PTRL6001 RESERVOIR ENGINEERING I
by Val Pinczewski School of Petroleum Engineering University of New South Wales Sydney NSW 2052. AUSTRALIA
March, 2002
Prepared for
PETROLEUM ENGINEERING DISTANCE LEARNING 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 accordance with copyright laws for the sole use of students enrolled in a course at the University of New South Wales. It is illegal to reproduce any of this material or to use 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 ORIGINAL OIL-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 BALANCE METHOD . . . . . . . . . . . . . . . . . . . . . . . . . .
67
3.5 ESTIMATING RESERVES . . . . . . . . . . . . . . . . . . . . .
68
3.6 ESTIMATING DECLINE IN OIL PRODUCTION RATES . . . .
69
4 HYDROSTATIC PRESSURE DISTRIBUTION IN RESERVOIRS 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 RESERVOIR CONTAINING OIL, WATER AND GAS . . . . . . . . . .
74
2
4.3 GEOTHERMAL GRADIENT . . . . . . . . . . . . . . . . . . . . 5 FLUID PROPERTIES
84 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 LABORATORY EXPERIMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 laboratory data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Drive Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . 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 ANALYSIS PROCEDURES 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
7.4.2
Measurement of Formation Compressibility . . . . . . . . . 190
7.4.3
Use of Rock Compressibilities . . . . . . . . . . . . . . . . 192
8 FLUID FLOW
. . . . . . . . . . . . . . . . 189
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
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Chapter 1 INTRODUCTION For those unfamiliar with reservoir engineering or reservoir geology the following section gives a brief summary of some basic concepts. If you are familiar with this material skip this section.
1.1 1.1.1
BASIC CONCEPTS Accumulation of Sediments
The accumulation of sediments in a basin depends on the balance between the energy of the depositional environment (water velocity) and the sedimentation velocity of the particles. The sedimentation velocity depends on the size and density of the particles. Sediments carried by high velocity streams may be deposited 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 other locations where the environmental energy is insufficient to carry the particles further. This process leads to sorting of sediments with the accumulation of sand grains in one area and clay and silt particles in another area. The depositional energy at a particular location varies with time. This results in a sequential deposition of sand (large particles) and shale (fine particles) producing sequences of sand and shale. (layering). This layering or vertical heterogeneity is the single most important characteristic determining reservoir performance and recovery.
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, are deposited together with fine-grained sediments in shallow marine environments during low energy periods of basin formation. Quite water is deficient in oxygen leading to the creation of anaerobic conditions and preservation of the organic matter. Anaerobic bacteria decomposes the organic matter to produce compounds of carbon, hydrogen (hydrocarbons) and oxygen. The conversion of the organic matter occurs over geological time as sediments become progressively buried and temperature and pressure increase. The minimum temperature for oil and gas formation is about 150o F and the maximum is about 320o F. Since temperature increases with depth, this results in a burial depth window of between 7,000 ft to 23,000 ft for fine grained sediments containing organic matter (source rocks) to produce oil and gas. Increasing burial causes the fine-grained sediments to undergo compaction and the source rocks eventually become effectively impermeable. As the particles compact the generated hydrocarbon particles (oil or gas) are squeezed from the source rock. This process is called primary migration. The expelled hydrocarbon particles, in the form of colloidal solutions (micelles) or individual drops or bubbles, enter the overlying and underlying water saturated permeable sand layers which have retained their porosity and permeability because the coarse sand grains are stronger than the fine-grained silt and clay particles and therefore better 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 pore space of the permeable sand layers and moves upwards as a result of buoyancy. This is called secondary migration. The upward migration of generated hydrocarbon continues until it is halted by an impermeable barrier or trap. As hydrocarbon accumulates under the trap a reservoir is formed. The characteristics of hydrocarbon traps are illustrated by considering a porous permeable formation between two impermeable layers which has been folded by tectonic action into an anticline (see the attached figure). This is called an anticlinal trap. The hydrocarbon contained in the reservoir may be oil or gas or both. Here are some commonly used terms to describe petroleum reservoirs. You will probably 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 limestone 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/or gas) 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 communication 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 layers after 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 unequal tectonic 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 formation such as pinchouts and lenticular sand bodies surrounded by impermeable shales. The processes involved in the formation of stratigraphic traps are complex because they involve changes in the depositional environment. Stratigraphic traps may be associated with unconformities. An uncomformity forms when a site of sediments is uplifted, eroded and buried again under new layers of sediments that may form the trap. Unconformities generally separate formations formed under very different depositional conditions. Study the attached figures in the context of the above discussion. This will improve your understanding of the basic concepts. Note the following important points in the last of the figures: (i) The oil and gas zones in the central block are not in direct pressure communication with the oil and gas zones in the left and right blocks. The oil and gas zones have different contacts. (ii) However, it cannot be concluded from the section alone that the central block is completely isolated from the rest of the formation. The blocks may be 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 and different 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 (natural water drive, gas cap expansion, solution gas drive and pressure depletion drive). Secondary Recovery - water or gas injection to maintain reservoir pressure (waterflooding and immiscible gas injection to supplement natural reservoir energy). Tertiary Recovery - enhanced oil recovery processes (EOR). An EOR process is any process which does a better job of recovering oil than conventional technology (primary and secondary recovery processes). In an EOR process 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 the world 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 oil resource.
1.2.2
Residual Oil is Trapped or By-passed
Oil which cannot be recovered using existing facilities or infrastructure (existing investment) is trapped (microscale) or by-passed (macroscale) - trapping of oil occurs 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 bypassing. 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 and immobile or residual oil. - Unrecovered mobile oil can be recovered by conventional processes by improved 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 recovery processes such as waterflooding and immiscible gas injection. This oil can only 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 be made 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 a solution-gas drive mechanism. (ii) Reservoirs having high closure and high permeability may experience the beneficial effects of gravity segregation of oil and gas. This may supplement the effectiveness of the original gas cap, or form a secondary gas cap. (iii) High permeability reservoirs in contact with an extensive aquifer will benefit to some degree from natural water influx (water drive). Material balance calculations may be used to determine the relative importance of the various natural drive mechanisms for a particular reservoir. Material balance calculations 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 possible in the development and production life of a reservoir cannot be over emphasized. A good data gathering program is a key component of an effective reservoir management 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. Useful estimates 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 on wells. Drill-stem tests on dry holes and structurally low wells (wells penetrating the aquifer) may yield important indications of aquifer permeability and continuity. Sometimes it is useful to complete dry holes as aquifer observation wells. Pressure data 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 used to assess the need for tertiary developments. Decisions regarding the actual implementation of EOR processes are not made until sufficient field data is available to 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 additional laboratory fluid and core data and the use of considerably more complex analysis techniques.
1.3
WHAT IS RESERVOIR ENGINEERING?
Reservoir Engineering is the science of understanding the production characteristics of oil and gas fields under primary (natural pressure depletion), secondary (water flooding, immiscible gas flooding) and tertiary (EOR) drive mechanisms. A basic understanding of reservoir engineering concepts is necessary for the planning, 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 formation or reservoir and vital physical parameters describing the reservoir internal structure and initial distribution of fluids. The reservoir engineer utilizes this data together with well production rates and measured reservoir pressures to analyze 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 initially contain(initial hydrocarbon in-place)?
2.
How much of the hydrocarbon initially in-place can ultimately be recovered?
3.
How will the production rates of wells depend on the physical parameters of the reservoir, reservoir geometry (shape), well number and development pattern and how will the rate decline with time?
The first question may be answered if the shape and size of the pay zone is known and 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 reservoir engineer may assist by estimating the positions of gas-oil and water-oil contacts if the positions of these contacts is unknown. The answer to the second question is very complex. It requires the use of sophisticated mathematical models (usually computer based reservoir simulators). These are required to model the many alternative developments which result in different ultimate recoveries and profitability. The process of producing production forecasts for the different recovery mechanisms and development plans will always be associated with some degree of uncertainty because geologic formations are highly heterogeneous and the geological description, no matter how sophisticated, can never capture all the geologic variability. The answer to the third question relies on detailed analysis of pressure and individual well test data.
22
Chapter 2 RESERVOIR DESCRIPTION The basis for a sound understanding of reservoir performance is a good description of reservoir geology (geological model) and the distribution of fluids (oil, water and 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 be represented 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 the reservoir, 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 depositional environment. Depositional conditions at any instant also vary from one location to another which results in lateral as well as vertical changes within the reservoir and 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 a measure 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 determining reservoir layering. Reservoir cross-sections based on logs can show if permeable zones are continuous throughout the reservoir or if they are lenses having limited areal extent. Logs also show the net sand at each well. Net sand is the rock that contains recoverable hydrocarbons and possesses permeability. A geologist should be consulted when analyzing logs for reservoir zonation and net sand thickness. Generally, logs are available for all wells drilled and represent the most complete set of reservoir descriptive data. Methods of analyzing logs and techniques of using logs for determining net sand thickness and evaluating sand continuity are beyond the scope of this course. Core analysis Core analysis data provide an additional basis for determining net sand thickness and reservoir zonation. Core analysis data is more quantitative than logs in describing 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. The logs can be calibrated by core data to make them more useful from a quantitative standpoint. Under some conditions, the calibrated logs can be used to obtain fairly good estimates of the porosity and permeability profiles at all wells in a reservoir. Stratification Stratification or heterogeneity occurs on all length scales in a reservoir. Although reservoir stratification is usually considered only in terms of net sand layers and impermeable streaks of sand or shale, stratification also exists within the net sand layers themselves and the degree of porosity and permeability can vary greatly within 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 and dolomite reservoirs locally may have porosities as high as 60 %. The porosity in reservoir rocks usually occurs in the spaces between rock particles and this is called primary porosity or intergranular porosity. Primary porosity can also exist in fractures and vugs. Primary porosity may be modified by post depositional events. For example, small mineral particles may be deposited in the space be25
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 greatly reduce the original porosity and permeability of a rock. Typically, the permeabilities of the net sand portion of a reservoir will vary from 10 md or less to 500 md or more. The ability of a rock to allow fluid flow is proportional to permeability, so fluid flow profiles in the reservoir may be very uneven. The process of reservoir description is sometimes referred to as formation evaluation 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 of fluids within these rocks.
(iii)
Interpretation of the above data for accuracy and reliability. These data are used to evaluate initial volumes of oil and gas in the reservoir
(iv)
Evaluation of potential sources of reservoir production energy. e.g. extent of 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.) and the collection of poor quality or inaccurate data is contradictory and wasteful. It is therefore necessary to place considerable effort into preparing an effective data gathering program.
2.1
RESERVOIR DESCRIPTION PROGRAM
The primary objectives of a data gathering program are to answer the following questions: (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 to supplement this energy by water or gas injection?
(iii)
What are the oil and gas reserves (production volumes for a particular field development)?
(iv)
How will individual well rates decline, and how can the decline in production rate 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 discovery 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, and this is integrated with the existing data to reduce the level of uncertainty associated with the initial estimates. The overall data gathering program is thus a continuing process over the life of the field. It is important to recognize this when developing the initial data gathering program.
2.2
SOURCES OF DATA
The following is a guide to the type of data which may be collected with the drilling 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 well should 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. The presence of multiple wells allows interference testing between wells to determine average permeabilities between wells and to test the continuity of individual sand units. Not all of the above data will be collected from each well drilled. The engineer must continually assess the data gathering process and only collect that data which materially reduces the level of uncertainty in estimating the important reservoir parameters. When areal variations in rock properties are large, it may be necessary to collect all 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 with the only opportunity to physically inspect a piece of the reservoir. It provides the 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 and reservoir 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 analysis data. This is because even in a heavily cored reservoir, the total cored volume constitutes 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 one well in the reservoir should be cored over the entire producing interval. The data obtained provides valuable information for describing vertical variations in reservoir 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 must be 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 purpose of logging is to identify a suitable completion interval for the well. A logging program should be designed to designed to provide the data required for reservoir description at a minimum cost. Lithology and borehole conditions must be considered in the selection of a suitable suite of logs. This will usually require a trial-and-error process of selecting the most effective combination of logs for a particular reservoir. Prior experience is a major fracture in the design of 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 valuable information 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 a constant rate for a predetermined period of time and then shutting-in the well and observing the rate at which the wellbore pressure rises or falls with time. An analysis of the pressure response allows an estimate of formation productivity and 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 pressure decline at nearby wells. Drill-stem tests are usually run at the time the well is drilled to determine if the well should be completed over the interval identified by the well log. The DST tool allows us to make a temporary completion to conduct pressure build-up and draw-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 the oil is produced to the surface the gas comes out of solution and the oil volume consequently shrinks. The oil that fills a barrel at surface conditions will have occupied 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 barrels occupied by a reservoir barrel of oil. Other important properties of oil, such as density and viscosity, also change when solution gas is released. These properties are very important in reservoir fluid flow calculations and must be accurately known. In large reservoirs having high closure, fluid properties may vary significantly both areally and vertically. Enough samples should be taken to adequately describe these variations. Reservoir fluid may be sampled in two ways: 1.
Subsurface — wireline samplers in the wellbore at the perforations or as part of the initial open-hole formation evaluation program.
2.
Recombined surface samples — oil and gas samples from the test separator recombined in the ratio of the produced gas-oil ratio.
2.3
INTEGRATED FORMATION EVALUATION PROGRAM
No single wireline tool or procedure is capable of providing all the data required to characterize a reservoir and its contents. It is therefore necessary to develop an integrated formation evaluation program which consists of a carefully considered mix 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 are taken in the development of such a program: 1.
A number of Key wells are selected to provide a representative coverage over the reservoir. A rule of thumb is that at least one well is needed for each 640 acres. For heterogeneous reservoirs this is reduced to 320 acres or less. 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 include logging, coring and well testing to provide complete reservoir coverage over the reservoir.
3.
Determining which logs provide the best quantitative data by comparing log interpretations with core data. Logs run in non-key wells are then interpreted according to the correlations developed for the key wells.
4.
Any available production data from key wells is also used to aide log interpretation.
5.
The above data gathering procedures are continuously monitored with the objective of eliminating any unnecessary or ineffective procedures.
The above steps ensure a data gathering procedure which is both simple and cost effective. A reservoir description program is needed during all phases of the producing life of a field. 1.
Early in the development of the field this information is required to achieve proper well spacing and productive completions.
2.
After field development the information is used for reservoir energy control 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 these processes it is necessary to have an accurate reservoir description and an estimate 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 4D high-resolution interwell seismic.
2.4
AQUIFER DESCRIPTION
The aquifer is the total volume of porous water-bearing rock in pressure communication with a hydrocarbon reservoir. The size and permeability of an aquifer will 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 by liquid or gas expansion within the reservoir. Reservoir drive mechanisms will be discussed in detail later in the course. At the present time, we are concerned only with 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 drilled outside the reservoir area. Logs and drill-stem tests on dry holes are about the only source of aquifer data. Logs will show the thickness of water-bearing rock that 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 aquifer permeability 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 the portion of the course entitled fluid flow. The aquifer can have a major effect on the production-pressure performance of a reservoir. This topic will be covered in detail in the section on water drive reservoirs.
47
Chapter 3 VOLUMETRICS AND INITIAL HYDROCARBON VOLUME The first step in a reservoir study is to accurately determine the initial hydrocarbon volume. This requires data which will allow us to calculate the size and geometry of the reservoir and the fluid volumes which the reservoir contains. In this section we will briefly look at how maps and cross-sections are used to describe the geometry of a reservoir and calculate the hydrocarbon volume in-place. Contour maps are commonly used to show reservoir geometry and the distribution of important reservoir parameters.
3.1
STRUCTURE MAPS
Structure maps show the geometric shape of a reservoir or formation. The maps may show the top or the bottom of a structure or reservoir unit. Examples of top of structure maps are attached. These maps also show the positions of fluid contacts in the reservoir. Structure maps are prepared by geologists. The data on which the maps are drawn 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 and the 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, most of the wells are not drilled to the base of the reservoir so a base structure map cannot be drawn. In this case, the gross sand interval must be estimated from reservoir cross-sections based on logs from the well which penetrated the entire interval. The gross pay isopach map for an oil reservoir is more descriptive of the hydrocarbon reservoir geometry than the gross thickness isopach.
3.2
ISOPACH MAPS
Isopach maps show the distribution and thickness of reservoir properties of interest. The contour lines connect points of equal vertical interval. Examples of common isopach maps are: Gross oil thickness isopach map: contours gross pay - the depth of the top of the oil column minus the top of the bottom of the oil column. Net oil thickness isopach map: contours net pay - gross pay minus nonreservoir intervals such as shales. Net oil isopach maps are commonly used to calculate volumes of hydrocarbons in-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 contributes to recovery. Iso-porosity map: contours average porosity over net-pay portions of the desired formation. Iso-water saturation map: contours average water saturation over net-pay portions 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 DETERMINING ORIGINAL OIL-IN-PLACE
The original oil-in-place (OOIP) contained in a reservoir expressed in Stock Tank Barrels (STB) is designated by the symbol N and is given by the equation (field units): 7758Vb φ(1 − S wi ) N= B oi where,
Vb φ S wi B oi
= = = =
reservoir bulk volume (acre-feet) average porosity (fraction) average initial water saturation (fraction) average initial oil formation volume factor (RB/STB)
or, in any set of self consistent units: N=
Vb φ(1 − S wi ) B oi
In this and the following reservoir engineering courses it is assumed that you are familiar with units and unit conversions i.e., starting with the above equation for any set of self consistent units, you should be able to calculate the constant 7758 in the preceeding equation. If you wish to review the topic of units go to the section Units - unit conversions on page 108 of Dake, Fundamentals of reservoir engineering. The original gas-in-place (OGIP) contained in a reservoir, expressed in Standard Cubic Feet (SCF), is designated by the symbol G and is given by the equation (in field units), 7758Vb φ(1 − S wi ) G= B gi where,
B gi
= average initial gas formation volume factor (RB/SCF)
or, in any set of self consistent units: G=
Vb φ(1 − S wi ) B gi 53
Figure 3.6: Schematic of the microstructure of a sandstone showing sand grains and 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 structure maps 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 the bottom of the reservoir zone. 2.
Construct a gross sand isopach map by removing intervals containing only 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 rock intervals (shale, siltstone, coal seam etc.). This procedure will normally involve estimates of minimum porosity and permeability cut-offs. The actual intervals to be excluded are picked from porosity 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 porosity values. These are determined from core data and from sonic and radioactivity logs 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 saturation. This requires a knowledge of the wettability state of the reservoir and special core test data.
2.
Coring with oil-based muds. Provided that invasion of drilling fluids has not changed reservoir wettability, coring with oil-based mud can result in core which gives a good indication of the irreducible water saturation. This corresponds to the water saturation in the reservoir above the transition zone. In the transition zone the core will indicate low water saturation because the oil from the oil-based mud will have displaced some of the mobile water from the core during the coring operation.
3.
Restored state tests. These tests attempt to restore the wettability of core in the laboratory to that in the actual reservoir. The restored state core is then used in the laboratory to duplicate the displacement processes by which the reservoir water saturation was initially established.
55
The data gathering program should include water saturation data determined by all 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 calibration of log resistivity. Laboratory measurements on restored state core also provide capillary pressure data which can be used to calculate water saturations in the transition zone. This data is also used to calibrate logs. The overall objective is to collect sufficient data to develop meaningful correlations between log measured resistivity and water 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 or gas filled hydrocarbon volume to the equivalent volumes at surface or stock tank conditions. Oil and gas formation volume factors are determined in laboratory PVT tests conducted with representative samples of reservoir oil and gas. Estimates of oil formation volume factors may be obtained from empirical correlations 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, we consider the very simple case of a homogeneous reservoir with unity net-to-gross ratio where the bottom of the reservoir is the horizontal surface formed by the water-oil contact (WOC). For these conditions a top-of-structure map and average porosity and water saturation are all that is needed to calculate the initial or original oil-in-place.
56
Figure 3.7: Schematic of interstitual, irreducible or connate water in a waterwet porous medium
57
Figure 3.8: Reservoir pressure survey map used to determine average reservoir pressure
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-ofstructure map attached). The average porosity, average connate water saturation and average oil formation volume factor have been determined and these values are given below. Data: φ S wi B oi
= 0.27 = 0.38 = 1.2715 (RB/STB)
We use this exercise to introduce you to Mathcad. It is assumed that you have installed Mathcad on your PC and that you have worked through the basics in the on-line tutorial. In order to see the solution to this exercise you will need to change some of the inputs to the spreadsheets. The areas of the spreadsheet where 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. The contour 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 example I have digitized each of the contours by hand and used Simpson’s rule to numerically calculate the area. The use of Simpson’s rule for numerical integration is shown in the attached figures. Study the three figures to be sure that you understand what is being done. To determine the area of a depth contour I first printed an enlarged copy of 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 the attached figure and read-off the (x, y) points defining each curve. These points 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 points are entered for the lower and upper contours. Note that the end-points for the lower and upper contours are the same. [Contour.mcd] , also located in the [Contours] folder, reads the contour files and calculates the areas, and plots the contour. This is done in the manner outlined in the course notes 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 as a numerical input. This makes it easy to annotate or ”pretty-up” the input file so that it is understandable to the casual reader. You can use as many data points as you like to define a contour - the more points you use, the smoother the contour and the more accurate the area calculation. The only limitation is that the total number of points defining the lower and upper contours must be the same. This makes it a little easier to program the calculations. When you open [Contour.mcd] you will see that I have left it doing one contour over and over. You will need to change the other input contour file names to see the areas and plots for each of the contours. The areas within each contour, A0 , A1 , A2 , etc. are multiplied by the map scale-factor squared to convert the map areas to actual field values.
60
2. The bulk volume between two successive contours is given by the pyramidal formula: µ ¶ q h Ai + Ai+1 + Ai Ai+1 ∆Vi = 3 where, ∆Vi Ai Ai+1 h
= = = =
bulk volume between contours i and i + 1. the area enclosed by the lower contour. the area enclosed by the upper contour. the vertical height between the contours, or the contour interval.
3. The initial volume of oil-in-place in volume element ∆Vi ,in any consistent set of units (as used by Mathcad), is given by, ∆Ni =
∆Vi φi (1 − S wc,i ) B oi,i
4. The initial oil-in-place is the sum of the values for each volume element. N=
X
∆Vi
i
See the attached figures which show how Simpson’s rule is used to calculate contour 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
yi
yi+ 1
yi
AU
yi+ 1
AL ∆x
∆x A=
X
yi − yi+ 1 ∆x 2
Figure 3.11: Computer calculation of contour areas
62
Contour Area
AU
AC = AU − AL
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 was 1500 psia, the initial reservoir temperature was 200o F, and the gas compressibility factor was 0.877 ? All other data as for Exercise 3.1. [Answer: 223 BSCF]
Solution hint The gas formation volume factor is given by, B g = 0.00504
zT P
[RB/SCF]
where T - the average reservoir temperature - is in o R (o R=o F +460) and P the average reservoir pressure - is in psia. Use this equation to calculate the gas formation factor. You can use the areas, volumes etc., calculated in the previous exercise to help you 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 Mathcad you can modify the files used in the previous example to do the calculation. If you choose to do this make sure that you have looked at the Units section of the on-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-structure map attached). The average reservoir and fluid properties are given below. Data: φ S wi T P z
= = = = =
0.21 0.19 210o F 2300 psia 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 commercial or in-house software to which you may have access. If you have been following the Mathcad route - I recommend this strongly - this problem will bring you right up the learning curve. Print a full-size (or larger) copy of the Marlin structure map. Create a full copy of the folder for Example 3.1 and in order to modify the files to solbe the present problem. Follow the outline for Example 3.1 and construct the data files for contours (there are 9 of these). Pay special attention to the particular manner in which these files are constructed! Modify the [Contour.mcd] by copy-and-paste to calculate the areas for all nine contours. Modify the final graph which plots the contours to plot all nine contours. You will probably need to revise the ’graphs’ section of the on-line manual to do this. You will need to make the same changes to [Contour.mcd] made for Problem 3.1 to convert it to a gas in-place calculation. Note that for the Marlin contours the contour depth intervals are not all the same size. The best way to handle this is to read-in individual values of h in the same way as we already read-in individual 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 appropriate changes 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 the array elements hi . Describing these changes is a lot more difficult than making them! If you succede in making the above changes you will have learnt all you need to know 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 hydrocarbon content of a reservoir. Material balance calculations: 1.
Require some production history (reservoir pressure and fluid production rates with time). The more production history available, the greater the accuracy of the calculation.
2.
The method is most precise for reservoirs where water influx is not significant. The method is most accurate for gas-cap-drive reservoirs if the pressure drop is at least 100 psi.
The material balance method is the preferred method for determining the original gas-in-place (OGIP) for gas reservoirs with no significant water influx. For reservoirs where water influx is significant the analysis procedure is as follows: 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 to characterize the size and strength of the aquifer.
4.
The Material Balance Equation is then used to predict future water influx and 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 the discovery well.
2.
Initiate a program for periodic reservoir pressure measurement immediately after commencement of production. This program (on an annual basis) should be continued throughout the producing life of the field. 67
3.
Accurately measure and diligently report the production volumes for all the fluids produced. Although produced oil volumes are accurately measured as a matter of routine, material balance calculations require equally accurate water and gas volumes.
4.
Accurately determine fluid PVT data. This requires representative oil and gas samples at reservoir conditions. The best oil samples are obtained before the reservoir is produced because the pressure drop which accompanies production may result in the release of solution gas and this may preclude the possibility of obtaining a representative fluid sample. Gas sampling does not have this problem.
3.5
ESTIMATING RESERVES
Reserves for a particular field are estimated from an analysis of the available engineering data. The implication of this is that the more reliable the data (production history data), the more accurate will be the estimate of reserves. 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 estimate reserves. This will usually involve the use of numerical reservoir simulators. Rock property data (porosity and permeability), fluid property data, reservoir description and production performance data (pressure decline and fluid production rates) are the basis for the analysis. Collecting all production data from early in the life of a field is ,again, of critical importance.
68
3.6
ESTIMATING DECLINE IN OIL PRODUCTION RATES
The decline in field and individual well production rates is estimated on the basis of: 1.
Decline curve analysis. Decline curve extrapolation is satisfactory only for wells which are produced continuously at maximum or near maximum rates. For wells producing at limited draw-down (below their maximum rate) extrapolated decline rates are not meaningful. Decline curve analysis are studied in detail in course PTRL6007 - Reservoir Engineering II.
2.
Productivity index method. The productivity index for a well is the well production rate per psi of pressure draw-down. This is determined from well flow tests. The productivity index is usually a function of flow rate and reservoir pressure (oil relative permeability, oil viscosity). Well flow tests must therefore be conducted at different rates and the productivity index must be adjusted for falling reservoir pressure. For wells not subject to severe water and/or gas coning this method provides reliable estimates of production decline rates.
3.
Reservoir simulation. For wells subject to severe water and/or gas coning reservoir simulation offers the most reliable method for predicting well decline rates and overall reservoir performance. Reservoir engineering is studied in detail in course PTRL6004 - Reservoir Simulation.
69
Chapter 4 HYDROSTATIC PRESSURE DISTRIBUTION IN RESERVOIRS This section deals with the hydrostatic or initial pressure distribution in a reservoir - 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 pressure data.
4.
Calculation of pressure kicks on penetrating a hydrocarbon zone.
5.
Locating exploration wells searching for oil in a water- or gas-bearing formation.
70
Symbols and Units g D p ρ Dowc Dgoc Dgwc
gravity constant depth to subsurface pressure density depth of initial oil-water contact depth of initial gas-oil contact depth of initial gas-water contact
g 1 Pascal (Pa) 1 bar 1 Atmosphere
m/s2 m N/m2 kg/m3 m m m
= = = =
9.81 m/s2 1 N/m2 105 N/m2 1.01325 bar or 14.695 psia
× × × ×
16.02 6.9 0.069 3.28
Conversion factors lb/ft3 psi psi ft
= = = =
kg/m3 kPa bar m
All of the above units may be embedded into working Mathcad files and can be used interchangeably or you can even mix units if you want. Normally we would not elect to mix units!
4.1 4.1.1
SUBSURFACE PRESSURES Water zone pressures
In normally pressured sedimentary basins water contained in the pore space of a reservoir (connate water) is in pressure communication with the atmosphere through the oceans or outcrops. Since the pore spaces which comprise the effective porosity of the reservoir are all interconnected, and the reservoir and associated aquifer are initially in static equilibrium, pressure varies only with depth. The difference in hydrostatic pressure between any two depth points in a formation containing water is, p2 − p1 = ρw g(D2 − D1 ) 71
where, ρw is the average density of water over the interval (D2 − D1 ) and g is the gravitational acceleration. ρw varies with depth (pressure) and salinity. Since water is only slightly compressible, it is customary to neglect the overbar symbol and simply write, p2 − p1 = ρw g(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, the equation may be written as, pwD − p0 = ρw gD or, pwD = p0 + ρw gD The term ρw g(D2 − D1 ) is the hydrostatic pressure difference which has units of pressure. If we divide this pressure difference by (D2 − D1 ), we obtain the water gradient, γw , which is equal to ρw g and has units of pressure per unit depth. In general we define a phase pressure gradient as γi = ρi g 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 difference between 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 of the oil pressure at the initial OWC (a commonly used datum point for reservoir engineering calculations) as, poD − poDOW C = ρo g(D − DOW C ) 72
Figure 4.1: Hydrostatic pressure distribution in a reservoir
which may be rearranged to give, poD = poDOW C − ρo g(DOW C − D) or, poD = poDOW C − γo (DOW C − 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 terms of the gas pressure at the initial GOC as, pgD − pgDGOC = ρg g(D − DGOC ) or, pgD = pgDGOC − ρg g(DGOC − D) or, pgD = pgDGOC − γg (DGOC − D) where ρg is the average density of gas at reservoir conditions. 73
4.2
HYDROSTATIC PRESSURE DISTRIBUTION IN A RESERVOIR CONTAINING OIL, WATER AND GAS
In the absence of capillary pressure, the WOC and GOC are sharp and the fluid pressures at the contacts are equal. The resulting hydrostatic pressure distribution 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 pressuredepth data.
74
Figure 4.2: Hydrostatic pressure distribution in a reservoir assuming negligible capillary pressure
75
Example 4.1 [REGPRES.mcd]
-
Calculation
of
regional
hydrostatic
pressures
Five widely separated exploration wells were drilled in a large sedimentary basin. While penetrating the aquifer, the following pressure measurements were made in each of the wells: Well A B C Rotary table elevation (ft above MSL) 2133 1312 2707 Measured depth (ft) 9744 13993 8235 Gauge Pressure (psia) 4340 5656 2480
D 3937 17388 6005
E 197 9383 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 DT V D , measured depth DM D , and depth above mean sea level DM SL is given by DT V D = DM D − DM SL The average water gradient (sea-level to depth of measurement), γw , is given by γw =
∆p DT V D
76
Example 4.2 - Calculation of fluid contacts and fluid gradients from pressure data [CONTACTS.mcd] The following formation tester pressure measurements were made in an exploration 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 2912 8202 2920 8284 2927 8366 2950 8448 2978 8530 3005 8612 3038 8694 3075 8776 3114
Solution hints Plot the data on a depth-pressure plot. Identify lines of constant slope, these show zones of different fluid saturation. Determine the gradients of depth-pressure lines and from these calculate the fluid densities. The densities identify the fluids. [CONTACTS.mcd] is a general spreadsheet which I have found very useful in analysing a wide range of formation pressure-depth data. If you have access to commercial or in-house software which does similar things, feel free to use it. [Answer: DGOC = 8302 ft ss., DOW C = 8573 ft ss., ρg = 13.2 lb/cubic ft., ρo = 48.3 lb/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 vertical thickness of 200 ft between 6560 and 6760 ft-subsea. A wireline formation tester recovered 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-ss and 7980 ft-ss but found it to be only water-bearing. Mechanical problems with the downhole gauge prevented a pressure measurement from being taken. The two wells are 12,000 ft apart. On the basis of previous experience and sampling of reservoir fluids we have determined that the regional water gradient is 0.524 psi/ft. The formation is normally pressured. The pressure gradient for the gas phase at reservoir conditions is calculated to be 0.084 psi/ft. The oil gradient, at reservoir conditions, is estimated to be 0.296 psi/ft. Questions (i)
Is the above data consistent with the assumption that the formation could be oil-bearing between 6660 and 7880 ft-ss ?
(ii)
Where would you locate an additional exploration well which would definitely 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 a mechanical problem prevented the measurement of a water phase pressure, we can 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 the depth of highest known water. If an oil zone exists, it must be between these depths. Knowing the gas and water gradients, and a fluid pressure and corresponding depth 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 be found from the intersection of the gas and water gradient lines. If a drop of oil is present it must gravitate to this depth splitting the GWC into a WOC and a GOC. The third well must intersect this depth to prove if any oil is present in the reservoir. The maximum possible thickness of the oil zone occurs when either the lowest known gas or the highest known water corresponds to an oil contact. How do we select which of the two is a possible contact? Simple, draw the oil gradient line through 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 known water. If you understand the principles involved and the steps required to solve the problem you should have little problem in entering the correct data into [EXPLOR.mcd] and manipulating the graph by setting appropriate contact and oil gradient lines to produce the correct answer. 80
Problem 4.2 - Calculation of a Pressure Kick when Penetrating a Hydrocarbon Zone [PKICK.mcd] Calculate the pressure kick to be expected when a well first penetrates a reservoir at 5000 ft subsea if the reservoir is known to have an OWC at 5500 ft and an GOC at 5200 ft. The formation is known to be normally pressured with the following phase gradients; water gradient 0.45 psi/ft oil gradient 0.35 psi/ft gas 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 bit experiences a water hydrostatic pressure gradient because the seal is saturated with water and in a normally pressured system this water is in hydraulic communication with the aquifer. Just before the bit penetrates the reservoir the bit pressure is the of the water gradient and depth. On penetrating the hydrocarbon column the reservoir pressure is equal to the pressure at the contact minus the gravity head of hydrocarbon column from the contact to the top of the reservoir. Since a hydrocarbon column is lighter than a water column of equivalent height, the reservoir pressure at the top of the reservoir is higher than the pressure at the base of the seal. This difference gives rise to a pressure kick. You do not really need a spreadsheet to solve this problem. If you know what you are doing you can solve the problem faster with a hand calculator! I have prepared the spreadsheet [PKICK.mcd] as an example of how a simple calculation can be layed 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 made in a single trip in an open hole. This allows the engineer to determine pressuredepth profiles across reservoir sections of interest. Comparing pressure-depth profiles in different wells can provide valuable information on field fluid contacts and 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 the structure and encountered only the aquifer. The operator had the foresight to measure water pressures over a 160 foot interval to determine the water pressure regime in this new area. The second Well-B, was drilled up-structure several kilometers from Well-A and discovered a 50 foot oil column with good porosity and permeability. Six RFT pressures were measured over this interval.
82
The pressures recorded for the two wells are summarized below. Note that in order to meaningfully compare the pressures in the wells it is necessary to convert measured depths to a common datum (depth sub-sea) and to express pressures in absolute units (psia). Well-A Depth (ft.ss) 6075 6091 6108 6220 6232
Well-B Depth (ft.ss) Pressure (psia) 5771 2602 5778 2604 5790 2608 5800 2610 5806 2612 5816 2615
Pressure (psia) 2662 2669 2677 2725 2731
What conclusions can you draw from this data with regard to the fluids, the reservoir 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. On the average it is about 1o F/100 ft. We can use this to estimate subsurface formation (reservoir) temperatures, Tf , as, Tf = Ts + GT Df 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 reservoir pressure. When sufficient production has taken place for the pressure to fall below the oil bubble point pressure, gas initially dissolved in the oil is released from solution. The volume of solution gas released and the associated changes in oil properties have a significant influence on the production characteristics of the reservoir. The attached 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 representative reservoir fluid samples. The properties are called PVT (pressurevolume-temperature) properties of the reservoir fluids. Before considering the specific PVT parameters of interest to reservoir engineering it is necessary to review 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 substance exists as a gas, a liquid or a solid (or a mixture of liquid and gas or a mixture of all three) depends on: -
Pressure.
-
Temperature.
-
Composition
5.1.1
Pure Hydrocarbons
For a pure hydrocarbon such as ethane, propane, butane, methane etc., temperature and pressure fully define the state of the hydrocarbon. The state of a pure substance can be represented on a P-T phase diagram. The P-T phase diagram can be constructed by experimentally observing the behavior of the pure substance in a windowed high pressure test cell as shown in the attached figure. Over the pressure and temperature range shown in the attached figure, the vapor pressure curve, which terminates in a critical point (C), divides the P-T space into liquid and gas regions. To the right of the critical point lies a single phase super critical region. A gas at a temperature above it’s critical temperature, may be compressed indefinitely without 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 temperature can result in a phase change from liquid to vapor when the pressure falls 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 of the line AF and the vapour pressure curve) both liquid and vapour phases can 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 constant temperature can result in a phase change from vapor to liquid when the pressure increases to the vapor pressure.
-
For pure substances at temperatures above the critical temperature changing pressure (states E to D) at constant temperature does not result in a phase change.
-
For a gas state at temperatures below the critical temperature, state A, increasing temperature at constant pressure (states A to B) does not result in 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 behavior of more complex mixtures is similar. The attached figure shows the P -T phase diagram for a 50:50 mixture of propane and heptane. Also plotted are the pure component vapor pressure curves. The major differences between the P -T phase diagrams for a single component and 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 pressures over which two phases can coexist at constant temperature.
2.
The two-phase region is bound by a bubble point line and a dew point line which meet at the mixture critical point. The bubble point pressure is the pressure at which the first bubble of gas appears as the pressure of a liquid is reduced at constant temperature. The dew point pressure is the pressure at which the first drop of liquid appears 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 properties of the liquid and vapor phases are identical. We note that for mixtures, liquid and gas phases mixtures may exist at conditions above the critical point.
5.
The size and shape of the two-phase region and the location of the critical point 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 in the 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 along the line A’ABC. The reservoir is said to be undersaturated because the initial reservoir pressure is higher than the bubble point pressure which is at point A. As the pressure continues to fall below the bubble point pressure along the line ABC, 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 liquid volume 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 critical point.
Retrograde Condensate Gas Reservoir The phase diagram for a typical retrograde condensate gas reservoir having a temperature T is shown in the attached figure. When pressure falls this reservoir enters the two-phase region through a dew point producing a condensate phase in the reservoir. This behavior is termed retrograde. For a retrograde condensate gas reservoir T , the reservoir temperature, lies to the right 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 shown in 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 falls in 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 engineering to characterize oil and gas reservoirs. Parameter
Symbol Definition
Oil Formation Volume Factor
Bo
The reservoir volume in barrels (RB) that is occupied by one stock tank barrel (STB) of oil and its dissolved gas.
Gas Formation Volume Factor
Bg
The reservoir volume in barrels (RB) that is occupied by one standard cubic foot (SCF) of gas.
Solution Gas-Oil Ratio
Rs
The volume of gas in SCF dissolved in one STB of oil at a specific reservoir pressure.
Oil Compressibility
co
The fractional reduction in oil volume that results from a pressure increase of one psi.
Water Compressibility
cw
The fractional reduction in water volume that results from a pressure increase of one psi.
Bubble Point Pressure
pb
The pressure at which the first bubble of gas 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 volumes. Calculation of underground withdrawal rates from surface measurements Oil and gas production rates, Qo (STB/D)and Qg (SCF/D), are measured at the surface at a time when the reservoir pressure is P . This pressure is below the bubble point pressure Pb . What are the corresponding underground withdrawal rates in (RB/D)?
The producing gas-oil ratio is R = Qg /Qo (SCF/ST B) This means that for every STB of oil which is produced at this pressure, we produce 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 rates to reservoir withdrawal volumes Each STB of oil produced to the surface requires the withdrawal of Bo RB of oil and 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 2.125 MMSCF/D. If the reservoir pressure at this time is 2400 psia, what are the underground withdrawal rates? If the density of stock-tank oil is 52.8 lb/ft3 and the gas gravity is 0.67 (density of air at standard conditions is 0.0763 lb/ft3 ) calculate the oil density and gas density 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 97
Bg =0.00119 RB/SCF
[Answer: qo = 2956 RB/day, qg = 1482 RB/day, ρo =47.4 lb/cubic ft., ρg =7.65 lb/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. Use the PVT properties and the surface densities to calculate the corresponding fluid masses at reservoir conditions. Enter the correct numbers into the spreadsheet [SUBSURV.mcd] and see the solution. Study the spreadsheet carefully in order to fully understand the calculation of densities at reservoir conditions.
5.2.1
Pressure Dependence of PVT Properties
Typical relationships between PVT properties and pressure are shown schematically in the attached figure.
Solution GOR The volume of dissolved gas decreases from Rsi , the initial solution GOR at the bubble 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 greater than the standard reference temperature of 60o F (oil expands with increasing temperature at constant pressure). Bo increases with pressure to Pb as a result of the increasing solubility of gas (the mass 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 increases.
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 real gas law. P V = znRT where, P V z n R T
= = = = = =
pressure, psia volume, ft3 gas deviation factor, dimensionless number of moles, lb mole gas constant, 10.73 psia ft3 /o R lb mole temperature, o R (o R =o F + 460)
Taking one mole of gas at standard conditions (14.65 psia and 60 o F) and compressing it to reservoir conditions (PR psia and TR o R) gives, 14.65V1 = z1 nR(60 + 460) PR VR = zR nR(TR ) Dividing the above equations we obtain, VR 14.65 zR TR = V1 520 z1 PR Since Bg has the units RB/SCF (1 bbl = 5.615 ft3 ) and z1 = 1.0, Bg =
14.65 VR zR TR = 5.615 5.615 × 520 1 PR
Bg = 5.02 × 10−3
zR T R PR
The compressibility factor, zR , for natural gas may be estimated from the attached figure where the compressibility factor is plotted as a function of pseudo-reduced temperature (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-critical pressure respectively. The pseudo-critical properties of a gas are a function of gas gravity and may be estimated from the attached figure.
Gas gravity Gas gravity is the ratio of the density of a gas to the density of air at standard conditions. At standard conditions gases obey the ideal gas law and gas density may be expressed as; P V = nRT The number of moles n, may be expressed in terms of the mass of gas m, and the molecular weight of the gas M , m n= M Combining the equations gives, PV =
m RT M
or, ρg =
m P Mg = V RT
ρair =
m P Mair = V RT
Similarly, the density of air is,
The gas gravity, γg is thus simply, γg =
ρg Mg = ρair 29
The molecular weight of the gas is determined from a laboratory analysis of a gas sample.
101
Figure 5.13: Pseudo-critical properties for natural gas and gas condensates
102
5.3 5.3.1
CALCULATION OF GAS PROPERTIES Single Gas Component
For a single gas component, the reduced temperature, Tr , and reduced pressure, Pr , are defined as follows: T Tr = Tc P Pr = Pc where, T Tc P Pc
= = = =
temperature of interest, o R critical temperature of the gas, o R pressure of interest, psia critical pressure of the gas, psia
Tc and Pc are constants for a single gas component and may be obtained from standard tables of the properties of pure gas components. Once Tr and Pr are known, the compressibility factor, z, may be read from compressibility charts and Bg in units of RB/SCF is calculated from: Bg = 5.02 × 10−3 where T is in o R and P is in psia.
103
zT P
CRITICAL PROPERTIES OF GASES Component
Molecular Weight
Tc R
Pc psia 673 712 617 551 485 435 397 362
o
Methane Ethane Propane n-Butane n-Pentane n-Hexane n-Heptane n-Octane
16.04 30.07 44.09 58.12 72.15 86.17 100.2 114.2
344 550 666 766 847 914 972 1025
Water Carbon dioxide Nitrogen Hydrogen sulphide Air
18.02 44.01 28.02 34.08 28.90
1365 3206 548 1073 227 492 673 1306 239 547
104
5.3.2
Multi-Component Gas Mixtures
A multi-component gas mixture may be treated as a single pseudo-component gas by defining appropriate pseudo critical temperature, Tpc , and pseudo critical pressure, Ppc : Tpc = Ppc =
n X 1 n X
yi Tci yi Pci
1
where, yi Tpc Ppc
= mole fraction of component i in the mixture = pseudo critical temperature of the gas, o R = pseudo critical pressure of the gas, psia
The pseudo reduced temperature, Tpr , and pseudo reduced pressure, Ppr , are defined as follows: T Tpr = Tpc P Ppr = Ppc Once Tpr and Ppr are known, the compressibility factor, z, may be read from compressibility charts and Bg calculated from: Bg = 5.02 × 10−3
zT P
where T is in o R and P is in psia. Gas gravity known If the gas gravity is known, the chart may be used to estimate the gas pseudo critical properties. Note: the gas gravity may be calculated from gas composition. Gas molecular weight, Mg Mg =
n X
yi Mi
1
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 of 225o F and 2500 psia. The gas composition is: Component Mole Fraction Methane Ethane Propane
0.80 0.15 0.05
Solution Component Mole Fraction Tc (o R) Tc (o F ) Pc (psia)
o
Methane Ethane Propane
0.80 0.15 0.05
TOTAL
1.00
-117 90 206
343 550 666
673 708 617
yTc
yPc
274.8 82.5 33.3
539 106 31
390.6
676
R =o F + 460 1. Tpc = 391o R Ppc = 676psia 2. Tpr =
T 460 + 225 = = 1.75 Tpc 391
Ppr =
P 2500 = = 3.70 Ppc 676
3. From the generalized compressibility chart, read z = 0.87 4. Bg = 5.02×10−3
zT (0.87)(225 + 460) = 5.02×10−3 = 1.2×10−3 RB/SCF P 2500 106
Example Calculation - Mg , γg and ρg Calculate the gas molecular weight, gas gravity and the gas density at reservoir conditions of 225o F and 2500 psia. The gas composition is: Component Mole Fraction Methane Ethane Propane
0.80 0.15 0.05
Solution Component Mole Fraction Mi Methane Ethane Propane
0.80 0.15 0.05
TOTAL
1.00
yi Mi (o F )
16 30 44
12.83 4.51 2.20 19.54
1. Mg =
3 X
yi Mi = 19.54
1
2. γg =
Mg 19.54 = = 0.675 29 29
3. ρg =
P Mg 2500 × 19.54 = = 7.64 lb/f t3 zRT 0.87 × 10.73 × (460 + 225)
107
Calculation of Gas Properties Calculate γg , Bg and ρg at reservoir conditions of 150o F and 1500 psia. The gas composition is: Component Mole Fraction Methane Ethane Propane n-Butane
0.70 0.12 0.10 0.08
Solution Component Mole Fraction Tc (o R) Tc (o F ) Pc (psia)
o
Methane Ethane Propane n-Butane
0.70 0.12 0.10 0.08
TOTAL
1.00
-117 90 206 306
343 550 666 766
673 708 617 551
yTc
yPc
240.1 471.1 66.0 85.0 66.6 61.7 61.3 44.1 434.0 661.8
R =o F + 460 1. Tpc = 434o R Ppc = 662psia 2. Tpr =
T 460 + 150 = = 1.406 Tpc 434
Ppr =
P 1500 = = 2.266 Ppc 662
3. From the generalized compressibility chart, read z = 0.75 108
4. Bg = 5.02×10−3
zT (0.75)(150 + 460) = 5.02×10−3 = 1.53×10−3 RB/SCF P 1500
Component Mole Fraction Mi Methane Ethane Propane n-Butane
0.70 0.12 0.10 0.08
TOTAL
1.00
16 30 44 58
yi Mi (o F ) 11.2 3.6 4.4 4.6 23.8
5. Mg =
3 X
yi Mi = 23.8
1
6. γg =
Mg 23.8 = = 0.82 29 29
7. ρg =
P Mg 1500 × 23.8 = = 7.29 lb/f t3 zRT 0.75 × 10.73 × (460 + 150)
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 reservoir oil at a constant reservoir temperature. The laboratory tests are conducted in high pressure windowed cells. These cells allow accurate visual measurement of oil and gas volumes in the cell at different pressures. The pressure in the cell is lowered by increasing the cell volume. This is achieved by removing accurately measured volumes of mercury from the cell using a calibrated high precision piston pump. The attached figure shows the essential features 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 raise the cell pressure above the initial reservoir pressure, pi . The cell pressure is lowered in small steps by removing a small volume of mercury at each step. The total hydrocarbon volume, vt , in the cell is measured at each step. All cell volumes are reported relative to the volume in the cell at the bubble point pressure, pb . The following table shows a set of typical laboratory data for this test. Pressure (psia)
Relative Total Volume vt = v/vb (RB/RBb )
5000 4500 4000 3500 3330 3290 3000 2700 2400 2100
0.9810 0.9850 0.9925 0.9975 1.0000 1.0025 1.027 1.0603 1.1060 1.1680
pi pb
This data is used to determine the bubble point pressure, pb , by noting that the total compressibility of the cell increases dramatically when gas is released from solution.
5.4.2
Differential Liberation Test
This test is designed to approximate the conditions in a reservoir where gas is released from solution as the pressure falls and the released gas moves away from the oil as a result of gravity segregation. experiment starts with the cell at the bubble point pressure. The pressure in the cell is reduced in small steps by removing mercury from the cell. 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 constant pressure. The displaced gas volume is measured at standard conditions and the volume, 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 table below: Pressure (psia) 4000 3500 3330 3000 2700 2400 2100 1800 1500 1200 900 600 300 14.7 14.7
Relative Gas Vol. Relative Gas Vol. Relative Oil Vol. at p and T , vg at sc, Vg at p and T , vo
pb 0.0460 0.0417 0.0466 0.0535 0.0597 0.0687 0.0923 0.1220 0.1818 0.3728
8.5221 6.9731 6.9457 6.9457 6.5859 6.2333 6.5895 6.4114 6.2369 6.2297
200o F 60o F
0.9925 0.9975 1.0000 0.9769 0.9609 0.9449 0.9298 0.9152 0.9022 0.8884 0.8744 0.8603 0.8459 0.8296 0.7794
In addition to the above data, the laboratory also reports the Cumulative Relative Gas Volumes collected at atmospheric pressure and reservoir temperature and at STC: Cumulative Relative Gas Volume at 14.7 psia and 200o F=74.9557 Cumulative Relative Gas Volume at 14.7 psia and 60o F=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 the separator system which may be a single separator or a series of staged separators with staged pressures and temperatures to simulate the actual production system. The cell contents are then flashed through the separator system to standard conditions. The resulting volumes of produced oil and gas are measured. The data from such tests is used to determine the oil shrinkage factor, cbf , and the initial solution GOR, Rsif . Separator Stock Tank Shrinkage GOR Pressure Pressure Factor Rsif (psia) (psia) cbf (STB/RBb )) (SCF/STB) 200 150 100 50
14.7 14.7 14.7 14.7
0.7983 0.7993 0.7932 0.7834
116
512 510 515 526
5.4.4
Procedure for calculating PVT parameters from laboratory data
The above data can be used to compute the PVT properties for the tested oil and gas. The procedure involves the following steps: 1.
Plot the flash expansion data, p vs vt , to determine the bubble point pressure.
2.
Use the differential flash data to compute Bg . Eqns.(4.4—4.6) may be manipulated to yield, 1 vg Bg = 5.615 Vg
3.
Use the differential flash data to compute the cumulative relative gas volume, F , liberated from the cell at every pressure step; Fp =
pb X
Vg
p
4.
Use the differential flash data and the oil shrinkage factor, cbf , to compute Bo , " # ∙ ¸ vo RB RB/RBb Bo = = STB cbf STB/RBb
5.
Use the differential flash data, the oil shrinkage factor, cbf and the cumulative gas liberated, F , to compute Rs ,
Cumulative Gas Liberated at pressure p ∙
¸
= Total Gas in - Gas in solution solution at pressure p ∙
¸
∙
¸
∙
STB SCF SCF 1 RBb F × 5.615 × = (Rsif − Rs ) RBb STB cbf STB STB
¸
The following example illustrates the application of the above procedure.
117
Example 5.2 - Calculation of PVT properties from laboratory PVT data [PVT.mcd] Using the Flash expansion, differential liberation and separator flash expansion data discussed in the previous sections, determine the following as a function pressure 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. The reservoir 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 truly representative sample of reservoir oil are the same. If the reservoir is undersaturated (reservoir pressure is above the bubble point pressure) and the well production rate is sufficiently low for the pressure in the vicinity of the wellbore to be everywhere above the bubble point pressure, the oil flowing into the wellbore is reservoir oil and the sample is representative. If the reservoir is saturated (initial reservoir pressure is equal to the bubble point pressure), the pressure in the vicinity of the wellbore will be below the bubble point 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 have the same solution GOR as the reservoir oil away from the well. When gas is first released in a reservoir, the gas remains immobile until the gas saturation increases to a value called the critical gas saturation. During the period that the gas saturation is below the critical value, oil flowing into the wellbore will be deficient in gas. On the other hand, when sufficient gas has been released for gas to become mobile, the very high mobility (low viscosity) of the gas will result in a 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 in the reservoir. Gas condensate tests are designed to measure this volume of liquid as a function of pressure - liquid dropout curve. The liquid volume is usually very small compared to the total volume of the system and in order to measure it accurately it is necessary to use special condensate PVT 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 cells are made out of a solid block of sapphire to allow un obstructed viewing of the liquid phase. The most common type of test used to characterize a gas condensate is the constant mass depletion test. Since the liquid volume produced in the reservoir is small, it is immobile and remains in contact with the gas from which it was produced. The constant volume depletion test simulates this behavior. In the case of very rich condensates where the volume of liquid produced is relatively large, the liquid phase in the reservoir may become mobile. The liquid will move to the bottom of the reservoir under the action of gravity and a constant volume depletion test is more appropriate. Typical liquid dropout curves for constant mass and constant volume depletion tests are shown in the attached figure. The richer the gas condensate, the more liquid is produced and the greater the difference between the liquid dropout curves.
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 gas flowlines, processing and measuring equipment, and pipelines. Hydrates consist of geometric lattices of water molecules containing cavities filled with gas - a solid foam. Gas hydrates form when hydrocarbon gases come into contact with water at temperatures below 35o C. A typical phase diagram for a hydrate forming gas-water system is shown in the attached figure. The temperature for hydrate formation is reduced by the addition of inhibitors such as methanol, ethanol, ethylene glycol diethylene glycol and triethylene glycol. The attached sheet provides a guide to temperature reduction which may be expected from the use of the different additives. More accurate predictions require the use of special equations of state designed specifically for this purpose.
121
Figure 5.21: Liquid drop-out curves for a gas condensate
Gas Hydrates Solid, semi-stable compounds which sometimes plug natural gas flowlines, processing equipment and pipelines Geometric lattices of water molecules containing cavities filled with gas Formed when hydrocarbon gases are in contact with water at temperatures below 35C Temperature for hydrate formation is reduced by the addition of inhibitors such as Methanol, Ethanol, Ethylene Glycol, Diethylene Glycol and Triethylene Glycol
Figure 5.22: A simple summary of gas hydrates
122
Figure 5.23: Gas hydrate co-existence curves. H-hydrate, G-gas, L1-water, L2liquid hydrocarbon, I-ice
Figure 5.24: Comparison between measured and calculated hydrate curves for a gas 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 analysis because it determines the manner in which fluids move through reservoir rock. Surface tension may exist between water and oil phases, water and gas phases and between gas and oil phases. Surface tension may be measured at reservoir conditions using the pendant drop method. A typical test setup is shown in the attached figure. The surface tension is given by, σ=
∆ρgDe H
where, ∆ρ g De , Ds H
5.8.1
= = = =
density difference for the two fluids gravitational acceleration drop diameters as shown in the attached figure tabulated function of Ds /De
Estimating Surface Tension
The surface tension between two fluids is a function of pressure, temperature and composition 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’s vapor, ρL − ρV σ 1/4 = Pch M where, Pch ρL , ρV σ M
= = = =
parachor phase densities, gm/cm3 surface tension, dynes/cm molecular weight
Parachors account for intermolecular forces and are predicted from the structures of molecules. A correlation for the parachor with molecular weight is given in the attached figure. For mixtures the surface tension is given by, σ 1/4 =
X
µ
Pchi xi
ρL ρV − yi ML MV
¶
where, Pchi xi , yi ρL , ρV σ ML , MV
5.9
= = = = =
parachor for component i in the mixture mole fraction of component i in the liquid and vapor phases phase densities, gm/cm3 mixture surface tension, dynes/cm phase molecular weights
CORRELATIONS FOR PROPERTIES OF RESERVOIR FLUIDS
Several useful correlations exist for estimating reservoir fluid properties in the absence 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 computerized which makes their use particularly easy. An example of such a program is 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 from a DST on an exploration well. What are the reservoir properties of the oil, water and gas? TR PR PS TS AP I GOR γg C N aCl
= = = = = = = = =
200 3330 115 60 30 510 0.75 100 6
o
F psia psi o F o API SCF/STB (air= 1) (water salinity in units of kg/m3 ) (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 atmospheric pressure for hydrocarbon gases
131
Problem 5.1 - Calculation of PVT properties from laboratory PVT data A laboratory PVT cell contained 280.0 cm3 of reservoir liquid at its bubble point of 2000 psia at 140o F. A volume of 18.8 cm3 of mercury was withdrawn from the cell and the pressure fell to 1600 psia. Mercury was then reinjected at constant temperature 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 temperature to 60o F. Then 0.388 SCF of gas were removed from the cell and 205.9 cm3 of liquid remained 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 good understanding of PVT properties.
132
Chapter 6 MATERIAL BALANCE EQUATIONS When a volume of oil is produced from a reservoir, the space originally occupied by the produced oil volume must now be occupied by something else. Unless fluid is injected, the production of oil must result in a decline in reservoir 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 of decline of reservoir pressure. Material balance equations express this dependence in mathematical form. The material balance equation is the basic reservoir engineering analysis tool used to examine past reservoir performance and to predict future performance.
The diagrams in this section are best viewed on the electronic version of the course 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 occupied 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 bubble point).
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 purposes and is not intended to imply that the materials in the reservoir are segregated in the manner shown. Under some conditions, however, gas released from solution will segregate to the top of the structure. Some of the released solution gas will always be evenly distributed throughout the reservoir, as are the expansions of rock 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 expand to replace some of the volume initially occupied by the produced oil. Gas cap expansion = |
where, G Gpc Bg Bgi
= = = =
{z
[RB]
}
(G − Gpc )Bg |
{z
}
−
gas volume at lower pressure
GBgi
| {z }
initial gas volume
original gas cap gas volume, [SCF] cumulative gas production from the gas cap, [SCF] gas formation volume factor at current pressure, [RB/SCF] 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.Gas cap shrinkage should never be allowed because it always results in a loss of oil recovery. 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 has been produced while the reservoir pressure had dropped from 1225 psig to 1100 psig. The gas cap originally contained 21.3 × 109 SCF. Data: G = 21.3 ×109 Bgc at 1225 psig = 0.002125 Bgc at 1100 psig = 0.002370
[SCF] [RB/SCF] [RB/SCF]
Solution 1. At 1100 psig the gas cap containing 0.8 times the original gas cap volume i.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 RB
i.e., the gas cap has actually shrunk. This would allow about 5 MMRB oil to migrate into the gas cap, a zone originally free of oil. This would result in 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 reservoir pressure had dropped from 1225 psig to 900 psig at the time that 20% of the gas was produced. Data: G = 21.3 ×109 Bgc at 1225 psig = 0.002125 Bgc at 900 psig = 0.002905
[SCF] [RB/SCF] [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 RB
i.e., the gas cap has expanded into the oil zone and no oil is lost to the gas zone.
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 released from solution. At any time during the production of a reservoir, the gas originally in solution can 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]
where, N Np Rsi Rs Bg
= = = = =
}
N Rsi Bg
− (N − Np )Rs Bg −
Gps Bg
gas originally in solution
gas still in solution
produced solution gas
|
{z
}
|
{z
}
| {z }
original oil volume, [STB] cumulative oil produced, [STB] original solution gas-oil ratio, [SCF/STB] solution gas-oil ratio at current pressure, [SCF/STB] gas formation volume factor at current pressure, [RB/SCF]
The above equation is usually written as, ∆VRSG = {N Rsi − (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 at the time when reservoir pressure was 900 psig. At the same time cumulative production of solution gas was 4.05 ×109 SCF. Calculate the reservoir volume occupied by released gas. Data: N Rsi at 1225 psig Rs at 900 psig Bg at 900 psig
= = = =
90.46 ×106 230 169 0.002905
[STB] [SCF/STB] [SCF/STB] [RB/SCF]
Solution ∆VRSG = {N Rsi − (N − Np )Rs − Gps }Bg ∆VRSG = {(90.46 × 230 − (90.46 − 14.73) × 169 − 4050) × 106 } 0.002905 ∆VRSG = 11.5 × 106 RB
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} = (N − Np )Bo | {z |
[RB]
{z
}
oil still in reservoir
where, N Np Bo
= original oil volume, [STB] = cumulative oil produced, [STB] = 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 × 106 RB
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 expressed as the formation compressibility cf . cf is defined as the fractional change in pore volume per psi change in reservoir pressure. The pore volume can be expressed in terms of the original oil volume, as is required for material balance calculations. original oil volume = N Boi = Vp Soi = Vp (1 − Swi ) where, N Boi Vp Soi Swi
= = = = =
original oil volume, [STB] oil formation volume factor at initial pressure, [RB/STB] reservoir pore volume, [RB] initial oil saturation initial or connate water saturation
or, Vp =
µ
N Boi 1 − Swi
142
¶
The total amount of rock expansion is calculated by utilizing the definition of compressibility; 1 ∆V c= V ∆P from which we have ∆V = cV ∆P Rock expansion = formation compressibility × pore volume × pressure change
|
{z
}
[RB]
Rock expansion = cf |
where, cf p1 p
{z
}
[RB]
µ
¶
N Boi (pi − p) 1 − Swi
= formation compressibility, [vol/vol/psi] = initial reservoir pressure, [psig] = current reservoir pressure, [psig]
The total amount of connate water expansion is given by, Water expansion = water compressibility × water volume × pressure change
|
{z
[RB]
}
The initial connate water volume is given by, original connate water volume = Swi Vp = Swi where, cw Swi
µ
N Boi 1 − Swi
¶
= water compressibility, [vol/vol/psi] = initial water saturation
The connate water expansion is, Water expansion = cw Swi
|
{z
[RB]
}
µ
¶
N Boi (pi − p) 1 − Swi
Combining the above equations, we have for rock and water expansion, ∆VRW E
µ
¶
N Boi = (cf + cw Swi ) (pi − p) 1 − Swi
For reservoirs where a gas phase is present, the rock and water expansion term is so small that it may safely be neglected. The rock and connate water term is usually only important for oil reservoirs when the oil pressure remains above the bubble point. 143
Example 6.5 - Calculation of rock and connate water expansion volume MBE.mcd] What is the rock and water expansion volume for the previous example when the pressure falls from 1225 psig to 900 psig. Data: Swi cf cw
= 0.205 = 3.0×10−6 = 3.0×10−6
[psi−1 ] [psi−1 ]
Solution ∆VRW E
µ
¶
N Boi = (cf + cw Swi ) (pi − p) 1 − Swi
∆VRW E = (cf + cw Swi ) (Vp ) (pi − p) ³
´
∆VRW E = (3.0 + 3.0(0.205)) 10−6 127.84(106 ) (1225 − 900) ∆VRW E = 0.15 × 106 RB
We see that rock and water expansion amounts to about 1% of the released gas volume. This is usually the case in practice and rock and connate water expansion can 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 water influx cannot be calculated directly. This is because the calculation of water influx requires information characterizing the size and strength of the aquifer and this information is not generally known with any certainty during the early production life of a reservoir. However, since we can calculate all the other terms in the material balance equation, we can determine the net water influx into the reservoir — total water entering the reservoir less water produced — by difference. |Net water {z influx} = [RB]
where, We Wp Bw
W
e |{z}
cumulative water influx
−
Wp Bw
| {z }
cumulative produced water
= cumulative water influx, [RB] = cumulative water produced, [STB] = water formation volume factor at current pressure, [RB/STB]
The above equation may be written as, ∆VN W I = We − Wp Bw
145
6.1.6
General Material Balance Equation
Having analyzed all the individual terms in the material balance equation, we can write, Original oil volume = Gas cap expansion + Released gas volume + Oil volume + Rock and water expansion + Net water influx
∆VOOIP = ∆VGCE + ∆VRGV + ∆VROV + ∆VRW E + ∆VN W I
N Boi = (G − Gpc )Bg − GBgi + {N Rsi − (N − Np )Rs − Gps }Bg + µ ¶ N Boi (N − Np )Bo + (cf + Swi cw ) (pi − p) + 1 − Swi (6.1) We − Wp Bw Note that the two gas production terms are additive and equal to the total gas production at the surface ie., (Gpc + Gps )Bg = Gp Bg The material balance equation may be re-written as, N Boi = G(Bg − Bgi ) + {N Rsi − (N − Np )Rs }Bg − Gp Bg + µ ¶ N Boi (N − Np )Bo + (cf + Swi cw ) (pi − p) + 1 − Swi We − Wp Bw
(6.2)
One of the most important uses of the material balance equation is to calculate water influx into a reservoir. This only requires that we know the volume of initial fluids in-place, reservoir pressure and cumulative oil, gas and water production. Cumulative water influx is, We
µ
N Boi = N Boi − (N − Np )Bo − (cf + Swi cw ) (pi − p) 1 − Swi −GBg + GBgi + Gp Bg − {N Rsi − (N − Np )Rs }Bg +Wp Bw
146
¶
(6.3)
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 the pressure falls from 1225 psig to 900 psig. Cumulative water production at this time is 620,000 STB. Data: Wp Bw
= 0.62×106 = 1.0
[STB] [RB/STB]
Solution The following items were calculated in the previous examples. Ex. 2 3 4 5
Item Calculated Gas Cap Exp. Released Gas Remaining Oil Rock & Water Exp.
(G − Gpc )Bg − GBgi {N Rsi − (N − Np )Rs − Gps }Bg (N − Np )Bo ³ ´ N Boi (cf + Swi cw ) 1−S (p − p) i wi
×106 RB 4.24 11.50 83.63 0.15
Re-arranging the material balance equation to calculate We ,
We
µ
N Boi = N Boi − (N − Np )Bo − (cf + Swi cw ) (pi − p) 1 − Swi −GBg + GBgi + Gp Bg − {N Rsi − (N − Np )Rs }Bg +Wp Bw
¶
(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 mechanism which contributes to the total energy required to produce oil from the reservoir. If we divide the overall material balance equation by N Boi we obtain,
1 =
(G − Gp )Bg − GBgi {N Rsi − (N − Np )Rs − Gps }Bg + + N B N B oi oi | {z } | {z } Igc
µ
Isg
¶
1 (N − Np )Bo + (cf + Swi cw ) (pi − p) + N Boi 1 − Swi
|
{z
Ipd
We − Wp Bw N{z Boi } |
}
(6.5)
Iw
The above equation may be written as, 1 = Igc + Isg + Ipd + Iw where, Igc Isg Ipd Iw
= = = =
gas cap drive index solution gas drive index pressure depletion drive index water drive index
The above indices represent the effectiveness of each drive mechanism for a particular reservoir. A reservoir for which the dominant drive mechanism is, Iw , is called a water drive reservoir. A reservoir for which the dominant drive mechanism is, Igc , is called a gas cap drive reservoir. A reservoir for which the dominant drive mechanism is, Isg , is called a solution gas 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 depletion drive reservoir. A reservoir for which more than one drive mechanism is important, is called a combination drive reservoir.
6.2.1
Typical Performance Characteristics for the Different Drive Mechanisms
The dominant drive mechanism for a particular reservoir can often be deduced from 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 summarized 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 straight line pressure response and may be expected to recover less than 5% of the original oil-in-place.
2.
Solution gas drive — pressure drops slowly at first. As the producing GOR increases, 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 decreases as water influx from the aquifer increases. The producing GOR remains 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
6.3.1
USING MATERIAL BALANCE EQUATIONS Average Reservoir Pressure
The material balance equation describes the whole reservoir in terms of average reservoir 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 equation it 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 wells are shut-in and that the bottom-hole pressure is measured at a time after shut-in which 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 in considerable loss in revenue because of lost production. The direct measurement approach is therefore impractical. Pressure data obtained from a pressure buildup test can be used to estimate the average pressure in the volume or area drained by the tested well. These tests require short test times (hours to days) and allow reservoir pressure to be mapped over the field. The mapped pressures may be averaged to give the average reservoir pressure as, P pj Vp j p= P (6.6) Vp j
where pj and Vp j are the drainage area pressure and pore volume drained by well j.
6.3.2
Knowns and Unknowns
When solving the material balance equation the parameters may be classified into the following groups of knowns and unknowns. knowns unknowns Np N Gp G Wp We Swc p (Bo , Bg , Rs ) cw cf Bw 153
The PVT properties (Bo , Bg , Rs ) may be considered to be known if the average reservoir pressure is known and representative fluid samples have obtained and analysed. This reduces the number of unknowns to five but we have only one equation. This is the central problem with the use of the material balance equation. Before considering specific examples of the application of the material balance equation, 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 is sold to generate revenue. Gp , Wp - cumulative gas and water production be unknown for older oil and gas fields because they had little or no value at the time of production. When these are unknown it is not possible to perform material balance calculations for the reservoir. Swc - is usually considered to be accurately known from petrophysical evaluation. cw , Bw - these are known, or may be estimated, from laboratory tests on formation brine. Unknowns N, G - volumetric estimates of the original oil and gas-cap volumes in-place are always known from the field appraisal stage. These are disregarded as soon as production-pressure data becomes available and an attempt is made to estimate in-place volumes from material balance calculations. This is because volumetric estimates include all the mapped hydrocarbon volume, whereas material balance calculation provides the effective volume or the volume which contributes to actual production. This will usually be smaller than the volumetric estimate due to compartmentilization of the reservoir by faults or low permeability zones. We - this is probably the greatest unknown in reservoir development - whether there has been any water influx or not. One of the most important uses of material balance calculations is to estimate water influx. p - although we usually consider reservoir pressure to be known, problems with the interpretation of pressure buildup test data may introduce serious errors and there may be considerable uncertainty in estimates of average reservoir pressure. cf - pore volume or formation compressibility is usually considered to be small and 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 on land in remote areas but may cause serious problems for operations offshore.
Problem 6.1 Calculating water influx from material balance for history matching aquifer performance The first step in characterizing or history matching an aquifer is to calculate water influx from available production and pressure data. You will be performing this history match in the reservoir engineering course which follows from the present course. After the aquifer has been characterized, the resultant aquifer model is attached to a numerical reservoir simulation model which is used to predict future reservoir performance. The reservoir for this exercise is a small offshore oil field which has been on-stream for only 700 days. The field was shut-in for a 250 day period for upgrading of the gathering system and production facilities. There was a strong and clear aquifer response to the shut-in (use Mathcad to plot the field pressure as a function of time to see the response). The field production history is given in the table below. PRODUCTION HISTORY Time P Np Gp (day) (psia) (MMSTB) (BSCF) 0 50 100 150 200 250 300 350 400 450 500 550 600 650 700
4217 3915 3725 3570 3450 3390 3365 3600 3740 3850 3900 3920 3720 3650 3640
0.00000 0.34270 0.63405 1.06898 1.36788 1.64397 1.91465 1.91465 1.91465 1.91465 1.91465 1.91465 2.11402 2.24921 2.33133
155
0.00000 0.17821 0.32970 0.55587 0.71130 0.85487 0.99562 0.99562 0.99562 0.99562 0.99562 0.99562 1.09929 1.16959 1.21229
Wp (MMSTB) 0.00000 0.00000 0.00121 0.02636 0.07041 0.12218 0.17304 0.17304 0.17304 0.17304 0.17304 0.17304 0.20368 0.25094 0.31690
The initial oil in-place is estimated (volumetrically) to be 33.2 MMSTB. Reservoir system properties were estimated to be: cf = 5×10−6 psi−1 cw = 3×10−6 psi−1 Swc = 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 3465 3565 3665 3765 3865 3965 4065 4165 4265
1.2511 1.2497 1.2483 1.2469 1.2455 1.2441 1.2427 1.2413 1.2399 1.2385
510 510 510 510 510 510 510 510 510 510
-
1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0
Reservoir temperature is 210o F and the gas gravity is 0.69. Determine the cumulative water influx for each of the 50 day time periods in the production 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 calculations performed in Examples 6-1 to 6-6. You can simply use [MBE.mcd] to do the calculation for each period. Alternatively, you could modify [MBE.mcd] to readin the production data and do the 14 calculations in one pass or prepare your owm 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 aquifer response is small) there is negligible water production and the material balance equation reduces to, µ
¶
N Boi 0 = N Boi − (N − Np )Bo − (cf + Swi cw ) (pi − p) 1 − Swi −GBg + GBgi + Gp Bg − {N Rsi − (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 is available, the only unknown is the original oil in-place, N . This provides a very valuable check on the volumetrically determined value. As discussed previously, the material balance derived value of N may be more reliable than the volumetrically determined volume.
Prediction of future reservoir pressure After the original oil in-place has been established, the material balance equation may be used to predict future reservoir pressure if future cumulative production can be estimated. The main difficulty with this procedure is estimating the volumes 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 balance equation [OIPMBE.mcd] The discovery pressure of a reservoir containing a gas cap was 3330 psia. Geological evidence suggests that aquifer quality is poor and that water influx into the reservoir is likely to be negligible. The reservoir was produced until the reservoir fell to 2700 psia. Cumulative oil production at this time is 11.503 MMSTB and cumulative gas production is 14.206 BSCF. The volumetric estimates of the original oil-in-place, (N ), is 105 MMSTB and the initial gas cap volume is 81.14 BSCF. Does the material balance calculation confirm this estimate? Data: Swi cf cw
= 0.21 = 3.0×10−6 = 3.0×10−6
[psi−1 ] [psi−1 ]
[Answer: 113 MMSTB]
Solution hints The material balance equation is explicit in N , so we could rearrange it and solve directly for N - you can do this if you like. An alternative way is to solve the material balance equation by trial and error ie., guess a value of N which makes both sides of the material balance equation equal. If you use [OIPMBE.mcd] you will see the I have simply rearranged [MBE.mcd] to allow me to enter a value of N 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 sum to 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 may be solved graphically. The graphical solution method is commonly referred to as the P/z-plot. The plot can be used to estimate the original gas in-place and to predict 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
zT P
we can write for the above material balance equation (where reservoir temperature is assumed to remain constant), µ
G 5.02 × 10−3
zi T Pi
¶
µ
= (G − Gp ) 5.02 × 10−3
zT P
¶
which may be rearranged to, G or,
µ
P G z or,
zi z = (G − Gp ) Pi P ¶ µ
µ
¶
µ
¶
P = (G − Gp ) z
P P = z z
¶
1 P − Gp G z i
i
i
The above equation results in a straight line relationship between (P/z) and Gp . The straight line on a (P/z)—Gp plot passes through (P/z)i at Gp = 0 and through G at (P/z) = 0. The straight-line relationship is very useful in estimating the initial volume of gas-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 reservoir [GIP.mcd] Estimate the initial gas content, G, for a closed gas reservoir having the following production history. Pressure Cumulative Gas Gas Deviation (psig) Production (MMSCF) Factor (z) 3500 3350 3200 3050 2750
0 46.7 125.0 203.5 380.0
160
0.84 0.82 0.81 0.80 0.78
Solution
1. Calculate P/z for each pressure point and plot against cumulative production, Gp . Pressure Cumulative Gas Gas Deviation (psig) Production (MMSCF) Factor (z) 3500 3350 3200 3050 2750
0 46.7 125.0 203.5 380.0
0.84 0.82 0.81 0.80 0.78
P/z (psig) 4166.7 4085.3 3950.6 3812.5 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 plot open [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 will occur as the reservoir pressure declines with production. The material balance for such a reservoir may be written as, GBgi = (G − Gp )Bg + We − Wp Bw This equation can be solved for water influx, We = GBgi − (G − Gp )Bg + Wp Bw If G can be evaluated volumetrically and if the field production data is known (Gp and Wp ) at corresponding reservoir pressures, the material balance equation can 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. Can we estimate G from this data? Pressure Cumulative Gas Gas Deviation (psig) Production (MMSCF) Factor (z) 3500 3350 3200 3050 2950
0 52 155 425 880
0.84 0.82 0.81 0.80 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 decrease in reservoir pressure. This means that early in the life of the reservoir will behaive like 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 realistic estimate of the original gas-in-place. I have setup the spreadsheet [GIP.mcd] to allow 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 cumulative water influx corresponding to the times at which cumulative gas production is given. Cumulative water production is zero for the period and the reservoir temperature is TR = 275 o F. [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 added a calculation for Bg from z and the material balance equation for a water drive gas reservoir.
164
Chapter 7 RESERVOIR ROCK PROPERTIES AND CORE ANALYSIS PROCEDURES Reservoir engineering calculations require a knowledge of certain basic rock properties. 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 compressibility. The saturation dependent properties will be considered in the next chapter.
7.1
POROSITY
Parameter Symbol Porosity
φ
Definition The total fraction of the bulk volume which is void space
165
well sorted sandstone 21.5%
poorly sorted sandstone 1%
Figure 7.1: Effect of grain size distribution on porosity
Porosity is important in determining the volume of oil and/or gas which will 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 weak function of grain size itself. — for sandstones sedimentological processes are important. For limestones porosity is determined by changes after deposition - dissolution and fracturing. Porosity for both sandstones and limestones can be greatly modified by postdepositional events or digenesis. Cementation, chemical action, and fracturing are events which can modify the original porosity of rocks. The original porosity of a formation is usually refereed to as primary porosity. Porosity modified by post-depositional processes is referred to as secondary porosity. The attached figures show the effect dispersed fines (particulate kaolinite, porelining chlorite and pore-bridging illite) affect primary porosity. Effective porosity and permeability may be greatly reduced by deposition of fine particulate minerals in 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 no interest because it is not accessible to flow of fluids and therefore does not contribute to hydrocarbon production. Total porosity is the sum of effective and disconnected porosity. The total porosity, φt , is given by, φt =
total pore volume bulk volume − grain volume = bulk 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 on porosity
Conventional laboratory experiments used to determine porosity are all based on introducing fluids into the pore space. They therefore all measure effective porosity.
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 schematically in the attached figure. The apparatus and measurement procedure consists: 1.
Two cells of accurately known volume, V1 and V2 , are connected and at atmospheric pressure, pa . The clean and dry core sample is placed in the second cell.
2.
Valve B is closes and cell 1 is pressured with gas G through valve A to an absolute 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 equilibrium absolute pressure, (p2 + pa ), is recorded. This is called Condition II.
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 2 at Condition I at Condition II Assume an isothermal expansion process, and pressures sufficiently low so that the ideal gas law applies. The grain volume of the sample is Vs . The ideal gas law is written as, pV n= RT where, p T V n R
= = = = =
absolute pressure absolute temperature volume moles of gas 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 )V1 pa (V2 − Vs ) (p2 + pa )V1 (p2 + pa )(V2 − Vs ) + = + RT RT RT RT Solving for the grain volume, Vs , gives, Vs = V1 + V2 −
p1 V1 p2
Note that the pressures p1 and p2 are gauge pressures as read in the actual experiment. Porosity is calculated from the equation, φ=
Vb − Vs Vb
where the bulk volume of the sample, Vb , is either determined from measurements of the sample dimensions (core plugs) or by measuring the displaced liquid volume when 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 helium porosimetry [HEPOR.mcd] Given the diameter and length of a cylindrical core plug; D L
= 2.5 cm = 4 cm
and the following measured porosimeter pressures and parameters; V1 V2 p1 p2
= = = =
25 cm3 50 cm3 100.03 psig 41.79 psig
Calculate the porosity of the plug. [Answer: 0.23]
173
q=
kA ∆P µ L ∆P
A q
q
L
Figure 7.11: Datcy’s law and permeability
7.2
PERMEABILITY
Parameter Permeability
Symbol Definition k
A measure of the ease with which fluid flows through a porous rock
Permeability is important in determining the rate at which oil and gas will flow to the wellbore Permeability will usually vary over several orders of magnitude in the same reservoir. The variation in permeability (heterogeneity) has a great effect on oil recovery 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 function 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 to measure the permeability of cylindrical core samples or plugs. The steady-state flow rate (q) of a fluid which completely saturates the core is directly proportional to the x-sectional area (A) over which the flow occurs, the imposed pressure gradient over the core (∆p/L) and inversely proportional to the viscosity of the fluid. µ ¶ ∆p 1 q∝A ∆L µ The constant of proportionality is called permeability and is an intrinsic property of the rock. kA∆p q= µ∆L This equation is known as Darcy’s law and is the basic equation for calculating fluid 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 computed permeability has the units of darcys. A rock sample has a permeability of one darcy when; ∆p ∆L A q µ k
= = = = = =
pressure difference, 1 atmosphere flow length, 1 cm x-sectional area, 1 cm2 volumetric flow rate, 1 cm3 /sec viscosity, 1 cp permeability, 1 darcy
The darcy unit is too large for most reservoir rocks and it is common practice to express 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 incompressible fluid is written as, kA∆p q = 0.001127 µ∆L where, ∆p ∆L A q µ k
= = = = = =
pressure difference, psi flow length, ft x-sectional area, ft2 volumetric flow rate, res. bbl/day (RB/D) viscosity, cp permeability, md
176
Example 7.2 - Laboratory determination of the liquid permeability of a core plug [LPERM.mcd] The laboratory has conducted a steady-state liquid permeability test on a core plug from a reservoir of interest to you. Check the laboratory calculation which suggests that the permeability of the plug is 200 md. The laboratory measured data is given below: plug diameter plug length test oil viscosity outlet pressure inlet pressure flow rate
= = = = = =
2.52 cm 2 cm 1.82 cp 1 atm 2 atm 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 permeability is 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 a horizontal line drive over a flow length of 10,000 feet. Production well spacing is 1000 feet and the sand thickness is 30 feet. The reservoir oil has a viscosity of 0.8 cp. What is the well rate when the pressure difference between injectors and producers is 500 psi? [Answer: 425 RB/day]
Solution hint Use q = 0.001127
kA∆p µ∆L
to calculate the rate in RB/day if you are not using Mathcad which handles mixed systems of units automatically. The constant 0.001127 arises from unit conversions. 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 a gas as the working fluid. Liquids may adsorb onto the rock surface and potentially change wettability which could adversely affect later experiments. Gas does not have this problem. The experiment consists of flowing a gas through the core plug. Since gas is compressible, the previous equations, which are valid only for incompressible fluids, must be modified. The average gas pressure (¯ p) in the core is given by, p¯ =
(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 are low, the product qp (equivalent to V p for a static system) is constant. We can therefore express the volumetric flow rate at the average pressure (¯ q ), in terms of the measured volumetric flow rate at atmospheric conditions (qa ), qa pa = q¯(p1 + p2 )/2 qa pa q¯ = (p1 + p2 )/2 where pa is atmospheric pressure. Darcy’s law may be written as, q¯ =
kA(p1 − p2 ) µ∆L
Substituting for q¯ gives, qa pa kA(p1 − p2 ) = (p1 + p2 )/2 µ∆L which on re-arrangement gives, kA(p1 2 − p2 2 ) qa = 2pa µ∆L or 2qa pa µ∆L k= A(p1 2 − p2 2 ) We use this equation to calculate the gas permeability of the test plug. When his done at a number of different average pressures we find that the calculated permeability is not constant but that it decreases with increasing pressure. This may appear to be a problem, particularly if we expect the permeability to be constant! 179
7.2.3
The Klinkenberg Effect
The apparent decrease in measured gas permeability which occurs with increasing average flowing pressure is known as the Klinkenberg effect. At very low pressures, the mean free path of gas molecules is similar to the dimensions of the pore space. Under these conditions the solid pore walls do not significantly retard gas flow close to the pore walls because molecules spend little time at the walls (free-slip condition). For liquids and gases at very high pressure, molecules are always present at the pore 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 pressures, p¯.
2.
Plot kgas vs 1/¯ p.
3.
Extrapolate the linear plot to infinite pressure i.e. to 1/¯ p = 0. The extrapolated permeability is equivalent to the liquid permeability.
The variation of gas permeability with pressure is given by the equation, kgas = kliquid
Ã
b 1+ p¯
!
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 of a 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 with increasing inlet gas pressure. The test data is given below. Calculate the permeability of the plug. The laboratory data is plug diameter plug length atmospheric pressure gas viscosity inlet pressure flow rate
as follows: = 1.953 cm = 3 cm = 0.979 atm = 0.0182 cp = 2 atm = 0.275 cm3 /sec
Run Inlet Pressure Outlet Pressure Discharge Rate p1 (atm) p2 (atm) qa (cm3 /sec) 1 1.146 0.979 0.348 1 1.481 0.979 1.168 3 1.815 0.979 2.169 4 3.486 0.979 9.737 [Answer: 27.5 md]
Solution hint 1.
Calculate the mean flowing pressure, p¯, for each test. p¯ =
2.
(p1 + p2 ) 2
Calculate kgas for each test using Darcy’s law. k=
2qa pa µ∆L A(p1 2 − p2 2 )
3.
Plot kgas vs 1/¯ p.
4.
Extrapolate the linear plot to infinite pressure i.e. to 1/¯ p = 0. The intercept at 1/¯ p = 0 is the equivalent liquid permeability.
Open [KLINK.mcd] , check the input data and read the answer. 181
kA ∆P πR4 ∆P q = nq c = n = 8µ L µ L
πR4 ∆P qc = 8µ L
Figure 7.13: Simple capillary tube model for a porous medium
7.3
7.3.1
POROSITY-PERMEABILITY TIONSHIPS
RELA-
Capillary Tube Model
In this section we examine the relationship between porosity and permeability using simple models to represent the porous medium. The simplest possible model 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 bundle of capillaries is a good model for porous medium made up of packed particles such as a sandstones. The matrix consists of a solid block having dimensions L × L × L. The pore space is 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 Poiseuille equation πR4 ∆P qc = 8µ 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 = n
πR4 ∆P 8µ L
According to Darcy’s law, the flow q is given by, q=
kA ∆P µ L
where A is the x-sectional area of the porous medium perpendicular to the direction of flow. Equating the two equations for, q, gives kA ∆P πR4 ∆P =n µ L 8µ L Eliminating the common parameters and rearranging to solve for permeability, k=n
πR4 8L2
The porosity of the simple porous medium is, φ=
pore volume bulk volume
πR2 L L3 πR2 φ= 2 L We use this equation to express L in terms of φ and R. Rearranging gives, φ=
L2 = n
πR2 φ
Substituting for L2 in the equation for k gives, k=
R2 φ 8
It has been experimentally verified that permeability is well correlated to the square 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 pore surf ace area ΣV p = pore volume n2πRL ΣV p = nπR2 L 2 ΣV p = R or 2 R= 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 pore surf ace area ΣV g = grain volume ΣV g = ΣV p
pore volume grain volume
ΣV g = ΣV p
φ 1−φ
ΣV p = ΣV g
1−φ φ
or
Substituting into the equation for k, gives k=
1 φ3 2ΣV g (1 − φ)2
This is similar to the empirical Kozeny relation for unconsolidated porous media and forms the basis for power-law correlations or transforms between porosity and permeability for actual reservoir rocks. Actual relationships follow a power law 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 convert log measured porosities to permeabilities. This is very important in reservoir characterization. 184
∆P
H q h
h3 W ∆P q= 12µ L
W L q=
kA ∆P kHW ∆P = µ L µ L
k=
1 1 2h h = h 2φ f 12 H 12
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 of height h as shown in the attached figure. The average velocity for laminar (viscous) flow in the fracture is given by, v=
h2 ∆P 12µ L
The volumetric flow rate through the fracture (q = vAf = vhW ) is, q=
h3 W ∆P 12µ L
From Darcy’s law we have q=
kA ∆P kHW ∆P = µ L µ L
Equating the two equations for q gives, k=
h2 h3 ³ ´ = 12H 12 Hh
We can define a fracture porosity as
φf =
hW L h = HW L H 185
This gives k=
1 2 h φf 12
Actual fractured media confirm that fracture permeability is a strong function of fracture 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 in the 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 different in different directions.
The resultant stress, σ−P , causes a strain or deformation of the rock. The overall result 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 the rock or matrix compressibility, cr . cr =
1 dVr Vr d(σ − P )
where, Vr is the grain or matrix volume. The fractional change in total or bulk volume per unit change in rock stress is called the bulk volume compressibility, cb . cb =
1 dVr Vb d(σ − P )
Bulk volume compressibility may be important in determining levels of ground subsidence. The fractional change in pore volume per unit change in rock stress is called the pore volume or formation compressibility, cp . cp = −
1 dVp Vp d(σ − P ) 186
Figure 7.15: Relationships between porosity, permeability and rock microstructure
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 sandstones The matrix, bulk and pore volume compressibilities are measured in special laboratory experiments which will be described in the next section. For reservoir engineering calculations the primary interest is in the change in pore volume as reservoir or pore volume pressure changes.
7.4.1
Pore Volume Compressibility
Experimental measurements for sandstones and shales, in which the external rock stress (σ) is constant, show that cr << cb . This means that, dVb ≈ dVp and since Vp = φVb , we have that cp =
cb φ
with cp = cf = −
1 dVp Vp dP
where cf is the formation compressibility and is commonly used interchangeably with pore volume compressibility. The above relationships are useful in interpreting laboratory measurements. 189
Figure 7.19: Hydrostatic load cell for measuring pore volume compressibility
For actual reservoir conditions only the vertical component of external stress is constant with the stress components in the horizontal plane characterized by boundary conditions that there is no bulk deformation in these directions. Geersma (1966) showed that for these conditions, 1 cf ≈ cp 2 this shows that the effective pore volume compressibility for reservoir rocks is only 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 measure rock compressibility (pore volume compressibility). The apparatus allows both the internal or fluid pressure in the core and the external or overburden pressure to be varied independently. A small bore capillary tube is used to measure the changes in pore volume by measuring the volume of fluid expelled from the pore space.
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 increased with the internal pressure held at atmospheric. The volume of liquid squeezed out into the capillary tube is directly related to the change in bulk volume of the sample.
2.
Pore volume compressibility — The external pressure is maintained constant. The internal pressure is reduced in steps and the liquid produced at each pressure is noted. The volume of liquid which would have been produced had the pore space volume remained constant is readily calculated from a knowledge of the liquid compressibility. The difference between the measured 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 pressure may be written as, ct = So co + Sw cw + Sg cg + cf where, ct cf co , cw , cg So , Sw , Sg
= = = =
total compressibility formation or pore space compressibility oil, water and gas phase compressibilities oil, water and gas phase saturations
The total compressibility is used to evaluate the diffusivity factor for unsteadystate flow calculations. Estimates of total compressibility are important in well pressure 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 cw co cg
= = = =
3 × 10−6 psi−1 3 × 10−6 psi−1 15 × 10−6 psi−1 500 × 10−6 psi−1 @ 2000 psia
for, Sw So Sg
= 0.2 = 0.7 = 0.1 ct = So co + Sw cw + Sg cg + cf ct = 64 × 10−6 psi−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 on Darcy’s law. Darcy (1856), a French engineer, investigated the flow of water through sand filters for water purification. He observed the following relationship between superficial velocity and the properties of the sand bed and fluid, q k ∆p =− A µ ∆L where, q A k µ ∆p ∆L
= = = = = =
flow rate, cross-sectional area open to flow, permeability, fluid viscosity, outlet pressure - inlet pressure, distance between inlet and outlet,
cc/sec cm2 darcy cp atmospheres 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. The linear dependence of flow rate on the pressure gradient implies laminar or viscous flow in the pore-spaces of the porous medium. This is generally true for most reservoir conditions where actual flow velocities are low. A common exception is flow 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 k dp =− A µ dx where dp/dx is the pressure gradient in the direction of flow The negative sign in the above equations indicates that pressure decreases in the direction of flow. The sign convention is therefore that distance is measured positive in the direction of flow. This is the basic equation used to analyze single phase flow in reservoir engineering.
8.1.1
Pressure Potential
The above form of Darcy’s law applies to horizontal flow where there are no gravity effects. In reservoir analysis we are often interested in vertical or inclined flow where the flow occurs in the presence of a hydrostatic gradient. In such cases it is convenient to define a pressure potential as, Φ=
Z
p
pr
dp − gdz ρ
where z is the vertical coordinate measured positive downward and g is the gravitational acceleration. pr is an arbitrary reference pressure, usually taken as pr = 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 sin α and the pressure potential is, Φ=
Z
p
pr
dp − gdx sin α ρ
If the reservoir fluid is considered to be incompressible the potential may be written as p Φ = − gx sin α ρ 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 k dΨ =− A µ dx or, q k =− A µ
Ã
dp − ρg sin α dx 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 the problem shown in Figure 8.3. If you are not using Mathcad be careful with units.
8.2 8.2.1
STEADY-STATE FLOW Horizontal Linear Flow of an Incompressible Fluid
The linear flow system is depicted in the attached figure. The medium has uniform permeability, k, and is saturated with an incompressible (ρ is constant) fluid having constant viscosity, µ. Darcy’s law is written as, q=−
kA ∆p µ L
This equation is commonly used to evaluate flow tests in reservoir core samples and 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 different thicknesses, hi , and permeabilities, ki . For a common inlet face pressure, P1 , and outlet face pressure, P2 , the pressure drop across each layer is the same and the flow rate for each layer is, qi = −
ki (hi w) (P2 − P1 ) µ L
The total flow rate is q=
n X
qi
i=1
or q=−
n w(P2 − P1 ) X ki hi µL i=1
Darcy’s law for the total flow may also be written as, q=−
¯ k(hw) (P2 − P1 ) µ L
where k¯ is the average system permeability and h is the total thickness. Equating the two expressions for q gives, Pn
ki h i k¯ = Pi=1 n i=1 hi This equation is used to determine the average horizontal permeability of a layered system. 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 following thicknesses and permeabilities. Layer Thickness Horizontal Permeability (ft) (md) 1 20 100 2 15 200 3 10 300 4 5 400 [Answer: 200 md]
Horizontal flow in series Consider the arrangement where the different permeabilities are arranged in series. For this arrangement the flow through each layer is the same, q = qi where, q=−
ki (hw) ∆Pi µ Li 199
where ∆Pi is the pressure drop across slab i. This may be rewritten as, ∆Pi =
qµ Li hw ki
and the total pressure drop is the sum of the pressure drops in each layer, (P1 − P2 ) =
n X
∆Pi =
i=1
n qµ X Li hw i=1 ki
Darcy’s law for the series arrangement may be written as, (P1 − P2 ) = where, L =
P
qµ L hw k¯
Li , is the total length for flow.
Equating the above equations gives, Pn
Li k¯ = Pn i=1 i=1 Li /ki This equation is used to calculate the effective vertical permeability for a layered reservoir system.
Example 8.2 - Average permeability for beds in series [SERK.mcd] Determine the average permeability of four beds in series having the following lengths and permeabilities. Layer Length Permeability (ft) (md) 1 250 25 2 250 50 3 500 100 4 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 basic model for flow into a well and is used in reservoir engineering to evaluate well performance. The following nomenclature employed, rw re h
= wellbore radius = well drainage radius = formation thickness
Flow enters the system across the outer perimeter at radius re , and leaves at the wellbore of radius rw . The volumetric flow rate, q, is constant and the flow area, A, is 2πrh. Substituting into Darcy’s law gives, q k dp = 2πrh µ dr Note that for radial flow pressure decreases with decreasing r.
201
Separating variables and integrating gives: q
Z
re
rw
dr 2πkh Z pe = dp r µ pw
q(ln rw − ln re ) =
2πkh(pe − pw ) µ
Solving for q, q=
2πkh (pe − pw ) µ ln rrwe
This equation is widely used in reservoir engineering to calculate production and injection rates for individual wells. The equation may be written in field units as; q = 0.00708
kh(pe − pw ) µ ln rrwe
where, q k h µ pe pw re rw
= = = = = = = =
well rate, RB/D permeability of well drainage area, md average reservoir thickness, feet fluid viscosity, cp pressure at external radius, psi wellbore pressure , psi external drainage radius, feet 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 where the pressure is being maintained at 2000 psia at the drainage radius. The following data apply: k h µ rw re
= = = = =
230 10 3.5 0.5, 2000,
md feet cp feet feet
[Answer: 1114 RB/day] Solution hint Study [RADSSF.mcd] carefully - you will be using it a lot. It solves a great variety of 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 this equation and ignore the other equations and results. The theoretical minimum flowing 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 permeability as that for linear flow.
Radial flow in series Using the same reasoning as for linear flow in series yields the result, M X 1 log(ri )/(ri−1 ) 1 = ¯ log(re /rw ) i=1 ki k
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 the following radial distances to bed boundaries and bed permeabilities. The wellbore radius is 0.5 ft and the well drainage radius is 2000 ft. Radial Layer Radius (ft) r1 250 r2 250 500 r3 r4 1000
Permeability (md) 25 50 100 200
[Answer: 30.4 md]
8.2.3
Wellbore Damage
Wells are always drilled with overpressure (mud pressure greater than formation pressure) to prevent inflow of reservoir fluids which could lighten the mud column and lead to a blowout. As a result of overpressure some of the drilling mud flows into the formation and the particles suspended in the mud partially plug the pore spaces through which fluid flows. This creates a lower permeability zone or a damaged zone close to the wellbore. Wellbore damage is depicted in the attached figure, where ra is the radius of the damaged zone. The damaged zone results in an additional pressure ∆pskin at the wellbore. 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 (pe − pw ) µ ln rrwe
or pe − pw =
µ
qµ re ln 2πkh rw 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 total pressure drop for a damaged well may be written as, µ
¶
qµ re pe − pw = ln + ∆pskin 2πkh rw If we write,
qµ (s) 2πkh Than, for a damaged well we have a modified radial flow equation, ∆pskin =
pe − pw = or q=
µ
¶
qµ re ln +s 2πkh rw
2πkh (pe − pw ) µ ln rrwe + s
where s is the skin-factor for the well. In general, the skin-factor may be positive (a damaged well), negative (a stimulated well) or zero (undamaged well). In practice, s is determined from the interpretation of transient pressure well tests (pressure build-ups and drawdown). In the above equations the rate, q, is the reservoir or sandface rate. In application 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 (pe − pw ) µB ln rrwe + s
where B is the formation volume factor which relates surface rates to reservoir rates. It is understood that when an equation contains B the rate, q, is in surface volumes. 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 reservoir where 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 h µ rw re
= = = = =
230 10 3.5 0.5, 2000,
md feet cp feet feet
(ii) What would the well rate be if the wellbore is cleaned with an acid. A successful acid job may be expected to reduce the skin factor to approximately zero. (iii) What would the rate be for a stimulated well after a frac job which is expected 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 altered zone
The skin factor for a stimulated well ,where the skin factor is negative, has an upper 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 (pe − pw ) ³ ´ µB ln re + s rw
If −s is equal to, or greater than, ln(re /rw ), the equation produces answers which are physically impossible. This problem is removed by the introduction of the thick 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 if the bulk formation permeability (from ra to re ) is k, we can use the steady-state radial flow equation to calculate the pressure drops across the altered zone and the formation, µ ¶ qBµ ra ln pa − pw = 2πkh rw where ra is the radius of the altered zone and pa is the pressure at ra . Also,
µ
¶
qBµ re pe − pa = ln 2πkh ra 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 =
½
µ
¶
µ
qBµ 1 re 1 ra ln + ln 2πh k ra ka rw
The thin skin model expresses the above pressure drop as, pe − pw =
½
µ
¶
¾
qBµ re ln +s 2πkh rw 211
¶¾
where k is the permeability of the bulk formation and s is the skin factor. This may be re-written as, ½
µ
¶
¾
qBµ 1 1 re pe − pw = ln + s 2πh k rw k Equating the two expressions for (pe − pw ) gives, ½
µ
¶
µ
1 re 1 ra + ln ln k ra ka rw
¶¾
½
=
µ
¶
¾
1 re 1 + s ln k rw k
This may be simplified to give, "
#
µ
k ra s= − 1 ln ka rw If ka → ∞, k/ka → 0 and the equation gives µ
ra s = − ln rw
¶
¶
which is the upper physical limit on the value of s for a stimulated well. The skin factor for a stimulated well depends both on the permeability of the altered zone, 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 effective wellbore radius. Again, using the steady-state radial flow equation, we have for a well with skin, 2πkh (pe − pw ) ³ ´ q= Bµ ln re + s rw
We can also write this as,
q=
2πkh (pe − pw ) ³ ´ Bµ ln r0e rw
where
rw0
is the effective wellbore radius.
Equating both expressions for q gives, Ã
re ln 0 rw which can be written as
!
= ln
Ã
!
r0 ln w rw
212
µ
¶
re +s rw
= −s
This may be simplified to, rw0 = rw e−s The effective wellbore radius concept is often used to characterize fractured and horizontal wells. For example, the effective wellbore radius for a vertical well with an infinite conductivity fracture is, rw0 = xf where xf is the fracture half-length. The effective wellbore radius for a long horizontal well in a thin reservoir approaches that of a fractured well with a fracture half-length equal to one-quarter the length of the horizontal well, rw0 =
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 to determine the well flow efficiency F E. This is defined as the ratio of the pressure drop at the damaged wellbore to that for an undamaged well, FE =
pi − pwf − ∆ps qactual = pi − pwf qundamaged
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 fluid properties apply: k = 30 md h = 40 ft. rw = 0.5 ft. re = 2000 ft. µ = 0.4 cp B = 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 a skin-factor of s = −3, what would the above parameters be for the stimulated well? [Answer: -94 psi, 1.63, 10 ft.] If the fracture is estimated to affect a zone extending 15 ft. into the formation estimate 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 radial system is the radial diffusivity equation. 1 ∂ r ∂r
Ã
kρ ∂p r µ ∂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 dependent variable 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 the following assumptions: - the reservoir is homogeneous and isotropic (k is constant). - the well is fully penetrating ie., it is completed over the entire vertical section 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 = (2πrhφρdr) ∂t This may be expressed as, qρ |r+dr − qρ |r = 2πrhφdr The above may be simplified, Ã
qρ |r+dr − qρ |r dr
!
216
= 2πrhφ
∂ρ ∂t ∂(ρ) ∂t
Figure 8.15: Differential material balance for the radial diffusivity equation
∂(qρ) ∂ρ = 2πrhφ ∂r ∂t this is called the continuity equation.
Darcy’s law for a radial system may be written as, q=
2πkhr ∂p µ ∂r
Substituting into the continuity equation gives, ∂ ∂r or
Ã
2πkhr ∂p ρ µ ∂r
1 ∂ r ∂r
Ã
!
kρ ∂p r µ ∂r
= 2πrhφ
!
=φ
∂ρ ∂t
∂ρ ∂t
This is the general form of the radial diffusivity equation.
217
The term ∂ρ/∂t may be expressed in terms of pressure by introducing an equationof-state for the fluid. Defining the isothermal compressibility of the fluid as c=−
1 ∂V 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
∂ρ ∂p = ∂t ∂t
Substituting into the previous equation gives 1 ∂ r ∂r
Ã
kρ ∂p r µ ∂r
!
= φcρ
∂p ∂t
This is the general form of the radial diffusivity equation for the flow of a single phase in a porous medium. The equation is non-linear because density, ρ, which is 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 constant and small compressibility is a good model for these fluids. Since we have assumed that k, φ, µ and c are constant, the diffusivity equation becomes Ã
1 ∂ ∂p ρr r ∂r ∂r
!
=
φcρµ ∂p k ∂t
Using the chain rule for differentiation, we expand the LHS of the equation, 1 r Ã
Ã
∂(ρr) ∂p ∂ 2p + ρr 2 ∂r ∂r ∂r
!
=
φcρµ ∂p k ∂t
1 ∂(ρ) ∂p ∂(r) ∂p ∂2p r +ρ + ρr 2 r ∂r ∂r ∂r ∂r ∂r Ã
1 ∂ρ ∂p ∂p ∂2p r + ρ + ρr 2 r ∂r ∂r ∂r ∂r
!
!
=
=
φcρµ ∂p k ∂t
φcρµ ∂p k ∂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 Ã
1 ∂p ∂p ∂p ∂2p rcρ + ρ + ρr 2 r ∂r ∂r ∂r ∂r
!
=
φcρµ ∂p k ∂t
Finally, dividing throughout by ρ, gives
Ã
1 ∂p rc r ∂r
!2
∂p ∂2p φcµ ∂p + + r 2 = ∂r ∂r k ∂t
This is simplified to Ã
∂p c ∂r
!2
+
1 ∂p ∂ 2 p φµc ∂p + 2 = r ∂r ∂r k ∂t
219
∂p/∂r is small and since c << 1, the leading term on the LHS of the above equation may be neglected and the diffusivity equation becomes φcµ ∂p 1 ∂p ∂ 2 p + 2 = r ∂r ∂r k ∂t which may be written as
Ã
1 ∂ ∂p r r ∂r ∂r
!
=
φµc ∂p k ∂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 the basis for many important reservoir analysis techniques which include transient pressure well test analysis and aquifer modeling. The equation is also written as,
Ã
1 ∂ ∂p r r ∂r ∂r
!
=
1 ∂p η ∂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 the formation contains connate water and the formation is itself compressible, the compressibility, c, is the total compressibility of the system. This is determined using the following equation c = co So + cw Sw + cf where co , cw and cf are the oil, water and pore volume or formation compressibilities, respectively. So and Sw are the oil and water saturations. Typical values may be co cw cf Sw p
= = = = =
10 × 10−6 psi−1 3 × 10−6 psi−1 6 × 10−6 psi−1 0.20 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 steadystate solution) is for pseudo-steady-state flow. Pseudo-steady-state flow is a special form of stabilized unsteady-state flow in which the shape of the pressure profile from the wellbore to the drainage radius is independent of time, but the average drainage area pressure declines with production. The pseudo-steady state solution is important because it describes the situation of a reservoir producing by natural pressure depletion where the production for the well originates from within the well’s own drainage area. No flow occurs across the drainage radius. For a reservoir producing at a constant production rate, q, by natural pressure depletion, the average pressure, p¯, can be calculated from a simple material balance (total compressibility of the pore-fluid system). cVp (pi − p¯) = qt where pi is the initial pressure and, t is time, c is the compressibility and Vp is the pore volume. The boundary conditions are; (i) no flow across the external boundary, r
∂p = 0 at r = re ∂r
(ii) pseudo-steady-state flow condition, ∂p = constant ∂t The value for the constant is obtained from the material balance equation written above. Differentiating this equation we obtain, cVp (−d¯ p) = qdt or
d¯ p dp q = =− dt dt cVp
For the radial model, Vp = πre2 hφ, thus ∂p q =− ∂t cπre2 hφ 221
The radial diffusivity equation is Ã
1 ∂ ∂p r r ∂r ∂r
!
=
φµc ∂p k ∂t
substituting for ∂p/∂t, gives Ã
1 ∂ ∂p r r ∂r ∂r which simplifies to
!
=−
Ã
1 ∂ ∂p r r ∂r ∂r
!
φµc q k cπre2 hφ
=−
qµ πre2 kh
This equation may be written as Ã
∂p ∂ r ∂r
!
=−
qµ r∂r πre2 kh
Integrating the above equation gives, ∂p qµr2 r =− + C1 ∂r 2πre2 kh The constant of integration, C1 , is easily evaluated from the first BC i.e., r∂p/∂r = 0 at r = re qµ C1 = 2πkh Substituting for C1 , and simplifying gives ∂p qµ = ∂r 2πkh Separating variables, qµ ∂p = 2πkh Integrating,
Z
pr pw
1 r − 2 r re
!
!
1 r − 2 ∂r r re
qµ Z r ∂p = 2πkh rw
[p]pprw or
Ã
Ã
"
Ã
qµ r2 = ln r − 2 2πkh 2re Ã
!
1 r − 2 ∂r r re #r
rw
qµ r r2 r2 pr − pw = ln − 2 + w2 2πkh rw 2re 2re 222
!
The term rw2 /re2 is clearly negligible and the equation is written as, Ã
qµ r r2 pr − pw = ln − 2πkh rw 2re2
!
The above equation is a general expression for pressure as a function of radius. The particular case of practical interest is when r = re µ
qµ re 1 pe − pw = ln − 2πkh rw 2 or q=
2πkh (pe − pw ) µB ln(re /rw ) −
¶
1 2
With the inclusion of the skin-factor the equation is q=
2πkh (pe − pw ) µB ln(re /rw ) − 12 + s
This equation is useful for determining well production rates. It is interesting to compare this equation to the previously derived steady-state equation, 2πkh (pe − pw ) q= µB ln(re /rw ) + s The term ln(re /rw ) has a value typically between 6 and 7. The rates predicted by the steady-state and the pseudo steady-state equations are usually very similar (within 10%).
8.3.4
Flow Equations in terms of Average Reservoir Pressure
A problem with using the above radial flow equations is that although pw and q can 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 therefore more convenient to express the pressure drawdown in terms of the average reservoir, p¯. The average reservoir pressure is given by R re
pdVp rw dVp
p¯ = Rrwre
223
Now, dVp = 2πrhφdr, so
R re
p2πrhφdr rw 2πrhφdr
p¯ = Rrwre
R re
prdr rw rdr
p¯ = Rrwre
or
Z re 2 p¯ = 2 prdr re − rw2 rw
Since re2 − rw2 ≈ re2
2 Z re p¯ = 2 prdr re rw
For pseudo-steady-state flow, Ã
!
Ã
!
qµ r r2 ln − 2 pr − pw = 2πkh rw 2re or
qµ r r2 p = pw + ln − 2 2πkh rw 2re
Therefore, Ã ! 2 Z re 2 Z re qµ r r2 p¯ = 2 pw rdr + 2 ln − rdr re rw re rw 2πkh rw 2re2 Ã
2 re2 rw2 − p¯ = 2 pw re 2 2
!
Ã
!
2 qµ Z re r r2 + 2 r ln − dr re 2πkh rw rw 2re2
Again, rw2 << re2 , so Ã
!
2 qµ Z re r r2 p¯ = pw + 2 r ln − dr re 2πkh rw rw 2re2 The first term in the integrand must be evaluated by parts; Z
re
rw
Z
µ
"
¶
r r r2 r ln dr = ln rw 2 rw
re
rw
µ
¶
"
#re
rw
r r r2 r ln dr = ln rw 2 rw
224
−
#re
rw
Z
re rw
"
1 r2 dr r 2
r2 − 4
#re
rw
Again, neglecting small terms, Z
re
rw
µ
¶
r r2 re r2 r ln dr = e ln − e rw 2 rw 4
Integration of the second term gives; Z
re rw
"
r3 r4 dr = 2re2 8re2
#re
rw
=
re2 8
The equation for average pressure becomes, 2 qµ p¯ = pw + 2 re 2πkh Simplifying,
Ã
re2 re r2 r2 − e − e ln 2 rw 4 8
µ
qµ re 1 1 p¯ = pw + ln − − 2πkh rw 2 4 µ ¶ qµ re 3 p¯ − pw = ln − 2πkh rw 4
!
¶
Which is the required inflow equation. In the presence of formation damage or stimulation (positive or negative skin) the equation becomes µ
¶
qµ re 3 p¯ − pw = ln − +s 2πkh rw 4 or q=
2πkh (¯ p − pw ) µB ln(re /rw ) − 34 + s
The equation is similar to the previously derived equations of pseudo-steady-state and 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 (¯ p − pw ) µB ln(re /rw ) − 34 + s
may be generalized for wells located anywhere in an arbitrary drainage area q=
2πkh (¯ p − pw ) q µB ln( 2.2458A/(CA r2 ) + s w
225
where CA is the Dietz shape factor which depends on the geometry of the drainage area. 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 = πre 2 the pseudo steady-state solution may be written as 2πkh (¯ p − pw ) h q i q= µB ln(re /rw ) − 0.75 + ln( 2.2458π/CA ) + 0.75 + s
This equation may be written as 2πkh (¯ p − pw ) q= µB ln(re /rw ) − 0.75 + sCA + s where sCA is a shape related skin-factor and is given by, q
sCA = ln( 2.2458π/CA ) + 0.75
Problem 8.3 - Pseudo-steady-state radial flow and drainage area shape factors [RADSSF.mcd] Calculate the production rate of an oil well in a 160 acre drainage area when the average pressure is 1850 psia. The following data apply: k h µ rw pw s Bo
= = = = = = =
180 16 2.2 0.5 1230 0 1.1
md feet cp feet psia 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
CA
Shape of drainage area
CA 10.8
1
31.6 2
4.86 30.9 2.07 31.6 2.72
1 4
0.232
27.6
0.115 27.1 60o 3.39 21.9 1/3
3.13
1
1 2
22.6
1
0.607
2
5.38
1
1
4 1
0.111
2
2.36
5
0.098 12.9 In water drive reservoirs 19.1 4.57 In reservoirs of unknown production character 25
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 by the Dietz shape factors, it is usually possible to approximate the system using simple idealizations. An example of this is shown in the attached figure for a well completed close to the intersection of two linear sealing faults. If the angle between the faults is θ and the flow is assumed to be steady-state, the well rate is given by, 2πkh (pe − pw ) ³ ´ q=f µB ln re rw
where,
f=
θ 360o
The above equation is based on the fact that every streamline in a radial flow system can represent a no-flow boundary.
228
8.4 8.4.1
WELL PRODUCTIVITY Productivity Index
The productivity index, J, of a well is defined as the production rate, q, in STB/D divided by the pressure drawdown, ∆p, in psi. J=
q ∆p
The productivity index is a simple measure of the productivity of a production well. The productivity index for a well is not constant and may vary considerably over the life of the well. To see why this is so, consider the pseudo-steady-state radial flow equation, 2πkh(P¯ − Pw ) q= µB(ln( rrwe ) − 34 + s) So, J=
q 2πkkro h = (∆p) µBo (ln( rrwe ) − 34 + s)
The above equation may be grouped into those parameters which remain constant with time and those which change with time, J=
Ã
2πkh re (ln( rw ) − 34 + s)
!Ã
kro µBo
!
The terms kro /µBo vary with time because reservoir pressure ( which affect µ and Bo ) and saturation ( which affects kro ) change with time. Moreover, if the well is stimulated or if it gradually cleans-up with production, s will also change with 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 entire section of the reservoir. Many oil and gas wells are completed over only part of the producing sand interval. These wells are referred to as partially penetrating wells. When comparing a partially penetrating well to a fully penetrating well, the flow in the vicinity of a partially penetrating well must deviate from pure radial flow as streamlines converge towards the well. This results in an additional pressure drop close to the wellbore which can be treated as a pseudo-skin due to partial penetration.
230
A number of correlations have been published which account for the effect of partial penetration on well performance. All are based on potential flow solutions or numerical simulations for the two-dimensional cylindrical diffusivity equation in two-dimensions. The correlation by Papatzacos (1988) is typical. For this correlation the pseudo-skin factor, sp , is given by (see attached figure for definition of variables), sp =
Ã
!
Ã
1 πhD − 1 ln hpD 2
!
"
1 hpD + ln hpD 2 + hpD
µ
A−1 B−1
¶0.5 #
where,
hpD = h hD = rw
hp h s
kh kv
A=
h h1 + 0.25hp
B=
h h1 + 0.75hp
Initially a partially penetrating well behaves as if it is producing from a formation of thickness hp . After some time a transition occurs to flow from the entire formation and the establishment of the pseudo-skin factor, sp . If the productivity index (J = q/∆p) for a fully penetrating well is, J=
2πkh h 1 µB ln re /rw − 0.75
and that for a partially penetrating well is, Jpp =
2πkh h 1 µB ln re /rw − 0.75 + sp
than the ratio J/Jpp is readily calculated. As a rule of thumb, this ratio is approximately 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 for the 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; 100 md kh = kv = 5 md rw = 0.5 ft. µ= 1 cp B = 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 the continuity equation for a radial element, ∂ ∂ρ (ρq) = 2πrhφ ∂r ∂t We then substituted Darcy’s law for the volumetric flow rate, q, q=
2πkhr ∂p µ ∂r
to eliminate q. The result is, Ã
k ∂p 1 ∂ ρ r r ∂r µ ∂r
!
=φ
∂ρ ∂t
Assuming that permeability, k, is constant, we write, 1 ∂ r ∂r
Ã
ρ ∂p r µ ∂r
!
=
φ ∂ρ k ∂t
This is the basic material balance for a radial volume element of the reservoir and it applies to all flows. For the case of a slightly compressible fluid with constant viscosity, we have previously written the radial diffusivity equation as, Ã
1 ∂ ∂p r r ∂r ∂r
!
=
φµc ∂p k ∂t
Gases are highly compressible and we cannot expect that the above diffusivity equation will apply for gases.
Equation of state for a compressible fluid The pressure-volume-temperature (PVT) behavior of a gas is given by the real gas law, pv = znRT where z is the compressibility factor, v is the volume of the gas, T is the temperature (assumed to be constant), n is the number of moles of gas and R is the universal gas law constant. 233
Since
m M where m is the mass of the gas and M is the gas molecular weight, the real gas law may be written as, m M p ρ= = v RT z We may use this expression together with the definition of compressibility to write an expression for the compressibility of a real gas, n=
c=
1 ∂ρ ρ ∂p
which gives c=
z ∂(p/z) p ∂p
In performing the integration we note that z = z(p). z c= p
(
z c= p
(
)
∂ (pz −1 ) ∂p
1 p ∂z − 2 z z ∂p
)
Finally, c=
1 1 ∂z − p 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
Ã
ρ ∂p r µ ∂r
!
=
φ ∂ρ k ∂t
For a gas, µ = µ(p) and we cannot assume a constant viscosity. Since ρ=
M p RT z
234
we have
1 ∂ r ∂r
Ã
M p 1 ∂p r RT z µ ∂r
!
φ ∂p = k ∂t
Ã
∂ρ ∂p
!
Now, from our definition of compressibility, ∂ρ = cρ ∂p ∂ρ M p =c ∂p RT z which gives 1 ∂ r ∂r
Ã
M p 1 ∂p r RT z µ ∂r
!
=
φ ∂p M p c k ∂t RT z
dividing both sides by the constants we finally obtain, 1 ∂ r ∂r
Ã
p ∂p r µz ∂r
!
=
φcp ∂p kz ∂t
This is a highly non-linear equation because the coefficients on both sides of the equations are highly non-linear functions of pressure. In order to solve the equation it is necessary to first linearize it.
Pseudo-pressure Introduce a new variable called the pseudo pressure, ψ(p), defined as ψ(p) = 2
Z
p
a
p dp µz
where a is an arbitrarily low reference pressure. This transformation gives
Ã
!
p dψ = 2 dp µz Re-writing the non-linear diffusivity equation in therm of pseudo-pressure we have, Ã ! 1 ∂ p ∂ψ dp φcp ∂ψ dp r = r ∂r µz ∂r dψ kz ∂t dψ Now, ∂p µz = ∂ψ 2p
235
therefore, 1 ∂ r ∂r
Ã
p ∂ψ µz r µz ∂r 2p
!
=
φcp ∂ψ µz kz ∂t 2p
Simplifying this expression gives Ã
∂ψ 1 ∂ r r ∂r ∂r
!
=
φµc ∂ψ k ∂t
This has eliminated the non-linearity on the LHS and made the equation look like the diffusivity equation for a slightly compressible liquid except that the coefficient on the RHS is still non-linear (c is a function of pressure).
8.5.1
Low pressure approximation — p < 3, 000psi and ∆p small
For low pressure and small pressure drop, µz ≈ constant = µ ¯z¯ The definition of pseudo pressure, ψ(p) = 2
Z
p
a
p ∂p µz
with the reference pressure to zero, is now written as, ψ(p) =
2 Zp p∂p µ ¯z¯ 0
ψ(p) =
2 p2 µ ¯z¯ 2
p2 µ ¯z¯ 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 for the coefficient terms, c¯ and µ ¯, resulting in a linear diffusivity equation in terms of the pseudo pressure. Ã ! 1 ∂ ∂ψ φ¯ µc¯ ∂ψ r = r ∂r ∂r k ∂t ψ(p) =
If we are doing the analysis by hand and wish to eliminate the need to evaluate pseudo-pressures we may simply write the equation in terms of p2 . ∂ψ 1 = ∂(p2 ) µ ¯z¯ 236
and
Ã
1 ∂ ∂(p2 ) ∂ψ r r ∂r ∂r ∂(p2 )
which gives
Ã
1 ∂ ∂(p2 ) r r ∂r ∂r
!
φ¯ µc¯ ∂(p2 ) ∂ψ = k ∂t ∂(p2 )
!
=
φ¯ µc¯ ∂(p2 ) k ∂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 ∆p small
At high pressures and small pressure drops, p p¯ ≈ constant = µz µ ¯z¯ From the definition of pseudo pressure we have, ψ(p) = 2
Z
p
a
which can be written as or
p ∂p µz
p¯ Z p ∂p ψ(p) = µ ¯z¯ a p¯ p µ ¯z¯
ψ(p) =
Again the pseudo-pressure is easily evaluated, and as for the case of low pressure we can show that the diffusivity in terms of ψ is linearized to Ã
1 ∂ ∂(p) r r ∂r ∂r
!
=
φ¯ µc¯ ∂(p) k ∂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 average values for these parameters over the flow conditions. In terms of ψ the non-linear diffusivity equation is Ã
1 ∂ ∂ψ r r ∂r ∂r We may write this as
Ã
1 ∂ ∂ψ r r ∂r ∂r
!
!
Defining a pseudo-time as, τ=
=
=
Z
t
0
or
φµc ∂ψ k ∂t
φ ∂ψ k {∂t/µc} dt µc
dt µc Pseudo-time is easily evaluated since pressure, as a function of time, is known. dτ =
The diffusivity equation may now be written as, Ã
∂ψ 1 ∂ r r ∂r ∂r
!
=
φ ∂ψ k ∂τ
A problem with the pseudo-time it does not have units of time and this is undesirable in actual practice where the engineer likes to relate time to distance via the depth of investigation concept. The situation is readily remedied by the introduction of a normalized pseudotime, τn , defined as Z t dt τn = (µc)i 0 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, but convention is to take the value at the initial test pressure. The diffusivity equation in terms of pseudo-pressure and pseudo-time is, Ã
1 ∂ ∂ψ r r ∂r ∂r
!
=
φ(µc)i ∂ψ k ∂τn
Again, this equation has exact ally the same form as the diffusivity equation for a slightly compressible fluid. 238
8.5.4
Steady and Pseudo-Steady State Radial Gas Flow
The radial diffusivity equation for gas is Ã
1 ∂ ∂ψ r r ∂r ∂r
!
φµc ∂ψ k ∂t
=
For steady-state flow, ∂ψ/∂t = 0, and we can write Ã
∂ψ 1 ∂ r r ∂r ∂r or,
Ã
∂ ∂ψ r ∂r ∂r
!
!
=0
=0
Integrating gives r
∂ψ = C1 ∂r
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 for radial flow, 2πkh ∂p q= r µ ∂r where q is the rate at reservoir conditions. The rate at surface conditions is given by 2πkh ∂p r qsc = µBg ∂r where Bg is the gas formation volume factor which is given by µ
psc Bg = Tsc
¶
zT p
Substituting for Bg gives qsc =
Ã
2πTsc psc
!
kh p ∂p r µ zT ∂r
The equation is written in terms of pseudo-pressure as qsc =
Ã
!
2πTsc psc
kh p ∂ψ dp r µ zT ∂r dψ
From the definition of pseudo-pressure dp µz = dψ 2p 239
Substituting for dp/dψ gives qsc =
Ã
or, r
!
πTsc psc µ
kh ∂ψ r T ∂r
∂ψ psc = ∂r πTsc
¶
qsc T kh
This is the required constant of integration C1 . Separating variables,
Integrating,
gives
Z
rw re
∂r = r
Ã
∂r = r
Ã
=
Ã
qsc =
Ã
πTsc psc
re ln rw
kh ∂ψ qsc T
!
kh Z ψw ∂ψ qsc T ψe
!
kh (ψe − ψw ) qsc T
πTsc psc
¶
µ
!
πTsc psc
πTsc psc
or, !
kh (ψe − ψw ) ³ ´ T ln re rw
As before, for pseudo-steady state flow in terms of average pseudo-pressure and in the presence of skin, we can write qsc =
Ã
πTsc psc
!
kh (ψ¯e − ψw ) h ³ ´ i T ln re − 0.75 + s rw
The above equation is valid for all conditions and may be used to estimate the stabilized rate for gas wells under Darcy flow conditions. It requires evaluation of pseudo-pressures. Pseudo-pressures are easily calculated by numerically integrating P , µ and z data for the gas. An example of such a calculation is given in Exercise-8.11.
8.5.5
Non-Darcy flow
In high rate gas wells flow velocities close to the wellbore may sufficiently high for 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 be included 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. The absolute value of rate indicates that there is a positive non-Darcy flow loss for both production and injection operations. D and s are determined from multirate well pressure test data. Pseudo-steady state flow in terms of average pseudo-pressure and in the presence of skin and non-Darcy flow effects, can now be written as qsc = or, qsc =
8.5.6
Ã
Ã
πTsc psc
πTsc psc
!
!
kh (ψ¯e − ψw ) h ³ ´ i T ln re − 0.75 + s0 rw
kh (ψ¯e − ψw ) h ³ ´ i T ln re − 0.75 + s + D|qsc | rw
Pressure-squared approximation
For pressures below about 3000 psia the product µz is approximately constant and pseudo-pressure, defined as Z p p ψ=2 dp 0 µz may be written as
2 Zp p2 ψ= pdp = µ ¯z¯ 0 µ ¯z¯ where µ ¯ and z¯ are evaluated at average flowing conditions. Substituting for ψ in the rate equation gives qsc =
8.5.7
Ã
πTsc psc
!
kh (pe 2 − pw 2 ) h ³ ´ i µ ¯z¯T ln re − 0.75 + s + D|qsc | rw
High Pressure approximation
For pressures above about 3000 psia the product p/µz is approximately constant and pseudo-pressure, defined as Z p p ψ=2 dp 0 µz 241
may be written as
2¯ p 2¯ pZp dp = ψ= p µ ¯z¯ 0 µ ¯z¯
where µ ¯, z¯ and p¯ are evaluated at average flowing conditions. Substituting for ψ in the rate equation gives qsc =
Ã
2πTsc psc
!
(pe − pw ) kh¯ p h ³ ´ i r e µ ¯z¯T ln − 0.75 + s + D|qsc | rw
The average gas formation factor may be written as ¯g = B
Ã
Tsc psc
!
z¯T¯ p¯
The gas rate equation becomes qsc =
2πkh (pe − pw ) h ³ ´ i ¯ µ ¯Bg ln re − 0.75 + s + D|qsc | rw
With the non-Darcy flow term set to zero (D = 0), this equation is identical to the 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 pressure is 2723 psia and the flowing wellbore pressure is maintained at 2655 psia. The following reservoir and fluid properties are known: k= h= rw = re = µ= z= T = s= D=
5 md 35 ft. 0.5 ft. 2640 ft. 0.018 cp 0.87 200o F -0.3 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 is 2723 psia and the flowing wellbore pressure is maintained at 2655 psia. Reservoir temperature is 200o F and the gas gravity is 0.7. Use pseudo-pressure and the pressure-squared method to calculate the gas rate. The following reservoir and fluid properties are known: k= h= rw = re = T = s= D=
5 md 35 ft. 0.5 ft. 2640 ft. 200o F -0.3 0.015 (MMSCF/day)−1 GAS PROPERTIES p (psia)
µ (cp)
z
150 300 450 600 750 900 1050 1200 1350 1500 1650 1800 1950 2100 2250 2400 2550 2700 2850 3000 3150
0.01238 0.01254 0.01274 0.01303 0.01329 0.01360 0.01387 0.01428 0.01451 0.01485 0.01520 0.01554 0.01589 0.01630 0.01676 0.01721 0.01767 0.01813 0.01862 0.01911 0.01961
0.9856 0.9717 0.9582 0.9453 0.9332 0.9218 0.9112 0.9016 0.8931 0.8857 0.8795 0.8745 0.8708 0.8684 0.8671 0.8671 0.8683 0.8705 0.8738 0.8780 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(¯ p2R − 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. The equation is frequently used to determine well deliverability.
Flow after flow tests The constants C and n are determined from flow after flow or conventional backpressure tests. The well is flowed at a constant rate (qsc ) until the flowing wellbore pressure (pw ) has stabilized. The rate is then changed (increased) and the process repeated. After a number of rate changes (usually four), the well is shutin until the shutin wellbore has stabilized. This is the average reservoir pressure (¯ pR ).
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(¯ p2R − p2w ) against log qsc gives a straight line of slope equal to 1/n and intercept −(1/n) log C. The test and interpretation procedure is illustrated in 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 point flow-after-flow test on a gas well: Flow period qsc (MMSCF/day) pw (psia) 1 2190 3387 2 2570 3268 3 3160 3092 4 3400 3015 shutin 0 3884
247
Determine the exponent n and the performance coefficient C. Calculate the Absolute Open Flow (AOF) for the well (the AOF is a commonly used figure-ofmerit for a gas well and corresponds to the well rate at a flowing well pressure of one atmosphere). [Answer: n = 0.874, C = 4.035 × 10−6 MMSCF/day − psia2n , AOF=7.6 MMSCF/day]
Problem 8.7 - Well deliverability using back pressure equation [DELGAS.mcd] A four point flow-after-flow test produced the following back pressure equation constants, n = 0.69 C = 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 maintained 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 recovery operations. The major advantages over conventional vertical wells are in overcoming: Reservoir heterogeneity - Horizontal wells contact a larger area of the reservoir than vertical wells. In reservoirs which display complex and difficult to predict heterogeneity (carbonates, channel point bar sands, braided stream systems and fractured reservoirs), horizontal wells can connect regions of high permeability rock which are separated by regions of low permeability. Reservoir flow problems - The greater length of horizontal wells may result in smaller pressure gradients (increased productivity) when compared to vertical wells flowing at the same rate. This can be important in allowing production of heavier crudes at economical rates and reducing or completely elimination the detrimental effects of water and gas coning. 249
Figure 8.24: Comparison between effective reservoir contact areas for horizontal and vertical wells - In sands where water coning is a problem horizontal wells can be placed near the top of the sand increasing the vertical distance between the perforations and the water-oil contact and therefore increasing recovery when compared with a vertical well in the same sand. Recovery is also increased because the cone for a similar situation exists in a sand subject to gas coning with the horizontal well being placed at the base of the sand. - In low permeability reservoirs it is possible to place multiple transverse fractures along the length of the well and greatly increase productivity over that possible with a vertical well.
Completion options The attached figure shows a number of different techniques used to complete horizontal wells. Open hole - Inexpensive but limited to competent formations. Difficult to control rate along length of well and difficult to stimulate. Rarely used except for formations such as Austin Chalk. Slotted liner - Three types commonly used; perforated, slotted, and prepacked liners. Liners are used to prevent hole collapse and to control sand production.
250
Figure 8.25: Horizontal wells with multiple transverse fractures play a vital role in the production of tight gas reservoirs
Liners with partial isolation allow the horizontal section to be divided into several isolated sections. This allows production control along the well length and selective stimulation or multiple fracking. Cemented and perforated liners provide the best completions provided that care 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. The area drained by a single well in the pattern is the drainage area for the well. A vertical well drains a cylindrical volume and a horizontal well drains an ellipsoid (three-dimensional ellipse). A horizontal well will usually drain a larger area than a vertical well. Areally isotropic reservoir - kx = ky A horizontal well can be viewed as a number of vertical wells drilled on a line and 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 in Alaska the drainage area of a horizontal well may be estimated from that for a vertical well as shown in the attached figure. If the drainage area for a vertical well is Av , the drainage radius is rev =
q
Av /π
The drainage area for a vertical well may also be approximated as a square of dimensions 2xe by 2ye where xe = ye Av = 4xe or
s
Av 4 A horizontal well can be viewed as a number of vertical wells drilled on a line and completed in a limited pay thickness. Under similar operating conditions, the drainage area of a horizontal well may be estimated from that for a vertical well as shown in the attached figure. xe = ye =
Ah = 2ye L + Av where L is the length of the horizontal section. The effective drainage radius of a horizontal well is q reh = Ah /π
As a rule of thumb, a 1000 ft horizontal well will drain twice the area of a vertical well 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). Assuming steady-state flow in two dimensions, we can write the following conservation equation à ! à ! ∂ ∂p ∂ ∂p kx + ky =0 ∂x ∂x ∂y ∂y or, since kx and ky are constant kx
∂2p ∂2p + k =0 y ∂x2 ∂y 2
For an isotropic reservoir, kx = ky = kh , where k−h is the horizontal permeability, we can write " # ∂2p ∂2p kh + =0 ∂x2 ∂y 2 For an anisotropic q reservoir we multiply and divide the conservation equation throughout by kx ky v s u 2 u kx ∂ 2 p k ∂ p y =0 kx ky t + 2 2
q
or
ky ∂x
q
kx ky
"
kx ∂y #
∂ 2 p ky ∂ 2 p + =0 ∂x2 kx ∂y 2
This equation may be written as q
kx ky
"
#
∂2p ∂ 2p + =0 ∂x2 ∂Y 2
where y=Y
s
ky kx
Comparing this equation with that for the isotropic system, kh
"
#
∂2p ∂2p + =0 ∂x2 ∂y 2
shows that in an anisotropic reservoir a vertical q well drains a rectangle which has a length along the high permeability direction ky /kx the length along the low permeability direction with an effective permeability
256
q
kx ky .
Figure 8.30: Drainage areas for horizontal wells in fractured or anisotropic reservoirs
257
Figure 8.31: Drainage areas for vertical and horizontal wells in anisotropic reservoirs If ky > kx , ye =
s
ky xe kx
and the drainage area for a vertical well is Ah = 4ye xe and the effective horizontal permeability is, kh =
q
kx ky
The above analysis shows that it is difficult to drain a reservoir in the low permeability direction using vertical wells. If a horizontal well is drilled in the direction of low permeability, the drainage area for the well is Ah = 2ye L + Av The number of horizontal wells required for adequate reservoir drainage is considerably smaller than that for a conventional vertical well development.
258
Example 8.7 - Drainage area for horizontal wells in an isotropic reservoir [HORWELL.mcd] A 600 acre lease is to be developed with 10 vertical wells. An alternative is to drill 500 ft, 1000 ft, or 2000 ft long horizontal wells. Estimate the number of horizontal wells required to drain the lease effectively assuming that the reservoir is 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 fractured reservoir. The permeability along the direction of the fractures is ky =4.5 md. The permeability in the direction perpendicular to the fractures is only kx =0.5md. What are the drainage areas for a 2000 ft horizontal well drilled along the direction of the fractures and one drilled in a direction perpendicular to the fractures. [Answer: 75 acres, 145 acres.]
259
Figure 8.32: Complexity of flow regimes for horizontal wells. a-early time radial flow, 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 horizontal wells with two two-dimensional solutions
8.6.2
Productivity of Horizontal Wells
The steady-state rate for a vertical is given by qv =
2πkh h ∆p µB ln(rev /rw )
The productivity index, Jv , is defined as the rate, qv , divided by the pressure difference, ∆p, 2πkh h/(µB) Jv = 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 solution techniques employed and the details of the assumptions made in developing the solution. The differences are generally small. Joshi (JPT June 1988)divided the three-dimensional steady-state flow problem into separate two-dimensional vertical and horizontal problems and summed the solutions. The solution obtained was shown to be in excellent agreement with laboratory simulations using electrical networks. For a horizontal well completed in the middle of sand of thickness, h, with constant pressure boundary conditions at the drainage area boundary, Jh =
qh = ∆p
ln
∙
√
a+
2πkh h/(µB)
a2 −(L/2)2 L/2
261
¸
+
³ ´ h L
ln
³
h 2rw
´
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,
L a = 0.5 + 2
s
µ
2reh 0.25 + L
reh =
s
¶4
0.5
Ah π
for L > h and (L/2) < 0.9reh . The above equation shows that as reservoir thickness, h, increases, the effectiveness of the horizontal well, Jh , decreases. Horizontal wells are most effective in thin reservoirs. Another equation for horizontal wells was developed by Borisov(Nedra, Moscow 1964 - translated into English in 1984) Jh =
2πkh h/(µB) ln (4reh /L) + (h/L) ln (h/2πrw )
This equation gives results very similar to those using the Joshi equation. In the limit L >> h, the second term in the denominator of the equation is negligible and the result can be rewritten as Jh =
2πkh h/(µB) ln (reh /[L/4]) 262
Comparing this equation with the corresponding equation for a vertical well shows that in the limit L >> h, a horizontal well is equivalent to a vertical well with 0 an effective wellbore radius rw = L/4 (a highly stimulated vertical well). We will later show that the above equation is identical to that for a vertical well with an infinite-conductivity vertical fracture of length L.
Isotropic reservoirs - kv 6= kh When the horizontal permeability is different from the vertical permeability, in a manner similar to that previously shown, the flow field is equivalent to that for √ an average permeability of kh kv with the z-axis modified, Z=z
s
kh kv
The influence of anisotropy is therefore to change the effective reservoir thickness, H, s kh H=h kv 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 simply substitute the effective thickness of the reservoir for h in the equation for an isotropic reservoir. The result is, Jh = ln
∙
√
a+
2πkh h/(µB)
a2 −(L/2)2 L/2
where β=
s
263
¸
+
kh kv
³
βh L
´
ln
³
βh 2rw
´
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 with respect to the sand in which the well is completed. A schematic of an off-centered horizontal well is shown in the attached figure. We define well eccentricity, δ, as δ=
h/2 − zw h/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 = ln
∙
a+
√
2πkh h/(µB)
a2 −(L/2)2 L/2
¸
+
³
βh L
´
ln
³
βh [1 2rw
+ 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 well can be located anywhere in the vertical plane without significant loss in productivity. This is strictly true only for bound reservoirs where the top and bottom boundaries 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 160 acre spacing. Other data reservoir and fluid parameters are; kh =50 md µ =0.4 cp rw =0.33 ft B =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 located 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, 13.2 STB/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. The horizontal permeability averages 200 md and the vertical permeability averages 20 md. Other reservoir and fluid parameters are; µ =0.2 cp rw =0.33 ft B =1.35 RB/STB What is the improvement in productivity which may be expected over that for a vertical 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 on horizontal well productivity. Consider the case of a horizontal well completed at the center of a homogeneous anisotropic sand. Renard and Dupuy (SPE19414, 1990) developed the following equation for the productivity index, Jh =
−1
cosh
³
2πkh h/(µB) 2a L
´
+
³
βh L
´
ln
³
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 Joshi produce similar estimates for horizontal well productivity. For a damaged well, Renard and Dupuy wrote, Jh,d =
cosh−1
³
2a L
´
2πkh h/(µB) +
³
βh L
´
ln
³
βh (1+β)πrw
´
+ sh
where sh is the horizontal well skin factor which must be determined from well tests. The ratio of productivities for a damaged and undamaged horizontal well is therefore, ³ ´ ³ ´ ³ ´ βh βh cosh−1 2a + ln Jhd L L (1+β)πrw ³ ´ ³ ´ ³ ´ = βh βh −1 2a Jh cosh + ln + sh L L (1+β)πrw or Jhd B = Jh B + sh where B = cosh
−1
µ
¶
Ã
!
Ã
2a βh βh + ln L L (1 + β)πrw
267
!
Figure 8.38: Comparison between formation damage for horizonral and vertical wells
268
