater than Pb, Bo decreases due to compression as pressure in-creases.

Gas Formation Volume Factor

Bg decreases with increasing pressure as a result of the compressibility of the gas.

At reservoir conditions the volumetric behavior of gases is governed by the realgas law.

PV = znRT

where,

P = pressure, psiaV = volume, ft3

z = gas deviation factor, dimensionlessn = number of moles, lb moleR = gas constant, 10.73 psia ft3/oR lb moleT = temperature, oR (oR =o F + 460)

Taking one mole of gas at standard conditions (14.65 psia and 60 oF) and com-pressing it to reservoir conditions (PR psia and TR

oR) gives,

14.65V1 = z1nR(60 + 460)

PRVR = zRnR(TR)

Dividing the above equations we obtain,

VRV1=14.65

520

zRz1

TRPR

Since Bg has the units RB/SCF (1 bbl = 5.615 ft3) and z1 = 1.0,

Bg =VR5.615

=14.65

5.615× 520zR1

TRPR

Bg = 5.02× 10−3 zRTRPR

The compressibility factor, zR, for natural gas may be estimated from the attachedfigure where the compressibility factor is plotted as a function of pseudo-reducedtemperature (Tpr) and pseudo-reduced pressure (Ppr).

Tpr =T

Tpc

100

Ppr =P

Ppc

where Tpc and Ppc are the gas pseudo-critical temperature and pseudo-criticalpressure respectively.

The pseudo-critical properties of a gas are a function of gas gravity and may beestimated from the attached figure.

Gas gravity

Gas gravity is the ratio of the density of a gas to the density of air at standardconditions.

At standard conditions gases obey the ideal gas law and gas density may beexpressed as;

PV = nRT

The number of moles n, may be expressed in terms of the mass of gas m, and themolecular weight of the gas M ,

n =m

M

Combining the equations gives,

PV =m

MRT

or,

ρg =m

V=PMg

RT

Similarly, the density of air is,

ρair =m

V=PMair

RT

The gas gravity, γg is thus simply,

γg =ρgρair

=Mg

29

The molecular weight of the gas is determined from a laboratory analysis of a gassample.

101

Figure 5.13: Pseudo-critical properties for natural gas and gas condensates

102

5.3 CALCULATION OF GAS PROPERTIES

5.3.1 Single Gas Component

For a single gas component, the reduced temperature, Tr, and reduced pressure,Pr, are defined as follows:

Tr =T

Tc

Pr =P

Pc

where,

T = temperature of interest, oRTc = critical temperature of the gas, oRP = pressure of interest, psiaPc = critical pressure of the gas, psia

Tc and Pc are constants for a single gas component and may be obtained fromstandard tables of the properties of pure gas components.

Once Tr and Pr are known, the compressibility factor, z, may be read fromcompressibility charts and Bg in units of RB/SCF is calculated from:

Bg = 5.02× 10−3 zTP

where T is in oR and P is in psia.

103

CRITICAL PROPERTIES OF GASES

Component Molecular Tc PcWeight oR psia

Methane 16.04 344 673Ethane 30.07 550 712Propane 44.09 666 617n-Butane 58.12 766 551n-Pentane 72.15 847 485n-Hexane 86.17 914 435n-Heptane 100.2 972 397n-Octane 114.2 1025 362

Water 18.02 1365 3206Carbon dioxide 44.01 548 1073Nitrogen 28.02 227 492Hydrogen sulphide 34.08 673 1306Air 28.90 239 547

104

5.3.2 Multi-Component Gas Mixtures

A multi-component gas mixture may be treated as a single pseudo-componentgas by defining appropriate pseudo critical temperature, Tpc, and pseudo criticalpressure, Ppc:

Tpc =nX1

yiTci

Ppc =nX1

yiPci

where,

yi = mole fraction of component i in the mixtureTpc = pseudo critical temperature of the gas, oRPpc = pseudo critical pressure of the gas, psia

The pseudo reduced temperature, Tpr, and pseudo reduced pressure, Ppr, aredefined as follows:

Tpr =T

Tpc

Ppr =P

Ppc

Once Tpr and Ppr are known, the compressibility factor, z, may be read fromcompressibility charts and Bg calculated from:

Bg = 5.02× 10−3 zTP

where T is in oR and P is in psia.

Gas gravity known

If the gas gravity is known, the chart may be used to estimate the gas pseudocritical properties.

Note: the gas gravity may be calculated from gas composition.

Gas molecular weight, Mg

Mg =nX1

yiMi

where Mi is the molecular weight of component i.

γg =Mg

29

105

Example Calculation — Bg

Calculate the gas formation volume and gas density at reservoir conditions of225oF and 2500 psia. The gas composition is:

Component Mole Fraction

Methane 0.80Ethane 0.15Propane 0.05

Solution

Component Mole Fraction Tc (oR) Tc (

oF ) Pc (psia) yTc yPc

Methane 0.80 -117 343 673 274.8 539Ethane 0.15 90 550 708 82.5 106Propane 0.05 206 666 617 33.3 31

TOTAL 1.00 390.6 676

oR =o F + 460

1.

Tpc = 391oR Ppc = 676psia

2.

Tpr =T

Tpc=460 + 225

391= 1.75

Ppr =P

Ppc=2500

676= 3.70

3. From the generalized compressibility chart, read

z = 0.87

4.

Bg = 5.02×10−3 zTP= 5.02×10−3 (0.87)(225 + 460)

2500= 1.2×10−3 RB/SCF

106

Example Calculation - Mg, γg and ρg

Calculate the gas molecular weight, gas gravity and the gas density at reservoirconditions of 225oF and 2500 psia. The gas composition is:

Component Mole Fraction

Methane 0.80Ethane 0.15Propane 0.05

Solution

Component Mole Fraction Mi yiMi (oF )

Methane 0.80 16 12.83Ethane 0.15 30 4.51Propane 0.05 44 2.20

TOTAL 1.00 19.54

1.

Mg =3X1

yiMi = 19.54

2.

γg =Mg

29=19.54

29= 0.675

3.

ρg =PMg

zRT=

2500× 19.540.87× 10.73× (460 + 225) = 7.64 lb/ft

3

107

Calculation of Gas Properties

Calculate γg, Bg and ρg at reservoir conditions of 150oF and 1500 psia. The gas

composition is:

Component Mole Fraction

Methane 0.70Ethane 0.12Propane 0.10n-Butane 0.08

Solution

Component Mole Fraction Tc (oR) Tc (

oF ) Pc (psia) yTc yPc

Methane 0.70 -117 343 673 240.1 471.1Ethane 0.12 90 550 708 66.0 85.0Propane 0.10 206 666 617 66.6 61.7n-Butane 0.08 306 766 551 61.3 44.1

TOTAL 1.00 434.0 661.8

oR =o F + 460

1.

Tpc = 434oR Ppc = 662psia

2.

Tpr =T

Tpc=460 + 150

434= 1.406

Ppr =P

Ppc=1500

662= 2.266

3. From the generalized compressibility chart, read

z = 0.75

108

4.

Bg = 5.02×10−3 zTP= 5.02×10−3 (0.75)(150 + 460)

1500= 1.53×10−3 RB/SCF

Component Mole Fraction Mi yiMi (oF )

Methane 0.70 16 11.2Ethane 0.12 30 3.6Propane 0.10 44 4.4n-Butane 0.08 58 4.6

TOTAL 1.00 23.8

5.

Mg =3X1

yiMi = 23.8

6.

γg =Mg

29=23.8

29= 0.82

7.

ρg =PMg

zRT=

1500× 23.80.75× 10.73× (460 + 150) = 7.29 lb/ft

3

109

Figure 5.14: Schematic of a laboratory PVT test cell

5.4 DETERMINATION OF OIL PVT DATA

FROM LABORATORY EXPERIMENTS

Laboratory tests to measure PVT parameters are carried using samples of reser-voir oil at a constant reservoir temperature.

The laboratory tests are conducted in high pressure windowed cells. These cellsallow accurate visual measurement of oil and gas volumes in the cell at differentpressures.

The pressure in the cell is lowered by increasing the cell volume. This is achievedby removing accurately measured volumes of mercury from the cell using a cal-ibrated high precision piston pump. The attached figure shows the essentialfeatures of the PVT test apparatus.

Three specific tests are performed to characterize an oil and it’s associated gas:

1. Flash expansion test .

2. Differential liberation test.

3. Separator flash expansion test.

110

Figure 5.15: Windowed PVT test cell

111

Figure 5.16: PVT laboratory at UNSW

112

5.4.1 Flash Expansion Test

A reservoir oil sample is introduced into the cell and mercury is injected to raisethe cell pressure above the initial reservoir pressure, pi.

The cell pressure is lowered in small steps by removing a small volume of mercuryat each step. The total hydrocarbon volume, vt, in the cell is measured at eachstep.

All cell volumes are reported relative to the volume in the cell at the bubble pointpressure, pb.

The following table shows a set of typical laboratory data for this test.

Pressure Relative Total Volume(psia) vt = v/vb (RB/RBb)

5000 0.98104500 0.98504000 pi 0.99253500 0.99753330 pb 1.00003290 1.00253000 1.0272700 1.06032400 1.10602100 1.1680

This data is used to determine the bubble point pressure, pb, by noting that thetotal compressibility of the cell increases dramatically when gas is released fromsolution.

5.4.2 Differential Liberation Test

This test is designed to approximate the conditions in a reservoir where gas isreleased from solution as the pressure falls and the released gas moves away fromthe oil as a result of gravity segregation. experiment starts with the cell at thebubble point pressure.

The pressure in the cell is reduced in small steps by removing mercury from thecell.

113

Figure 5.17: Schematic representation of flash and differential liberation tests

114

At each step the cell oil volume, vo, and gas volume, vg, are measured.

The gas in the cell is then slowly displaced from the cell with mercury at constantpressure. The displaced gas volume is measured at standard conditions and thevolume, Vg, is noted.

The above procedure is repeated for small steps down to atmospheric pressure.

Typical data produced from a differential liberation test is shown in the tablebelow:

Pressure Relative Gas Vol. Relative Gas Vol. Relative Oil Vol.(psia) at p and T , vg at sc, Vg at p and T , vo

4000 0.99253500 0.99753330 pb 1.00003000 0.0460 8.5221 0.97692700 0.0417 6.9731 0.96092400 0.0466 6.9457 0.94492100 0.0535 6.9457 0.92981800 0.0597 6.5859 0.91521500 0.0687 6.2333 0.90221200 0.0923 6.5895 0.8884900 0.1220 6.4114 0.8744600 0.1818 6.2369 0.8603300 0.3728 6.2297 0.845914.7 200oF 0.829614.7 60oF 0.7794

In addition to the above data, the laboratory also reports the Cumulative RelativeGas Volumes collected at atmospheric pressure and reservoir temperature and atSTC:

Cumulative Relative Gas Volume at 14.7 psia and 200oF=74.9557Cumulative Relative Gas Volume at 14.7 psia and 60oF=74.9557

115

Figure 5.18: Laboratory test separator

5.4.3 Separator Flash Expansion Test

This test is designed to approximate the surface production system.

The PVT cell is brought to the bubble point pressure and connected to theseparator system which may be a single separator or a series of staged separatorswith staged pressures and temperatures to simulate the actual production system.

The cell contents are then flashed through the separator system to standardconditions. The resulting volumes of produced oil and gas are measured.

The data from such tests is used to determine the oil shrinkage factor, cbf , andthe initial solution GOR, Rsif .

Separator Stock Tank Shrinkage GORPressure Pressure Factor Rsif(psia) (psia) cbf (STB/RBb)) (SCF/STB)

200 14.7 0.7983 512150 14.7 0.7993 510100 14.7 0.7932 51550 14.7 0.7834 526

116

5.4.4 Procedure for calculating PVT parameters fromlaboratory data

The above data can be used to compute the PVT properties for the tested oiland gas. The procedure involves the following steps:

1. Plot the flash expansion data, p vs vt, to determine the bubble point pres-sure.

2. Use the differential flash data to compute Bg. Eqns.(4.4—4.6) may be ma-nipulated to yield,

Bg =1

5.615

vgVg

3. Use the differential flash data to compute the cumulative relative gas vol-ume, F , liberated from the cell at every pressure step;

Fp =pbXp

Vg

4. Use the differential flash data and the oil shrinkage factor, cbf , to computeBo,

Bo

∙RB

STB

¸=vocbf

=

"RB/RBbSTB/RBb

#

5. Use the differential flash data, the oil shrinkage factor, cbf and the cumu-lative gas liberated, F , to compute Rs,

Cumulative Gas = Total Gas in - Gas in solutionLiberated at pressure p solution at pressure p

F∙STB

RBb

¸× 5.615

∙SCF

STB

¸× 1

cbf

∙RBbSTB

¸= (Rsif − Rs)

∙SCF

STB

¸

The following example illustrates the application of the above procedure.

117

Example 5.2 - Calculation of PVT properties from laboratory PVTdata [PVT.mcd]

Using the Flash expansion, differential liberation and separator flash expansiondata discussed in the previous sections, determine the following as a functionpressure at reservoir temperature:

1. Bubble point pressure

2. Oil compressibility above the bubble point

3. Oil formation volume factor

4. Solution gas-oil ratio

5. Gas formation volume factor

6. Gas compressibility

Determine the above properties for the optimum surface operating condition.

[Answer: Open PVT.mcd]

118

5.5 FLUID SAMPLING

Fluid samples for PVT analysis are collected early in the life of a reservoir. Thereservoir is usually sampled in one of two ways;

1. sub-surface sampling

2. surface sampling.

Irrespective of the sampling method used, the problems of obtaining a trulyrepresentative sample of reservoir oil are the same. If the reservoir is under-saturated (reservoir pressure is above the bubble point pressure) and the wellproduction rate is sufficiently low for the pressure in the vicinity of the wellboreto be everywhere above the bubble point pressure, the oil flowing into the wellboreis reservoir oil and the sample is representative.

If the reservoir is saturated (initial reservoir pressure is equal to the bubble pointpressure), the pressure in the vicinity of the wellbore will be below the bubblepoint pressure and gas will come out of solution as the oil flows to the wellbore.Under these conditions the oil sample may not be representative.

If the sample is to be representative, the oil flowing into the wellbore must havethe same solution GOR as the reservoir oil away from the well. When gas isfirst released in a reservoir, the gas remains immobile until the gas saturationincreases to a value called the critical gas saturation. During the period that thegas saturation is below the critical value, oil flowing into the wellbore will bedeficient in gas. On the other hand, when sufficient gas has been released for gasto become mobile, the very high mobility (low viscosity) of the gas will result ina disproportion ally high volume of gas entering the wellbore.

5.6 PVT TESTS FOR GAS CONDENSATE

FIELDS

Gas condensate exist as a single gaseous phase at initial reservoir conditions.When reservoir pressure falls below the dew-point, a liquid phase is produced inthe reservoir. Gas condensate tests are designed to measure this volume of liquidas a function of pressure - liquid dropout curve.

The liquid volume is usually very small compared to the total volume of the sys-tem and in order to measure it accurately it is necessary to use special condensatePVT cells. These cells are designed to allow light to pass through the liquid and

119

Figure 5.19: Constant mass expansion test for a gas condensate

are fitted with front and back windows for this purpose. Some condensate cellsare made out of a solid block of sapphire to allow un obstructed viewing of theliquid phase.

The most common type of test used to characterize a gas condensate is theconstant mass depletion test. Since the liquid volume produced in the reservoiris small, it is immobile and remains in contact with the gas from which it wasproduced. The constant volume depletion test simulates this behavior.

In the case of very rich condensates where the volume of liquid produced isrelatively large, the liquid phase in the reservoir may become mobile. The liquidwill move to the bottom of the reservoir under the action of gravity and a constantvolume depletion test is more appropriate.

Typical liquid dropout curves for constant mass and constant volume depletiontests are shown in the attached figure. The richer the gas condensate, the moreliquid is produced and the greater the difference between the liquid dropoutcurves.

120

Figure 5.20: Constant volume expansion test for a gas condensate

5.7 GAS HYDRATES

Gas hydrates are solid, semi-stable compounds which sometimes plug natural gasflowlines, processing and measuring equipment, and pipelines. Hydrates consistof geometric lattices of water molecules containing cavities filled with gas - asolid foam. Gas hydrates form when hydrocarbon gases come into contact withwater at temperatures below 35oC. A typical phase diagram for a hydrate forminggas-water system is shown in the attached figure.

The temperature for hydrate formation is reduced by the addition of inhibitorssuch as methanol, ethanol, ethylene glycol diethylene glycol and triethylene glycol.The attached sheet provides a guide to temperature reduction which may beexpected from the use of the different additives. More accurate predictions requirethe use of special equations of state designed specifically for this purpose.

121

Figure 5.21: Liquid drop-out curves for a gas condensate

Solid, semi-stable compounds which sometimes plug naturalgas flowlines, processing equipment and pipelines

Geometric lattices of water molecules containing cavities filledwith gas

Formed when hydrocarbon gases are in contact with water attemperatures below 35C

Temperature for hydrate formation is reduced by the additionof inhibitors such as Methanol, Ethanol, Ethylene Glycol, DiethyleneGlycol and Triethylene Glycol

Gas Hydrates

Figure 5.22: A simple summary of gas hydrates

122

Figure 5.23: Gas hydrate co-existence curves. H-hydrate, G-gas, L1-water, L2-liquid hydrocarbon, I-ice

Figure 5.24: Comparison between measured and calculated hydrate curves for agas with MEOH inhibitor

123

Figure 5.25: Effect of different inhibitors on hydrate formation

5.8 SURFACE TENSION

Surface tension between two fluids is an important parameter in reservoir analysisbecause it determines the manner in which fluids move through reservoir rock.Surface tension may exist between water and oil phases, water and gas phasesand between gas and oil phases. Surface tension may be measured at reservoirconditions using the pendant drop method. A typical test setup is shown in theattached figure. The surface tension is given by,

σ =∆ρgDeH

where,

∆ρ = density difference for the two fluidsg = gravitational accelerationDe, Ds = drop diameters as shown in the attached figureH = tabulated function of Ds/De

5.8.1 Estimating Surface Tension

The surface tension between two fluids is a function of pressure, temperature andcomposition of the phases. The surface tension of hydrocarbon mixtures may be

124

Figure 5.26: Pendant drop tensiometer

125

estimated using the parachor method. For a pure liquid in equilibrium with it’svapor,

σ1/4 = PchρL − ρVM

where,

Pch = parachorρL, ρV = phase densities, gm/cm3

σ = surface tension, dynes/cmM = molecular weight

Parachors account for intermolecular forces and are predicted from the structuresof molecules. A correlation for the parachor with molecular weight is given in theattached figure. For mixtures the surface tension is given by,

σ1/4 =XPchi

µxiρLML− yi ρV

MV

¶where,

Pchi = parachor for component i in the mixturexi, yi = mole fraction of component i in the liquid and vapor phasesρL, ρV = phase densities, gm/cm3

σ = mixture surface tension, dynes/cmML, MV = phase molecular weights

5.9 CORRELATIONS FOR PROPERTIES OF

RESERVOIR FLUIDS

Several useful correlations exist for estimating reservoir fluid properties in theabsence of laboratory data.

Examples of these correlations are given in attached figures.

Computer based correlations

Many of the available graphical correlations of fluid properties have been com-puterized which makes their use particularly easy. An example of such a programis given in the file PropEst.mcd.

126

Figure 5.27: Surface tensions for hydrocarbons at atmospheric pressure

Example 5.3 - Estimation of reservoir fluid properties [PropEst.mcd]

The following estimates are made of reservoir and produced fluid properties froma DST on an exploration well. What are the reservoir properties of the oil, waterand gas?

TR = 200 oFPR = 3330 psiaPS = 115 psiTS = 60 oFAPI = 30 oAPIGOR = 510 SCF/STBγg = 0.75 (air= 1)C = 100 (water salinity in units of kg/m3)NaCl = 6 (water salinity in units of 10,000 ppm)

[Answer: open the spread-sheet, enter the appropriate input values and see]

127

Figure 5.28: Parachors for hydrocarbons

128

Figure 5.29: Standing correlation for reservoir oils

Figure 5.30: Viscosity of saturated crude oil

129

Figure 5.31: Viscosity of reservoir brine

130

Figure 5.32: Viscosity of hydrocarbon gases at atmospheric pressure

Figure 5.33: Ratio of viscosity at reservoir pressure to viscosity at atmosphericpressure for hydrocarbon gases

131

Problem 5.1 - Calculation of PVT properties from laboratory PVTdata

A laboratory PVT cell contained 280.0 cm3 of reservoir liquid at its bubble pointof 2000 psia at 140oF. A volume of 18.8 cm3 of mercury was withdrawn from thecell and the pressure fell to 1600 psia. Mercury was then reinjected at constanttemperature and pressure to displace the gas from the cell. This left 263.5 cm3

of liquid in the cell and resulted in the collection of 0.129 SCF of gas.

The process was repeated, reducing the pressure to 14.7 psia and the temperatureto 60oF. Then 0.388 SCF of gas were removed from the cell and 205.9 cm3 of liquidremained in the cell.

Determine the following:

1. Bob and Rsi.

2. Bo, Bg, Rs and z at 1600 psia.

[Answer: Bob= 1.3599 RB/STB, Rsi= 399.2 SCF/STB, Bo= 1.2797 RB/STB,Bg= 0.001721 RB/SCF, Rs= 299.6 SCF/STB,]

solution hints

If you can do this without using the PVT.mcd spread-sheet you have a goodunderstanding of PVT properties.

132

Chapter 6

MATERIAL BALANCEEQUATIONS

When a volume of oil is produced from a reservoir, the space originallyoccupied by the produced oil volume must now be occupied by somethingelse.

Unless fluid is injected, the production of oil must result in a decline inreservoir pressure.

The decline in reservoir pressure can cause;

— influx of fluid from an adjoining aquifer or gas cap.

— expansion of fluids in the reservoir.

— expansion of reservoir rock grains - reduction of pore volume.

There is therefore a relationship between the rate of production and the rate ofdecline of reservoir pressure. Material balance equations express this dependencein mathematical form.

The material balance equation is the basic reservoir engineering analy-sis tool used to examine past reservoir performance and to predictfuture performance.

The diagrams in this section are best viewed on the electronic version of thecourse notes. You may wish to make colour prints of these figures.

133

GAS CAP

OIL ZONE

AQUIFER

Figure 6.1: Schematic of a combination reservoir

6.1 ORIGINAL OIL VOLUME BALANCE

Most material balance equations are written in terms of the original volume oc-cupied by the oil.

The volume originally occupied by produced oil may now be occupied by;

1. expansion of an adjoining gas cap (if one is present).

2. volume of gas released by oil (if reservoir pressure falls below the bubblepoint).

3. volume of oil remaining in the reservoir.

4. volume occupied by expansion of rock grains - pore space compressibility.

5. volume occupied by expansion of connate water.

6. influx of water from an adjoining aquifer (if the aquifer is strong enough).

It should be understood that the attached diagram is for illustrative purposesand is not intended to imply that the materials in the reservoir are segregated inthe manner shown. Under some conditions, however, gas released from solutionwill segregate to the top of the structure. Some of the released solution gas willalways be evenly distributed throughout the reservoir, as are the expansions ofrock and connate water.

134

Figure 6.2: Material balance on the pore space occupied by the original oil volume

6.1.1 Gas Cap Expansion

If a gas cap is present, and the reservoir pressure drops, the gas cap will expandto replace some of the volume initially occupied by the produced oil.

Gas cap expansion| {z }[RB]

= (G−Gpc)Bg| {z }gas volume at lower pressure

− GBgi| {z }initial gas volume

where,

G = original gas cap gas volume, [SCF]Gpc = cumulative gas production from the gas cap, [SCF]Bg = gas formation volume factor at current pressure, [RB/SCF]Bgi = gas formation volume factor at original reservoir pressure, [RB/SCF]

Note that if Gpc is large, the gas cap may actually shrink rather than expand.Gascap shrinkage should never be allowed because it always results in a loss of oilrecovery.

The above equation may be written as,

∆VGCE = (G−Gpc)Bg −GBgi

135

Figure 6.3: Gas cap expansion volume

Example 6.1 - Calculation of gas cap expansion [MBE.mcd]

Calculate the change in the size of the gas cap after 20% of the gas cap gas hasbeen produced while the reservoir pressure had dropped from 1225 psig to 1100psig. The gas cap originally contained 21.3× 109 SCF.Data:

G = 21.3 ×109 [SCF]Bgc at 1225 psig = 0.002125 [RB/SCF]Bgc at 1100 psig = 0.002370 [RB/SCF]

Solution

1. At 1100 psig the gas cap containing 0.8 times the original gas cap volumei.e., G = 0.8Gi

2.

∆VGCE = (G−Gpc)Bg −GBgi∆VGCE = (0.8× 21.3× 109 − 0) 0.002370− 21.3× 109 × 0.002125

∆VGCE = −4.88× 106 RBi.e., the gas cap has actually shrunk. This would allow about 5 MMRB oilto migrate into the gas cap, a zone originally free of oil. This would resultin a loss of about 1MMRB of oil.

We would not allow this to actually happen in the actual reservoir.

136

Example 6.2 - Calculation of gas cap expansion [MBE.mcd]

Calculate the change in gas cap size for the reservoir of Example-1 if the reservoirpressure had dropped from 1225 psig to 900 psig at the time that 20% of the gaswas produced.

Data:

G = 21.3 ×109 [SCF]Bgc at 1225 psig = 0.002125 [RB/SCF]Bgc at 900 psig = 0.002905 [RB/SCF]

Solution

1. At 900 psig the gas cap cumulative production is i.e., Gp = 0.2Gi

2.

∆VGCE = (G−Gpc)Bg −GBgi∆VGCE = (0.8× 21.3× 109 − 0) 0.002905− 21.3× 109 × 0.002125

∆VGCE = 4.24× 106 RBi.e., the gas cap has expanded into the oil zone and no oil is lost to the gaszone.

137

Figure 6.4: Released solution gas volume

6.1.2 Released Gas Volume

If the reservoir pressure falls below the bubble point pressure gas will be releasedfrom solution.

At any time during the production of a reservoir, the gas originally in solutioncan be placed into three categories;

1. gas remaining in solution

2. gas released from solution and produced from the reservoir

3. gas released from solution but remaining in the reservoir.

On this basis we write,

Released gas volume| {z }[RB]

= NRsiBg| {z }gas originally in solution

− (N −Np)RsBg| {z }gas still in solution

− GpsBg| {z }produced solution gas

where,

N = original oil volume, [STB]Np = cumulative oil produced, [STB]Rsi = original solution gas-oil ratio, [SCF/STB]Rs = solution gas-oil ratio at current pressure, [SCF/STB]Bg = gas formation volume factor at current pressure, [RB/SCF]

The above equation is usually written as,

∆VRSG = {NRsi − (N −Np)Rs −Gps}Bg

138

Example 6.3 - Calculation of released solution gas volume [MBE.mcd]

Cumulative oil production for our example reservoir was 14.73 ×106 STB atthe time when reservoir pressure was 900 psig. At the same time cumulativeproduction of solution gas was 4.05 ×109 SCF. Calculate the reservoir volumeoccupied by released gas.

Data:

N = 90.46 ×106 [STB]Rsi at 1225 psig = 230 [SCF/STB]Rs at 900 psig = 169 [SCF/STB]Bg at 900 psig = 0.002905 [RB/SCF]

Solution

∆VRSG = {NRsi − (N −Np)Rs −Gps}Bg∆VRSG = {(90.46× 230− (90.46− 14.73)× 169− 4050)× 106} 0.002905

∆VRSG = 11.5× 106RB

139

Figure 6.5: Remaining oil volume

6.1.3 Remaining Oil Volume

The oil volume remaining in the reservoir is simply,

Reservoir oil volume| {z }[RB]

= (N −Np)Bo| {z }oil still in reservoir

where,

N = original oil volume, [STB]Np = cumulative oil produced, [STB]Bo = oil formation volume factor at current pressure, [RB/STB]

The above equation is written as,

∆VROV = (N −Np)Bo

140

Example 6.4 - Calculation of remaining oil volume [MBE.mcd]

What is the remaining reservoir oil volume for the previous example at 900 psig.

Data:

Bo at 900 psig = 1.104 [RB/STB]

Solution

∆VROV = (N −Np)Bo∆VROV = (90.46× 106 − 14.73× 106) 1.104

∆VROV = 83.63× 106RB

141

Figure 6.6: Rock and connate water expansion volume

6.1.4 Rock and Connate Water Expansion

The expansion of rock and connate water are combined into one term and ex-pressed as the formation compressibility cf .

cf is defined as the fractional change in pore volume per psi change in reservoirpressure.

The pore volume can be expressed in terms of the original oil volume, as isrequired for material balance calculations.

original oil volume = NBoi = VpSoi = Vp(1− Swi)

where,

N = original oil volume, [STB]Boi = oil formation volume factor at initial pressure, [RB/STB]Vp = reservoir pore volume, [RB]Soi = initial oil saturationSwi = initial or connate water saturation

or,

Vp =µNBoi1− Swi

¶

142

The total amount of rock expansion is calculated by utilizing the definition ofcompressibility;

c =1

V

∆V

∆Pfrom which we have

∆V = cV∆P

Rock expansion| {z }[RB]

= formation compressibility× pore volume× pressure change

Rock expansion| {z }[RB]

= cf

µNBoi1− Swi

¶(pi − p)

where,

cf = formation compressibility, [vol/vol/psi]p1 = initial reservoir pressure, [psig]p = current reservoir pressure, [psig]

The total amount of connate water expansion is given by,

Water expansion| {z }[RB]

= water compressibility × water volume× pressure change

The initial connate water volume is given by,

original connate water volume = SwiVp = Swi

µNBoi1− Swi

¶where,

cw = water compressibility, [vol/vol/psi]Swi = initial water saturation

The connate water expansion is,

Water expansion| {z }[RB]

= cwSwi

µNBoi1− Swi

¶(pi − p)

Combining the above equations, we have for rock and water expansion,

∆VRWE = (cf + cwSwi)µNBoi1− Swi

¶(pi − p)

For reservoirs where a gas phase is present, the rock and water expansion term isso small that it may safely be neglected.

The rock and connate water term is usually only important for oil reservoirs whenthe oil pressure remains above the bubble point.

143

Example 6.5 - Calculation of rock and connate water expansion volumeMBE.mcd]

What is the rock and water expansion volume for the previous example when thepressure falls from 1225 psig to 900 psig.

Data:

Swi = 0.205cf = 3.0×10−6 [psi−1]cw = 3.0×10−6 [psi−1]

Solution

∆VRWE = (cf + cwSwi)µNBoi1− Swi

¶(pi − p)

∆VRWE = (cf + cwSwi) (Vp) (pi − p)∆VRWE = (3.0 + 3.0(0.205)) 10

−6 ³127.84(106)´ (1225− 900)∆VRWE = 0.15× 106 RB

We see that rock and water expansion amounts to about 1% of the released gasvolume. This is usually the case in practice and rock and connate water expansioncan be neglected for calculations below the bubble point.

144

Figure 6.7: Water influx volume

6.1.5 Water Influx

Unlike the above items in the material balance equation, the volume of waterinflux cannot be calculated directly.

This is because the calculation of water influx requires information characterizingthe size and strength of the aquifer and this information is not generally knownwith any certainty during the early production life of a reservoir.

However, since we can calculate all the other terms in the material balance equa-tion, we can determine the net water influx into the reservoir — total waterentering the reservoir less water produced — by difference.

Net water influx| {z }[RB]

= We|{z}cumulative water influx

− WpBw| {z }cumulative produced water

where,

We = cumulative water influx, [RB]Wp = cumulative water produced, [STB]Bw = water formation volume factor at current pressure, [RB/STB]

The above equation may be written as,

∆VNWI =We −WpBw

145

6.1.6 General Material Balance Equation

Having analyzed all the individual terms in the material balance equation, wecan write,

Original oil volume = Gas cap expansion + Released gas volume +Oil volume + Rock and water expansion +Net water influx

∆VOOIP = ∆VGCE +∆VRGV +∆VROV +∆VRWE +∆VNWI

NBoi = (G−Gpc)Bg −GBgi + {NRsi − (N −Np)Rs −Gps}Bg +(N −Np)Bo + (cf + Swicw)

µNBoi1− Swi (pi − p)

¶+

We −WpBw (6.1)

Note that the two gas production terms are additive and equal to the total gasproduction at the surface ie.,

(Gpc +Gps)Bg = GpBg

The material balance equation may be re-written as,

NBoi = G(Bg −Bgi) + {NRsi − (N −Np)Rs}Bg −GpBg +(N −Np)Bo + (cf + Swicw)

µNBoi1− Swi (pi − p)

¶+

We −WpBw (6.2)

One of the most important uses of the material balance equation is to calculatewater influx into a reservoir. This only requires that we know the volume of initialfluids in-place, reservoir pressure and cumulative oil, gas and water production.Cumulative water influx is,

We = NBoi − (N −Np)Bo − (cf + Swicw)µNBoi1− Swi (pi − p)

¶−GBg +GBgi +GpBg − {NRsi − (N −Np)Rs}Bg+WpBw (6.3)

146

Example 6.6 - Calculation of water influx from material balance[MBE.mcd]

Using the results of the Examples 2,3,4 and 5, find the water influx when thepressure falls from 1225 psig to 900 psig. Cumulative water production at thistime is 620,000 STB.

Data:

Wp = 0.62×106 [STB]Bw = 1.0 [RB/STB]

Solution

The following items were calculated in the previous examples.

Ex. Item Calculated ×106 RB2 Gas Cap Exp. (G−Gpc)Bg −GBgi 4.243 Released Gas {NRsi − (N −Np)Rs −Gps}Bg 11.504 Remaining Oil (N −Np)Bo 83.63

5 Rock & Water Exp. (cf + Swicw)³NBoi1−Swi (pi − p)

´0.15

Re-arranging the material balance equation to calculate We,

We = NBoi − (N −Np)Bo − (cf + Swicw)µNBoi1− Swi (pi − p)

¶−GBg +GBgi +GpBg − {NRsi − (N −Np)Rs}Bg+WpBw

(6.4)

We = (101.64− 83.63− 4.24− 11.50− 0.15 + 0.62× 1.0)× 106 = 2.74× 106 RB

147

6.2 PRIMARY RECOVERY MECHANISMS

Each of the terms of the material balance equation represents a drive mech-anism which contributes to the total energy required to produce oil from thereservoir.

If we divide the overall material balance equation by NBoi we obtain,

1 =(G−Gp)Bg −GBgi

NBoi| {z }Igc

+{NRsi − (N −Np)Rs −Gps}Bg

NBoi| {z }Isg

+

(N −Np)BoNBoi

+ (cf + Swicw)µ

1

1− Swi (pi − p)¶

| {z }Ipd

+

We −WpBwNBoi| {z }Iw

(6.5)

The above equation may be written as,

1 = Igc + Isg + Ipd + Iw

where,

Igc = gas cap drive indexIsg = solution gas drive indexIpd = pressure depletion drive indexIw = water drive index

The above indices represent the effectiveness of each drive mechanism for a par-ticular reservoir.

A reservoir for which the dominant drive mechanism is, Iw, is called a water drivereservoir.

A reservoir for which the dominant drive mechanism is, Igc, is called a gas capdrive reservoir.

A reservoir for which the dominant drive mechanism is, Isg, is called a solutiongas drive reservoir.

148

Figure 6.8: Reservoir drive index as a function of time

A reservoir for which the dominant drive mechanism is, Ipd, is called a depletiondrive reservoir.

A reservoir for which more than one drive mechanism is important, is called acombination drive reservoir.

6.2.1 Typical Performance Characteristics for the Differ-ent Drive Mechanisms

The dominant drive mechanism for a particular reservoir can often be deducedfrom the rate of pressure decline and the trend of the producing gas-oil ratio.

The trends which are observed when one of the mechanisms dominates are sum-marized in the following figures.

Ultimate recovery is strongly influenced by the type of drive mechanism.

1. Pressure depletion drive (fluid and rock expansion) results in a rapid straightline pressure response and may be expected to recover less than 5% of theoriginal oil-in-place.

2. Solution gas drive — pressure drops slowly at first. As the producing GORincreases, pressure falls rapidly. Ultimate recoveries are in the range 15-30%.

149

Figure 6.9: Reservoir pressure as a function of cumulative oil production

Figure 6.10: Producing GOR as a function of cumulative oil production

150

Figure 6.11: Producing characteristics of gas cap drive reservoirs

3. Water drive — pressure declines rapidly at first, but the decline rate de-creases as water influx from the aquifer increases. The producing GORremains approximately constant and ultimate recoveries are 50% or more.

151

Figure 6.12: Producing characteristics of solution gas drive reservoirs

Figure 6.13: Producing characteristics of water drive reservoirs

152

6.3 USING MATERIAL BALANCE EQUA-

TIONS

6.3.1 Average Reservoir Pressure

The material balance equation describes the whole reservoir in terms of averagereservoir pressure, initial volumes of oil, gas and water in-place and cumulative oil,water and gas production volumes. In order to use the material balance equationit is necessary to determine average reservoir pressure for the time (cumulative oil,water and gas production) when the material balance equation is to be applied.

Direct measurement of average reservoir pressure would require that all the wellsare shut-in and that the bottom-hole pressure is measured at a time after shut-inwhich is sufficiently long for all the pressure gradients in the reservoir to equalize.This may take months to years (depending on reservoir permeability) and result inconsiderable loss in revenue because of lost production. The direct measurementapproach is therefore impractical.

Pressure data obtained from a pressure buildup test can be used to estimatethe average pressure in the volume or area drained by the tested well. Thesetests require short test times (hours to days) and allow reservoir pressure to bemapped over the field. The mapped pressures may be averaged to give the averagereservoir pressure as,

p =

PpjVpjPVpj

(6.6)

where pj and Vpj are the drainage area pressure and pore volume drained by wellj.

6.3.2 Knowns and Unknowns

When solving the material balance equation the parameters may be classified intothe following groups of knowns and unknowns.

knowns unknownsNp NGp GWp We

Swc p (Bo, Bg, Rs)cw cfBw

153

The PVT properties (Bo, Bg, Rs) may be considered to be known if the aver-age reservoir pressure is known and representative fluid samples have obtainedand analysed. This reduces the number of unknowns to five but we have onlyone equation. This is the central problem with the use of the material balanceequation.

Before considering specific examples of the application of the material balanceequation, lets consider some of the parameters in the material balance equation.

Knowns

Np - this is usually known because oil is the primary production target which issold to generate revenue.

Gp, Wp - cumulative gas and water production be unknown for older oil and gasfields because they had little or no value at the time of production. When theseare unknown it is not possible to perform material balance calculations for thereservoir.

Swc - is usually considered to be accurately known from petrophysical evaluation.

cw, Bw - these are known, or may be estimated, from laboratory tests on forma-tion brine.

Unknowns

N, G - volumetric estimates of the original oil and gas-cap volumes in-place arealways known from the field appraisal stage. These are disregarded as soon asproduction-pressure data becomes available and an attempt is made to estimatein-place volumes from material balance calculations. This is because volumetricestimates include all the mapped hydrocarbon volume, whereas material balancecalculation provides the effective volume or the volume which contributes to ac-tual production. This will usually be smaller than the volumetric estimate dueto compartmentilization of the reservoir by faults or low permeability zones.

We - this is probably the greatest unknown in reservoir development - whetherthere has been any water influx or not. One of the most important uses of materialbalance calculations is to estimate water influx.

p - although we usually consider reservoir pressure to be known, problems withthe interpretation of pressure buildup test data may introduce serious errors andthere may be considerable uncertainty in estimates of average reservoir pressure.

cf - pore volume or formation compressibility is usually considered to be smalland constant. In some cases it may be large and variable. When it is large,compaction may form a major part of the reservoir drive energy and usually leads

154

to considerable subsidence at the surface. This may be of little consequence onland in remote areas but may cause serious problems for operations offshore.

Problem 6.1 Calculating water influx from material balance for history matchingaquifer performance

The first step in characterizing or history matching an aquifer is to calculate waterinflux from available production and pressure data. You will be performing thishistory match in the reservoir engineering course which follows from the presentcourse. After the aquifer has been characterized, the resultant aquifer model isattached to a numerical reservoir simulation model which is used to predict futurereservoir performance.

The reservoir for this exercise is a small offshore oil field which has been on-streamfor only 700 days. The field was shut-in for a 250 day period for upgrading of thegathering system and production facilities. There was a strong and clear aquiferresponse to the shut-in (use Mathcad to plot the field pressure as a function oftime to see the response).

The field production history is given in the table below.

PRODUCTION HISTORY

Time P Np Gp Wp(day) (psia) (MMSTB) (BSCF) (MMSTB)

0 4217 0.00000 0.00000 0.0000050 3915 0.34270 0.17821 0.00000100 3725 0.63405 0.32970 0.00121150 3570 1.06898 0.55587 0.02636200 3450 1.36788 0.71130 0.07041250 3390 1.64397 0.85487 0.12218300 3365 1.91465 0.99562 0.17304350 3600 1.91465 0.99562 0.17304400 3740 1.91465 0.99562 0.17304450 3850 1.91465 0.99562 0.17304500 3900 1.91465 0.99562 0.17304550 3920 1.91465 0.99562 0.17304600 3720 2.11402 1.09929 0.20368650 3650 2.24921 1.16959 0.25094700 3640 2.33133 1.21229 0.31690

155

The initial oil in-place is estimated (volumetrically) to be 33.2 MMSTB.

Reservoir system properties were estimated to be:cf = 5×10−6psi−1cw = 3×10−6psi−1Swc = 0.248

PVT data for the reservoir oil is given in the following table.

PVT DATA

P Bo Rs Bg Bw(psia) (RB/STB) (SCF/STB) (RB/SCF) (RB/STB)

3365 1.2511 510 - 1.03465 1.2497 510 - 1.03565 1.2483 510 - 1.03665 1.2469 510 - 1.03765 1.2455 510 - 1.03865 1.2441 510 - 1.03965 1.2427 510 - 1.04065 1.2413 510 - 1.04165 1.2399 510 - 1.04265 1.2385 510 - 1.0

Reservoir temperature is 210oF and the gas gravity is 0.69.

Determine the cumulative water influx for each of the 50 day time periods in theproduction history table.

Solution hint

You need to calculate a cumulative water influx volume for each 50 day period(a total of 14 calculations). Each calculation is identical to the calculationsperformed in Examples 6-1 to 6-6. You can simply use [MBE.mcd] to do thecalculation for each period. Alternatively, you could modify [MBE.mcd] to read-in the production data and do the 14 calculations in one pass or prepare yourowm spreadsheet using whatever software you are most comfortable with.

[Answer: We at 700 days is 2.778 MMRB]

156

6.4 MATERIAL BALANCE FOR A CLOSED

OIL RESERVOIR

When water influx from the aquifer is small or negligible (small ineffective aquifer,weak to moderate aquifers during early stages of production when the aquiferresponse is small) there is negligible water production and the material balanceequation reduces to,

0 = NBoi − (N −Np)Bo − (cf + Swicw)µNBoi1− Swi (pi − p)

¶−GBg +GBgi +GpBg − {NRsi − (N −Np)Rs}Bg

(6.7)

If the original gas cap volume may be estimated from a volumetric calculation,the formation compressibility is known, and reliable production-pressure data isavailable, the only unknown is the original oil in-place, N . This provides a veryvaluable check on the volumetrically determined value. As discussed previously,the material balance derived value of N may be more reliable than the volumet-rically determined volume.

Prediction of future reservoir pressure

After the original oil in-place has been established, the material balance equationmay be used to predict future reservoir pressure if future cumulative productioncan be estimated. The main difficulty with this procedure is estimating the vol-umes of gas and water which will accompany the specified oil production volume.If all the produced oil and water is reinjected, this difficulty is eliminated.

157

Problem 6.2 - Calculation of original oil in-place using material balanceequation [OIPMBE.mcd]

The discovery pressure of a reservoir containing a gas cap was 3330 psia. Geo-logical evidence suggests that aquifer quality is poor and that water influx intothe reservoir is likely to be negligible. The reservoir was produced until the reser-voir fell to 2700 psia. Cumulative oil production at this time is 11.503 MMSTBand cumulative gas production is 14.206 BSCF. The volumetric estimates of theoriginal oil-in-place, (N), is 105 MMSTB and the initial gas cap volume is 81.14BSCF. Does the material balance calculation confirm this estimate?

Data:

Swi = 0.21cf = 3.0×10−6 [psi−1]cw = 3.0×10−6 [psi−1]

[Answer: 113 MMSTB]

Solution hints

The material balance equation is explicit in N , so we could rearrange it and solvedirectly for N - you can do this if you like. An alternative way is to solve thematerial balance equation by trial and error ie., guess a value of N which makesboth sides of the material balance equation equal. If you use [OIPMBE.mcd] youwill see the I have simply rearranged [MBE.mcd] to allow me to enter a value ofN and see the sum of the drive indices with the drive index for water set to zero(closed reservoir condition). The value of N which makes the drive indices sumto one is the correct oil-in-place.

158

6.5 MATERIAL BALANCE FOR A CLOSED

GAS RESERVOIR

The material balance equation for a closed gas reservoir is so simple that it maybe solved graphically. The graphical solution method is commonly referred to asthe P/z-plot. The plot can be used to estimate the original gas in-place and topredict future reservoir pressure given a production forecast.

The general material balance equation for a closed gas reservoir reduces to;

GBgi = (G−Gpc)Bg

Since Bg is given by,

Bg = 5.02× 10−3 zTP

we can write for the above material balance equation (where reservoir temperatureis assumed to remain constant),

Gµ5.02× 10−3 ziT

Pi

¶= (G−Gp)

µ5.02× 10−3 zT

P

¶

which may be rearranged to,

GziPi= (G−Gp) z

P

or,

GµP

z

¶= (G−Gp)

µP

z

¶i

or,P

z=µP

z

¶i−Gp 1

G

µP

z

¶i

The above equation results in a straight line relationship between (P/z) andGp. The straight line on a (P/z)—Gp plot passes through (P/z)i at Gp = 0 andthrough G at (P/z) = 0.

The straight-line relationship is very useful in estimating the initial volume ofgas-in-place (G) from limited production history.

159

Figure 6.14: p/z plot for a closed gas reservoir

Example 6.7 - Calculation of initial gas in-place for a closed gas reser-voir [GIP.mcd]

Estimate the initial gas content, G, for a closed gas reservoir having the followingproduction history.

Pressure Cumulative Gas Gas Deviation(psig) Production (MMSCF) Factor (z)

3500 0 0.843350 46.7 0.823200 125.0 0.813050 203.5 0.802750 380.0 0.78

160

Solution

1. Calculate P/z for each pressure point and plot against cumulative produc-tion, Gp.

Pressure Cumulative Gas Gas Deviation P/z(psig) Production (MMSCF) Factor (z) (psig)

3500 0 0.84 4166.73350 46.7 0.82 4085.33200 125.0 0.81 3950.63050 203.5 0.80 3812.52750 380.0 0.78 3525.6

2. Draw the best-fit straight line through the data and extrapolate to P/z = 0.The intercept at P/z = 0 is the initial gas in place, G. To see this plotopen [GIP.mcd].

G = 2400 MMSCF.

161

Figure 6.15: Material balance for a gas reservoir

6.5.1 Water Drive Gas Reservoirs

When a gas reservoir is in contact with a significant aquifer, water influx willoccur as the reservoir pressure declines with production. The material balancefor such a reservoir may be written as,

GBgi = (G−Gp)Bg +We −WpBw

This equation can be solved for water influx,

We = GBgi − (G−Gp)Bg +WpBw

If G can be evaluated volumetrically and if the field production data is known(Gp and Wp) at corresponding reservoir pressures, the material balance equationcan be used to calculate the cumulative water influx at these times.

162

Figure 6.16: p/z plot for a gas reservoir experiencing water influx

Example 6.8 - Production history of a water drive gas reservoir[GIP.mcd]

The following pressure production data were recorded for a gas reservoir. Canwe estimate G from this data?

Pressure Cumulative Gas Gas Deviation(psig) Production (MMSCF) Factor (z)

3500 0 0.843350 52 0.823200 155 0.813050 425 0.802950 880 0.79

[Answer: yes, G= 2.67 TSCF]

Solution hint

Even if the aquifer is active, it takes time for the aquifer to respond to the decreasein reservoir pressure. This means that early in the life of the reservoir will behaivelike a closed system. The early data should therefore plot as a straight line on a

163

P/z-plot. Extrapolating this plot to zero pressure will therefore give a realisticestimate of the original gas-in-place. I have setup the spreadsheet [GIP.mcd] toallow you to select the points through which to fit the straight line.

Problem 6.3 - Calculation of water influx for a gas reservoir [GWMTB.mcd]

For the water drive gas reservoir of the previous example calculate the cumulativewater influx corresponding to the times at which cumulative gas production isgiven. Cumulative water production is zero for the period and the reservoirtemperature is TR = 275

oF.

[Answer: 0, 0, 0.016, 0.192, 0.569 MMRB]

Solution hint

Do this any way you like. I used [GIP.mcd] from the previous example and addeda calculation for Bg from z and the material balance equation for a water drivegas reservoir.

164

Chapter 7

RESERVOIR ROCKPROPERTIES AND COREANALYSIS PROCEDURES

Reservoir engineering calculations require a knowledge of certain basic rock prop-erties. These include:

1. Porosity.

2. Permeability.

3. Rock compressibility.

4. Saturation dependent properties.

— Relative permeability.

— Capillary pressure.

— Wettability.

In this section we look at the basic properties - porosity, permeability and com-pressibility. The saturation dependent properties will be considered in the nextchapter.

7.1 POROSITY

Parameter Symbol Definition

Porosity φ The total fraction of the bulk volume which is void space

165

well sorted sandstone21.5%

poorly sorted sandstone1%

Figure 7.1: Effect of grain size distribution on porosity

Porosity is important in determining the volume of oil and/or gas whichwill be contained in a reservoir

— for reservoirs of interest, φ is in the range 0.05 — 0.40.

— porosity is a strong function of grain size distribution but only a weakfunction of grain size itself.

— for sandstones sedimentological processes are important. For limestonesporosity is determined by changes after deposition - dissolution and frac-turing.

Porosity for both sandstones and limestones can be greatly modified by post-depositional events or digenesis. Cementation, chemical action, and fracturingare events which can modify the original porosity of rocks. The original porosityof a formation is usually refereed to as primary porosity. Porosity modified bypost-depositional processes is referred to as secondary porosity.

The attached figures show the effect dispersed fines (particulate kaolinite, pore-lining chlorite and pore-bridging illite) affect primary porosity. Effective porosityand permeability may be greatly reduced by deposition of fine particulate mineralsin the pore space.

166

Figure 7.2: Schematic of a clean well-sorted sandstone

7.1.1 Effective and Total Porosity

Porosity is divided into two categories;

— effective porosity — interconnected pore space.

— disconnected porosity — this porosity is inaccessible to flow and is of nointerest because it is not accessible to flow of fluids and therefore does notcontribute to hydrocarbon production.

Total porosity is the sum of effective and disconnected porosity. The totalporosity, φt, is given by,

φt =bulk volume− grain volume

bulk volume=total pore volume

bulk volume

Effective porosity, φ, is given by,

φ =effective or connected pore volume

bulk volume

167

Discrete- particle kaolinite Pore- lining chlorite Pore- bridging illite

Figure 7.3: Common types of dispersed shales, clays and fines and their effect onporosity

Conventional laboratory experiments used to determine porosity are all basedon introducing fluids into the pore space. They therefore all measure effectiveporosity.

7.1.2 Laboratory Measurement of Porosity

The most common method employed to measure porosity is based on Boyle’s law.The equipment used is called a Boyle’s law porosimeter and is shown schematicallyin the attached figure. The apparatus and measurement procedure consists:

1. Two cells of accurately known volume, V1 and V2, are connected and atatmospheric pressure, pa. The clean and dry core sample is placed in thesecond cell.

2. Valve B is closes and cell 1 is pressured with gas G through valve A to anabsolute pressure (p1+ pa). This pressure is recorded and valve A is closed.This is referred to as Condition I.

3. Valve B is opened and the two cells are allowed to equilibrate. The equi-librium absolute pressure, (p2 + pa), is recorded. This is called ConditionII.

168

Figure 7.4: SEM micrograph of pore-filling authigenic kaolinite (x1000)

A mole balance at Condition I and Condition II gives,

moles of gas in cells 1 and 2 = moles of gas in cells 1 and 2at Condition I at Condition II

Assume an isothermal expansion process, and pressures sufficiently low so thatthe ideal gas law applies. The grain volume of the sample is Vs. The ideal gaslaw is written as,

n =pV

RT

where,

p = absolute pressureT = absolute temperatureV = volumen = moles of gasR = universal gas constant

169

Figure 7.5: SEM micrograph of chlorite (feldspar) lined pore (x1000)

Figure 7.6: SEM micrograph of delicate fibers of illite bridging pores (x1000).These fibers may break-off at high flow rates and block pores

170

Figure 7.7: SEM micrograph of pore filling kaolinite plates (x1000).

Using the ideal gas law we write, for the above word equation,

(p1 + pa)V1RT

+pa(V2 − Vs)

RT=(p2 + pa)V1

RT+(p2 + pa)(V2 − Vs)

RT

Solving for the grain volume, Vs, gives,

Vs = V1 + V2 − p1p2V1

Note that the pressures p1 and p2 are gauge pressures as read in the actualexperiment.

Porosity is calculated from the equation,

φ =Vb − VsVb

where the bulk volume of the sample, Vb, is either determined from measurementsof the sample dimensions (core plugs) or by measuring the displaced liquid volumewhen the sample is immersed in a non-wetting liquid such as mercury.

171

Figure 7.8: Solution and fracture porosity in limestones

Figure 7.9: Boyle’s law porisometer

172

Figure 7.10: Schematic of Boyle’s law porisometer test procedure

Example 7.1 - Determination of the porosity of a core plug using heliumporosimetry [HEPOR.mcd]

Given the diameter and length of a cylindrical core plug;

D = 2.5 cmL = 4 cm

and the following measured porosimeter pressures and parameters;

V1 = 25 cm3

V2 = 50 cm3

p1 = 100.03 psigp2 = 41.79 psig

Calculate the porosity of the plug.

[Answer: 0.23]

173

q q

q =kA

µ

∆P

L

∆P

L

A

Figure 7.11: Datcy’s law and permeability

7.2 PERMEABILITY

Parameter Symbol Definition

Permeability k A measure of the ease with which fluid flowsthrough a porous rock

Permeability is important in determining the rate at which oil and gaswill flow to the wellbore

Permeability will usually vary over several orders of magnitude in the same reser-voir. The variation in permeability (heterogeneity) has a great effect on oil recov-ery for both conventional and EOR processes.

— reservoir rocks typically have permeabilities in the range 1 — 10,000 md.

— permeability is a weak function of grain size distribution but a strong func-tion of pore size (throat size).

174

Figure 7.12: Schematic of permeameter

7.2.1 Measurement of Permeability

The attached figure shows a schematic of the experimental apparatus used tomeasure the permeability of cylindrical core samples or plugs.

The steady-state flow rate (q) of a fluid which completely saturates the core isdirectly proportional to the x-sectional area (A) over which the flow occurs, theimposed pressure gradient over the core (∆p/L) and inversely proportional to theviscosity of the fluid.

q ∝ Aµ∆p

∆L

¶1

µ

The constant of proportionality is called permeability and is an intrinsic propertyof the rock.

q =kA∆p

µ∆L

This equation is known as Darcy’s law and is the basic equation for calculatingfluid flow rates in reservoir engineering.

Darcy’s law defines permeability as,

k =qµ∆L

A∆p

175

When the variables in the above equation are expressed in Darcy units, the com-puted permeability has the units of darcys. A rock sample has a permeability ofone darcy when;

∆p = pressure difference, 1 atmosphere∆L = flow length, 1 cmA = x-sectional area, 1 cm2

q = volumetric flow rate, 1 cm3/secµ = viscosity, 1 cpk = permeability, 1 darcy

The darcy unit is too large for most reservoir rocks and it is common practice toexpress permeability in terms of millidarcys.

1darcy = 1000 millidarcy = 1000 md

Field Units

In oil field units the linear form of Darcy’s Law for the flow of an incompressiblefluid is written as,

q = 0.001127kA∆p

µ∆L

where,

∆p = pressure difference, psi∆L = flow length, ftA = x-sectional area, ft2

q = volumetric flow rate, res. bbl/day (RB/D)µ = viscosity, cpk = permeability, md

176

Example 7.2 - Laboratory determination of the liquid permeability ofa core plug [LPERM.mcd]

The laboratory has conducted a steady-state liquid permeability test on a coreplug from a reservoir of interest to you. Check the laboratory calculation whichsuggests that the permeability of the plug is 200 md.

The laboratory measured data is given below:

plug diameter = 2.52 cmplug length = 2 cmtest oil viscosity = 1.82 cpoutlet pressure = 1 atminlet pressure = 2 atmflow rate = 0.275 cm3/sec

[Answer: 201 md]

Solution hint

Use

k =qµ∆L

A∆p

to calculate the permeability. Since a liquid is used to measure it, the permeabilityis called the liquid permeability.

177

Example 7.3 - Calculation of field flow rates [LINQ.mcd]

The core plug in the previous example was taken from a reservoir undergoing ahorizontal line drive over a flow length of 10,000 feet. Production well spacing is1000 feet and the sand thickness is 30 feet. The reservoir oil has a viscosity of0.8 cp. What is the well rate when the pressure difference between injectors andproducers is 500 psi?

[Answer: 425 RB/day]

Solution hint

Use

q = 0.001127kA∆p

µ∆L

to calculate the rate in RB/day if you are not using Mathcad which handles mixedsystems of units automatically. The constant 0.001127 arises from unit conver-sions. Study [LINQ.mcd] to see how Mathcad handles the unit conversions.

178

7.2.2 Laboratory Measurement of Permeability

It is usually more convenient to measure permeability in the laboratory using agas as the working fluid. Liquids may adsorb onto the rock surface and potentiallychange wettability which could adversely affect later experiments. Gas does nothave this problem.

The experiment consists of flowing a gas through the core plug. Since gas iscompressible, the previous equations, which are valid only for incompressiblefluids, must be modified. The average gas pressure (p̄) in the core is given by,

p̄ =(p1 + p2)

2

where, p1 and p2 are the inlet and outlet pressures, respectively.

Assuming that the gas is ideal - a reasonable assumption if the pressures arelow, the product qp (equivalent to V p for a static system) is constant. We cantherefore express the volumetric flow rate at the average pressure (q̄), in terms ofthe measured volumetric flow rate at atmospheric conditions (qa),

qapa = q̄(p1 + p2)/2

q̄ =qapa

(p1 + p2)/2

where pa is atmospheric pressure.

Darcy’s law may be written as,

q̄ =kA(p1 − p2)µ∆L

Substituting for q̄ gives,

qapa(p1 + p2)/2

=kA(p1 − p2)µ∆L

which on re-arrangement gives,

qa =kA(p1

2 − p22)2paµ∆L

or

k =2qapaµ∆L

A(p12 − p22)We use this equation to calculate the gas permeability of the test plug. Whenhis done at a number of different average pressures we find that the calculatedpermeability is not constant but that it decreases with increasing pressure. Thismay appear to be a problem, particularly if we expect the permeability to beconstant!

179

7.2.3 The Klinkenberg Effect

The apparent decrease in measured gas permeability which occurs with increasingaverage flowing pressure is known as the Klinkenberg effect.

At very low pressures, the mean free path of gas molecules is similar to the di-mensions of the pore space. Under these conditions the solid pore walls do notsignificantly retard gas flow close to the pore walls because molecules spend littletime at the walls (free-slip condition).

For liquids and gases at very high pressure, molecules are always present at thepore walls and the velocity of the fluid at the pore walls is zero (no-slip condition).The measured permeability is therefore lower than that for gases at low pressure.

To correct gas measured permeabilities for the Klinkenberg effect:

1. Measure several gas permeabilities (kgas) at different mean flowing pres-sures, p̄.

2. Plot kgas vs 1/p̄.

3. Extrapolate the linear plot to infinite pressure i.e. to 1/p̄ = 0. The extrap-olated permeability is equivalent to the liquid permeability.

The variation of gas permeability with pressure is given by the equation,

kgas = kliquid

Ã1 +

b

p̄

!

where b is the Klinkenberg factor.

kliquid and b are determined from the slope and intercept of the plot.

kliquid = intercept

b =slope

kliquid

The Klinkenberg correction may be quite significant for gas flow in:

- low permeability rocks and quite high pressures.

- laboratory data because pressures are low.

180

Problem 7.1 - Laboratory determination of the liquid permeability ofa core plug using gas [KLINK.mcd]

A clean core plug was mounted in a gas permeameter to measure permeability.The core was discharged to atmospheric pressure. A series of tests was made withincreasing inlet gas pressure. The test data is given below.

Calculate the permeability of the plug.

The laboratory data is as follows:plug diameter = 1.953 cmplug length = 3 cmatmospheric pressure = 0.979 atmgas viscosity = 0.0182 cpinlet pressure = 2 atmflow rate = 0.275 cm3/sec

Run Inlet Pressure Outlet Pressure Discharge Ratep1 (atm) p2 (atm) qa (cm

3/sec)1 1.146 0.979 0.3481 1.481 0.979 1.1683 1.815 0.979 2.1694 3.486 0.979 9.737

[Answer: 27.5 md]

Solution hint

1. Calculate the mean flowing pressure, p̄, for each test.

p̄ =(p1 + p2)

2

2. Calculate kgas for each test using Darcy’s law.

k =2qapaµ∆L

A(p12 − p22)3. Plot kgas vs 1/p̄.

4. Extrapolate the linear plot to infinite pressure i.e. to 1/p̄ = 0. The interceptat 1/p̄ = 0 is the equivalent liquid permeability.

Open [KLINK.mcd] , check the input data and read the answer.

181

qc =πR4

8µ

∆P

Lq = nqc = n

πR4

8µ

∆P

L=kA

µ

∆P

L

Figure 7.13: Simple capillary tube model for a porous medium

7.3 POROSITY-PERMEABILITY RELA-

TIONSHIPS

7.3.1 Capillary Tube Model

In this section we examine the relationship between porosity and permeabilityusing simple models to represent the porous medium. The simplest possiblemodel for a porous medium is to consider a bundle of equal sized capillary tubes.This model is illustrated in the attached diagram. Although simple, the bundleof capillaries is a good model for porous medium made up of packed particlessuch as a sandstones.

The matrix consists of a solid block having dimensions L×L×L. The pore spaceis made up of a total of n uniform capillaries of radius R and length L.

Laminar (creeping) flow in a single capillary, qc, is described by the Poiseuilleequation

qc =πR4

8µ

∆P

L

where µ is the viscosity of the fluid and ∆P is the imposed pressure gradient.

182

The total steady-state flow rate, q, is simply the sum of the flows in each capillary,

q = nqc = nπR4

8µ

∆P

L

According to Darcy’s law, the flow q is given by,

q =kA

µ

∆P

L

where A is the x-sectional area of the porous medium perpendicular to the direc-tion of flow.

Equating the two equations for, q, gives

kA

µ

∆P

L= n

πR4

8µ

∆P

L

Eliminating the common parameters and rearranging to solve for permeability,

k = nπR4

8L2

The porosity of the simple porous medium is,

φ =pore volume

bulk volume

φ =πR2L

L3

φ =πR2

L2

We use this equation to express L in terms of φ and R. Rearranging gives,

L2 = nπR2

φ

Substituting for L2 in the equation for k gives,

k =R2φ

8

It has been experimentally verified that permeability is well correlated to thesquare of particle diameter for porous media consisting of spherical particles.The unit of permeability is clearly length2.

183

Defining the specific surface area per unit pore volume of the porous medium,ΣV p, as

ΣV p =pore surface area

pore volume

ΣV p =n2πRL

nπR2L

ΣV p =2

Ror

R =2

ΣV p

Substituting for R in the equation for k yields,

k =φ

2Σ2V p

Defining the specific surface area per unit grain volume of the porous medium,ΣV g, as

ΣV g =pore surface area

grain volume

ΣV g = ΣV ppore volume

grain volume

ΣV g = ΣV pφ

1− φor

ΣV p = ΣV g1− φ

φ

Substituting into the equation for k, gives

k =1

2ΣV g

φ3

(1− φ)2

This is similar to the empirical Kozeny relation for unconsolidated porous mediaand forms the basis for power-law correlations or transforms between porosityand permeability for actual reservoir rocks. Actual relationships follow a powerlaw because in real porous media ΣV g depends on φ and k in a complex way.

Porosity-permeability transforms are important because they can be used to con-vert log measured porosities to permeabilities. This is very important in reservoircharacterization.

184

H

q

W

L

h

∆P

q =h3W

12µ

∆P

L

q =kA

µ

∆P

L=kHW

µ

∆P

Lk =

1

12h2h

H=1

12h2φf

Figure 7.14: Simple model for a fractured porous medium

7.3.2 Fractured medium model

Consider a block of length L, height H and width W containing a fracture ofheight h as shown in the attached figure. The average velocity for laminar (vis-cous) flow in the fracture is given by,

v =h2

12µ

∆P

L

The volumetric flow rate through the fracture (q = vAf = vhW ) is,

q =h3W

12µ

∆P

L

From Darcy’s law we have

q =kA

µ

∆P

L=kHW

µ

∆P

L

Equating the two equations for q gives,

k =h3

12H=

h2

12³Hh

´We can define a fracture porosity as

φf =hWL

HWL=h

H

185

This gives

k =1

12h2φf

Actual fractured media confirm that fracture permeability is a strong function offracture height.

7.4 ROCK COMPRESSIBILITY

Reservoir rocks (porous rocks) are subject to two stresses:

1. Internal stress — reservoir or pore pressure, P , exerted by the fluids inthe pore space.This stress is isotropic i.e. the same in all directions.

2. External stress — exerted by the pressure of the overburden, σ.This stress increases with burial depth and is anisotropic i.e. it is differentin different directions.

The resultant stress, σ−P , causes a strain or deformation of the rock. The overallresult is that grain, pore and bulk volumes decrease with increasing stress.

The fractional change in grain volume per unit change in rock stress is called therock or matrix compressibility, cr.

cr =1

Vr

dVrd(σ − P )

where, Vr is the grain or matrix volume.

The fractional change in total or bulk volume per unit change in rock stress iscalled the bulk volume compressibility, cb.

cb =1

Vb

dVrd(σ − P )

Bulk volume compressibility may be important in determining levels of groundsubsidence.

The fractional change in pore volume per unit change in rock stress is called thepore volume or formation compressibility, cp.

cp = − 1Vp

dVpd(σ − P )

186

Figure 7.15: Relationships between porosity, permeability and rock microstruc-ture

187

Figure 7.16: Measured porosity-permeability relationships for different rock types

Figure 7.17: Measured porosity-permeability relationships for different limestones

188

Figure 7.18: Measured permeability-specific area relationships for different sand-stones

The matrix, bulk and pore volume compressibilities are measured in special lab-oratory experiments which will be described in the next section. For reservoirengineering calculations the primary interest is in the change in pore volume asreservoir or pore volume pressure changes.

7.4.1 Pore Volume Compressibility

Experimental measurements for sandstones and shales, in which the external rockstress (σ) is constant, show that cr << cb. This means that,

dVb ≈ dVpand since Vp = φVb, we have that

cp =cbφ

with

cp = cf = − 1Vp

dVpdP

where cf is the formation compressibility and is commonly used interchangeablywith pore volume compressibility. The above relationships are useful in interpret-ing laboratory measurements.

189

Figure 7.19: Hydrostatic load cell for measuring pore volume compressibility

For actual reservoir conditions only the vertical component of external stressis constant with the stress components in the horizontal plane characterizedby boundary conditions that there is no bulk deformation in these directions.Geersma (1966) showed that for these conditions,

cf ≈ 12cp

this shows that the effective pore volume compressibility for reservoir rocks isonly one-half of the value measured in laboratory tests.

7.4.2 Measurement of Formation Compressibility

The apparatus shown schematically in the attached figure is used to measurerock compressibility (pore volume compressibility). The apparatus allows boththe internal or fluid pressure in the core and the external or overburden pressureto be varied independently. A small bore capillary tube is used to measure thechanges in pore volume by measuring the volume of fluid expelled from the porespace.

190

Figure 7.20: Porosity and permeability as a function of stress

Two types of measurements may be made:

1. Rock compaction — the external pressure on the sample sheath is in-creased with the internal pressure held at atmospheric. The volume ofliquid squeezed out into the capillary tube is directly related to the changein bulk volume of the sample.

2. Pore volume compressibility — The external pressure is maintained con-stant. The internal pressure is reduced in steps and the liquid produced ateach pressure is noted. The volume of liquid which would have been pro-duced had the pore space volume remained constant is readily calculatedfrom a knowledge of the liquid compressibility. The difference between themeasured and calculated liquid volumes is the change in pore space volume.

Typical results for various reservoir rock types are shown in attached figure.

191

7.4.3 Use of Rock Compressibilities

When oil, water and gas all occupy the pore space, (φ), in a unit bulk volume,the volume of liquid which is expelled from the rock per unit drop in pressuremay be written as,

ct = Soco + Swcw + Sgcg + cf

where,

ct = total compressibilitycf = formation or pore space compressibilityco, cw, cg = oil, water and gas phase compressibilitiesSo, Sw, Sg = oil, water and gas phase saturations

The total compressibility is used to evaluate the diffusivity factor for unsteady-state flow calculations. Estimates of total compressibility are important in wellpressure test analysis.

When there is no oil or free gas present (aquifer zone), So = Sg = 0 and Sw = 1,we have,

ct = cw + cf

Typical Values

Typical values of the individual compressibilities at reservoir conditions are:

cf = 3× 10−6 psi−1cw = 3× 10−6 psi−1co = 15× 10−6 psi−1cg = 500× 10−6 psi−1 @ 2000 psiafor,

Sw = 0.2So = 0.7Sg = 0.1

ct = Soco + Swcw + Sgcg + cf

ct = 64× 10−6psi−1

192

Figure 7.21: Porosity variation with depth in a sedimentary basin

Figure 7.22: Anisotropic stress and directional permeability

193

Chapter 8

FLUID FLOW

8.1 DARCY’S LAW

All the equations used to describe the flow of fluids in reservoirs are based onDarcy’s law. Darcy (1856), a French engineer, investigated the flow of waterthrough sand filters for water purification. He observed the following relationshipbetween superficial velocity and the properties of the sand bed and fluid,

q

A= −k

µ

∆p

∆L

where,

q = flow rate, cc/secA = cross-sectional area open to flow, cm2

k = permeability, darcyµ = fluid viscosity, cp∆p = outlet pressure - inlet pressure, atmospheres∆L = distance between inlet and outlet, cm

The permeability, k, is an intrinsic property of the porous medium.

In the above equation the porous medium is saturated with a single fluid. Thelinear dependence of flow rate on the pressure gradient implies laminar or viscousflow in the pore-spaces of the porous medium. This is generally true for mostreservoir conditions where actual flow velocities are low. A common exception isflow close to a wellbore where, as a result of the converging nature of the flow,velocities may be high enough for turbulent or non-linear flow.

194

Figure 8.1: Darcy’s law and linear flow

The above equation is general and may be written in differential form as,

q

A= −k

µ

dp

dx

where dp/dx is the pressure gradient in the direction of flow

The negative sign in the above equations indicates that pressure decreases inthe direction of flow. The sign convention is therefore that distance is measuredpositive in the direction of flow. This is the basic equation used to analyze singlephase flow in reservoir engineering.

8.1.1 Pressure Potential

The above form of Darcy’s law applies to horizontal flow where there are nogravity effects. In reservoir analysis we are often interested in vertical or inclinedflow where the flow occurs in the presence of a hydrostatic gradient. In such casesit is convenient to define a pressure potential as,

Φ =Z p

pr

dp

ρ− gdz

where z is the vertical coordinate measured positive downward and g is the grav-itational acceleration. pr is an arbitrary reference pressure, usually taken aspr = 0.

195

Figure 8.2: Pressure potential and inclined linear flow

For a reservoir dipping upward in the direction of flow with dip angle α,

z = x sinα

and the pressure potential is,

Φ =Z p

pr

dp

ρ− gdx sinα

If the reservoir fluid is considered to be incompressible the potential may bewritten as

Φ =p

ρ− gx sinα

An alternate definition of fluid potential is

Ψ = ρΦ = p− ρgz

With this definition, potential has the units of pressure.

Darcy’s law may be written as,

q

A= −k

µ

dΨ

dx

or,q

A= −k

µ

Ãdp

dx− ρg sinα

!

196

Figure 8.3: Linear vertical flow problem

Problem 8.1 - Fluid potential and Darcy’s law

In order to test your understanding of Darcy’s law and fluid potential solve theproblem shown in Figure 8.3. If you are not using Mathcad be careful with units.

8.2 STEADY-STATE FLOW

8.2.1 Horizontal Linear Flow of an Incompressible Fluid

The linear flow system is depicted in the attached figure. The medium has uniformpermeability, k, and is saturated with an incompressible (ρ is constant) fluidhaving constant viscosity, µ. Darcy’s law is written as,

q = −kAµ

∆p

L

This equation is commonly used to evaluate flow tests in reservoir core samplesand calculate rates in line drives.

197

Figure 8.4: Linear flow in parallel

Horizontal flow in parallel

Consider the case where the flow system consists of a n layers having differentthicknesses, hi, and permeabilities, ki. For a common inlet face pressure, P1, andoutlet face pressure, P2, the pressure drop across each layer is the same and theflow rate for each layer is,

qi = −ki(hiw)µ

(P2 − P1)L

The total flow rate is

q =nXi=1

qi

or

q = −w(P2 − P1)µL

nXi=1

kihi

Darcy’s law for the total flow may also be written as,

q = − k̄(hw)µ

(P2 − P1)L

where k̄ is the average system permeability and h is the total thickness.

Equating the two expressions for q gives,

k̄ =

Pni=1 kihiPni=1 hi

This equation is used to determine the average horizontal permeability of a layeredsystem.

198

Figure 8.5: Linear flow in series

Example 8.1 - Average permeability for beds in parallel [PARK.mcd]

Determine the average permeability of four parallel beds having the followingthicknesses and permeabilities.

Layer Thickness Horizontal Permeability(ft) (md)

1 20 1002 15 2003 10 3004 5 400

[Answer: 200 md]

Horizontal flow in series

Consider the arrangement where the different permeabilities are arranged in se-ries. For this arrangement the flow through each layer is the same,

q = qi

where,

q = −ki(hw)µ

∆PiLi

199

where ∆Pi is the pressure drop across slab i. This may be rewritten as,

∆Pi =qµ

hw

Liki

and the total pressure drop is the sum of the pressure drops in each layer,

(P1 − P2) =nXi=1

∆Pi =qµ

hw

nXi=1

Liki

Darcy’s law for the series arrangement may be written as,

(P1 − P2) = qµ

hw

L

k̄

where, L =PLi, is the total length for flow.

Equating the above equations gives,

k̄ =

Pni=1 LiPn

i=1 Li/ki

This equation is used to calculate the effective vertical permeability for a layeredreservoir system.

Example 8.2 - Average permeability for beds in series [SERK.mcd]

Determine the average permeability of four beds in series having the followinglengths and permeabilities.

Layer Length Permeability(ft) (md)

1 250 252 250 503 500 1004 1000 200

[Answer: 80 md]

200

Figure 8.6: Radial flow model

8.2.2 Radial Flow of an Incompressible Fluid

The radial flow system is depicted in the attached figure. This is the basicmodel for flow into a well and is used in reservoir engineering to evaluate wellperformance. The following nomenclature employed,

rw = wellbore radiusre = well drainage radiush = formation thickness

Flow enters the system across the outer perimeter at radius re, and leaves at thewellbore of radius rw. The volumetric flow rate, q, is constant and the flow area,A, is 2πrh.

Substituting into Darcy’s law gives,

q

2πrh=k

µ

dp

dr

Note that for radial flow pressure decreases with decreasing r.

201

Separating variables and integrating gives:

qZ re

rw

dr

r=2πkh

µ

Z pe

pwdp

q(ln rw − ln re) = 2πkh(pe − pw)µ

Solving for q,

q =2πkh

µ

(pe − pw)ln re

rw

This equation is widely used in reservoir engineering to calculate production andinjection rates for individual wells.

The equation may be written in field units as;

q = 0.00708kh(pe − pw)µ ln re

rw

where,

q = well rate, RB/Dk = permeability of well drainage area, mdh = average reservoir thickness, feetµ = fluid viscosity, cppe = pressure at external radius, psipw = wellbore pressure , psire = external drainage radius, feetrw = wellbore radius, feet

202

Figure 8.7: Pressure distribution in steady-state radial flow

Example 8.3 - Steady-state radial flow [RADSSF.mcd]

Calculate the maximum possible pumping rate for an oil well in a reservoir wherethe pressure is being maintained at 2000 psia at the drainage radius. The followingdata apply:

k = 230 mdh = 10 feetµ = 3.5 cprw = 0.5, feetre = 2000, feet

[Answer: 1114 RB/day]

Solution hint

Study [RADSSF.mcd] carefully - you will be using it a lot. It solves a great varietyof radial flow problems and does many vertical well performance calculations.Since we have only studied steady-state flow to this point, take the answer for thisequation and ignore the other equations and results. The theoretical minimumflowing bottom hole pressure for a pumped well is one atmosphere.

203

Figure 8.8: Radial flow in series

Radial flow in parallel

This arrangement results in the same expression for effective or average perme-ability as that for linear flow.

Radial flow in series

Using the same reasoning as for linear flow in series yields the result,

1

k̄=

1

log(re/rw)

MXi=1

log(ri)/(ri−1)ki

204

Figure 8.9: Radial flow in parallel

205

Example 8.4 - Average permeability for radial beds in series[RADSERK.mcd]

Determine the average permeability of four radial layers in series having thefollowing radial distances to bed boundaries and bed permeabilities. The wellboreradius is 0.5 ft and the well drainage radius is 2000 ft.

Radial Layer Radius Permeability(ft) (md)

r1 250 25r2 250 50r3 500 100r4 1000 200

[Answer: 30.4 md]

8.2.3 Wellbore Damage

Wells are always drilled with overpressure (mud pressure greater than formationpressure) to prevent inflow of reservoir fluids which could lighten the mud columnand lead to a blowout.

As a result of overpressure some of the drilling mud flows into the formation andthe particles suspended in the mud partially plug the pore spaces through whichfluid flows. This creates a lower permeability zone or a damaged zone close tothe wellbore.

Wellbore damage is depicted in the attached figure, where ra is the radius of thedamaged zone. The damaged zone results in an additional pressure ∆pskin at thewellbore. This pressure drop simply adds to that for the case of no damage.

For steady-state radial flow, the pressure drop for an undamaged well is given by,

q =2πkh

µ

(pe − pw)ln re

rw

or

pe − pw = qµ

2πkh

µlnrerw

¶

206

Figure 8.10: Radial flow model with an altered or damaged zone

If the additional pressure drop due to the damaged zone is ∆pskin, then the totalpressure drop for a damaged well may be written as,

pe − pw = qµ

2πkh

µlnrerw

¶+∆pskin

If we write,

∆pskin =qµ

2πkh(s)

Than, for a damaged well we have a modified radial flow equation,

pe − pw = qµ

2πkh

µlnrerw+ s

¶or

q =2πkh

µ

(pe − pw)ln re

rw+ s

where s is the skin-factor for the well.

In general, the skin-factor may be positive (a damaged well), negative (a stim-ulated well) or zero (undamaged well). In practice, s is determined from theinterpretation of transient pressure well tests (pressure build-ups and drawdown).

In the above equations the rate, q, is the reservoir or sandface rate. In appli-cation of the radial flow equation to wells we usually specify the rate at surface

207

conditions. When this is done we write the radial flow equation as,

q =2πkh

µB

(pe − pw)ln re

rw+ s

where B is the formation volume factor which relates surface rates to reservoirrates. It is understood that when an equation contains B the rate, q, is in surfacevolumes. In the absence of B, the rate is understood to be in reservoir volumes.

208

Figure 8.11: Skin pressure drop

209

Example 8.5 - Steady-state radial flow with skin [RADSSF.mcd]

(i) Calculate the maximum possible pumping rate for an oil well in a reservoirwhere the pressure is being maintained at 2000 psia at the drainage radius.

A well test has shown the well to be highly damaged with a skin factor of+7.

The following data apply:

k = 230 mdh = 10 feetµ = 3.5 cprw = 0.5, feetre = 2000, feet

(ii) What would the well rate be if the wellbore is cleaned with an acid. A suc-cessful acid job may be expected to reduce the skin factor to approximatelyzero.

(iii) What would the rate be for a stimulated well after a frac job which isexpected to achieve a skin factor of -3.

[Answer: (i) 604 RB/day, (ii) 1114 RB/day, (iii) 1745 RB/day]

8.2.4 Relationship between s and the size of the alteredzone

The skin factor for a stimulated well ,where the skin factor is negative, has anupper limit on the magnitude of the actual value which the skin factor can take.This may be demonstrated by considering the steady-state radial flow equation,

q =2πkh

µB

(pe − pw)ln³rerw

´+ s

If −s is equal to, or greater than, ln(re/rw), the equation produces answers whichare physically impossible. This problem is removed by the introduction of thethick skin model.

210

Figure 8.12: Realatinship between skin-factor and the size of the altered zone

If the altered zone extends from rw to ra and the altered permeability is ka, and ifthe bulk formation permeability (from ra to re) is k, we can use the steady-stateradial flow equation to calculate the pressure drops across the altered zone andthe formation,

pa − pw = qBµ

2πkhlnµrarw

¶where ra is the radius of the altered zone and pa is the pressure at ra.

Also,

pe − pa = qBµ

2πkhlnµrera

¶where re is the drainage radius and pe is the pressure at re.

Adding the two equations gives the pressure across the formation (pe − pw)

pe − pw = qBµ

2πh

½1

klnµrera

¶+1

kalnµrarw

¶¾

The thin skin model expresses the above pressure drop as,

pe − pw = qBµ

2πkh

½lnµrerw

¶+ s

¾

211

where k is the permeability of the bulk formation and s is the skin factor. Thismay be re-written as,

pe − pw = qBµ

2πh

½1

klnµrerw

¶+1

ks¾

Equating the two expressions for (pe − pw) gives,½1

klnµrera

¶+1

kalnµrarw

¶¾=½1

klnµrerw

¶+1

ks¾

This may be simplified to give,

s =

"k

ka− 1

#lnµrarw

¶

If ka →∞, k/ka → 0 and the equation gives

s = − lnµrarw

¶which is the upper physical limit on the value of s for a stimulated well. Theskin factor for a stimulated well depends both on the permeability of the alteredzone, ka, and the size of the altered zone, ra.

8.2.5 Effective Wellbore Radius

Another common way of expressing the skin factor is in terms of an effectivewellbore radius. Again, using the steady-state radial flow equation, we have for awell with skin,

q =2πkh

Bµ

(pe − pw)ln³rerw

´+ s

We can also write this as,

q =2πkh

Bµ

(pe − pw)ln³rer0w

´where r0w is the effective wellbore radius.

Equating both expressions for q gives,

ln

Ãrer0w

!= ln

µrerw

¶+ s

which can be written as

ln

Ãr0wrw

!= −s

212

This may be simplified to,r0w = rwe

−s

The effective wellbore radius concept is often used to characterize fractured andhorizontal wells. For example, the effective wellbore radius for a vertical well withan infinite conductivity fracture is,

r0w = xf

where xf is the fracture half-length.

The effective wellbore radius for a long horizontal well in a thin reservoir ap-proaches that of a fractured well with a fracture half-length equal to one-quarterthe length of the horizontal well,

r0w =L

4

where L is the length of the horizontal well.

8.2.6 Flow Efficiency

Once a numerical value for the skin factor has been determined it can be used todetermine the well flow efficiency FE. This is defined as the ratio of the pressuredrop at the damaged wellbore to that for an undamaged well,

FE =pi − pwf −∆pspi − pwf =

qactualqundamaged

When ∆ps = 0, the well has 100% efficiency.

213

Problem 8.2 - Skin-factor [SKIN.mcd]

A pressure test conducted on a production well indicated a skin-factor of s = +3.The last flowing pressure was pwf=2650 psi. The following information and fluidproperties apply:

rw = 0.5 ft. k = 30 md h = 40 ft.re = 2000 ft. µ = 0.4 cpB = 1.12 RB/STB. pi = 2800 psi

Calculate,

1. The pressure drop due to skin-effect. [Answer: 42 psi]

2. The flow efficiency for the well. [Answer: 0.72]

3. The effective wellbore radius. [Answer: 0.025 ft.]

The well is a candidate for stimulation by fracturing. If fracturing results in askin-factor of s = −3, what would the above parameters be for the stimulatedwell? [Answer: -94 psi, 1.63, 10 ft.]

If the fracture is estimated to affect a zone extending 15 ft. into the formationestimate the permeability of the affected zone. [Answer: 255 md.]

214

Figure 8.13: Transient radial flow

8.3 UNSTEADY STATE FLOW

The basic equation which describes single phase unsteady-state flow in a radialsystem is the radial diffusivity equation.

1

r

∂

∂r

Ãkρ

µr∂p

∂r

!= φcρ

∂p

∂t

where c is the total system compressibility, t is time and ρ is the fluid density.

The independent variables are time (t) and spatial position (r). The dependentvariable is pressure (p). Pressure is a function of both space and time i.e., p =p(r, t). The partial differential equation is non-linear because the coefficients(particularly the density, ρ) are functions of pressure.

215

Figure 8.14: Pressure distribution in transient radial flow

8.3.1 Radial Diffusivity Equation

The radial system is similar to that described for steady-state flow. We make thefollowing assumptions:

- the reservoir is homogeneous and isotropic (k is constant).

- the well is fully penetrating ie., it is completed over the entire verticalsection so that the flow is strictly radial.

Consider the flow through a volume element of thickness dr.

mass rate in −mass rate out = rate of accumulation of mass

qρ |r+dr −qρ |r =∂

∂t(2πrhφρdr)

This may be expressed as,

qρ |r+dr − qρ |r = 2πrhφdr∂ρ

∂t

The above may be simplified,Ãqρ |r+dr − qρ |r

dr

!= 2πrhφ

∂(ρ)

∂t

216

Figure 8.15: Differential material balance for the radial diffusivity equation

∂(qρ)

∂r= 2πrhφ

∂ρ

∂t

this is called the continuity equation.

Darcy’s law for a radial system may be written as,

q =2πkhr

µ

∂p

∂r

Substituting into the continuity equation gives,

∂

∂r

Ã2πkhr

µρ∂p

∂r

!= 2πrhφ

∂ρ

∂t

or1

r

∂

∂r

Ãkρ

µr∂p

∂r

!= φ

∂ρ

∂t

This is the general form of the radial diffusivity equation.

217

The term ∂ρ/∂tmay be expressed in terms of pressure by introducing an equation-of-state for the fluid. Defining the isothermal compressibility of the fluid as

c = − 1V

∂V

∂p

Since ρ = m/V , and mass m is constant,

c =1

ρ

∂ρ

∂p

This may be written as,cρ∂p = ∂pρ

dividing both sides of the equation by ∂t, gives

cρ∂p

∂t=∂ρ

∂t

Substituting into the previous equation gives

1

r

∂

∂r

Ãkρ

µr∂p

∂r

!= φcρ

∂p

∂t

This is the general form of the radial diffusivity equation for the flow of a singlephase in a porous medium. The equation is non-linear because density, ρ, whichis one of the coefficients is a function of pressure p.

218

8.3.2 Liquids Having Small and Constant Compressibility

Reservoir oil and water are relatively incompressible and the assumption of con-stant and small compressibility is a good model for these fluids. Since we haveassumed that k, φ, µ and c are constant, the diffusivity equation becomes

1

r

∂

∂r

Ãρr∂p

∂r

!=φcρµ

k

∂p

∂t

Using the chain rule for differentiation, we expand the LHS of the equation,

1

r

Ã∂(ρr)

∂r

∂p

∂r+ ρr

∂2p

∂r2

!=φcρµ

k

∂p

∂t

1

r

Ãr∂(ρ)

∂r

∂p

∂r+ ρ

∂(r)

∂r

∂p

∂r+ ρr

∂2p

∂r2

!=φcρµ

k

∂p

∂t

1

r

Ãr∂ρ

∂r

∂p

∂r+ ρ

∂p

∂r+ ρr

∂2p

∂r2

!=φcρµ

k

∂p

∂t

An alternate definition for compressibility, in terms of fluid density,

c =1

ρ

∂ρ

∂p

or,∂ρ = cρ∂p

Substituting in the differential equation

1

r

Ãrcρ

∂p

∂r

∂p

∂r+ ρ

∂p

∂r+ ρr

∂2p

∂r2

!=φcρµ

k

∂p

∂t

Finally, dividing throughout by ρ, gives

1

r

rcÃ∂p∂r

!2+∂p

∂r+ r

∂2p

∂r2

= φcµ

k

∂p

∂t

This is simplified to

c

̶p

∂r

!2+1

r

∂p

∂r+∂2p

∂r2=φµc

k

∂p

∂t

219

∂p/∂r is small and since c << 1, the leading term on the LHS of the aboveequation may be neglected and the diffusivity equation becomes

1

r

∂p

∂r+∂2p

∂r2=φcµ

k

∂p

∂t

which may be written as

1

r

∂

∂r

Ãr∂p

∂r

!=φµc

k

∂p

∂t

This is the final form of the diffusivity equation for a slightly compressible fluid.The equation is linear and therefore amenable to analytical solution. It forms thebasis for many important reservoir analysis techniques which include transientpressure well test analysis and aquifer modeling.

The equation is also written as,

1

r

∂

∂r

Ãr∂p

∂r

!=1

η

∂p

∂t

where η = k/φµc is the hydraulic diffusivity.

Total system compressibility

In actual applications of the above equation to single phase oil flow where theformation contains connate water and the formation is itself compressible, thecompressibility, c, is the total compressibility of the system. This is determinedusing the following equation

c = coSo + cwSw + cf

where co, cw and cf are the oil, water and pore volume or formation compressibil-ities, respectively. So and Sw are the oil and water saturations. Typical valuesmay be

co = 10× 10−6 psi−1cw = 3× 10−6 psi−1cf = 6× 10−6 psi−1Sw = 0.20

p = 3000 psi

This results in

c = 14.6× 10−6 psi−1

220

8.3.3 Pseudo-Steady-State Radial Flow

The simplest solution to the radial diffusivity equation (other than the steady-state solution) is for pseudo-steady-state flow. Pseudo-steady-state flow is a spe-cial form of stabilized unsteady-state flow in which the shape of the pressureprofile from the wellbore to the drainage radius is independent of time, but theaverage drainage area pressure declines with production.

The pseudo-steady state solution is important because it describes the situationof a reservoir producing by natural pressure depletion where the production forthe well originates from within the well’s own drainage area. No flow occursacross the drainage radius.

For a reservoir producing at a constant production rate, q, by natural pressure de-pletion, the average pressure, p̄, can be calculated from a simple material balance(total compressibility of the pore-fluid system).

cVp(pi − p̄) = qtwhere pi is the initial pressure and, t is time, c is the compressibility and Vp isthe pore volume.

The boundary conditions are;

(i) no flow across the external boundary,

r∂p

∂r= 0 at r = re

(ii) pseudo-steady-state flow condition,

∂p

∂t= constant

The value for the constant is obtained from the material balance equation writtenabove. Differentiating this equation we obtain,

cVp(−dp̄) = qdtor

dp̄

dt=dp

dt= − q

cVp

For the radial model, Vp = πr2ehφ, thus

∂p

∂t= − q

cπr2ehφ

221

The radial diffusivity equation is

1

r

∂

∂r

Ãr∂p

∂r

!=φµc

k

∂p

∂t

substituting for ∂p/∂t, gives

1

r

∂

∂r

Ãr∂p

∂r

!= −φµc

k

q

cπr2ehφ

which simplifies to1

r

∂

∂r

Ãr∂p

∂r

!= − qµ

πr2ekh

This equation may be written as

∂

Ãr∂p

∂r

!= − qµ

πr2ekhr∂r

Integrating the above equation gives,

r∂p

∂r= − qµr2

2πr2ekh+ C1

The constant of integration, C1, is easily evaluated from the first BCi.e., r∂p/∂r = 0 at r = re

C1 =qµ

2πkh

Substituting for C1, and simplifying gives

∂p

∂r=

qµ

2πkh

Ã1

r− r

r2e

!

Separating variables,

∂p =qµ

2πkh

Ã1

r− r

r2e

!∂r

Integrating, Z pr

pw∂p =

qµ

2πkh

Z r

rw

Ã1

r− r

r2e

!∂r

[p]prpw =qµ

2πkh

"ln r − r2

2r2e

#rrw

or

pr − pw = qµ

2πkh

Ãlnr

rw− r2

2r2e+r2w2r2e

!

222

The term r2w/r2e is clearly negligible and the equation is written as,

pr − pw = qµ

2πkh

Ãlnr

rw− r2

2r2e

!

The above equation is a general expression for pressure as a function of radius.

The particular case of practical interest is when r = re

pe − pw = qµ

2πkh

µlnrerw− 12

¶or

q =2πkh

µB

(pe − pw)ln(re/rw)− 1

2

With the inclusion of the skin-factor the equation is

q =2πkh

µB

(pe − pw)ln(re/rw)− 1

2+ s

This equation is useful for determining well production rates.

It is interesting to compare this equation to the previously derived steady-stateequation,

q =2πkh

µB

(pe − pw)ln(re/rw) + s

The term ln(re/rw) has a value typically between 6 and 7. The rates predicted bythe steady-state and the pseudo steady-state equations are usually very similar(within 10%).

8.3.4 Flow Equations in terms of Average Reservoir Pres-sure

A problem with using the above radial flow equations is that although pw and qcan be measured directly, the pressure at the outer drainage radius, pe, cannot.Well tests provide a measure of the average drainage area pressure and it is there-fore more convenient to express the pressure drawdown in terms of the averagereservoir, p̄.

The average reservoir pressure is given by

p̄ =

R rerwpdVpR re

rwdVp

223

Now, dVp = 2πrhφdr, so

p̄ =

R rerwp2πrhφdrR re

rw2πrhφdr

p̄ =

R rerwprdrR re

rwrdr

or

p̄ =2

r2e − r2wZ re

rwprdr

Since r2e − r2w ≈ r2ep̄ =

2

r2e

Z re

rwprdr

For pseudo-steady-state flow,

pr − pw = qµ

2πkh

Ãlnr

rw− r2

2r2e

!

or

p = pw +qµ

2πkh

Ãlnr

rw− r2

2r2e

!Therefore,

p̄ =2

r2e

Z re

rwpwrdr +

2

r2e

Z re

rw

qµ

2πkh

Ãlnr

rw− r2

2r2e

!rdr

p̄ =2

r2epw

Ãr2e2− r

2w

2

!+2

r2e

qµ

2πkh

Z re

rwr

Ãlnr

rw− r2

2r2e

!dr

Again, r2w << r2e , so

p̄ = pw +2

r2e

qµ

2πkh

Z re

rwr

Ãlnr

rw− r2

2r2e

!dr

The first term in the integrand must be evaluated by parts;

Z re

rwrµlnr

rw

¶dr =

"r2

2lnr

rw

#rerw

−Z re

rw

1

r

r2

2dr

Z re

rwrµlnr

rw

¶dr =

"r2

2lnr

rw

#rerw

−"r2

4

#rerw

224

Again, neglecting small terms,Z re

rwrµlnr

rw

¶dr =

r2e2lnrerw− r

2e

4

Integration of the second term gives;

Z re

rw

r3

2r2edr =

"r4

8r2e

#rerw

=r2e8

The equation for average pressure becomes,

p̄ = pw +2

r2e

qµ

2πkh

Ãr2e2lnrerw− r

2e

4− r

2e

8

!

Simplifying,

p̄ = pw +qµ

2πkh

µlnrerw− 12− 14

¶

p̄− pw = qµ

2πkh

µlnrerw− 34

¶

Which is the required inflow equation. In the presence of formation damage orstimulation (positive or negative skin) the equation becomes

p̄− pw = qµ

2πkh

µlnrerw− 34+ s

¶or

q =2πkh

µB

(p̄− pw)ln(re/rw)− 3

4+ s

The equation is similar to the previously derived equations of pseudo-steady-stateand steady-state flow and predicts similar well rates.

8.3.5 Dietz Shape Factors for Vertical Wells

The pseudo steady-state equation for a well centered in a circular drainage area,

q =2πkh

µB

(p̄− pw)ln(re/rw)− 3

4+ s

may be generalized for wells located anywhere in an arbitrary drainage area

q =2πkh

µB

(p̄− pw)ln(q2.2458A/(CAr2w) + s

225

where CA is the Dietz shape factor which depends on the geometry of the drainagearea. Values for the shape factor are given in various handbooks and texts.

If re is the effective drainage radius, the drainage area, A, may be expressed as

A = πre2

the pseudo steady-state solution may be written as

q =2πkh

µB

(p̄− pw)ln(re/rw)− 0.75 +

hln(q2.2458π/CA) + 0.75

i+ s

This equation may be written as

q =2πkh

µB

(p̄− pw)ln(re/rw)− 0.75 + sCA + s

where sCA is a shape related skin-factor and is given by,

sCA = ln(q2.2458π/CA) + 0.75

Problem 8.3 - Pseudo-steady-state radial flow and drainage area shapefactors [RADSSF.mcd]

Calculate the production rate of an oil well in a 160 acre drainage area when theaverage pressure is 1850 psia.

The following data apply:

k = 180 mdh = 16 feetµ = 2.2 cprw = 0.5 feetpw = 1230 psias = 0Bo = 1.1 RB/STB

What would the rate be using the steady-state solution? [Answer: 721 STB/day]

What would the rate be if the well was located,?

(i) - at the center of a square drainage area. [Answer: 652 STB/day]

(ii) - at the center of a rectangular (5:1). drainage area [Answer: 562 STB/day]

(iii) - at the apex of a triangular drainage area. [Answer: 480 STB/day]

226

31.6

30.9

31.6

27.6

27.1

21.9

22.6

5.38

2.36

12.9

4.57

10.8

4.86

2.07

2.72

0.232

0.115

3.39

3.13

0.607

0.111

0.098

19.1

25

Shape of drainage area CA CA

60o

1/3

1

1

2

1

4

1

5

1

2

1

4

1

2

1

2

In water drive reservoirs

In reservoirs of unknown production character

Figure 8.16: Dietz shape factors for vertical wells

227

Figure 8.17: Approximating complex geometries

8.3.6 Approximating Complex Geometries

In cases where the flow geometry is more complex than those considered bythe Dietz shape factors, it is usually possible to approximate the system usingsimple idealizations. An example of this is shown in the attached figure for awell completed close to the intersection of two linear sealing faults. If the anglebetween the faults is θ and the flow is assumed to be steady-state, the well rateis given by,

q = f2πkh

µB

(pe − pw)ln³rerw

´where,

f =θ

360o

The above equation is based on the fact that every streamline in a radial flowsystem can represent a no-flow boundary.

228

8.4 WELL PRODUCTIVITY

8.4.1 Productivity Index

The productivity index, J , of a well is defined as the production rate, q, in STB/Ddivided by the pressure drawdown, ∆p, in psi.

J =q

∆p

The productivity index is a simple measure of the productivity of a productionwell.

The productivity index for a well is not constant and may vary considerably overthe life of the well. To see why this is so, consider the pseudo-steady-state radialflow equation,

q =2πkh(P̄ − Pw)

µB(ln( rerw)− 3

4+ s)

So,

J =q

(∆p)=

2πkkroh

µBo(ln(rerw)− 3

4+ s)

The above equation may be grouped into those parameters which remain constantwith time and those which change with time,

J =

Ã2πkh

(ln( rerw)− 3

4+ s)

!ÃkroµBo

!

The terms kro/µBo vary with time because reservoir pressure ( which affect µand Bo) and saturation ( which affects kro) change with time. Moreover, if thewell is stimulated or if it gradually cleans-up with production, s will also changewith time.

Increasing Well PI

The above equation shows the options available to increase well productivity are:

- increase length of completion interval, h.

- stimulate or fracture to reduce skin, S.

- increase reservoir pressure to increase kro and reduce µo

229

Figure 8.18: Partial penetration

8.4.2 Partial Penetration

The well models discussed above considered a well completed over the entiresection of the reservoir. Many oil and gas wells are completed over only part ofthe producing sand interval. These wells are referred to as partially penetratingwells.

When comparing a partially penetrating well to a fully penetrating well, the flowin the vicinity of a partially penetrating well must deviate from pure radial flowas streamlines converge towards the well. This results in an additional pressuredrop close to the wellbore which can be treated as a pseudo-skin due to partialpenetration.

230

A number of correlations have been published which account for the effect ofpartial penetration on well performance. All are based on potential flow solutionsor numerical simulations for the two-dimensional cylindrical diffusivity equationin two-dimensions. The correlation by Papatzacos (1988) is typical. For thiscorrelation the pseudo-skin factor, sp, is given by (see attached figure for definitionof variables),

sp =

Ã1

hpD− 1

!ln

ÃπhD2

!+

1

hpDln

"hpD

2 + hpD

µA− 1B − 1

¶0.5#

where,

hpD =hph

hD =h

rw

skhkv

A =h

h1 + 0.25hp

B =h

h1 + 0.75hp

Initially a partially penetrating well behaves as if it is producing from a formationof thickness hp. After some time a transition occurs to flow from the entireformation and the establishment of the pseudo-skin factor, sp. If the productivityindex (J = q/∆p) for a fully penetrating well is,

J =2πkhh

µB

1

ln re/rw − 0.75and that for a partially penetrating well is,

Jpp =2πkhh

µB

1

ln re/rw − 0.75 + spthan the ratio J/Jpp is readily calculated. As a rule of thumb, this ratio isapproximately hp/h.

231

Problem 8.4 - Partially penetrating well [PARPEN.mcd]

A field is developed on a 40 acre spacing. The formation thickness is 100 ft.Calculate the productivity index for the well and the ratio of productivities forthe well and for a fully penetrating well;

(i) over the entire 100 ft of pay.

(ii) over the top 75 ft of pay.

(iii) over the top 50 ft of pay.

(iv) over the top 25 ft of pay.

(v) over the top 5 ft of pay.

(vi) over the middle 50 ft of pay.

Other reservoir properties are;

kh = 100 mdkv = 5 mdrw = 0.5 ft.µ = 1 cpB = 1.25 RB/STB

[Answer: (i) 7.8 STB/day-psi, 1.0 (ii) 6.2 STB/day-psi, 0.8, (iii) 4.4 STB/day-psi,0.57, (iv) 2.5 STB/day-psi, 0.32, (v) 0.69 STB/day-psi, 0.09, (vi)4.4 STB/day-psi,0.57]

232

8.5 GAS FLOW

In treating the flow of a slightly compressible fluid we began by writing thecontinuity equation for a radial element,

∂

∂r(ρq) = 2πrhφ

∂ρ

∂t

We then substituted Darcy’s law for the volumetric flow rate, q,

q =2πkhr

µ

∂p

∂r

to eliminate q. The result is,

1

r

∂

∂r

Ãρk

µr∂p

∂r

!= φ

∂ρ

∂t

Assuming that permeability, k, is constant, we write,

1

r

∂

∂r

Ãρ

µr∂p

∂r

!=φ

k

∂ρ

∂t

This is the basic material balance for a radial volume element of the reservoirand it applies to all flows.

For the case of a slightly compressible fluid with constant viscosity, we havepreviously written the radial diffusivity equation as,

1

r

∂

∂r

Ãr∂p

∂r

!=φµc

k

∂p

∂t

Gases are highly compressible and we cannot expect that the above diffusivityequation will apply for gases.

Equation of state for a compressible fluid

The pressure-volume-temperature (PVT) behavior of a gas is given by the realgas law,

pv = znRT

where z is the compressibility factor, v is the volume of the gas, T is the tem-perature (assumed to be constant), n is the number of moles of gas and R is theuniversal gas law constant.

233

Sincen =

m

M

where m is the mass of the gas and M is the gas molecular weight, the real gaslaw may be written as,

ρ =m

v=M

RT

p

z

We may use this expression together with the definition of compressibility towrite an expression for the compressibility of a real gas,

c =1

ρ

∂ρ

∂p

which gives

c =z

p

∂(p/z)

∂p

In performing the integration we note that z = z(p).

c =z

p

(∂

∂p(pz−1)

)

c =z

p

(1

z− p

z2∂z

∂p

)Finally,

c =1

p− 1z

∂z

∂p

i.e., the compressibility of a gas is a highly non-linear function of pressure.

A further complication for gases is that viscosity is also a function of pressure(µ = µ(p))

Diffusivity equation for a compressible fluid

As previously, we begin our analysis for a gas by writing the continuity equation,

1

r

∂

∂r

Ãρ

µr∂p

∂r

!=φ

k

∂ρ

∂t

For a gas, µ = µ(p) and we cannot assume a constant viscosity.

Since

ρ =M

RT

p

z

234

we have1

r

∂

∂r

ÃM

RT

p

z

1

µr∂p

∂r

!=φ

k

∂p

∂t

Ã∂ρ

∂p

!Now, from our definition of compressibility,

∂ρ

∂p= cρ

∂ρ

∂p= c

M

RT

p

z

which gives1

r

∂

∂r

ÃM

RT

p

z

1

µr∂p

∂r

!=φ

k

∂p

∂tcM

RT

p

z

dividing both sides by the constants we finally obtain,

1

r

∂

∂r

Ãp

µzr∂p

∂r

!=φcp

kz

∂p

∂t

This is a highly non-linear equation because the coefficients on both sides ofthe equations are highly non-linear functions of pressure. In order to solve theequation it is necessary to first linearize it.

Pseudo-pressure

Introduce a new variable called the pseudo pressure, ψ(p), defined as

ψ(p) = 2Z p

a

p

µzdp

where a is an arbitrarily low reference pressure.

This transformation gives

dψ = 2

Ãp

µz

!dp

Re-writing the non-linear diffusivity equation in therm of pseudo-pressure wehave,

1

r

∂

∂r

Ãp

µzr∂ψ

∂r

dp

dψ

!=φcp

kz

∂ψ

∂t

dp

dψ

Now,∂p

∂ψ=µz

2p

235

therefore,1

r

∂

∂r

Ãp

µzr∂ψ

∂r

µz

2p

!=φcp

kz

∂ψ

∂t

µz

2p

Simplifying this expression gives

1

r

∂

∂r

Ãr∂ψ

∂r

!=φµc

k

∂ψ

∂t

This has eliminated the non-linearity on the LHS and made the equation looklike the diffusivity equation for a slightly compressible liquid except that thecoefficient on the RHS is still non-linear (c is a function of pressure).

8.5.1 Low pressure approximation — p < 3, 000psi and ∆psmall

For low pressure and small pressure drop,

µz ≈ constant = µ̄z̄The definition of pseudo pressure,

ψ(p) = 2Z p

a

p

µz∂p

with the reference pressure to zero, is now written as,

ψ(p) =2

µ̄z̄

Z p

0p∂p

ψ(p) =2

µ̄z̄

p2

2

ψ(p) =p2

µ̄z̄

The pseudo pressure is easily evaluated on a spreadsheet or in a computer code.Since ∆p is small (p approximately constant) we may use average properties forthe coefficient terms, c̄ and µ̄, resulting in a linear diffusivity equation in termsof the pseudo pressure.

1

r

∂

∂r

Ãr∂ψ

∂r

!=φµ̄c̄

k

∂ψ

∂t

If we are doing the analysis by hand and wish to eliminate the need to evaluatepseudo-pressures we may simply write the equation in terms of p2.

∂ψ

∂(p2)=1

µ̄z̄

236

and1

r

∂

∂r

Ãr∂(p2)

∂r

∂ψ

∂(p2)

!=φµ̄c̄

k

∂(p2)

∂t

∂ψ

∂(p2)

which gives1

r

∂

∂r

Ãr∂(p2)

∂r

!=φµ̄c̄

k

∂(p2)

∂t

The result is identical to that which would be obtained by defining

ψ = p2

This is commonly referred to as the pressure squared method of analysis.

8.5.2 High pressure approximation — p > 3, 000psi and ∆psmall

At high pressures and small pressure drops,

p

µz≈ constant = p̄

µ̄z̄

From the definition of pseudo pressure we have,

ψ(p) = 2Z p

a

p

µz∂p

which can be written as

ψ(p) =p̄

µ̄z̄

Z p

a∂p

or

ψ(p) =p̄

µ̄z̄p

Again the pseudo-pressure is easily evaluated, and as for the case of low pressurewe can show that the diffusivity in terms of ψ is linearized to

1

r

∂

∂r

Ãr∂(p)

∂r

!=φµ̄c̄

k

∂(p)

∂t

which is identical to the case of a slightly compressible fluid.

237

8.5.3 ∆p is not small

When ∆p is not small, c and µ vary considerably and we cannot use averagevalues for these parameters over the flow conditions.

In terms of ψ the non-linear diffusivity equation is

1

r

∂

∂r

Ãr∂ψ

∂r

!=φµc

k

∂ψ

∂t

We may write this as1

r

∂

∂r

Ãr∂ψ

∂r

!=φ

k

∂ψ

{∂t/µc}

Defining a pseudo-time as,

τ =Z t

0

dt

µcor

dτ =dt

µc

Pseudo-time is easily evaluated since pressure, as a function of time, is known.

The diffusivity equation may now be written as,

1

r

∂

∂r

Ãr∂ψ

∂r

!=φ

k

∂ψ

∂τ

A problem with the pseudo-time it does not have units of time and this is unde-sirable in actual practice where the engineer likes to relate time to distance viathe depth of investigation concept.

The situation is readily remedied by the introduction of a normalized pseudo-time, τn, defined as

τn = (µc)i

Z t

0

dt

µc

(µc)i is the viscosity-compressibility product evaluated at the initial test pressure.The product may be defined at any pressure, the pressure being arbitrary, butconvention is to take the value at the initial test pressure.

The diffusivity equation in terms of pseudo-pressure and pseudo-time is,

1

r

∂

∂r

Ãr∂ψ

∂r

!=φ(µc)ik

∂ψ

∂τn

Again, this equation has exact ally the same form as the diffusivity equation fora slightly compressible fluid.

238

8.5.4 Steady and Pseudo-Steady State Radial Gas Flow

The radial diffusivity equation for gas is

1

r

∂

∂r

Ãr∂ψ

∂r

!=φµc

k

∂ψ

∂t

For steady-state flow, ∂ψ/∂t = 0, and we can write

1

r

∂

∂r

Ãr∂ψ

∂r

!= 0

or,∂

∂r

Ãr∂ψ

∂r

!= 0

Integrating gives

r∂ψ

∂r= C1

This is the solution for steady-state radial flow of a gas.

To evaluate the value of the constant of integration, C1, we write Darcy’s law forradial flow,

q =2πkh

µr∂p

∂r

where q is the rate at reservoir conditions. The rate at surface conditions is givenby

qsc =2πkh

µBgr∂p

∂r

where Bg is the gas formation volume factor which is given by

Bg =µpscTsc

¶zT

p

Substituting for Bg gives

qsc =

Ã2πTscpsc

!kh

µ

p

zTr∂p

∂r

The equation is written in terms of pseudo-pressure as

qsc =

Ã2πTscpsc

!kh

µ

p

zTr∂ψ

∂r

dp

dψ

From the definition of pseudo-pressure

dp

dψ=µz

2p

239

Substituting for dp/dψ gives

qsc =

ÃπTscpsc

!kh

Tr∂ψ

∂r

or,

r∂ψ

∂r=µpscπTsc

¶qscT

kh

This is the required constant of integration C1.

Separating variables,∂r

r=

ÃπTscpsc

!kh

qscT∂ψ

Integrating, Z rw

re

∂r

r=

ÃπTscpsc

!kh

qscT

Z ψw

ψe∂ψ

gives

lnµrerw

¶=

ÃπTscpsc

!kh

qscT(ψe − ψw)

or,

qsc =

ÃπTscpsc

!kh

T

(ψe − ψw)

ln³rerw

´

As before, for pseudo-steady state flow in terms of average pseudo-pressure andin the presence of skin, we can write

qsc =

ÃπTscpsc

!kh

T

(ψ̄e − ψw)hln³rerw

´− 0.75 + s

i

The above equation is valid for all conditions and may be used to estimate thestabilized rate for gas wells under Darcy flow conditions. It requires evaluationof pseudo-pressures. Pseudo-pressures are easily calculated by numerically inte-grating P , µ and z data for the gas. An example of such a calculation is given inExercise-8.11.

8.5.5 Non-Darcy flow

In high rate gas wells flow velocities close to the wellbore may sufficiently highfor turbulent or non-Darcy flow to occur. This introduces an additional pressure

240

drop which depends on the gas rate. The non-Darcy flow pressure drop may beincluded by defining an effective skin factor, s0, as

s0 = s+D|qsc|

where D is the non-Darcy flow constant and s is the Darcy flow skin factor. Theabsolute value of rate indicates that there is a positive non-Darcy flow loss forboth production and injection operations. D and s are determined from multi-rate well pressure test data.

Pseudo-steady state flow in terms of average pseudo-pressure and in the presenceof skin and non-Darcy flow effects, can now be written as

qsc =

ÃπTscpsc

!kh

T

(ψ̄e − ψw)hln³rerw

´− 0.75 + s0

ior,

qsc =

ÃπTscpsc

!kh

T

(ψ̄e − ψw)hln³rerw

´− 0.75 + s+D|qsc|

i

8.5.6 Pressure-squared approximation

For pressures below about 3000 psia the product µz is approximately constantand pseudo-pressure, defined as

ψ = 2Z p

0

p

µzdp

may be written as

ψ =2

µ̄z̄

Z p

0pdp =

p2

µ̄z̄

where µ̄ and z̄ are evaluated at average flowing conditions.

Substituting for ψ in the rate equation gives

qsc =

ÃπTscpsc

!kh

µ̄z̄T

(pe2 − pw2)h

ln³rerw

´− 0.75 + s+D|qsc|

i

8.5.7 High Pressure approximation

For pressures above about 3000 psia the product p/µz is approximately constantand pseudo-pressure, defined as

ψ = 2Z p

0

p

µzdp

241

may be written as

ψ =2p̄

µ̄z̄

Z p

0dp =

2p̄

µ̄z̄p

where µ̄, z̄ and p̄ are evaluated at average flowing conditions.

Substituting for ψ in the rate equation gives

qsc =

Ã2πTscpsc

!khp̄

µ̄z̄T

(pe − pw)hln³rerw

´− 0.75 + s+D|qsc|

iThe average gas formation factor may be written as

B̄g =

ÃTscpsc

!z̄T̄

p̄

The gas rate equation becomes

qsc =2πkh

µ̄B̄g

(pe − pw)hln³rerw

´− 0.75 + s+D|qsc|

i

With the non-Darcy flow term set to zero (D = 0), this equation is identical tothe equation previously derived for the flow of a slightly compressible fluid.

242

Problem 8.5 - Calculation of gas well rate [GASFLOW.mcd]

Estimate the production rate of a gas well when the average reservoir pressureis 2723 psia and the flowing wellbore pressure is maintained at 2655 psia. Thefollowing reservoir and fluid properties are known:

k = 5 mdh = 35 ft.rw = 0.5 ft.re = 2640 ft.µ = 0.018 cpz = 0.87T = 200oFs = -0.3D = 0.015 (MMSCF/day)−1

What is the effect of non-Darcy (turbulent) flow on well performance?

[Answer: 344 MSCF/day]

243

Problem 8.6 - Gas well rate using pseudo-pressures [PSUDOP.mcd]

Estimate the production rate of a gas well when the average reservoir pressure is2723 psia and the flowing wellbore pressure is maintained at 2655 psia. Reservoirtemperature is 200oF and the gas gravity is 0.7. Use pseudo-pressure and thepressure-squared method to calculate the gas rate. The following reservoir andfluid properties are known:

k = 5 mdh = 35 ft.rw = 0.5 ft.re = 2640 ft.T = 200oFs = -0.3D = 0.015 (MMSCF/day)−1

GAS PROPERTIES

p µ z(psia) (cp)

150 0.01238 0.9856300 0.01254 0.9717450 0.01274 0.9582600 0.01303 0.9453750 0.01329 0.9332900 0.01360 0.92181050 0.01387 0.91121200 0.01428 0.90161350 0.01451 0.89311500 0.01485 0.88571650 0.01520 0.87951800 0.01554 0.87451950 0.01589 0.87082100 0.01630 0.86842250 0.01676 0.86712400 0.01721 0.86712550 0.01767 0.86832700 0.01813 0.87052850 0.01862 0.87383000 0.01911 0.87803150 0.01961 0.8830

[Answer: 346 MSCF/day, 342 MSCF/day]

244

Figure 8.19: Pseudo-pressure as a function of pressure

8.5.8 Gas Well Back Pressure Equation

This is an empirical equation which relates flow rate and pressure for gas wells.The equation is written as,

qsc = C(p̄2R − p2w)n

where pR is the average reservoir pressure, pw is the flowing wellbore pressure,n is an exponent which has a value between 0.5 (non-Darcy or turbulent flow)and 1 (Darcy or laminar flow) and C is the well performance coefficient. Theequation is frequently used to determine well deliverability.

Flow after flow tests

The constants C and n are determined from flow after flow or conventional back-pressure tests. The well is flowed at a constant rate (qsc) until the flowing wellborepressure (pw) has stabilized. The rate is then changed (increased) and the processrepeated. After a number of rate changes (usually four), the well is shutin untilthe shutin wellbore has stabilized. This is the average reservoir pressure (p̄R).

245

Figure 8.20: Gas well deliverability plot

Figure 8.21: Four point flow-after-flow well deliverability test

246

Figure 8.22: Interpretation of four point flow-after-flow well deliverability test

A plot of log(p̄2R − p2w) against log qsc gives a straight line of slope equal to 1/nand intercept −(1/n) logC. The test and interpretation procedure is illustratedin the attached figures.

Example 8.6 - Flow after flow conventional back-pressure test[FAFGAS.mcd]

The following flow rate and wellbore pressure data was obtained from a four pointflow-after-flow test on a gas well:

Flow period qsc (MMSCF/day) pw (psia)1 2190 33872 2570 32683 3160 30924 3400 3015

shutin 0 3884

247

Determine the exponent n and the performance coefficient C. Calculate theAbsolute Open Flow (AOF) for the well (the AOF is a commonly used figure-of-merit for a gas well and corresponds to the well rate at a flowing well pressure ofone atmosphere).

[Answer: n = 0.874, C = 4.035 × 10−6MMSCF/day − psia2n, AOF=7.6 MM-SCF/day]

Problem 8.7 - Well deliverability using back pressure equation[DELGAS.mcd]

A four point flow-after-flow test produced the following back pressure equationconstants,

n = 0.69C = 0.01508 MSCF/day-psia2n

The average reservoir pressure is 408 psia.

(i) Determine the absolute open flow rate (AOF) for the well.

(ii) What would the well flow rate be if the flowing wellbore pressure was main-tained at 300 psia.

(iii) Make a well deliverability plot for the well.

[Answer: (i) 60.4 MMSCF/day, (ii) 35.4 MMSCF/day]

248

Figure 8.23: Horizontal well in a compartmentalized reservoir

8.6 HORIZONTAL WELLS

Horizontal wells are now routinely used in primary, secondary and tertiary re-covery operations. The major advantages over conventional vertical wells are inovercoming:

Reservoir heterogeneity

- Horizontal wells contact a larger area of the reservoir than verticalwells. In reservoirs which display complex and difficult to predictheterogeneity (carbonates, channel point bar sands, braided streamsystems and fractured reservoirs), horizontal wells can connect regionsof high permeability rock which are separated by regions of low per-meability.

Reservoir flow problems

- The greater length of horizontal wells may result in smaller pressuregradients (increased productivity) when compared to vertical wellsflowing at the same rate. This can be important in allowing produc-tion of heavier crudes at economical rates and reducing or completelyelimination the detrimental effects of water and gas coning.

249

Figure 8.24: Comparison between effective reservoir contact areas for horizontaland vertical wells

- In sands where water coning is a problem horizontal wells can be placednear the top of the sand increasing the vertical distance between theperforations and the water-oil contact and therefore increasing recov-ery when compared with a vertical well in the same sand. Recovery isalso increased because the cone for a similar situation exists in a sandsubject to gas coning with the horizontal well being placed at the baseof the sand.

- In low permeability reservoirs it is possible to place multiple transversefractures along the length of the well and greatly increase productivityover that possible with a vertical well.

Completion options

The attached figure shows a number of different techniques used to completehorizontal wells.

Open hole - Inexpensive but limited to competent formations. Difficultto control rate along length of well and difficult to stimulate. Rarely usedexcept for formations such as Austin Chalk.

Slotted liner - Three types commonly used; perforated, slotted, andprepacked liners. Liners are used to prevent hole collapse and to controlsand production.

250

Figure 8.25: Horizontal wells with multiple transverse fractures play a vital rolein the production of tight gas reservoirs

Liners with partial isolation allow the horizontal section to be divided intoseveral isolated sections. This allows production control along the welllength and selective stimulation or multiple fracking.

Cemented and perforated liners provide the best completions provided thatcare is taken to ensure a good cement job.

8.6.1 Drainage Area

Wells are drilled on a pattern to provide adequate drainage of the reservoir. Thearea drained by a single well in the pattern is the drainage area for the well. Avertical well drains a cylindrical volume and a horizontal well drains an ellipsoid(three-dimensional ellipse). A horizontal well will usually drain a larger area thana vertical well.

Areally isotropic reservoir - kx = ky

A horizontal well can be viewed as a number of vertical wells drilled on a lineand completed in a limited pay thickness. Under similar operating conditions,

251

Figure 8.26: Completion techeniques for horizontal wells

252

Figure 8.27: Comparison of horizontal and vertical well costs for Prudhoe Bay inAlaska

the drainage area of a horizontal well may be estimated from that for a verticalwell as shown in the attached figure.

If the drainage area for a vertical well is Av, the drainage radius is

rev =qAv/π

The drainage area for a vertical well may also be approximated as a square ofdimensions 2xe by 2ye where xe = ye

Av = 4xe

or

xe = ye =

sAv4

A horizontal well can be viewed as a number of vertical wells drilled on a lineand completed in a limited pay thickness. Under similar operating conditions,the drainage area of a horizontal well may be estimated from that for a verticalwell as shown in the attached figure.

Ah = 2yeL+Av

where L is the length of the horizontal section. The effective drainage radius ofa horizontal well is

reh =qAh/π

As a rule of thumb, a 1000 ft horizontal well will drain twice the area of a verticalwell and a 2000 ft horizontal well will drain three times the area of a vertical well.

253

Figure 8.28: Drainage areas for horizontal and vertical wells

254

Figure 8.29: Drainage areas for horizontal wells

255

Areally anisotropic reservoir - kx < ky

Consider an anisotropic (kx < ky) homogeneous (kx, ky constant). Assumingsteady-state flow in two dimensions, we can write the following conservation equa-tion

∂

∂x

Ãkx∂p

∂x

!+

∂

∂y

Ãky∂p

∂y

!= 0

or, since kx and ky are constant

kx∂2p

∂x2+ ky

∂2p

∂y2= 0

For an isotropic reservoir, kx = ky = kh, where k−h is the horizontal permeability,we can write

kh

"∂2p

∂x2+∂2p

∂y2

#= 0

For an anisotropic reservoir we multiply and divide the conservation equation

throughout byqkxky

qkxky

vuutkxky

∂2p

∂x2+

skykx

∂2p

∂y2

= 0or q

kxky

"∂2p

∂x2+kykx

∂2p

∂y2

#= 0

This equation may be written as

qkxky

"∂2p

∂x2+

∂2p

∂Y 2

#= 0

where

y = Y

skykx

Comparing this equation with that for the isotropic system,

kh

"∂2p

∂x2+∂2p

∂y2

#= 0

shows that in an anisotropic reservoir a vertical well drains a rectangle which has

a length along the high permeability directionqky/kx the length along the low

permeability direction with an effective permeabilityqkxky.

256

Figure 8.30: Drainage areas for horizontal wells in fractured or anisotropic reser-voirs

257

Figure 8.31: Drainage areas for vertical and horizontal wells in anisotropic reser-voirs

If ky > kx,

ye =

skykxxe

and the drainage area for a vertical well is

Ah = 4yexe

and the effective horizontal permeability is,

kh =qkxky

The above analysis shows that it is difficult to drain a reservoir in the low per-meability direction using vertical wells.

If a horizontal well is drilled in the direction of low permeability, the drainagearea for the well is

Ah = 2yeL+Av

The number of horizontal wells required for adequate reservoir drainage is con-siderably smaller than that for a conventional vertical well development.

258

Example 8.7 - Drainage area for horizontal wells in an isotropic reser-voir [HORWELL.mcd]

A 600 acre lease is to be developed with 10 vertical wells. An alternative is todrill 500 ft, 1000 ft, or 2000 ft long horizontal wells. Estimate the number ofhorizontal wells required to drain the lease effectively assuming that the reservoiris isotropic.

[Answer: 8 500ft wells, 7 1000ft wells, 5 2000ft wells]

Problem 8.8 - Drainage for a horizontal well in an anisotropic reservoir[HORWELL.mcd]

A vertical well is known to drain approximately 40 acres in a naturally frac-tured reservoir. The permeability along the direction of the fractures is ky =4.5md. The permeability in the direction perpendicular to the fractures is onlykx =0.5md. What are the drainage areas for a 2000 ft horizontal well drilledalong the direction of the fractures and one drilled in a direction perpendicularto the fractures.

[Answer: 75 acres, 145 acres.]

259

Figure 8.32: Complexity of flow regimes for horizontal wells. a-early time radialflow, b-early time linear flow, c-late time radial flow, d-late time linear flow

260

Figure 8.33: Approximating three-dimensional steady-state flow in horizontalwells with two two-dimensional solutions

8.6.2 Productivity of Horizontal Wells

The steady-state rate for a vertical is given by

qv =2πkhh

µB

∆p

ln(rev/rw)

The productivity index, Jv, is defined as the rate, qv, divided by the pressuredifference, ∆p,

Jv =2πkhh/(µB)

ln(rev/rw)

Homogeneous isotropic reservoirs - kv = kh

Numerous equations have been developed for steady-state flow of horizontal wells.The differences between the various equations result from the differences in solu-tion techniques employed and the details of the assumptions made in developingthe solution. The differences are generally small.

Joshi (JPT June 1988)divided the three-dimensional steady-state flow probleminto separate two-dimensional vertical and horizontal problems and summed thesolutions. The solution obtained was shown to be in excellent agreement withlaboratory simulations using electrical networks.

For a horizontal well completed in the middle of sand of thickness, h, with con-stant pressure boundary conditions at the drainage area boundary,

Jh =qh∆p

=2πkhh/(µB)

ln∙a+√a2−(L/2)2L/2

¸+³hL

´ln³h2rw

´261

Figure 8.34: Productivity ratio Jh/Jv for horizontal wells in isotropic reservoirs

where a is half the major axis of the drainage area ellipse defined as,

a =L

2

0.5 +s0.25 +

µ2rehL

¶40.5

reh =

sAhπ

for L > h and (L/2) < 0.9reh.

The above equation shows that as reservoir thickness, h, increases, the effective-ness of the horizontal well, Jh, decreases. Horizontal wells are most effective inthin reservoirs.

Another equation for horizontal wells was developed by Borisov(Nedra, Moscow1964 - translated into English in 1984)

Jh =2πkhh/(µB)

ln (4reh/L) + (h/L) ln (h/2πrw)

This equation gives results very similar to those using the Joshi equation. In thelimit L >> h, the second term in the denominator of the equation is negligibleand the result can be rewritten as

Jh =2πkhh/(µB)

ln (reh/[L/4])

262

Comparing this equation with the corresponding equation for a vertical well showsthat in the limit L >> h, a horizontal well is equivalent to a vertical well withan effective wellbore radius r

0w = L/4 (a highly stimulated vertical well). We will

later show that the above equation is identical to that for a vertical well with aninfinite-conductivity vertical fracture of length L.

Isotropic reservoirs - kv 6= kh

When the horizontal permeability is different from the vertical permeability, in amanner similar to that previously shown, the flow field is equivalent to that foran average permeability of

√khkv with the z-axis modified,

Z = z

skhkv

The influence of anisotropy is therefore to change the effective reservoir thickness,H,

H = h

skhkv

When kv < kh, the effect is to increase the effective thickness of the reservoir.This reduces the effectiveness of a horizontal well relative to a vertical well.

To obtain the productivity of a horizontal well in an isotropic reservoir we simplysubstitute the effective thickness of the reservoir for h in the equation for anisotropic reservoir. The result is,

Jh =2πkhh/(µB)

ln∙a+√a2−(L/2)2L/2

¸+³βhL

´ln³βh2rw

´where

β =

skhkv

263

Figure 8.35: Productivity ratio Jh/Jv for horizontal wells in anisotropic reservoirs

Horizontal well eccentricity

The above equations are for the case where the horizontal well is centered withrespect to the sand in which the well is completed. A schematic of an off-centeredhorizontal well is shown in the attached figure. We define well eccentricity, δ, as

δ =h/2− zwh/2

δ = 0 for a centered well.δ = 1 for a well at the base or top of the sand.

Joshi gives the following equation for this case,

Jh =2πkhh/(µB)

ln∙a+√a2−(L/2)2L/2

¸+³βhL

´ln³βh2rw[1 + δ2]

´which is valid for L > βh, δ < h/2 and L < 1.8reh.

For long horizontal wells, (L >> h), eccentricity has little effect and the wellcan be located anywhere in the vertical plane without significant loss in produc-tivity. This is strictly true only for bound reservoirs where the top and bottomboundaries are closed ie., there is no bottom water or top gas.

264

Figure 8.36: Horizontal well accentricity

Figure 8.37: Effect of eccentricity on the productivity of a horizontal well

265

Problem 8.9 - Horizontal well productivities I [HORWELL.mcd]

A 1000 ft. horizontal well is drilled through the centre of a 50 ft sand on a 160acre spacing. Other data reservoir and fluid parameters are;

kh =50 mdµ =0.4 cprw =0.33 ftB =1.35 RB/STB

(i) Compare the horizontal well productivities for kv/kh of 0.01, 0.1, 0.5 and 1.

(ii) For kv/kh=0.1 compare the horizontal well productivities if the well is lo-cated at distances of 1 ft, 10 ft and 20 ft from the base of the sand.

[Answer: Joshi equation (i) 7.2, 13.3, 16.1, 16.9 STB/day-psi (ii) 12.7, 13.0, 13.2STB/day-psi ]

Problem 8.10 - Horizontal well productivities II [HORWELL.mcd]

A 1000 ft. horizontal well is drilled in a reservoir on a 160 acre spacing. Thehorizontal permeability averages 200 md and the vertical permeability averages20 md. Other reservoir and fluid parameters are;

µ =0.2 cprw =0.33 ftB =1.35 RB/STB

What is the improvement in productivity which may be expected over that for avertical well if the sand thickness is,?

(i) 200 ft.

(ii) 20 ft.

[Joshi equation (i) 1.5, (ii) 4.1]

266

Effect of Formation Damage on Horizontal Well Productivity

As in the case of vertical wells, formation damage can have a severe effect onhorizontal well productivity. Consider the case of a horizontal well completed atthe center of a homogeneous anisotropic sand. Renard and Dupuy (SPE19414,1990) developed the following equation for the productivity index,

Jh =2πkhh/(µB)

cosh−1³2aL

´+³βhL

´ln³

βh(1+β)πrw

´where a is half the major axis of the drainage area ellipse as previously defined.

The above equation and the previously discussed equation developed by Joshiproduce similar estimates for horizontal well productivity.

For a damaged well, Renard and Dupuy wrote,

Jh,d =2πkhh/(µB)

cosh−1³2aL

´+³βhL

´ln³

βh(1+β)πrw

´+ sh

where sh is the horizontal well skin factor which must be determined from welltests.

The ratio of productivities for a damaged and undamaged horizontal well is there-fore,

JhdJh

=cosh−1

³2aL

´+³βhL

´ln³

βh(1+β)πrw

´cosh−1

³2aL

´+³βhL

´ln³

βh(1+β)πrw

´+ sh

or

JhdJh

=B

B + sh

where

B = cosh−1µ2a

L

¶+

Ãβh

L

!ln

Ãβh

(1 + β)πrw

!

267

Figure 8.38: Comparison between formation damage for horizonral and verticalwells

268