+peters ekwere j. - petrophysics
TRANSCRIPT
TABLE OF CONTENTS
Page 1 PETROLEUM RESERVOIR ROCKS ......................................... 1-1
1.1 PETROPHYSICS ............................................................................... 1-1 1.2 PETROLEUM RESERVOIR ROCKS ................................................1-2 1.3 MINERAL CONSTITUENTS OF ROCKS—A REVIEW ...................1-4
1.4 ROCKS............................................................................................... 1-5 1.4.1 Igneous Rocks.................................................................... 1-5 1.4.2 Metamorphic Rocks...........................................................1-6 1.4.3 Sedimentary Rocks............................................................1-6
1.5 CLASSIFICATION OF SEDIMENTARY ROCKS ............................. 1-7 1.5.1 Clastic Sedimentary Rocks ................................................ 1-7 1.5.2 Chemical Sedimentary Rocks............................................ 1-7 1.5.3 Organic Sedimentary Rocks ..............................................1-8
1.6 DISTRIBUTION OF SEDIMENTARY ROCK TYPES .................... 1-10
1.7 SANDSTONE RESERVOIRS (CLASTIC SEDIMENTARY ROCK) ............................................................................................. 1-10 1.7.1 Pore Space ....................................................................... 1-12 1.7.2 Compaction and Cementation......................................... 1-15 1.7.3 Classification ................................................................... 1-17
1.8 CARBONATE RESERVOIRS (LIMESTONES AND DOLOMITES) .................................................................................1-20 1.8.1 Classification ................................................................... 1-21 1.8.2 Pore Space .......................................................................1-22
1.9 FRACTURED RESERVOIRS ..........................................................1-28
1.10 RESEVOIR COLUMN.....................................................................1-29
REFRENCES...................................................................................1-32
2 POROSITY AND FLUID SATURATIONS..................................2-1
2.1 DEFINITION OF POROSITY ...........................................................2-1
2.2 FACTORS AFFECTING SANDSTONE POROSITY ........................ 2-2
2.3 FACTORS AFFECTING CARBONATE POROSITY ........................ 2-4
2.4 TYPICAL RESERVOIR POROSITY VALUES.................................. 2-5
2.5 LABORATORY MEASUREMENT OF POROSITY.......................... 2-6
2.5.1 Direct Porosity Measurement by Routine Core Analysis .................................................................... 2-6 2.5.2 Indirect Porosity Measurement by CT Imaging.............. 2-11
2.6 FLUID SATURATIONS ..................................................................2-16
2.7 INDIRECT POROSITY MEASUREMENTS FROM WELL LOGS................................................................................... 2-24 2.7.1 Introduction to Well Logging......................................... 2-24 2.7.2 Mud filtrate Invasion...................................................... 2-25 2.7.3 Porosity Logs .................................................................. 2-32 Density Log .................................................................. 2-32 Sonic Log (Acoustic Log) ............................................. 2-36 Neutron Log .................................................................2-41 Combination Porosity Logs ......................................... 2-45 2.7.4 Resistivity Log ................................................................ 2-46 Electric Log ................................................................. 2-54 Induction-Electric Log................................................ 2-56 Dual Induction Laterolog ........................................... 2-58 Focused Electric Log (Guard and Laterolog) ............. 2-62 Microresistivity Logs................................................... 2-65 2.7.5 Lithology Logs ................................................................ 2-68 Spontaneous Potential Log (SP)................................. 2-68 The Gamma Ray Log (GR)...........................................2-73 2.7.6 Nuclear Magnetic Resonance (NMR) Logs.................... 2-76 Nuclear Spins in a Magnetic Field.............................. 2-76 The Effect of Radiofrequency Pulses - Resonance Absorption ............................................... 2-79 Relaxation Processes...................................................2-80 Molecular Diffusion Effect.......................................... 2-84 NMR Signal and Corresponding T2 Spectrum ........... 2-84 Pore Size Distribution................................................. 2-89 Estimation of Permeability from NMR Relaxation Times............................................... 2-95 2.7.7 NMR Imaging of Laboratory Cores................................ 2-97 The Effect of Magnetic Field Gradients...................... 2-98 Slice-Selective Excitation............................................ 2-99 Frequency Encoding ................................................. 2-100 Phase Encoding..........................................................2-101 Image Reconstruction............................................... 2-102 Three-Dimensional NMR Imaging............................2-103 Signal-to-Noise Ratio and Image Contrast............... 2-104 Example NMR Images of Laboratory Cores..............2-105 2.7.8 A Comparison of Various Porosity Measurements for Shaly Sand....................................... 2-112 2.8 RESERVE ESTIMATION PROJECT ........................................... 2-113 2.8.1 Reserve Estimation........................................................ 2-114
2.8.2 Economic Evaluation..................................................... 2-115 2.8.3 Simulation Procedure.................................................... 2-116 2.8.4 Sampling Procedure ...................................................... 2-116 2.8.5 Simulation Output.........................................................2-124
2.9 PORE VOLUME COMPRESIBILITY............................................2-126
NOMENCLATURE .......................................................................2-135
REFRENCES AND SUGGESTED READINGS.............................2-138
3 PERMEABILITY .....................................................................3-1
3.1 DEFINITION ....................................................................................3-1
3.2 DIMENSIONS AND UNIT OF PERMEABILITY ............................ 3-6
3.3 LABORATORY DETERMINATION OF PERMEABILITY...............3-7
3.4 FIELD DETERMINATION OF PERMEABILITY ..........................3-14 3.4.1 Diffusivity Equation for Slightly Compressible Liquid........................................................3-15 3.4.2 Pressure Drawdown Equation ........................................3-19 3.4.3 Pressure Buildup Equation ............................................ 3-22 3.4.4 Diagnostic Plots.............................................................. 3-24 3.4.5 Skin Factor...................................................................... 3-30 3.4.6 Homogenous Reservoir Model with Wellbore Storage and Skin............................................. 3-33 3.4.7 Type Curve Matching ......................................................3-37 3.4.8 Radius of Investigation of a Well Test ........................... 3-40 3.4.9 Field Example of Well Test Analysis .............................. 3-40 3.4.10 Welltest Model for Dry Gas Reservoir .......................... 3-52
3.5 FACTORS AFFECTING PERMEABILITY..................................... 3-56 3.5.1 Compaction..................................................................... 3-56 3.5.2 Pore Size (Grain Size) ..................................................... 3-56 3.5.3 Sorting ............................................................................ 3-60 3.5.4 Cementation ................................................................... 3-60 3.5.5 Layering .......................................................................... 3-60 3.5.6 Clay Swelling....................................................................3-61
3.6 TYPICAL RESERVOIR PERMEABILITY VALUES .......................3-61
3.7 PERMEABILITY-POROSITY CORRELATIONS............................3-61
3.8 CAPILLARY TUBE MODELS OF POROUS MEDIA..................... 3-69 3.8.1 Carman-Kozeny Equation .............................................. 3-69 3.8.2 Tortuosity ........................................................................3-75 3.8.3 Calculation of Permeability from Pore
Size Distribution............................................................. 3-79 3.9 STEADY STATE FLOW THROUGH FRACTURES....................... 3-84
3.10 AVERAGING PERMEABILITY DATA .......................................... 3-85
3.11 DARCY’S LAW FOR INCLINED FLOW........................................ 3-88 3.12 VALIDITY OF DARCY’S LAW....................................................... 3-99
3.13 NON-DARCY FLOW..................................................................... 3-101
3.14 DARCY’S LAW FOR ANISOTROPIC POROUS MEDIA............. 3-106 3.14.1 Definition of Homogeneity and Anisotropy ................ 3-106 3.14.2 Darcy’s Law for Homogeneous and Anisotropic Medium.............................................3-107 3.14.3 Transformation of Permeability Tensor from One Coordinate system to Another .................... 3-114 3.14.4 Alternative Calculation of the Principal Values and the Principal Axes of the Permeability Anisotropy..............................................3-122 3.14.5 Directional Permeability...............................................3-124 3.14.6 Measurement of Transverse Permeability of a Cylindrical Core ....................................................3-137
3.15 EXAMPLE APPLICATIONS OF PERMEABILITY.......................3-140 3.15.1 Productivity of Horizontal Well ....................................3-140 Introduction............................................................... 3-141 Homogeneous and Isotropic Reservoirs ................... 3-141 Homogeneous and Anisotropic Reservoirs ...............3-145 3.15.2 Productivity of a Vertically Fractured Well..................3-152
NOMENCLATURE .......................................................................3-155
REFRENCES AND SUGGESTED READINGS.............................3-159
4 HETEROGENEITY ................................................................ 4-1
4.1 INTRODUCTION..............................................................................4-1
4.2 MEASURES OF CENTRAL TENDENCY AND VARIABILITY (HETEROGENEITY)............................................... 4-3 4.2.1 Measures of Central Tendency......................................... 4-3 Mean.............................................................................. 4-3 Geometric Mean............................................................ 4-3 Median .......................................................................... 4-3 Mode.............................................................................. 4-4 4.2.1 Measures of Variability (Heterogeneity or Spread) ............................................... 4-4 Variance ........................................................................ 4-4
Dykstra-Parsons Coefficient of Variation..................... 4-5 Lorenz Coefficient ......................................................... 4-8
4.3 MEASURES OF SPATIAL CONTINUITY ...................................... 4-11 4.3.1 Variogram........................................................................4-13 Definition .....................................................................4-13 How to Calculate the Variogram .................................4-16 Physical Meaning of the Variogram............................ 4-27 Variogram Models....................................................... 4-28 Fitting a Theoretical Variogram Model to an Experimental Variogram ....................................... 4-35 Variogram Anisotropy .................................................4-41 Example Experimental Variograms............................ 4-44 4.3.2 Covariance (Autocovariance) Function...........................4-51 Definition .....................................................................4-51 Physical Meaning of Covariance Function ................. 4-54 4.3.3 Correlation Coefficient Function (Autocorrelation Function) .............................................4-57
4.4 PROBABILITY DISTRIBUTIONS ................................................. 4-59 4.4.1 Normal (Gaussian) Distribution .................................... 4-60 4.4.2 Log Normal Distribution................................................ 4-72
4.5 ESTIMATION .................................................................................4-75 4.5.1 Introduction ....................................................................4-75 4.5.2 Ordinary Kriging Equations ........................................... 4-86 Derivation in Terms of the Covariance Function ................................................... 4-89 Derivation in Terms of the Variogram ....................... 4-94 Solution of the Kriging Equation in terms of the Covariance Function......................................... 4-98 Solution of the Kriging Equation in terms of Variogram ..............................................................4-103
4.6 CONDITIONAL SIMULATION....................................................4-132 4.6.1 Introduction ..................................................................4-132 4.6.2 Sequential Gaussian Simulation ...................................4-132 4.6.3 A Practical Application of Sequential Gaussian Simulation .....................................................4-136
NOMENCLATURE .......................................................................4-148
REFRENCES AND SUGGESTED READINGS.............................4-149
5 DISPERSION IN POROUS MEDIA ..........................................5-1
5.1 INTRODUCTION..............................................................................5-1
5.2 LABORATORY FIRST-CONTACT MISCIBLE DISPLACEMENTS........................................................................... 5-3
5.3 ORIGIN OF DISPERSION IN POROUS MEDIA.......................... 5-20 5.3.1 Molecular Diffusion.........................................................5-21 5.3.2 Mechanical Dispersion ....................................................5-21
5.4 CONVECTION-DISPERSION EQUATION................................... 5-23 5.4.1 Generalized Equation in Vector Notation...................... 5-23 5.4.2 One Dimensional Convection-Dispersion Equation ........................................................................ 5-25
5.4.2 Solution of the One-Dimensional Convection-Dispersion Equation ................................... 5-26
5.5 DISPERSION COEFFICENT AND DISPERSIVITY ..................... 5-42
5.6 MEASURMENT OF DISPERSION COEFFICENT AND DISPERSIVITY ......................................................................5-53
5.6.1 Traditional Laboratory Method with Breakthrough Curve .......................................................5-53 5.6.2 Laboratory Method of Peters et al. (1996) ..................... 5-56 5.6.3 Field Measurement of Dispersion Coefficient and Dispersivity ............................................ 5-71
5.7 FACTORS THAT COULD AFFECT DISPERSION COEFFICENT AND DISPERSIVITY ..............................................5-75
5.8 NUMERICAL MODELING OF FIRST-CONTACT MISCIBLE DISPLACEMENT .........................................................5-79 5.8.1 Introduction ....................................................................5-79 5.8.2 Mathematical Model of First-Contact Miscible Displacement ....................................................5-79 5.8.3 Numerical Modeling of Laboratory Experiments.......... 5-82 Experiment 1 ................................................................. 5-84 Experiment 2..................................................................5-91 Experiment 3................................................................. 5-99 Experiment 4................................................................5-106 Experiment 5................................................................ 5-116 Experiment 6................................................................ 5-121 NOMENCLATURE .......................................................................5-126
REFRENCES AND SUGGESTED READINGS.............................5-128 6 INTERFACIAL PHENOMENA AND WETTABILITY................ 6-1
6.1 INTRODUCTION..............................................................................6-1
6.2 SURFACE AND INTERFACIAL TENSIONS................................... 6-2 6.2.1 Surface Tension ................................................................ 6-2 6.2.2 Interfacial Tension .......................................................... 6-11
6.2.3 Measurement of Surface and Interfacial Tension......................................................... 6-20 Capillary Rise Experiment ............................................ 6-20 Sessile Drop Method ..................................................... 6-24 Pendant Drop Method .................................................. 6-26 Ring Method.................................................................. 6-27 Spinning Drop Method ................................................. 6-30
6.3 WETTABILITY................................................................................6-31 6.3.1 Definition.........................................................................6-31 6.3.2 Determination of Wettability ......................................... 6-36 Contact Angle Method ................................................. 6-37 Amott Wettability Test.................................................. 6-40 United State Bureau of Mines (USBM) Wettability Test............................................................. 6-42 6.3.3 Wettability of Petroleum Reservoirs.............................. 6-45 6.3.4 Effect of Wettability on Rock-Fluid Interactions........... 6-46 Microscopic Fluid Distribution at the Pore Scale ..................................................................... 6-47 Effect of Wettability on Irreducible Water Saturation .......................................................... 6-47 Effect of Wettability on Electrical Properties of Rocks ...................................................... 6-48 Effect of Wettability on the Efficiency of an Immiscible Displacement .............................................6-51
6.3 THERMODYMAMICS OF INTERFACES ..................................... 6-64 6.4.1 Characterization of Interfacial Tension as Specific Surface Energy.............................................. 6-64 6.4.2 Characterization of Microscopic Pore Level Fluid Displacements....................................................... 6-66 Case 1. Displacement of a Nonwetting Phase by a Wetting Phase ............................................ 6-67 Case 2. Displacement of a Wetting Phase by a Nonwetting Phase ................................................. 6-69
NOMENCLATURE .........................................................................6-71
REFRENCES AND SUGGESTED READINGS ..............................6-73 7 CAPILLARY PRESSURE .........................................................7-1
7.1 DEFINITION OF CAPILLARY PRESSURE ..................................... 7-1
7.2 CAPILLARY PRESSURE-SATURATION RELATIONSHIP FOR A POROUS MEDIUM.............................................................. 7-8 7.3 DRAINAGE CAPILLARY PRESSURE CURVE .............................. 7-17
7.4 CONVERSION OF LABORATORY CAPILLARY PRESSURE
DATA TO RESERVOIR CONDITIONS .......................................... 7-21 7.5 AVERAGING CAPILLARY PRESSURE DATA .............................. 7-21 7.6 DETERMINATION OF INITIAL STATIC RESERVOIR FLUID SATURATIONS BY USE OF DRAINAGE CAPILLARY PRESSURE CURVE.................................................. 7-28 7.7 CAPILLARY PRESSURE HYSTERESIS.........................................7-45 7.8 CAPILLARY IMBIBITION..............................................................7-54 7.9 CAPILLARY END EFFECT IN A LABORATORY CORE ...............7-57 7.9.1 Capillary End Effect.........................................................7-57 7.9.2 Mathematical Analysis of Capillary End Effect ..............7-59 7.9.3 Mathematical Model of Capillary End Effect During Steady State Relative Permeability Measurement.................................................................. 7-68 7.9.4 Experimental Evidence of Capillary End Effect...............7-70 7.10 CAPILLARY PRESSURE MEASUREMENTS ................................7-76 7.10.1 Restored State Method (Porous Plate Method)..............7-76 7.10.2 Mercury Injection Method .............................................7-77 7.10.3 Centrifuge Method..........................................................7-81 7.11 PORE SIZE DISTRIBUTION......................................................... 7-96 7.11.1 Introduction.................................................................... 7-96 7.11.2 Pore Volume Distribution .............................................. 7-98 7.11.3 Pore Size Distribution Based on Bundle of Capillary Tubes Model............................................7-103 7.11.4 Mercury Injection Porosimeter..................................... 7-115 7.12 CALCULATION OF PERMEABILITY FROM DRAINAGE CAPILLARY PRESSURE CURVE................................................. 7-118 7.12.1 Calculation of Absolute Permeability from Drainage Capillary Pressure Curve............................. 7-118 7.12.2 Calculation of Relative Permeabilities from Drainage Capillary Pressure Curve.............................7-132 7.13 EMPIRICAL CAPILLARY PRESSURE MODELS ........................ 7-133 7.13.1 Brooks-Corey Capillary Pressure Models ..................... 7-133 7.13.2 van Genuchten Capillary Pressure Model ....................7-143 7.14 CAPILLARY TRAPPING IN POROUS MEDIA........................... 7- 145 7.14.1 Pore Doublet Model of Capillary Trapping................... 7-145
7.14.2 Snap-Off Model of Capillary Trapping ......................... 7-152 7.14.3 Mobilization of Residual Non-Wetting Phase.............. 7-155 7.14.4 Oil Migration................................................................. 7-159
7.15 EFFECTS OF WETTABILITY AND INTERFACIAL TENSION ON CAPILLARY PRESSURE CURVES.......................7-162 NOMENCLATURE .......................................................................7-164
REFRENCES AND SUGGESTED READINGS ............................ 7-168 8 RELATIVE PERMEABILITY................................................... 8-1 8.1 DEFINITION OF RELATIVE PERMEABILITY...............................8-1
8.2 LABORATORY MEASUREMENT OF TWO-PHASE RELATIVE PERMEABILITY BY THE STEADY STATE METHOD ......................................................................................... 8-6 8.3 THEORY OF ONE DIMENSIONAL IMMISCIBLE DISPLACEMENT IN A POROUS MEDIUM..................................8-15 8.3.1 Mathematical Model of Two-Phase Immiscible Displacement................................................8-15 8.3.2 Buckley-Leverett Approximate Solution of the Immiscible Displacement Equation................................8-21 8.3.3 Waterflood Performance Calculations from Buckley-Leverett Theory .................................................8-31 Oil Recovery at any Time ...............................................8-31 Oil Recovery Before Water Breakthrough .....................8-31 Oil Recovery at Water Breakthrough............................ 8-32 Oil Recovery After Water Breakthrough ...................... 8-36 Water Production...........................................................8-41 8.4 LABORATORY MEASUREMENT OF TWO-PHASE RELATIVE PERMEABILITY BY THE UNSTEADY STATE METHOD ........................................................................................8-51 8.5 FACTORS AFFECTING RELATIVE PERMEABILITIES.............. 8-65 8.5.1 Fluid Saturation.............................................................. 8-65 8.5.2 Saturation History.......................................................... 8-66 8.5.3 Wettability ...................................................................... 8-67 8.5.4 Injection Rate ................................................................. 8-70 8.5.5 Viscosity Ratio ................................................................ 8-73 8.5.6 Interfacial Tension ......................................................... 8-74 8.5.7Pore Structure .................................................................. 8-75 8.5.8 Temperature ................................................................... 8-76 8.5.9 Heterogeneity ................................................................. 8-78
8.6 THREE-PHASE RELATIVE PERMEABILITIES .......................... 8-79 8.4 CALCULATION OF RELATIVE PERMEABILITIES FROM DRAINAGE CAPILLARY PRESSURE CURVE .............................8-82 NOMENCLATURE .........................................................................8-91
REFERENCES AND SUGGESTED READINGS ............................8-94 APPENDIX A: A Systematic Approach To Dimensional Analysis .. A-1
Summary.......................................................................................... A-1 Introduction..................................................................................... A-1 Algebraic Theory of Dimensional Analysis......................................A-2
Transformation of the Dimensionless π Groups .............................A-9 Example Problem.............................................................................A-9 Procedure ....................................................................................... A-10
Transformation of the Dimensionless π Groups for Example Problem........................................................................... A-21 Some Practical Considerations ......................................................A-28 Concluding Remarks...................................................................... A-31 Nomenclature ................................................................................ A-31 References ......................................................................................A-32
CHAPTER 1
INTRODUCTION
1.1 PETROPHYSICS
Petrophysics is the study of rock properties and their interactions
with fluids (gases, liquid hydrocarbons and aqueous solutions). Because
petroleum reservoir rocks must have porosity and permeability, we are
most interested in the properties of porous and permeable rocks. The
purpose of this text is to provide a basic understanding of the physical
properties of permeable geologic rocks and the interactions of the various
fluids with their interstitial surfaces. Particular emphasis is placed on
the transport properties of the rocks for single phase and multiphase
flow.
The petrophysical properties that are discussed in this text
include:
• Porosity
• Absolute permeability
• Effective and relative permeabilities
• Water saturation
1-1
• Irreducible water saturation
• Hydrocarbon saturation
• Residual oil saturation
• Capillary pressure
• Wettability
• Pore size
• Pore size distribution
• Pore structure
• Net pay thickness
• Isothermal coefficient of compressibility
• Mineralogy
• Specific pore surface area
• Dispersivity
1.2 PETROLEUM RESERVOIR ROCKS
A petroleum reservoir rock is a porous medium that is sufficiently
permeable to permit fluid flow through it. In the presence of
interconnected fluid phases of different density and viscosity, such as
water and hydrocarbons, the movement of the fluids is influenced by
gravity and capillary forces. The fluids separate, therefore, in order of
density when flow through a permeable stratum is arrested by a zone of
low permeability, and, in time, a petroleum reservoir is formed in such a
trap. Porosity and permeability are two fundamental petrophysical
properties of petroleum reservoir rocks.
1-2
Geologically, a petroleum reservoir is a complex of porous and
permeable rock that contains an accumulation of hydrocarbons under a
set of geological conditions that prevent escape by gravitational and
capillary forces. The initial fluid distribution in the reservoir rock, which
is determined by the balance of gravitational and capillary forces, is of
significant interest at the time of discovery.
A rock capable of producing oil, gas and water is called a reservoir
rock. In general, to be of commercial value, a reservoir rock must have
sufficient thickness, areal extent and pore space to contain a large
volume of hydrocarbons and must yield the contained fluids at a
satisfactory rate when the reservoir is penetrated by a well. Any buried
rock, be it sedimentary, igneous or metamorphic, that meets these
conditions may be used as a reservoir rock by migrating hydrocarbons.
However, most reservoir rocks are sedimentary rocks.
Sandstones and carbonates (limestones and dolomites) are the
most common reservoir rocks. They contain most of the world’s
petroleum reserves in about equal proportions even though carbonates
make up only about 25% of sedimentary rocks. The reservoir character
of a rock may be primary such as the intergranular porosity of a
sandstone, or secondary, resulting from chemical or physical changes
such as dolomitization, solution and fracturing. Shales frequently form
the impermeable cap rocks for petroleum traps.
The distribution of reservoirs and the trend of pore space are the
end product of numerous natural processes, some depositional and some
post-depositional. Their prediction, and the explanation and prediction of
their performance involve the recognition of the genesis of the ancient
sediments, the interpretation of which depends upon an understanding
of sedimentary and diagenetic processes. Diagenesis is the process of
1-3
physical and chemical changes in sediments after deposition that convert
them to consolidated rock such as compaction, cementation,
recrystallization and perhaps replacement as in the development of
dolomite.
1.3 MINERAL CONSTITUENTS OF ROCKS - A REVIEW
The physical and chemical properties of rocks are the consequence
of their mineral composition. A mineral is a naturally occurring
crystalline inorganic material that has specific physical and chemical
properties, which are either constant or vary within certain limits. Rock-
forming minerals of interest in petroleum engineering can be classified
into the following families: silicates, carbonates, oxides, sulfates
(sulphates), sulfides (sulphides) and chorides. These are summarized in
Table 1.1. Silicates are the most abundant rock-forming minerals in the
Earth’s crust.
Table 1.1 Rock - Forming Minerals
Name Chemical Formula Specific Gravity Silicates
Quartz Orthoclase Plagioclase Clay
SiO2
KAlSi2O8 NaAlSi3O8 CaAl2Si2O8
Al2Si2O5(OH)
and many more
2.65 2.57
2.62 - 2.76
2.5
Carbonates Calcite Dolomite
CaCO3
CaMg(CO3)2
2.72 2.85
Oxides Magnetite Hematite
Fe3O4 Fe2O3
5.18
4.9 - 5.3
1-4
Sulfates Anhydrite Gypsum Barite
CaSO4
CaSO4.2H2O BaSO4
2.89 - 2.98
2.32 4.5
Sulfide Pyrite
FeS2
5.02
Chloride Halite
NaCl
2.16
1.4 ROCKS
A rock is an aggregate of one or more minerals. There are three
classes of rocks: igneous, metamorphic and sedimentary rocks .
1.4.1 Igneous Rocks
These are rocks formed from molten material (called magma) that
solidified upon cooling either:
1. At the earth’s surface to form volcanic or extrusive rocks, e.g.,
basaltic lava flows, volcanic glass and volcanic ash.
or
2. Below the surface, usually at great depths, to form plutonic or
intrusive rocks, e.g., granites and gabbros.
Igneous rocks are the most abundant rocks on the earth’s crust,
making up about 64.7% of the Earth’s crust.
1-5
1.4.2 Metamorphic Rocks
These are rocks formed by transformation, generally in the solid
state, of pre-existing rocks beneath the surface by heat, pressure and
chemically active fluids, e.g., quartz is transformed to quartzite and
limestone plus quartz plus heat gives marble and carbon dioxide.
Metamorphic rocks are the second most abundant rocks on the
earth’s crust, making up 27.4% of the Earth’s crust.
1.4.3 Sedimentary Rocks
These are rocks formed at the surface of the earth either by
1. Accumulation and consolidation of minerals, rocks and/or
organisms and vegetation, e.g., sandstone and limestone.
or
2. Precipitation from solution such as sea water or surface water,
e.g., salt and limestone.
Sedimentary rocks are the source of petroleum and provide the
reservoir rock and trap to hold the petroleum in the earth’s crust.
Sedimentary rocks are the least abundant rocks on the earth’s crust,
making up about 7.9% of the earth’s crust. Because most reservoir
rocks are sedimentary rocks, they are of particular interest to us in the
study of petrophysics. Therefore, we will examine sedimentary rocks in
more detail than igneous and metamorphic rocks.
1-6
1.5 CLASSIFICATION OF SEDIMENTARY ROCKS
Sedimentary rocks may be classified by origin and composition as
clastic, chemical or organic. Tables 1.2 to 1.4 show the various rock
types for each class.
1.5.1 Clastic Sedimentary Rocks
These rocks are composed of fragments or minerals broken from
any type of pre-existing rock. Clastic sedimentary rocks are usually
classified by grain size as boulder, cobble, gravel, sand, silt and clay.
Figure 1.1 shows such a classification known as the Wentworth scale.
Table 1.2 Clastic Sedimentary Rocks
1.5.2 Chemical Sedimentary Rocks
These rocks are formed by chemical precipitation as carbonates,
e.g., limestone (CaCO3) and dolomite (CaMg(CO3)2) or as evaporties, e.g.,
gypsum (CaSO4.2H2O), anhydrite (CaSO4) and salt (NaCl).
1-7
1.5.3 Organic Sedimentary Rocks
These rocks are formed by biologic precipitation and by
accumulation of organic (plant and animal) material, e.g., peat, coal,
diatomite and limestone.
Figure 1.1 Classification of clastic rocks according to texture.
1-8
Table 1.3 Chemical Sedimentary Rocks (Precipitates)
Table 1.4 Organic Sedimentary Rocks
1-9
1.6 DISTRIBUTION OF SEDIMENTARY ROCK TYPES
Table 1.5 shows the approximate distribution of sedimentary rocks
in the earth’s crust. Shales make up over 50% of total sedimentary rock
volume in the earth’s crust.
Table 1.5 Distribution of Sedimentary Rocks Type % Earth’s Crust % Sedimentary Rock Shale 4.2 53
Sandstone 1.7 22 Limestone and
Dolomite 2.0 25
Total 7.9 100
1.7 SANDSTONE RESERVOIRS (CLASTIC SEDIMENTARY ROCK)
Sandstones are composed of fragmented materials, which have
been transported to the site of deposition by water currents and which
have been subjected to varying degrees of wave and current action
during transport and during deposition. Consequently, sandstone
reservoirs vary from clean, well sorted quartz sandstone with well
rounded grains (Figure 1.2a) through more angular feldspathic
sandstone containing varying amounts of clay (Figure 1.2b), to
argillaceous, very poorly sorted sandstone containing varying amounts of
rock fragments (Figure 1.2c) all of which may be affected by varying
degrees of compaction, cementation, solution and replacement.
1-10
Figure 1.2: Examples of sandstone reservoir rocks. (A) clean, well sorted sandstone, (B) angular, feldspathic sandstone and (C) argillacious, very
poorly sorted sandstone.
1-11
1.7.1 Pore Space
The basic framework of a sandstone reservoir is formed by the
sand grains between which the pore space may or may not contain
interstitial fine material and/or cement (Figure 1.3). The amount of this
intergranular pore space or porosity is controlled primarily by sorting of
the sediment and to a lesser extent by the packing of the grains. Sorting
is a measure of the spread of distribution of grain size on either side of
an average in a sediment. Theoretically, grain size has no effect on
porosity. This is true only for spherical grains of the same size.
However, the arrangement of such spheres has a large effect on the
porosity of the pack.
Figure 1.3: Framework of reservoir sand with interstitial clay and
cement.
1-12
Porosity is at its maximum for spherical grains but becomes
progressively less as the angularity of the grains increases because such
grains pack together more closely. Experimental data from artificial
sands confirm that the grain size essentially has no influence on porosity
for well sorted sand. However, porosities of wet packed sands show a
general decrease as sorting becomes poorer. This is because with a
mixture of sizes, the smaller grains partially fill the interstices between
the larger grains.
Permeability, being a measure of the ease with which a material
transmits fluids, depends primarily upon the size, shapes and extent of
the interconnections between individual pores rather than the size of the
pores themselves. However, since the interconnections are directly
related to the pore size which in turn is related to grain size, there are
relationships between these factors and permeability. Krumbein and
Monk (1942) have shown that permeability varies as the square of the
mean grain diameter and a complex function of sorting (other factors
being equal). Experimental data show a marked decrease in permeability
as grain size decreases and as sorting becomes poorer. Experience has
also shown that the permeability measured normal to the bedding is
usually less than permeability measured parallel to the bedding and that
large variations in permeability occur in different directions in the
bedding plane.
Clay in the pore space of a reservoir may affect the performance of
a reservoir very adversely. The amount and kind of clay, as well as
distribution throughout the reservoir rock, has an important bearing on
liquid permeability, whereas a small amount has little effect on porosity.
Figures 1.4 and 1.5 show how the dispersed clay morphology affects the
permeability and capillary pressure of sandstones.
1-13
Figure1.4: Three general types of dispersed clay in sandstone reservoir
rocks and their effects on permeability: (a) discrete particles of kaolinite; (b) pore lining by chlorite; (c ) pore bridging by illite (from Neasham,
1977).
1-14
Figure1.5: Three general types of dispersed clay in sandstone reservoir rocks and their effects on capillary pressure (from Neasham, 1977).
If fresh water, for example drilling fluid filtrate, invades a reservoir,
certain clays, such as montmorillonite, will swell and plug some of the
pore interconnections, drastically reducing the permeability, whereas
saline water may have no effect.
1.7.2 Compaction and Cementation
The pore space of the original unconsolidated sediment is reduced
in ancient rocks by many factors. Compaction and cementation are the
most important of these factors but they in turn are affected by
1-15
composition of the sediment, and the contained fluids and their
pressures. Compaction by the weight of the overburden commences as
soon as a sediment is deposited. It produces reduction of pore space as
a result of:
1. Closer packing of the grains which causes smaller pores and
connected passages.
2. Crushing and fracturing of grains, and dissolution at points of
contact sometimes accompanied by precipitation of silica in the
pore space nearby. Alkaline interstitial water seems to provide
conditions more conducive to dissolution than saline water.
3. Plastic deformation of the softer grains which tend to mold around
the harder grains thus destroying pore space. The softer grains
may be composed of limestone, shale, siltstone and other rock
fragments, and when the amount of such soft material exceeds 10
to 12%, the permeability may be largely destroyed even though
some porosity usually remains.
Early cementation of sand may produce a rigid framework which
will inhibit compaction until the depth of burial exceeds that at which
fracturing of the grains and cement is initiated. Abnormal fluid pressure
in sandstone reservoirs also inhibits compaction because the overburden
is partly supported by the fluid pressure.
Reduction of reservoir pressure by production of fluid can lead to
compaction of the producing zone by rearrangement of the sand grains.
This can produce serious and expensive subsidence of surface facilities
such as occurred in the Wilmington field in California, Bolivar District
fields in Venezuela and Ekofisk field in the North Sea. This form of
1-16
compaction leads to porosity and permeability reduction which is
irreversible and which may affect the producing characteristics of the
reservoir very adversely.
Cementation is the result of recrystallization from solution of silica,
carbonates and other soluble materials in the pores of clastic rocks. The
most common cementing materials in sandstone reservoirs are silica and
calcite but many others do occur. It is not uncommon to find both silica
and calcite present in the cement and in such cases, the silica in the
form of quartz has precipitated first and the calcite later. Silica cement
usually occurs in the form of quartz and grows in optical continuity with
the sand grains until finally the crystals interfere with one another and
an interlocking network results. Calcite cement is often patchy but may
completely fill the pore space.
Silica cement appears to have two origins: (1) early cementation
before the sands were compacted appreciably, and (2) deposition of silica
predissolved by pressure solution during compaction. In the Eocene
Wilcox sandstones of the Gulf Coast, the distribution of silica cement can
be related to both primary texture and depth of burial. The amount of
silica cement tends to increase in coarser and better sorted sands and, to
a lesser extent, with depth. Much of this cement appears to be early and
unrelated to the compaction process.
1.7.3 Classification
Sandstones may be classified by mineral composition (Figure 1.6).
The principal types are: (a) quartz sandstones consisting of over 95%
detrital quartz, (b) feldspathic sandstone consisting of 5 - 25% feldspar,
(c) arkosic sandstone consisting of over 25% feldspar, (d) sublithic
sandstone consisting of 5 - 25% rock fragments and (e) lithic sandstone
consisting of over 25% rock fragments.
1-17
Sandstone grain texture consists of five components: (a) size, (b)
sorting, (c) shape (sphericity), (d) roundness, and (e) packing. Figure 1.7
shows a grain size comparator where some of these qualitative terms are
presented visually. Porosity is independent of grain size for uniform
grains but decreases as sorting gets poorer. Close packing reduces
porosity and permeability. The effects of shape and roundness on
porosity are less definite. Permeability increases with increasing grain
size, but decreases with poorer sorting. Permeability generally increases
with angularity and decreasing sphericity.
Figure 1.6: Classification of sandstones by composition.
1-18
Figure 1.7: Grain size comparator chart (from Stow, 2005).
1-19
1.8 CARBONATE RESERVOIRS (LIMESTONES AND DOLOMITES)
Most carbonate rocks, like clastics, are composed of particles of
clay to gravel size that were generally deposited in a marine environment.
However, they differ from terrigenous clastics in that they are deposited
as lime particles which are produced locally, whereas, sandstones are
composed of particles transported from an outside source by water
currents. They differ even more importantly from sandstones by being
subject to more post-depositional diagenesis ranging from simple
cementation of the original particles to complete recrystallization or
replacement by dolomite or chert. In addition, they are susceptible to
solution at any stage in their diagenesis. They are usually more poorly
sorted than clastics.
Components of carbonate rocks are usually (1) grains of various
kinds, (2) lime mud, and (3) carbonate cement precipitated later. There
are several kinds of grains, of which four are the most important. These
are (1) shell fragments, called “bio”, (2) fragments of previously deposited
limestone called “intraclasts”, (3) small round pellets - the excreta of
worms, and (4) ooliths - spheres formed by rolling lime particles along
the bottom. Lime mud consists of clay-sized particles of lime.
The material between the grains may be primary lime mud
deposited at the same time as the grains which would be grain-supported
rock, or the grains may be “floating” in lime mud which would be mud-
supported rock.
1.8.1 Classification
Carbonates are usually classified according to depositional texture
as shown in Figure 1.8. The presence or absence of lime mud and the
1-20
type and abundance of grains form the basis of the classification. A
boundstone consists of original skeletal components bound together
during deposition (Figure 1.9). Grainstones consist of packed carbonate
grains with the texture being grain-supported and very little lime mud
(Figure 1.10). Packstones are grain-supported but contain very
substantial amounts of lime mud. Wackestones have a larger amount of
lime muds, such that the grains effectively “float” in the mud. Mudstones
consist of essentially lime muds only. The presence of lime mud may be
most important in the development of porosity in carbonates because
under the right conditions, lime mud may be preferentially dolomitized
and may also be more readily leached out than the grains.
Good porosity in carbonate reservoirs is usually due to
dolomitization. The largest volume of carbonate petroleum reserves
comes from dolomites. Dolomitization occurs from the substitution of
magnesium for calcium in half of the sites in a carbonate crystal. A
volume loss of 12 to 13% due to dolomitization results in a corresponding
increase in porosity. Due to their larger surface area, mud-size grains
are more easily dolomitized than sand-sized grains. Thus, the best
carbonate reservoirs may have the lowest primary porosity.
Dolomitization also creates planar crystal surfaces and harder crystal
structures. Thus, dolomites retain more of their porosity during
compaction than limestones.
1-21
Figure 1.8: Classification of carbonates by texture.
1.8.2 Pore Space
The porosity, permeability and pore space distribution in carbonate
reservoir rocks are related to both the depositional environment and the
diagenesis of the sediment. They are most commonly of secondary
(diagenetic) origin although residual primary pore space does occur.
Carbonates have a large range of pore structures due to the
complex nature of carbonate constituents and diagenetic features. The
pore structures have been classified by Choquette and Pray, 1970) as
shown in Figure 1.10.
1-22
Figure 1.9: Examples of boundstone.
1-23
Figure 1.9: Examples of grainstones.
1-24
Fabric-selective porosity includes:
•Interparticle porosity.
•Intercrystalline porosity - typical of dolomites.
•Fenestral porosity - by solution along bedding planes or joint
surfaces.
•Skeletal, framework, molding, or shelter porosity - selective solution of,
within, or around fossil material.
•Oomoldic porosity - selective solution of ooliths.
•Non fabric-selective porosity includes:
•Fracture porosity - by stress or shrinkage.
•Channel porosity - widening and coalescence of fractures.
•Vuggy or cavernous porosity.
•Bioturbation porosity - from boring and burrowing.
•Breccia porosity - in some cases, really high fracture porosity.
In carbonates, porosity and permeability are not well-correlated
with grain size or sorting. Porosity and permeability are controlled
largely by the amount of fines and by diagenesis. Correlation of
petrophysical properties with rock type is thus very difficult.
1-25
Figure 1.10: Classification of pore systems in carbonate rocks (Choquette and Pray, 1970).
1-26
Table 1.6 shows a comparison of the pore space characteristics of
clastic and carbonate rocks (Choquette and Pray, 1970).
Table 1.6. Comparison of Pore-space Properties in Clastic and Carbonate Rocks (Choquette and Pray, 1970).
1-27
1.9 FRACTURED RESERVOIRS
Natural reservoir fractures are caused by brittle failure, usually
due to such factors as (a) folding, (b) faulting, (c) fluid pressure, (d)
release of lithostatic pressure, (e) pressure solution, (f) dehydration, (g)
weathering, (h) cooling and (i) impact craters. Natural fractures can exist
in essentially any type of rock although they are particularly common in
carbonates.
Naturally fractured reservoirs are usually treated by a dual
porosity approach to deal with their properties. The matrix rock
(between fractures) usually has reasonable porosity and extremely low
permeability. Fractures range in size from hair-size to several
millimeters in aperture. Fractures that have not been filled with cement
have very high permeabilities, even though they may be fairly widely-
spaced. However, the fracture system generally contains only a small
fraction of the reservoir pore space. Thus, the matrix contains the bulk
of the reservoir pore volume while the fractures contain the bulk of the
reservoir flow capacity.
Figure 1.11 shows a naturally fractured rock together with its
idealized dual porosity approximation.
1-28
Figure 1.11: Idealization of naturally fractured reservoir (Warren and Root, 1963)
1.10 RESERVOIR COLUMN
Figure 1.12 shows a reservoir column penetrated by a well. The
total thickness of the reservoir as determined from the spontaneous
potential (SP) log, discussed in Chapter 2, is H. This reservoir contains a
hydrocarbon bearing zone and a water bearing zone at the bottom. The
gross pay thickness, which is the thickness of the hydrocarbon bearing
portion of the reservoir as determined from the resistivity log (see
Chapter 2), is . However, this thickness contains shale breaks which
will not be productive and must be discounted to determine the net pay
to be used in reserves estimation. The net pay for this example is given
by
0h
1-29
6
1Net pay
i
ii
h=
=
= = h∑ (1.1)
The net to gross (NTG) pay is defined as
6
1
0 0
Net to gross (NTG)
i
ii
hhh h
=
== =∑
(1.2)
The net to gross is a number that is less than or equal to 1 or if
expressed as a percentage, is a number that is less than or equal to
100%. Notice that in this example, there is a gas oil contact (GOC) and
an oil water contact (OWC) in the reservoir. The thickness of the gas zone
is gash and that of the oil bearing zone is . Of course, not all petroleum
reservoirs have a gas oil contact or an oil water contact. Net pay is used
along with other petrophysical properties of the reservoir to estimate the
hydrocarbon reserve as discussed in Chapter 2.
oilh
1-30
Figure 1.12. Reservoir column showing gross and net pay.
1-31
1-32
REFERENCES
Archie, G.E. :“Introduction to Petrophysics of Reservoir Rocks,” AAPG Bull., Vol. 34, No. 5 (May 1950) 943-961.
Beard, D.C. and Weyl, P.K. : “Influence of Texture on Porosity and Permeability of Unconsolidated Sand,” AAPG Bull. (Feb. 1973) 57, 349-369.
Choquette, P.W. and Pray, L.C. : “Geologic Nomenclature and Classification of Porosity in Sedimentary Carbonates,” AAPG Bull., Vol. 54, No. 2 (1970) 207-250.
Krumbein, W.C. and Monk, G.D. : “Permeability as a Function of the Size Parameters of Unconsolidated Sand,” Amer. Int. Mining and Met. Tech. Pub. 1492, 1942.
Levorsen, A.I. : Geology of Petroleum, W.H. Freeman and Company, San Francisco, 1967.
Neasham, J.W.: "The Morphology of Dispersed Clay in Sandstone Reservoirs and Its Effect on Sandstone Shaliness, Pore Space and Fluid Flow Properties," SPE 6858, Presented at the 52nd Annual Technical Conference and Exhibition of the Society of Petroleum Engineers, Denver, Oct. 9-12, 1977.
Selley, R.C. : Elements of Petroleum Geology, W.H. Freeman, New York, 1985.
Stoneley, R. : An Introduction to Petroleum Exploration for Non-Geologists, Oxford University Press, New York, 1995.
Stow, D.A.V. : Sedimentary Rocks in the Field, Elsevier Academic Press, Burlington, 2005.
2-1
CHAPTER 2
POROSITY AND FLUID SATURATIONS
2.1 DEFINITION OF POROSITY
Porosity gives an indication of the rock’s ability to store fluids. It is
defined as the ratio of the pore volume to the bulk volume of the porous
medium as shown in the following equation:
p b s
b b
V V VV V
φ −= = (2.1)
Porosity may be classified as total or effective porosity. Total
porosity accounts for all the pores in the rock (interconnected and
isolated pores) whereas effective porosity only accounts for the
interconnected pores. Therefore, effective porosity will be less than or
equal to total porosity depending on the amount of isolated pores in the
rock. From a reservoir engineering standpoint, it is the effective porosity
that matters, not the total porosity.
2–2
Porosity may also be classified as primary or secondary. Primary
porosity is that which was formed at the time of deposition of the
sediments whereas secondary porosity was developed after deposition
and burial of the formation. Sandstone porosity is practically all primary
porosity whereas carbonate porosity tends to be secondary porosity.
2.2 FACTORS AFFECTING SANDSTONE POROSITY
Sandstone porosity is affected by packing, sorting and
cementation. Packing describes the arrangement of the sand grains
relative to one another. Figures 2.1 shows three idealized types of
packing for spherical sand grains and their theoretical porosities. The
cubic packing has a porosity of 47.6%; the hexagonal packing has a
porostiy of 39.5% and the rhombohedral packing has a porosity of
25.9%. As shown by the geometrical derivations in Figure 2.1, the
porosity of a pack of uniform spheres is independent of the grain size as
the grain diameter cancels out.
Well sorted sandstone consists of grains having approximately the
same size whereas poorly sorted sandstone consists of grains having a
wide range of different grain sizes. Poor sorting reduces the porosity of
the sandstone as may be seen in Figure 2.2 in which the small grains fit
into the pores created by the large grains, thereby reducing the porosity.
2-3
Figure 2.1. Effect of packing on porosity of uniform spheres.
Figure 2.2. Effect of sorting on porosity. (A) Irregular grains,
(B) Idealized spherical grains (from Tiab and Donaldson, 2004).
2–4
Table 2.1 shows experimentally measured porosities of various
artificial sandpacks. Note the general decrease of porosity with poor
sorting for all grain sizes and the approximately constant porosity of the
extremely well sorted sands for all grain sizes.
Table 2.1 Measured Porosities of Artificial Sandpacks (adapted from Beard and Weyl, 1973)
In consolidated rocks, the sand grains are cemented together
usually by quartz or carbonates. Cementation reduces the porosity of
the sand as shown in Figure 2.3.
Figure 2.3. Effect of cementation on porosity.
2-5
2.3 FACTORS AFFECTING CARBONATE POROSITY
In carbonates, secondary porosity is usually more important than
primary porosity. The major sources of secondary porosity are
fracturing, solution and chemical replacement.
Fractures are cracks in the rock. Figure 2.4 shows an idealized
fractured formation where the grains are bricks and the fractures
constitute the pore space. Although fracture porosity is generally small,
often 1-2%, the fractures are very useful in allowing fluids to flow more
easily through the rock. Therefore, they greatly enhance the flow
capacity of the rock.
Figure 2.4. Idealized fractured rock with low fracture porosity.
Solution is a chemical reaction in which water with dissolved
carbon dioxide reacts with calcium carbonate to form calcium
bicarbonate which is soluble. This reaction increases the porosity of the
limestone. The chemical reactions are
2–6
2 2 2 3CO H O H CO+ = (2.2)
( )2 3 3 3 2H CO CaCO Ca HCO+ = (2.3)
Chemical Replacement is a chemical reaction in which one type of
ion replaces another with a resulting shrinkage in the size of the new
compound. An example is dolomitization in which some of the calcium
ions in calcium carbonate are replaced by magnesium ions to form
calcium magnesium carbonate (dolomite). This replacement causes a
shrinkage of 12 to 13% in the grain volume, with a corresponding
increase in secondary porosity. The chemical reaction is
( )3 2 3 222CaCO MgCl CaMg CO CaCl+ = + (2.4)
2.4 TYPICAL RESERVOIR POROSITY VALUES
Sandstones have porosities that typically range from 8% to 38%,
with an average of 18%. About 95% of sandstone porosity is effective
porosity. Sandstone porosity is usually mostly intergranular porosity.
Carbonates have porosities that typically range from 3% to 15%, with an
average of about 8%. About 90% of carbonate porosity is effective
porosity. Carbonate porosities are much more difficult to characterize
and may consist of (1) intergranular, (2) intercrystalline, (3) fractures and
fissures, and (4) vugular porosities.
2-7
2.5 LABORATORY MEASUREMENT OF POROSITY
2.5.1 Direct Porosity Measurement by Routine Core Analysis
Direct measurement of porosity requires the measurements of two
of the three volumes Vb, Vs and Vp. In the laboratory, measurements
are usually performed on extracted cores, which have been cleaned and
dried.
Bulk volume can be determined by (1) caliper and (2) fluid
displacement. For well machined samples, the dimensions can be
measured with a caliper, from which the bulk volume can be calculated.
Two types of fluid displacements can be used to determine bulk
volume. In the first method, fluid that does not easily penetrate the
pores such as mercury is used. The apparatus, which is known as a
pycnometer, measures the volume of mercury displaced by the sample
(Figure 2.5a). Since mercury does not penetrate the pores at
atmospheric pressure, the volume of mercury displaced is equal to the
bulk volume of the sample.
In the second method, fluid which easily saturates the sample is
used. The sample is weighed in air, evacuated and then saturated with a
liquid (brine, kerosene, or toluene). The saturated sample is weighed in
air and then weighed fully immersed in the saturating liquid. The loss in
weight of the saturated sample when fully immersed in the saturating
liquid is proportional to the bulk volume of the sample (Archimedes
principle).
Grain volume can be determined by (1) fluid displacement and (2)
gas exapansion using Boyle’s law porosimeter. The loss in weight of the
2–8
dry sample and the sample fully immersed in a liquid is proportional to
the grain volume.
Figure 2.5b shows a schematic diagram of a Boyle’s law porosimeter for grain volume determination by gas expansion. The sample, which is confined in a vessel of known volume V1, is pressured
by gas (air, nitrogen or helium) to a pressure P1 (absolute units). The vessel of volume V1 is connected to a second vessel of known volume V2,
which is initially evacuated. The valve between the two vessels is opened
and the pressure in the two vessels is allowed to stabilize at P2 (absolute units). By Boyle’s law (PV = constant at a constant temperature),
( ) ( )1 1 1 2 2s sV V P V V V P− = − + (2.5)
Eq.(2.5) can be solved for the grain volume as
21 2
1 2s
PV V VP P
⎛ ⎞= − ⎜ ⎟−⎝ ⎠
(2.6)
The instrument can be calibrated with steel blanks of known volume.
Calibration consists of a plot of Vs versus P2/(P1-P2), which should be
linear with a slope, -V2, and an intercept, V1. At least three steel blanks
of different sizes should be used in the calibration to ensure reliability of
the calibration. The three data points should fall on the calibration line.
Also, the slope and the intercept should be checked against the known
volumes, V2 and V1. Once the calibration line has been established, it is
used to convert the measurements from core samples to grain volume.
2-9
Figure 2.5. Schematics of equipment for measurement of core plug porosity.(a) Bulk volume pycnometer; (b) Boyle’s law porosimeter; (c) Bulk and pore volume porosimeter.
2–10
Pore volume can be determined by (1) fluid saturation and (2)
mercury injection. The difference in the weight of the saturated sample
and the dry sample is proportional to the pore volume.
Mercury injection consists of forcing mercury under relatively high
pressure into the pores of the sample using a mercury porosimeter
(Figure 2.5c). Typically, the core is evacuated before mercury injection.
Because any air left in the pores is compressed to a negligibly small
value, the volume of mercury injected is essentially equal to the
connected pore volume of the sample. This is a destructive method
because after the test, the sample is no longer suitable for other
measurements. Mercury porosimetry is also used to determine capillary
pressure and pore size distribution of the sample (see Chapter 6).
The methods so far described determine the effective porosity of
the sample. To determine the total porosity, the sample is ground into a
fine powder after bulk volume measurement. The grain volume of the
ground sample can be determined either by liquid displacement or by
assuming an average grain density.
The measurement of porosity on consolidated samples in routine
core analysis might generally be expected to yield values of the true
fractional porosity plus or minus 0.005, i.e., a true value of 27% porosity
may be measured between 26.5% and 27.5% porosity.
Core porosities may differ from in-situ porosities for the following
reasons:
• The core may be altered during recovery.
2-11
• The core in the laboratory is no longer subjected to the overburden
and lateral stresses that it was subjected to in the reservoir.
• The porosities are measured on small plugs, which may not be
representative of the entire reservoir.
• The volume of the core analyzed is small and may not account for
the variability of the porosity in the reservoir.
Despite these limitations, core analysis provides the only direct
measurement of porosity. Frequently, the results of core analysis are
used to calibrate well logs.
Example 2.1
An experiment has been performed to determine the porosity of an
irregularly shaped core sample. The cleaned dry sample was weighed in
air. It was then evacuated and fully saturated with an oil with a density
of 0.85 gm/cc and then weighed again in air. Afterwards, the saturated
sample was weighed when it was fully immersed in the oil. Here are the
results of the experiment.
Weight of dry sample in air = 42.40 gm
Weight of the saturated sample in air = 45.49 gm
Weight of the saturated sample immersed in the oil = 28.80 gm
a. Calculate the porosity of the core.
2–12
b. Is there enough information from this experiment to determine the
mineralogy of the sample? If yes, what is it? Please justify your answer
with appropriate arguments.
Solution to Example 2.1
Wt of dry sample (Wdry) = 42.40 gm
Wt of saturated sample (Wsat) = 45.49 gm
Wt of sample immersed in oil (Wi) = 28.80 gm
Density of saturating oil (ρL) = 0.85 gm/cc
a. Required to calculate the porosity of the sample.
Pore volume (Vp) = (Wsat – Wdry)/ρL = (45.49–42.40)/0.85 = 3.64 cc
Bulk volume (Vb) = (Wsat – Wi)/ρL = (45.49–28.80)/0.85 = 19.64 cc
Porosity (φ) = Vp/Vb = 3.64/19.64 = 0.185 or 18.5%
b. Yes. There is enough information to determine the mineralogy of
the sample through the grain density.
Grain volume (Vs) = Vb–Vp =19.64–3.64 = 16.00 cc
Alternatively, Grain volume (Vs) = (Wdry–Wi)/ρL = (42.40–
28.80)/0.85 = 16.00 cc
2-13
Grain density (ρs) = Wdry/Vs = 42.40/16.00 = 2.65 gms/cc
Specific gravity of mineral (γm) = ρs /ρw = 2.65/1.00 = 2.65
Table 1.1 lists the specific gravities of common reservoir rock minerals.
From the table, quartz has a specific gravity of 2.65, which is the same
as the specific gravity of the sample matrix. Therefore, based on the
available information, the mineral of the sample is quartz.
2.5.2 Indirect Porosity Measurement by CT Imaging
With the availability of X-ray computed tomography (CT) imaging
systems in research laboratories, it is now possible to measure the
porosity distributions in core samples. Peters and Afzal (1992) have
made such measurements in an artificial sandpack and a Berea
sandstone approximately 60 cm long and 5 cm in diameter. CT imaging
gives rise to a very large data set, over 600,000 porosity values in some
cases. Therefore, it is convenient to present the results of the
measurements as images (Figures 2.7 and 2.8). It should be noted in
Figure 2.7 that a sandpack may not be as uniform as we always assume
it to be. The packing technique used in this test introduced significant
porosity variation into the pack. The packing history is clearly evident in
the image. The dominant feature in the porosity variation of the Berea
sandstone is layering which is clearly visible in Figure 2.8. The porosity
data also can be presented in histograms as shown Figures 2.9 and 2.10.
The porosity in each voxel (volume element) was obtained by
scanning the sample dry and then scanning it fully saturated with a
wetting fluid such as brine. The x-ray attenuation equations for the two
scans are
2–14
( )1m air dryψ φ φψ ψ− + = (2.7)
( )1m brine wetψ φ φψ ψ− + = (2.8)
Eqs.(2.7) and (2.8) can be solved simultaneously to obtain the porosity in
each voxel as
wet dry
brine air
ψ ψφ
ψ ψ−
=−
(2.9)
The x-ray attenuation coefficient of the brine in Eq.(2.9) is obtained by
scanning a sample of the brine in a test tube and the attenuation for air
is assumed to be zero.
Figure 2.7. Porosity image of a sandpack from CT imaging. L = 54.2 cm,
2-15
d = 4.8 cm. (a) Cross-sectional slice. (b) Longitudinal vertical slice. (Peters and Afzal, 1992).
Figure 2.8. Porosity image of a Berea sandstone from CT imaging. L = 60.2 cm, d = 5.1 cm. (a) Cross-sectional slice. (b) Longitudinal vertical
slice. (Peters and Afzal, 1992).
2–16
Figure 2.9. Porosity histogram for a sandpack from CT imaging. L = 54.2 cm, d = 4.8 cm. Mean = 29.7%, Standard deviation = 2.5%.(Peters
and Afzal, 1992).
2-17
Figure 2.10. Porosity histogram for a Berea sandstone from CT imaging. L = 60.2 cm, d = 5.1 cm. Mean = 17.3%, Standard
deviation = 2.0%.(Peters and Afzal, 1992).
2.6 FLUID SATURATIONS
In a petroleum reservoir, there is always more than one fluid phase
occupying the pore space. In an oil reservoir, oil and water occupy the
pore space. In a gas reservoir, gas and water occupy the pore space. At a
certain point in the production of an oil reservoir, oil, water and gas
could occupy the pore space. There is a need to keep track of the
quantity of each type of fluid occupying the pore space. The
petrophysical property that describes the amount of each fluid type in
2–18
the pore space is the fluid saturation. It is defined as the fraction of the
pore space occupied by a fluid phase. Thus, in general,
Fluid VolumeFluid Saturation = Effective Rock Pore Volume
(2.10)
If Sw = water saturation, So = oil saturation and Sg = gas saturation,
then Sw = Vw/Vp, So = Vo/Vp, and Sg = Vg/Vp, where Vw, Vo, Vg and
Vp are the volumes of water, oil, gas and pore space, respectively. For an
oil reservoir without a free gas saturation, So + Sw = 1.0. For a gas
reservoir without a liquid hydrocarbon saturation, Sg + Sw = 1.0. For an
oil reservoir with a free gas saturation, So + Sw + Sg = 1.0. Fluid
saturation may also be expressed in %.
There are two methods of determining the in-situ fluid saturations
in a petroleum reservoir rock. The direct approach is to measure the
fluid saturations from a core cut from the reservoir. The indirect
approach is to measure some other physical property of the rock that can
be related to fluid saturation. The direct approach is discussed here.
The indirect approach such as using electric logs or capillary pressure
measurements to estimate water saturation will be discussed later.
One method of direct measurement of fluid saturations is the
retort method. In this method, a core sample is heated so as to vaporize
the water and oil, which are condensed and collected in a small,
graduated receiving vessel (Figure 2.11). The volumes of oil and water
divided by the pore volume of the core sample give the oil and water
saturations. The gas saturation is obtained indirectly by the requirement
that saturations must sum to one.
2-19
Figure 2.11. Retort distillation apparatus.
There are two disadvantages to the retort method of saturation
determination. In order to remove all the oil, it is necessary to heat the
core to temperatures in the range of 1000 to 1100 °F. At these
2–20
temperatures, the water of crystallization (hydration) of the rock is driven
off, resulting in an estimated water saturation that is higher than the
true interstitial (connate) water saturation. The second disadvantage is
that the oil when heated to high temperatures has a tendency to crack
and coke. This cracking and coking tend to reduce the oil volume
resulting in an oil saturation that is less than the true value.
Corrections can be made to the retort measurements to make them more
accurate.
Another method of direct saturation measurement is by extraction
with a solvent. This is accomplished in a Dean-Stark distillation
apparatus (Figure 2.12). The core is placed in the apparatus in such a
way that the vapor from a solvent (e.g., toluene) rises through the core
and is condensed back over the sample. This process leaches out the oil
and water from the sample. The water and solvent are condensed and
trapped in a graduated receiver. The water settles to the bottom of the
receiver while the solvent refluxes back into the main heating vessel. The
extraction is continued until no more water is collected in the receiving
vessel. The water saturation is calculated directly from the volume of
water expelled from the sample. The oil saturation is calculated
indirectly from the weight of the saturated sample before distillation, the
weight of the dry sample after distillation and the weight of the water
expelled from the sample. Again, the gas saturation is calculated
indirectly from the requirement that the saturations must sum to one.
To ensure that all the oil has been removed from the sample, the
sample may be transferred from the Dean-Stark apparatus to a Soxhlet
extractor for further extraction (Figure 2.13). The Soxhlet extractor is
similar to the Dean-Stark apparatus except that there is no provision for
trapping the extracted liquids.
2-21
The saturations determined by direct measurements on cores
should be treated with caution because they may not represent the in-
situ fluid saturations for several reasons. If the core was cut with a
water-based drilling mud, it would have been flushed by mud filtrate
resulting in a higher water saturation than the original, undisturbed
formation water saturation. The measured oil saturation in this case
would be the residual oil saturation after waterflooding, which is less
than the original in-situ oil saturation. If the core was cut with an oil-
based mud, the water saturation obtained by direct measurement will be
essentially the correct original water saturation, if it was at irreducible
level. If the original in-situ water saturation was not at the irreducible
level, then the oil mud filtrate could potentially displace some of the
water making the laboratory measured water saturation to be too low.
Figure 2.12. Dean-Stark apparatus.
2–22
Figure 2.13. Soxhlet extractor.
As the core is brought from the high pressure and temperature of
the reservoir to the low pressure and temperature of the laboratory,
changes occur in the fluid saturations which can make them
considerably different from the original in-situ saturations. The free gas,
if present, will expand, expelling water and oil in the process. Solution
gas will evolve from the oil, expand and further reduce the oil and water
volumes. The evolution of solution gas causes the oil to "shrink". These
changes cause the saturations determined by core analysis to be
different from the in-situ saturations. In particular, the changes cause
gas saturation to be excessive even when there was no free gas
saturation at the original in-situ conditions.
2-23
Although the saturations determined by direct measurements on
cores may not reflect the true in-situ saturations, they do provide useful
information about the reservoir. The saturation measurements can be
used to approximately locate the gas-oil and water-oil contacts in the
reservoir if they are present.
Fluid saturations, So, Sw and Sg, only tell us the proportion of
each fluid type in the pore space. They do not tell us how the fluids are
distributed in the rock. To determine the fluid distribution, we need to
consider the interfacial forces and phenomena that arise when
immiscible fluids are confined in reservoir pores of capillary dimensions.
The important interfacial forces and phenomena include surface tension,
interfacial tension, wettability, capillarity and capillary pressure (see
Chapters 6 and 7).
Table 2.2 shows the data obtained in an example core analysis
from a hydrocarbon bearing formation from a depth of 4805.5 to 4851.5
feet. The table shows the depth, core permeability, core porosity, oil
saturation, water saturation and gas saturation as determined in the
laboratory. Although the fluid saturations are not the true in-situ
saturations, nevertheless they provide useful information. Figure 2.14
shows the saturation distributions from the core data. One can easily see
a water bearing zone at the bottom where the measured water saturation
is very high, an oil bearing zone above it, and a gas cap on top of the oil
zone. A gas oil contact exists at 4828.5 ft, and a water oil contact exists
at 4848.5 ft. Note the misleading gas saturation below the gas oil
contact. There was no free gas saturation below the gas oil contact at in-
situ conditions.
2–24
Table 2.2: Core Analysis Data
Depth k φ So Sw Sg (ft) (md
) (%) (%) (%) (%)
4805.5 0 7.5 0.0 68.0
32.0
4806.5 0 12.3
0.0 78.0
22.0
4807.5 2.5 17.0
0.0 43.0
57.0
4808.5 59 20.7
0.0 29.0
71.0
4809.5 221 19.1
0.0 31.4
68.6
4810.5 211 20.4
0.0 38.7
61.3
4811.5 275 23.3
0.0 34.7
65.3
4812.5 384 24.0
0.0 26.2
73.8
4813.5 108 23.3
0.0 30.9
69.1
4814.5 147 16.1
0.0 29.2
70.8
4815.5 290 17.2
0.0 34.3
65.7
4816.5 170 15.3
0.0 24.2
75.8
4817.5 278 15.9
0.0 26.4
73.6
4818.5 238 18.6
0.0 39.8
60.2
4819.5 167 16.2
0.0 39.5
60.5
4820.5 304 20.0
0.0 38.0
62.0
4821.5 98 16.9
0.0 34.3
65.7
4822.5 191 18.1
0.0 34.8
65.2
4823.5 266 20.3
0.0 31.1
68.9
4824.5 40 15.3
0.0 22.9
77.1
4825.5 260 15.1
0.0 13.9
86.1
4826.5 179 14.0
0.0 21.4
78.6
4827.5 312 15.6
0.0 28.8
71.2
2-25
4828.5 272 15.5
0.0 34.8
65.2
4829.5 395 19.4
6.2 25.3
68.5
4830.5 405 17.5
13.1
25.7
61.2
4831.5 275 16.4
17.7
22.5
59.8
4832.5 852 17.2
19.8
19.2
61.0
4833.5 610 15.5
21.9
21.3
56.8
4834.5 406 20.2
16.3
22.3
61.4
4835.5 535 18.3
19.7
24.6
55.7
4836.5 663 19.6
19.4
16.3
64.3
4837.5 597 17.7
17.5
19.8
62.7
4838.5 434 20.0
14.0
27.5
58.5
4839.5 339 16.8
20.8
19.7
59.5
4840.5 216 13.3
18.1
23.3
58.6
4841.5 332 18.0
15.6
15.6
68.8
4842.5 295 16.1
19.3
15.5
65.2
4843.5 882 15.1
19.2
21.2
59.6
4844.5 600 18.0
20.6
22.2
57.2
4845.5 407 15.7
15.3
13.4
71.3
4847.5 479 17.
8 20.8
14.6
64.6
4848.5 0 9.2 14.1
8.7 77.2
4849.5 139 20.5
0.0 77.1
22.9
4850.5 135 8.4 0.0 57.2
42.8
4851.5 0 1.1 0.0 63.6
36.4
2–26
Figure 2.14. Saturation distributions from core analysis data.
Other useful observations can be made from the core
analysis data. The low residual oil saturation of about 20% indicates a
light oil reservoir in contrast to a heavy (more viscous) oil reservoir in
which the residual oil saturation would be much higher than 20%. All of
2-27
the measured properties vary with depth, which is an indication that the
reservoir is heterogeneous in nature. The porosity and permeability
distributions are shown in Figure 2.15. Variability of reservoir properties
is pervasive. Not only do the properties vary along the well depth, they
also vary laterally away from the well. The characterization of this
variability and the estimation of the properties at unmeasured locations
are the subjects addressed by geostatistics (see Chapter 4).
2.7 INDIRECT POROSITY MEASUREMENT FROM WELL LOGS
2.7.1 Introduction to Well Logging
In-situ porosity cannot be measured directly in the field as in the
laboratory. Therefore, only indirect measurements are made through well
logging. These measurements use either sonic energy or some form of
induced or applied radiation. Most log evaluation is concerned primarily
with determining in-situ porosity and water saturation. Neither in-situ
water saturation nor hydrocarbon saturation can be measured directly in
the wellbore. However, it is possible to infer the water saturation if the
porosity is known by measuring the resistivity of the formation.
Therefore, in this section, porosity and resistivity logs are discussed.
Care should always be taken in comparing core versus log-derived
porosities, particularly in rocks that have been highly affected by
diagenesis. Logs measure average porosities over a much larger volume
than conventional laboratory core analysis. Also, a laboratory core has
been relieved of the overburden and lateral stresses and because it is an
elastic medium, it will expand. Since the minerals have very low
coefficient of expansion, the increase in volume must be due almost
solely to the increase in porosity. Thus, the porosity measured in the
2–28
laboratory at ambient conditions may be expected to be higher than at in
situ conditions.
Figure 2.15. Porosity and permeability distributions from core analysis
data.
2-29
2.7.2 Mud Filtrate Invasion
Well log measurements are made in the borehole after the well has
been drilled. The drilling operation alters the formation characteristics
near the wellbore where the log measurements are made. In order to
interpret the logs, it is necessary to understand the changes that have
occurred in the formation caused by the drilling mud.
Drilling mud is a complex liquid usually composed of mainly water
(for water-based muds) and suspended solids and various chemicals that
control the mud properties (viscosity, fluid loss, pH and others). Clays
(bentonite) are added to give the mud viscosity and weighting material
(barite) is added to increase the mud density above that of water.
The mud is circulated during drilling to lift the cuttings out of the
borehole. Another important function of the mud is to exert a
backpressure on the formation to prevent the well from "kicking" during
the drilling operations. In general, during drilling, the pressure in the
mud column in the borehole is higher than the formation pressure.
If we take a mud sample and place it in a mud press as is typically
done in mud testing, we can separate the mud into its two main
components: mud filtrate and mudcake. Mud filtrate is a clear liquid
whose salinity varies according to the source of the water used to mix the
mud and the chemical nature of the additives. Usually, the filtrate
salinity is lower than the formation water salinity. Since the filtrate is
clear (no suspended solids), it can invade the formation if the pressure in
the wellbore is greater than the formation pressure, which is the case
during overbalance drilling. The mud filtrate can displace some of the
original formation fluids away from the wellbore into the formation.
2–30
The mudcake seals off the formation from further invasion by the
mud filtrate. The presence of the mudcake can be detected by the logging
tool and will cause the borehole diameter to be smaller than the bit
diameter. It is an indication of invasion and, indirectly, of permeability.
Since the mudcake is a solid, it will not normally invade the formation.
Drilling mud itself usually cannot invade the formation because it
contains a lot of suspended solids. However, whole mud can be lost into
the formation if the formation is inadvertently fractured. This is lost
circulation, which should not be confused with mud filtrate invasion. The
mudcake has a very low permeability and as a result, controls the
volume of filtrate invasion. The depth of invasion is determined by the
porosity of the formation. The depth of invasion will be greater in a low
porosity formation than in a high porosity formation everything else
being equal.
Figure 2.16 shows a schematic diagram of the undisturbed
formation and the altered formation after it has been penetrated by the
bit. The left figure shows the undisturbed formation before it was drilled.
Let us assume that this is a sandstone formation bounded above and
below by impermeable shales. The bottom of the sand has water
saturation, Sw, of 100% and the upper section of the sand is at an
irreducible water saturation, Swirr. Irreducible water saturation is that
saturation at which the water cannot be produced during normal
production operations. The water occupies the lower portion of the
formation because it is denser than oil which floats to the top of the
formation. There is a transition zone in which the water saturation
decreases from 100% at the bottom to 25% at the top over a finite length
of the formation. This transition zone is caused by capillarity (see
2-31
Chapter 7). It should be pointed out that not all hydrocarbon bearing
zones contain irreducible water. Some can contain mobile water.
Figure 2.16. Schematic diagram of mud filtrate invasion. Undisturbed formation is at left and invaded formation is at right.
On the right is the same formation after it has been drilled. Here,
invasion has occurred. Because of the higher pressure in the borehole
than the formation, the formation acts as a mud press and separates the
mud into mudcake which plasters the borehole wall, and mud filtrate,
which invades the formation. In order to invade the formation, the fluids
that were originally there must be displaced. The filtrate flushes or
2–32
displaces the fluids deeper into the formation and takes their place near
the wellbore.
In the bottom of the formation where Sw = 100%, the flushing is
nearly complete because the formation water is being displaced by water
which is different only in salinity and is miscible with it. The salinity of
any formation water left behind will soon reach equilibrium with the
filtrate because of diffusion.
In the upper part of the formation, we had an initially large oil
saturation (Soi = 1-Swirr). Although most of the formation water has been
displaced by mud filtrate, residual oil remains in the flushed zone with a
saturation, Sor, because an immiscible displacement can never be
complete.
The filtrate has flushed out all the original fluids that it can flush
out to a certain depth. This depth is called the flushed zone. The flushed
zone water saturation is designated Sxo and the flushed zone diameter is
designated dxo. The resistivity of the water in the flushed zone is Rmf, the
resitivity of the mud filtrate, and the resistivity of the flushed zone
formation is Rxo. If we go a little deeper into the formation, we will find a
transition zone, which contains a mixture of formation fluids and mud
filtrate. The zone, from the borehole wall to the end of the mud filtrate is
the invaded zone and includes the flushed zone. The diameter of the
invaded zone is designated as di, the water saturation is Si, and the
formation resisitivity Ri. Note that in the flushed zone, Si = Sxo and Ri =
Rxo. Since the water between dxo and di is a mixture of formation water
and filtrate, it is not possible to measure a single value for Ri in this zone.
Finally, as we pass the invaded zone, we return to the undisturbed or
uncontaminated formation. This is the virgin zone. The conditions in this
2-33
zone are the same as at the left side of Figure 2.16. The resistivity of this
undisturbed zone is Rt, the true formation resisitivity, which we ideally
would like to measure for estimating the undisturbed water saturation,
Sw.
Figure 2.17 gives the same information about the fluid
distributions but in a different format. On the left is a plot of Sw versus
depth for the undisturbed formation. At the bottom, Sw = 100% and is
constant for about 30 ft from the bottom. We then enter the transition
zone, where Sw changes with depth until it reaches an irreducible water
saturation of about 25%. Sw is constant for the last 20 ft or so at Sw =
Swirr. Note that the hydrocarbon saturation, Sh is equal to (1-Sw).
Figure 2.17. Variation of water saturation with distance from the borehole.
2–34
On the right are three sections drawn horizontally through the
formation to show how Sw varies with distance from the borehole at three
depths. At the bottom section, Sxo = Si = Sw = 100%. This is because there
was never any hydrocarbon in this section of the formation and as a
result, Sh = 0.
The depth in the middle section was chosen in the transition zone
where Sw was about 40%. We can see a change in the various water
saturations because oil is present in this zone and some of it has been
displaced by mud filtrate. Because of the residual oil, Sxo is less than
100% (Sxo = 1-Sor). Si will be lower than Sxo because some of the
hydrocarbon that was originally in the flushed zone has been displaced
into this zone, and Sw in the uncontaminated zone will be 40%.
In the uppermost section, Sw is at its irreducible saturation value
for the undisturbed formation. We see the maximum variation in the
various saturations after invasion. Sxo will be lower than the Sxo in the
transition zone, and (1-Sxo) will be close to the residual oil saturation, Sor.
The water saturation will vary throughout between dxo and di. Beyond di,
Sw = Swirr = 25%. In this uppermost section, we clearly see the flushed
zone diameter, where Sxo is constant, and the end of the invaded zone,
where Sw becomes constant.
Most of the difficulties in log evaluation and the proliferation of
many tool configurations are caused by the presence of mud invasion,
usually of unknown depth, in the logging environment. Figure 2.18
shows a schematic diagram of the borehole condition for logging
measurements.
2-35
Figure 2.18. Borehole conditions for well logging (Courtesy of Schlumberger).
2–36
2.7.3 Porosity Logs
Conventional logging techniques for measuring porosity are the
Density, Neutron and Sonic logs. All of these logs provide an indication
of total porosity.
Density Log
The Density log measures the electron density of the formation by
using a pad mounted chemical source of gamma radiation and two
shielded gamma detectors (Figure 2.19). The medium-energy gamma rays
emitted into the formation collide with electrons in the formation. At each
collision, a gamma ray loses some, but not all, of its energy to the
electron and then continues with reduced energy. This type of interaction
is known as Compton scattering. The scattered gamma rays reaching the
detector, at a fixed distance from the source, are counted as an
indication of the formation density.
Figure 2.19. Schematic of density logging tool.
2-37
The number of Compton scattering collisions is related directly to
the number of electrons in the formation. Therefore, the response of the
density tool is determined essentially by the electron density (the number
of electrons per cubic centimeter) of the formation. Electron density is
related to the true bulk density in gm/cc, which in turn depends on the
density of the rock matrix, the formation porosity and the density of the
pore fluids.
For a pure element, the electron density index, which is
proportional to the electron density is defined as
2e b
ZA
ρ ρ ⎛ ⎞= ⎜ ⎟⎝ ⎠
(2.11)
where ρe is the electron density index, ρb is the bulk density, Z is the
atomic number of the element and A is the atomic weight of the element.
For a molecule, the electron density index is given by
2 i
e b
ZM
ρ ρ⎛ ⎞
= ⎜ ⎟⎜ ⎟⎝ ⎠
∑ (2.12)
where M is the molecular weight and iZ∑ is the sum of the atomic
numbers of the atoms making up the molecule, which is equal to the
number of electrons per molecule. For most materials encountered in the
formation, the quantities in brackets in Eqs.(2.11) and (2.12) are
approximately equal to unity as shown in Tables 2.3 and 2.4. The density
tool is calibrated in a fresh water filled limestone formation of high purity
to give an apparent density that is related to the electron density index
by
2–38
1.0704 0.1883a eρ ρ= − (2.13)
For liquid filled sandstones, limestones and dolomites, the apparent
density read by the tool is practically equal to bulk density of the
formation.
The bulk density of a clean formation is given by
( )1b f mρ φρ ρ φ= + − (2.14)
Eq.(2.14) can be solved for the porosity as
m b
m f
ρ ρφρ ρ
−=
− (2.15)
To calculate porosity from Eq.(2.15), the matrix and fluid densities must
be known or assumed. The depth of investigation of the density log is
relatively shallow. Therefore, in most permeable formations, the pore
fluid is the drilling mud filtrate, along with any residual hydrocarbons.
Usually, the fluid density is assumed to be 1.0 gm/cc. When residual
hydrocarbon saturations are fairly high, this can cause the calculated
porosity values to be greater than the true porosity, and should be
corrected for this effect. Table 2.4 gives the densities of various rock
matrices.
Figure 2.20 shows a typical presentation of a density log. Track 1
shows the Gamma Ray log, which measures the natural gamma
radiation of the formation. Radioactive elements such as uranium,
potassium and thorium tend to occur more in shales than in sands. As a
result, the Gamma Ray log is a lithology log that identifies shales from
sands. The caliper in the same track measures the borehole diameter.
2-39
The formation density and the porosity derived from it are presented in
Tracks 2 and 3. Also shown is the correction or compensation applied to
account for mud cake effect and small borehole irregularities.
Table 2.3: Atomic Properties of Common Elements in the Formation
Elemen
t
A
Z 2 Z
A⎛ ⎞⎜ ⎟⎝ ⎠
H 1.008 1 1.984
1
C 12.01
1
6 0.999
1
O 16.00
0
8 1.000
0
Na 22.99 11 0.956
9
Mg 24.32 12 0.986
8
Al 26.98 13 0.963
7
Si 28.09 14 0.996
8
S 32.07 16 0.997
8
Cl 35.46 17 0.958
8
K 39.10 19 0.971
9
Ca 40.08 20 0.998
0
2–40
Table 2.4: Densities of Rock Formations and Fluids
Compound
Formula
ρb (gm/cc)
2 iZM
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
∑ ρe
(gm/cc) ρa
(gm/cc)
Quartz SiO2 2.654 0.9985 2.650 2.648
Calcite CaCO3 2.710 0.9991 2.708 2.710
Dolomite CaMg(CO3)2 2.870 0.9977 2.863 2.876
Anhydrite CaSO4 2.960 0.9990 2.957 2.977
Sylvite KCl 1.984 0.9657 1.916 1.863
Halite NaCl 2.165 0.9581 2.074 2.032
Gypsum CaSO4.H2O 2.320 1.0222 2.372 2.351
Fresh Water H2O 1.000 1.1101 1.110 1.00
Salt Water 200,000
ppm
1.146 1.0797 1.237 1.135
“Oil” N(CH2) 0.850 1.1407 0.970 0.850
Methane CH4 ρmeth 1.247 1.247ρmet
h
1.335ρmeth-
0.188
“Gas” C1.1H4.2 ρgas 1.238 1.238ρgas 1.325ρgas -0.188
2-41
Figure 2.20. Presentation of density log.
2–42
Sonic Log (Acoustic Log)
The Sonic log measures the time, Δt, required for compressional
sound wave to traverse one foot of formation. Known as the interval
transit time, Δt is the reciprocal of the velocity of the compressional
sound wave. To avoid fractions, the interval transit time is scaled by 106
and reported in micro-seconds per ft (μsec/ft). Thus,
610t
vΔ = (2.16)
where Δt is the interval transit time in μsec/ft and v is the compressional
wave velocity in ft/s.
The sonic tool contains a transmitter and two receivers (Figure
2.21). When the transmitter is energized, the sound wave enters the
formation from the mud column, travels through the formation and back
to the receivers through the mud column. The difference between the
arrival times at the two receivers divided by the distance between the
receivers gives the interval transit time. The speed of sound in the tool
and the drilling mud is less than that in the formation. Accordingly, the
first arrival of sound energy at the receivers corresponds to sound travel
paths in the formation near the borehole. The logging tool has circuits to
compensate for hole size changes or any tilting of the tool in the hole.
The interval transit time in a formation depends upon lithology and
porosity. In general, the more dense or consolidated a formation, the
lower the interval transit time. An increase in travel time indicates an
increase in porosity. Based on laboratory measurements, Wyllie (1956)
concluded that in clean and consolidated formations with uniformly
2-43
distributed small pores, there is a linear relationship between porosity
and interval transit time as follows:
( )1f mt t tφ φΔ = Δ + − Δ (2.17)
where Δt is the interval transit time measured by the log, Δtf is the
interval transit time in the pore fluid, Δtm is the interval transit time in
the rock matrix, and φ is the formation porosity.
Figure 2.21. Schematic of sonic logging tool.
2–44
Eq.(2.17) can be solved for porosity as
m
f m
t tt t
φ Δ − Δ=
Δ − Δ (2.18)
To calculate porosity from Eq.(2.18), the transit times for the rock matrix
and the pore fluid must be known or assumed. Table 2.5 gives the sonic
speeds and interval transit times for common rock matrices. The depth of
investigation of the sonic log is relatively shallow. Thus, the pore fluid is
usually assumed to be mud filtrate with an interval transit time of 189
μsec/ft, corresponding to a fluid velocity of 5300 ft/sec.
If any shale laminae exist in the sandstone, the apparent sonic
porosity is increased by an amount proportional to the bulk volume
fraction of such laminae. The interval transit time is increased because
transit time for shale generally exceeds that of the matrix.
In carbonates having intergranular porosity, Wyllie’s average
formula still applies. But sometimes, the pore structure and pore size
distribution are significantly different from sandstones. Also, there is
often some secondary porosity such as vugs and fractures with much
larger dimensions than the pores of the primary porosity. In vuggy
formations, according to Wyllie, the velocity of sound depends mostly on
the primary porosity, and the porosity derived from the sonic reading
through the time average formula will tend to be too low by an amount
approaching the secondary porosity.
Direct application of the Wyllie formula to unconsolidated and
insufficiently compacted sands gives porosity values that are too high.
When shale transit time exceeds 100 μsec/ft, which is an indication of
2-45
undercompaction, then the compaction correction should be made to
obtain more accurate porosity values. This is accomplished by applying
an empirical correction factor as shown in Eq.(2.19):
100m
f m sh
t tt t t
φ⎛ ⎞⎛ ⎞Δ − Δ
= ⎜ ⎟⎜ ⎟⎜ ⎟Δ − Δ Δ⎝ ⎠⎝ ⎠ (2.19)
where Δtsh is the interval transit time in the adjacent shale. Figure 2.22
shows a typical presentation of the sonic log.
Table 2.5: Sonic Speed and Interval Transit Time for Rock Formations
vm (ft/sec)
Δtm (μsec/ft)
Δtm (μsec/ft)
(commonly used)
Sandstones 18,000 – 19,000 55.5 – 51.0 55.5 or 51.0
Limestones 21,000 – 23,000 47.6 – 43.5 47.5
Dolomites 23,000 43.5 43.5
Anhydrite 20,000 50.0 50.0
Salt 15,000 66.7 67.0
Casing (iron) 17,500 57.0 57.0
2–46
Figure 2.22. Presentation of sonic log.
Neutron Log
The Neutron log measures induced formation radiation produced
by bombarding the formation with fast moving neutrons (Figure 2.23).
2-47
The tool responds primarily to the hydrogen present in the formation.
Thus, in clean formations, whose pores are filled with water or oil, the
neutron log reflects the amount of liquid-filled porosity.
Neutrons are electrically neutral particles with a mass almost
identical to the mass of a hydrogen atom. High-energy (fast) neutrons are
continuously emitted from a radioactive source mounted in the logging
tool. These neutrons collide with the nuclei of the formation materials.
With each collision, a neutron loses some of its energy. The
Figure 2.23. Schematic of neutron logging tool.
amount of energy lost per collision depends on the relative mass of the
nucleus with which the neutron collides. The greatest energy loss occurs
when the neutron collides with a nucleus of practically equal mass, i.e.,
2–48
hydrogen. Collisions with heavy nuclei do not slow the neutrons down
very much. Thus, the slowing down of neutrons depends largely on the
amount of hydrogen in the formation.
Within a few microseconds, the neutrons have been slowed down
by successive collisions to thermal velocities, corresponding to energies
of around 0.025 electron volt (eV). They then diffuse randomly, without
losing any more energy, until they are captured by the nuclei of atoms
such as chlorine, hydrogen, silicon and others. The capturing nucleus
becomes intensely excited and emits a high-energy gamma ray of
capture. Depending on the type of Neutron logging tool, either these
capture gamma rays or the slow neutrons themselves are counted by a
detector in the tool.
When the hydrogen concentration of the material surrounding the
neutron source is large, most of the neutrons are slowed down and
captured within a short distance from the source. However, if the
hydrogen concentration is small, the neutrons travel farther from the
source before they are captured. Accordingly, the counting rate at the
detector increases for decreased hydrogen concentration and decreases
for increased hydrogen concentration. The porosity based on the neutron
count is given by
logN a b φ= − (2.20)
where N is the slow neutrons counted, a and b are empirical constants
determined by appropriate calibration and φ is the porosity.
Since there is very little difference in the concentration of hydrogen
in oil or water, neutron logs measure the liquid filled porosity. A high
2-49
neutron counting rate indicates low porosity and a low neutron counting
rate indicates high porosity. Figure 2.24 shows a typical presentation of
the Neutron log. The neutron count is presented in API (American
Petroleum Institute) units. The porosity is in neutron porosity units
based on calibration with limestone or sandstone.
Two additional factors should be considered in the interpretation of
neutron logs. First, shales and zones containing a significant amount of
shale, will indicate a high neutron porosity due to the bound water
associated with the shale. Secondly, because of the lower concentration
of hydrogen in gas than in oil or water, a zone containing gas will
indicate a neutron porosity that is lower than it should be. These
features are really an advantage since a comparison of the neutron
porosity to cores and other porosity logs provides a convenient method
for determining shale volumes and for distinguishing gas zones from oil
or water zones. In a gas zone, the fluid density is very much lower than
the 1.0 gm/cc used in Eq.(2.15) to calculate the density porosity. As a
result, the density porosity in a gas zone is higher than it should be.
Thus, in a gas zone, the neutron porosity is too low and the density
porosity is too high. When the two porosity logs are superimposed, the
two curves will agree is shales and in liquid zones and will cross over in
gas zones. This cross over of the two logs can be used to identify gas
bearing zones as shown in Figure 2.25.
2–50
Figure 2.24. Presentation of Neutron log.
2-51
Figure 2.25. A comparison of neutron and density porosities. Shaded areas indicate gas zones.
2–52
Combination Porosity Logs
In many areas, it is common practice to record more than one
porosity log on a well. Common combinations are Density-Neutron,
Density-Sonic and Sonic-Neutron. Sometimes, all three logs are run in
the same well. These logs are usually recorded along with a Gamma Ray
curve and a Caliper.
Combination porosity logs are used to (1) differentiate oil or water
from gas zones, (2) calculate quantitative values for lithology, and (3)
determine volume of shale in the rock matrix. Figure 2.26 shows a
section of the three porosity logs run in the same well.
2.7.4 Resistivity Log
Resistivity is one of the most useful physical properties measured
in the borehole. Formation resistivity measurements, in conjunction with
porosity and water resistivity, are used to obtain values of water
saturation and consequently, hydrocarbon saturation. They are also
used in conjunction with lithology logs to identify hydrocarbon bearing
intervals and to estimate the net pay thickness.
Resistivity is the degree to which a substance “resists” or impedes
the flow of electrical current. It is a physical property of the material,
independent of size and shape. In well logging, both resistivity and
conductivity are used frequently. One is the reciprocal of the other. Thus,
1ResistivityConductivity
= (2.21)
2-53
Figure 2.26. A comparison of the three porosity logs in the same formation.
2–54
Low resistivity corresponds to high conductivity and high resistivity
corresponds to low conductivity. The resistivity unit used in well logging
is ohm-meter2/meter, which is usually shortened to ohm-meter.
Electrical conductivity is expressed in mhos per meter. In order to avoid
decimal fractions, in electrical logging, it is expressed in millimhos per
meter.
In reservoir rocks, the sedimentary minerals that make up the
formation matrix are non-conductors. Also, hydrocarbons such as gas
and oil are non-conductors. Therefore, current flow in sedimentary rocks
is associated with the water in the pore space. Most of the waters
encountered in well logging contain some sodium chloride (NaCl) in
solution. The current then is carried by the ions of the salt, which is
dissolved in the water. Therefore, conductivity is proportional to the salt
concentration (salinity) of the water. Although each ion is capable of
carrying only a definite quantity of charge, as the formation temperature
is increased, these ions are capable of moving faster. This results in
increased conductivity. Figure 2.27 shows the variation of water
resistivity with temperature at various salinities.
The amount of water contained in the formation is directly related
to the porosity and, also, affects the formation resistivity. As the volume
of water increases, the capacity for ions increases and the conductivity
increases. Thus, the formation resistivity is affected by (1) salt
concentration in the water (salinity), (2) reservoir temperature, (3) water
volume (porosity) and (4) hydrocarbon content. Thus, although we
cannot directly measure the amount of hydrocarbon in a formation, we
can infer the hydrocarbon content from resistivity measurements.
2-55
Figure 2.27. Variation of water resistivity with temperature and salinity.
Let Rw be the resistivity of the formation water, Ro be the
resisitivity of the formation saturated 100% by the formation water of
resistivity Rw and Rt be the true resistivity of the formation partially
saturated with water of resistivity Rw and hydrocarbon. Based on
laboratory measurements, Archie (1953) found that Ro was directly
proportional to Rw for clean, consolidated sandstone cores for a fixed
porosity. Thus,
or oo w
w
RR FR FR
= = (2.22)
where F is a constant of proportionality at a given porosity known as the
formation resistivity factor. He further found that F could be related to
the porosity of the core by an equation of the form
2–56
m
aFφ
= (2.23)
where a is an empirical constant and m is a cementation factor that
varies from 1.3 for unconsolidated sands to 2.5 for consolidated
sandstones. Thus, for clean sandstone cores of varied porosity,
ln ln lnF a m φ= − (2.24)
Eq.(2.24) suggests that a graph of lnF versus lnφ should be linear. Figure
2.28 shows such a graph for clean sandstone cores. For these data, a =
1.10 and m = 1.73.
y = 1.1016x-1.7332
R2 = 0.8648
1
10
100
1000
0.01 0.1 1
Porosity
Form
atio
n R
esis
tivity
Fac
tor
Figure 2.28. Log-Log graph of formation resistivity factor versus porosity for various water resistivities.
2-57
For the cores used in his measurements, Archie found a to be
approximately 1.0. Therefore, he proposed the following relationship
between formation resistivity factor and porosity
1mF
φ= (2.25)
Others performed similar measurements using their own core samples. A
group in Humble Oil (Now ExxonMobil) performed similar measurements
and found that their data were best fitted by an equation of the form
2.15
0.62o
w
RFR φ
= = (2.26)
Eq. (2.26) is known as the Humble formula and is still widely used in the
petroleum industry.
Archie also conducted resistivity measurements in partially
saturated cores to measure Rt at various water saturations, Sw. He
defined formation resistivity index as
t
o
RIR
= (2.27)
He found that the formation resistivity index, I, was related to the water
saturation by an equation of the form
1t
no w
RIR S
= = (2.28)
2–58
where n is the water saturation exponent. Eq.(2.28) can be solved for the
water saturation as
1 o wn n nw
t t
R FRSI R R
= = = (2.29)
Figure 2.29 shows lnI versus lnSw for a Berea sandstone core. For this
sample, n = 2.27. Archie found n to be approximately 2 for the core
samples in his study. The hydrocarbon saturation is given by
1h wS S= − (2.30)
Figure 2.30 shows an example invasion profile for resistivity
measurements. This profile is analogous to the water saturation profile of
Figure 2.17.
2-59
Figure 2.29. Resistivity index for Berea sandstone core
2–60
Figure 2.30. Example invasion profile for resistivity logs.
Three types of logging tools are used to measure formation
resistivity: Induction tools, focused resistivity tools and unfocused
resistivity tools. These tools can be further subdivided into those that
measure a very small volume of the formation (microresistivity logs) and
those that measure a relatively large volume of the formation. Table 2.6
presents a summary of the various resistivity tools and their limitations.
2-61
Table 2.6: Resistivity Tools
Electric Log
The Electric log was the basic and most frequently used log until
the mid 1950's. This log was invented and developed by two French
brothers, Conrad and Marcel Schlumberger. Figure 2.31 shows the
presentation of an electric log. It consists of a Spontaneous Potential (SP)
curve in Track 1 and a combination of resistivity curves designated as
normal and lateral depending on the electrode arrangements.
The normal curve is produced by two effective electrodes downhole,
a current electrode and a pickup electrode as shown in Figure 2.32.
Resistivity values are measured by recording the voltage drop across
these electrodes. A short normal, with electrode spacing of 18 inches, is
used for correlation to define bed boundaries, and to measure the
2–62
resistivity near the wellbore. Normal curves have a radius of investigation
of approximately twice the electrode spacing.
Figure 2.31. Presentation of an electric log.
2-63
The lateral curve is produced by three effective electrodes, one
current and two pickup electrodes (Figure 2.32). The radius of
investigation is approximately equal to the electrode spacing, which is
the distance from the current electrode to the midpoint between the two
pickup electrodes. The spacing is usually in the range of 16 to 19 feet.
Lateral curves are nonsymmetrical and highly distorted by adjacent beds
and thin beds, but are effective in measuring true resistivity in thick
homogeneous formations.
Figure 2.32. Schematic of electric logging tool: (A) Normal curve, (B)
Lateral curve.
2–64
Induction-Electric Log
The Induction-Electric log is a combination of electric log curves
with induction curves. The induction tool was developed to provide a
means of logging wells drilled with oil-based (nonconductive) muds. All
the original electric logging tools used the mud column to conduct the
current into the formation, so they could not be run in nonconductive
muds or air-drilled holes. Although the induction tool was developed to
meet the need for a resistivity tool that could operate in a nonconductive
mud, it soon became recognized that the tool worked better than the
original electric log in freshwater muds. The induction curve was easier
to read than the electric log, and it read close to true formation resistivity
in formations where the resistivity was not over 200 ohm-meter and Rmf
was greater than Rw.
The induction tool works by the principle of electromagnetic
induction. A high-frequency alternating current flows through a
transmitter coil mounted on the logging tool (Figure 2.33). This current
sets up a high-frequency magnetic field around the tool, which extends
into the formation. The alternating magnetic field causes currents to flow
through the formation concentric with the axis of the induction tool. The
currents, called ground loops, are proportional to the conductivity of the
formation. They alternate at the same frequency as the magnetic field
and the transmitter current flowing through the induction coil. The
ground loop currents set up a magnetic field of their own. This secondary
magnetic field causes a current to flow in the receiver coil located in the
logging tool. The amount of current flowing in the receiver coil is
proportional to the ground loop currents and therefore to the
conductivity of the formation. The signal in the receiver coil is detected,
2-65
processed and recorded on the log as either a conductivity measurement
or a resistivity measurement.
The tool illustrated in Figure 2.33 is a simple two-coil device. In
practice, bucking coils are used to help focus the effects of the main
transmitter and receiver coils and to remove unwanted signals from the
borehole. One popular induction tool used today has six different coils.
The depth of investigation (the depth from which most of the
measurement is obtained) for a typical deep induction tool is about 10
feet. The vertical resolution (the thinnest bed that the tool will detect) is
40 inches. Both the depth of investigation and the vertical resolution are
affected by the spacing between the main transmitter and receiver coils
as well as by the placement of the focusing coils. By judicious selection of
these parameters, different depths of investigation can be designed into a
tool. Thus, one can measure the resistivity profile through the invaded
zone and correct the deep induction reading to move it close to the true
formation resistivity, Rt.
Figure 2.33. Schematic of induction logging tool.
2–66
Figure 2.34 shows a typical presentation of the induction electric
log (IEL). It includes an SP and/or Gamma Ray curve, 18" normal and
the induction curve on both the resistivity and conductivity scales. An
amplified 18" normal curve is often recorded in areas where low
resistivities are encountered.
For many years, the induction electric log was the most popular
induction tool in high-porosity formations such as in California, along
the US Gulf Coast and in other high-porosity, moderate-resisitivity
formations. A single induction curve with a vertical resolution of about 3
feet and a depth of investigation of about 10 feet was combined with
either a short normal curve or a shallow laterolog curve. Since mud
filtrate invasion is seldom deep in high-porosity formations, these two
curves, corrected for borehole and bed boundary effects, could be used to
determine Rt.
Dual Induction Laterolog
The Dual Induction Laterolog was developed for those areas that
had low porosities and deep invasion. The tool has two induction curves
(ILd and ILm) with a vertical resolution of about 40 inches. However, one
induction curve, the ILd, reads very deeply into the formation, while the
medium induction curve, ILm, reads only half as deep. A shallow-reading
laterolog combined with the two induction curves gives a good
description of the resistivity profile.
2-67
Figure 2.34. Presentation of induction-electric log.
2–68
Figure 2.35 shows a typical presentation of a dual induction
laterolog. An SP and/or Gamma Ray curve and three resistivity curves
having different depths of investigation are recorded. The shallow
laterolog measures the resistivity of the flushed zone, Rxo. The medium
induction curve, ILm, measures the combined resistivity of the flushed
and invaded zones, Ri. The deep induction curve responds primarily to
the resistivity of the uncontaminated zone, Rt. The resistivity curves may
be recorded on logarithmic or linear scales. The logarithmic presentation
permits a greater dynamic range for resistivities and is convenient for
determining ratios since the difference of two logarithms is equal to their
ratio.
The ratios of shallow to deep curves and medium to deep curves
are used to determine the diameter of invasion, di, the resistivity of the
flushed zone, Rxo, and the true formation resistivity, Rt. Figure 2.36
shows a typical "tornado chart" (so called because of its distinctive shape)
used to correct the dual induction log to obtain Rt.
2-69
Figure 2.35. Presentation of dual induction laterolog.
2–70
Figure 2.36. Tornado chart used to correct deep induction resistivity to true resistivity.
2-71
Focused Electric Log (Guard and Laterolog)
In boreholes which contain extremely saline drilling muds or very
high resistivity formations, the current that is emitted from a normal or
lateral electrode is almost entirely confined within the borehole and flows
up and down within the mud column. Very little, if any, of the current
penetrates surrounding resistive material. Under similar borehole
conditions, the induction logging tool is also adversely affected because
too much of the receiver voltage is derived from high conductivity of the
invaded zone. Focused-current logging tools have been designed to
overcome these problems in part.
There are two different focused-current logging systems, referred to
as the guard and laterolog, in use today (Figure 2.37). In the guard
system, guard electrodes are placed above and below a current electrode
and kept at the same potential to focus the formation current into a thin
disc, which flows perpendicularly to the borehole. The radius of
investigation is approximately three times the length of the guard
electrode.
The guard log defines bed boundaries very well and is affected very
little by adjacent bed resistivities. Shallow guard systems, utilizing short
guard electrodes (approximately 30 inches), are used with tools like the
Dual Induction for measuring the flushed zone resistivity, Rxo, or the
invaded zone resistivity, Ri. The longer guard (5 feet in length) systems
are used for measuring the true resistivity of the uncontaminated
formation, Rt. Figure 2.38 shows an example guard log presentation.
The laterolog electrode arrangement consists of a center current
electrode placed symmetrically between three short-circuited pairs of
electrodes. A controlled current is emitted from the short-circuited outer
2–72
pair of electrodes in such a manner that the voltage difference between
the two inner short-circuited pairs of electrodes is essentially zero. As in
the guard system, these electrode arrangements focus the formation
current into a thin disc, which flows perpendicularly to the borehole.
Figure 2.37. Focused-current electrode arrangements.
Various laterolog tools have been developed over the years. Among
the most commonly used tools, the dual laterolog is common. This tool,
similar to the dual induction tool, has both deep and shallow measuring
laterologs. It is often run in conjunction with a very shallow reading
laterolog tool, which is mounted on a pad pressed against the borehole.
2-73
This shallow reading curve, called the micro-spherically focused log,
measures the flushed zone resistivity (Rxo). This combination of
measurements can define the resistivity profile from the borehole,
through the invaded zone to the undisturbed formation. Since the
current path for these logs is through the mud to the borehole wall,
through the invaded zone, and then to the uncontaminated zone, the
resistivity readings are a combination of these different zones. However,
mud and the invaded zones affect the tool's resistivity measurement
much less than unfocused tools, a feature which minimizes corrections.
Microresistivity Logs
Microresistivity tools are designed to measure the resistivity of the
flushed zone (Rxo). Since the flushed zone could be only 3 or 4 inches
deep, Rxo tools have very shallow readings, with depths of investigation
approximately 1 to 4 inches. The electrodes are mounted on flexible pads
pressed against the borehole wall, thereby eliminating most of the effects
of the mud on the measurement (Figure 2.39). Microresistivity logs
include the microlog, microlaterologs and microspherically focused logs.
Collectively, these logs can be used to estimate
• Depth of invasion
• Flushed zone water saturation (Sxo)
• Moveable hydrocarbon saturation (Sxo-Sw)
• Corrections for deep induction and laterologs
• Permeability
• Hole diameter
• Pay zone thickness
2–74
• Porosity.
Figure 2.38. Presentation of a guard log.
2-75
Figure 2.39. Microresitivity logging tools.
2–76
2.7.5 Lithology Logs
Two lithology logs are commonly used in formation evaluation, the
Spontaneous Potential (SP) log and the Gamma Ray (GR) log. Both are
recordings of naturally occurring phenomena in the formation.
Spontaneous Potential Log (SP)
The SP curve records the electrical potential produced by the
interaction of formation water, conductive drilling mud, and certain ion-
selective rocks such as shale. It is a recording versus depth of the
difference between the electrical potential of a moveable electrode in the
borehole and the electrical potential of a fixed surface electrode. Opposite
shales, the SP curve usually defines a more or less straight line on the
log, called the shale baseline. Opposite permeable formations, the curve
shows deflections from the shale baseline. In thick beds, these
deflections tend to reach an essentially constant deflection defining a
sand line. The deflection may be to the left (negative) or to the right
(positive), depending primarily on the salinities of the formation water
and of the mud filtrate. If the formation water is more saline than the
mud filtrate, the deflection is to the left. If it is less saline than the mud
filtrate, the deflection is to the right.
The position of the shale baseline on the SP log is arbitrary as it is
set by the logging engineer so that the curve deflections remain in the SP
track of the log. The SP is measured in millivolts (mV).
An SP curve cannot be recorded in boreholes filled with
nonconductive muds, such as oil muds or air, because such muds do not
provide electrical continuity between the SP electrode and the formation.
Also, if the resistivities of the mud filtrate and formation water are about
2-77
equal, the SP deflections will be small and the curve will be rather
featureless and useless.
The deflections on the SP curve result from electric currents
flowing in the mud in the borehole. These currents are caused by
electromotive forces in the formation, which are of electrochemical and
electrokinetic origins. Consider a permeable formation with thick shales
above and below it. Assume that the two electrolytes present, formation
water and mud filtrate, contain sodium chloride (NaCl) only. Because of
the layered clay structure and the charges on the layers, shales are
permeable to the Na+ cations but impermeable to the Cl- anions. Only the
Na+ cations are able to move through the shale from the more saline to
the less saline NaCl solution. This movement of charged ions constitutes
an electric current, and the force causing them to move constitutes a
potential across the shale.
The curve arrow in the upper section of Figure 2.40 shows the
direction of the current corresponding to the flow of Na+ ions through the
adjacent shale from the more saline formation water to the less saline
drilling mud in the borehole. Since shales pass only the cations, shales
resemble ion-selective membranes, and the potential across the shale is
called the membrane potential.
A second component of the electrochemical potential is produced
at the edge of the invaded zone where the mud filtrate and formation
water (the electrolytes) are in direct contact. Here Na+ and Cl- can diffuse
from one electrolyte to the other. Since Cl- are more mobile than Na+
ions, the net result of the diffusion is the flow of negative Cl- ions from
the more saline to the less saline electrolyte. This is equivalent to a
conventional current flow in the opposite direction as shown by the
2–78
arrow A in the upper half of Figure 2.40. The current flowing across the
junction between solutions of different salinity is produced by an
electromotive force called liquid-junction potential. The magnitude of the
liquid-junction potential is much smaller than the membrane potential.
If the permeable formation is clean (not shaly), the total
electrochemical emf, Ec, corresponding to these two phenomena is given
by
log wc
mf
aE Ka
= − (2.31)
where aw and awf are the chemical activities of the two solutions at
formation temperature, K is a coefficient proportional to temperature,
and for NaCl formation water and mud filtrate is 71 at 25 ºC (77 ºF). The
chemical activity of a solution is roughly proportional to salinity and
hence to its conductivity. If the solutions contain substantial amounts of
salts other than NaCl, the value of K at 77 ºF may differ from 71. If the
permeable formation is shaly, or contains dispersed clay, the total
electrochemical emf will be reduced since the clay produces an
electrochemical membrane of opposite polarity to that of the adjacent
shale bed.
2-79
Figure 2.40. Schematic representation of potential and current distribution in and around a permeable bed.
2–80
An electrokinetic potential, Ek (also known as streaming potential
or electrofiltration potential), is produced when an electrolyte flows
through a permeable, nonmetallic, porous medium. The magnitude of the
electrokinetic potential is determined by several factors, among which are
the differential pressure producing the flow and the resistivity of the
electrolyte. In the borehole, an electrokinetic emf, Ekmc, is produced by
the flow of the mud filtrate through the mud cake. An electrokinetic emf,
Eksh, may also be produced across the shale, since it may have sufficient
permeability to permit a tiny amount of filtrate flow from the mud. Each
of these electrokinetic emfs contributes to a more negative SP reading
opposite the permeable bed and opposite the shale. The net contribution
to the SP deflection is, therefore, the difference between Ekmc and Eksh,
which is generally small and negligible.
The movement of ions, which causes the SP phenomenon, is
possible only in formations that have a certain minimum permeability.
However, there is no direct relationship between the magnitude of the SP
deflection and permeability, nor does the SP deflection have any direct
relationship with the porosity.
The lower portion of Figure 2.40 shows how the SP currents flow in
the borehole and formations. The current direction shown corresponds to
the more usual case where the salinity of the formation water is greater
than that of the mud filtrate. Thus, the potential observed over the
permeable bed is negative with respect to the potential opposite the
shale. This negative variation corresponds to an SP curve deflection to
the left of the SP log as shown in the figure.
As shown in Figure 2.40, the SP currents flow through four
different media: the borehole (mud), the invaded zone, the noninvaded
2-81
part of the permeable formation, and the surrounding shale. In each
medium, the potential along a line of current flow drops in proportion to
the resistance encountered. The total potential along a line of current
flow is equal to the total emf. The deflections on the SP curve are,
however, a measurement of only the potential drop in the borehole (mud)
resulting from the SP currents. This potential drop represents only a
fraction (although usually the major fraction) of the total emf. If the
currents could be prevented from flowing by means such as the
insulating plugs as shown in the upper part of Figure 2.40, the potential
differences observed in the mud would equal the total emf. The SP curve
recorded in such an idealized condition is called the static SP curve and
is shown in the lower part of Figure 2.40.
Figure 2.41 shows the presentation of an SP curve. In general,
shale-free permeable beds of moderate to low resistivity are sharply
defined by the SP curve. High resistivity beds distort the SP currents,
causing a change in the shape of the SP curve at bed boundaries and
thus poor boundary definitions. Also, the SP curve is depressed in
permeable zones that contain shale or hydrocarbon. The shape of the SP
curve is influenced by (1) the thickness (h) and resistivity (Rt) of the
permeable bed, (2) the resistivity (Ri) and the diameter (di) of the invaded
zone, (3) the resistivity (Rs) of the surrounding formation, and (4) the
resistivity of the mud (Rm) and the diameter (d) of the borehole.
2–82
Figure 2.41. Presentation of an SP curve in a sand-shale sequence.
The Gamma Ray Log (GR)
The GR log is a measurement of the natural radioactivity of the
formation. In sedimentary formations, the log normally reflects the shale
content of the formation because the radioactive elements tend to
concentrate in clays and shales. Clean formations usually have a very
low level of radioactivity, unless a radioactive contaminant such as
volcanic ash or granite wash is present or formation waters contain
dissolved radioactive salts.
The GR log can be recorded in cased holes, which makes it very
useful as a correlation curve in completion and workover operations. It is
2-83
frequently used to complement the SP log and as a substitute for the SP
curve in wells drilled with salt mud, air or oil-based muds. In each case,
it is useful for the location of shales and nonshaly beds and, most
importantly, for general correlation.
Gamma rays are bursts of high-energy electromagnetic waves that
are emitted spontaneously by some radioactive elements. Nearly all
gamma radiation encountered in the earth is emitted by the radioactive
potassium isotope of atomic weight 40 (K40) and by the radioactive
elements of the uranium and thorium series. These elements emit
gamma rays, the number and energies of which are characteristic of each
element. Figure 2.42 shows the energies of the emitted gamma rays.
Potassium emits gamma rays of a single energy at 1.46 MeV, whereas the
uranium and thorium series emit gamma rays of various energies.
In passing through matter, gamma rays experience successive
Compton-scattering collisions with atoms of the formation material,
losing energy with each collision. After the gamma ray has lost enough
energy, it is absorbed by means of photoelectric effect, by an atom of the
formation. Thus, natural gamma rays are gradually absorbed and their
energies reduced as they pass through the formation. Two formations
having the same amount of radioactive material per unit volume, but
having different densities, will show different levels of radioactivity. The
less dense formations will appear to be slightly more radioactive than the
more dense formations. The GR log response, after appropriate
corrections, is proportional to the weight concentrations of the
radioactive material in the formation:
i i i
b
V AGR
ρρ
= ∑ (2.32)
2–84
where ρi are the densities of the radioactive minerals, Vi are the bulk
volume factors of the minerals, Ai are proportionality factors
corresponding to the radioactivity of the mineral, and ρb is the bulk
density of the formation. In sedimentary formations, the depth of
investigation of the GR log is about 1 foot.
Figure 2.42. Gamma ray emission spectra of radioactive minerals.
The gamma ray logging tool contains a detector to measure the
gamma radiation originating in the volume of formation near the tool.
2-85
Scintillation counters are now generally used for this measurement. They
are much more efficient than the Geiger-Mueller counters used in the
past. Events resulting in gamma rays are random. For this reason,
gamma ray logs have a ragged appearance (see Figure 2.20).
The primary calibration for the gamma ray tools is in the API
(American Petroleum Institute) test facility in Houston. A field calibration
standard is used to normalize each tool to the API standard and the logs
are calibrated in API units. The radioactivities in sedimentary formations
generally range from a few API units in anhydrite or salt to 200 or more
units in shales.
The GR log is particularly useful for defining shale beds when the
SP is distorted (in very resistive formations), when the SP is featureless
(in fresh water-bearing formations or in salty muds, when Rmf = Rw), or
when the SP cannot be recorded (in nonconductive mud, empty or air-
drilled hole, cased holes). The bed boundary is picked at the point
midway between the maximum and minimum deflection of the anomaly.
The gamma ray reflects the proportion of shale and, in many
regions, can be used to quantitatively as a shale indicator. It is also used
for the detection and evaluation of radioactive minerals, such as potash
and uranium ore. The GR log is part of most logging programs in both
open hole and cased hole. It is readily combined with most other logging
tools and permits the accurate correlation of logs made on one trip into
the borehole with those that were made on another trip.
2.7.6 Nuclear Magnetic Resonance (NMR) Logs
NMR measurements of fluids in porous media can be used under
favorable conditions to estimate porosity, irreducible water saturation,
2–86
moveable fluid saturation, fluid viscosity, pore size distribution, surface
to volume ratios and permeability. In this section, the principles of NMR
and its use in petrophysical measurements are presented.
Nuclear Spins in a Magnetic Field
Most atomic nuclei possess a quantum mechanical property called
spin angular momentum. This means that the nucleus spins around an
axis. Since atomic nuclei are charged particles, the spinning motion
causes a magnetic moment that is co-linear with the direction of the spin
axis. The strength of this magnetic moment is a property of the type of
nucleus. 1H nuclei (protons) possess a strong magnetic moment (second
only to radioactive 3H), which, together with the high natural abundance
of hydrogen and prevalence in most fluids, makes it the ideal nucleus for
NMR logging and NMR imaging in radiology.
Consider a collection of 1H nuclei as in Figure 2.43. In the absence
of an externally applied magnetic field, the individual magnetic moments
have no preferred orientation and the net magnetization of the collection
of spins is zero. However, if an externally supplied magnetic field
(denoted as B0) is imposed, there is a tendency for the magnetic moments
to align with the external field. 1H nuclei with a quantum number of I = 12
in this situation may adopt one of two possible orientations: alignment
parallel or anti-parallel to B0 as shown in Figure 2.44. Thus, depending
on their orientation, we can define two groups or populations of spins.
Alignment parallel to B0 is the lower energy orientation and is thus
preferred, while the anti-parallel alignment is the higher energy state.
However, the energy difference between the two states is very small.
Thermal energy alone causes the two states to be almost equally
populated. The remaining population difference results in a net bulk
2-87
magnetization aligned parallel to B0. It is only this net magnetization
arising from a small population difference that is detectable by NMR
techniques.
The individual spins do not align exactly parallel or anti-parallel to
B0, but at an angle to B0. This is analogous to the case of a spinning top:
Figure 2.43. Random orientations of the nuclear magnetic moments in the absence of an externally applied magnetic field.
Bo
parallel
antiparrallel
Figure 2.44. Two orientations of the nuclear magnetic moments in the
2–88
presence of an externally applied magnetic field (Bo).
the top precesses about the axis defined by the pull of gravity. This
precession defines the surface of a cone. Figure 2.45 shows a model for a
large collection of spins at any given instant. Here, each of the vectors
(arrows) represents an individual spin. The vector M represents the bulk
net magnetization that results from the vector sum of the contributions
from each of the spins.
Bo M
parallel
antiparallel
Figure 2.45. The bulk net magnetization M
The individual magnetic moments precess at a certain frequency,
known as the Larmor frequency, which is determined by the strength of
the magnetic field and the type of nucleus. The Larmor equation gives
this frequency as
0 0Bω γ= (2.33)
where ω0 is the Larmor frequency (in radians per second), Bo is the
magnetic field strength, and γ is a constant for each nucleus known as
2-89
the gyromagnetic ratio. Eq.(2.33) is the fundamental equation for all
NMR methods. Table 2.7 gives γ values for some NMR active nuclei.
From Table 2.7, we can see that 1H nucleus has the highest
relative sensitivity and a high natural abundance. So 1H nuclei (protons)
possess a strong magnetic moment and are most commonly used in NMR
logging and NMR imaging.
Table 2.7: Magnetic Resonance Properties of Some Important Nuclei
Nucleus
Natural
Abundance %
γ
Hz / Gauss
Relative
Sensitivity
1H 13C 19F
23Na 31P
99.98
1.10
100.0
100.0
100.0
4257
1071
4005
1126
1723
1.0
0.016
0.830
0.093
0.066
The Effect of Radiofrequency Pulses - Resonance Absorption
In order to detect a signal, a condition of resonance needs to be
established. The term “resonance” implies alternating absorption and
dissipation of energy. Energy absorption is caused by radiofrequency (RF)
perturbation, and energy dissipation is mediated by relaxation processes.
RF radiation, like all electromagnetic radiation, possesses electric
and magnetic field components. We may consider the RF as another
magnetic field of strength B1 perpendicular to B0 as shown in Figure
2.46. At equilibrium, M (=Mo) is stationary and difficult to observe. By
2–90
applying an RF field B1 perpendicular to B0, M can be rotated so that it
has a component transverse to B0. A maximum transverse component is
obtained by applying B1 with an amplitude and duration that rotates M
by 90 degree and into the plane transverse to B0. This is called a 90-
degree pulse. Once in the transverse plane, M precesses (rotates) around
the B0 axis at the Larmor frequency. This rotating magnetization can
induce an alternating current (AC) in a receiver coil, and that current can
be used to record the action of the magnetization in the transverse plane.
The signal that is induced in the receiver coil decays over time. The
signal decay is due to a process known as relaxation.
Relaxation Processes
In resonance absorption, RF energy is absorbed by the nuclei when
it is broadcast at the Larmor frequency. Relaxation is the process by
which the nuclei release this energy and return to their original
configuration. There are two relaxation processes involved: transverse
relaxation and longitudinal relaxation.
Bo M
B1
Figure 2.46. A second magnetic field B1 generated by Radiofrequency (RF)
2-91
Given that at equilibrium, the net magnetization is longitudinal (B0
direction), the equilibrium magnetization in the transverse plane is zero.
So the signal that is induced in the receiver coil will decay over time and
reach zero some time after the 90º RF pulse is turned off. This decay
process is exponential and the decay envelop can be expressed by the
equation
( )0 *2exp /transverse transverseM M t T= − (2.34)
where 0transverseM is the initial transverse magnetization immediately after a
90º RF pulse, transverseM is the transverse magnetization at any given time t
after a 90º RF pulse, *2T is the apparent NMR transverse relaxation time
or spin-spin relaxation time, which characterizes the rate of signal decay.
The decay mechanism is that different components of the magnetization
may precess at slightly different rates, a process known as “dephasing”
in the transverse plane. Since the signal recorded is the vector sum of all
the transverse components, sufficient dephasing will lead to complete
cancellation of the signal.
One of the major causes of this dephasing is B0 inhomogeneity
(ΔB0) effects. Spins at different locations are not exposed to exactly the
same B0 field, which in turn yields a range of Larmor frequencies. ΔB0
effects are largely suppressed by correcting the magnetic field uniformity
as much as possible, and by employing NMR “spin-echo” techniques.
In spin-echo NMR (Hahn, 1950), the dephasing ΔB0 effects are
largely reversed by following a 90-degree RF pulse with a short delay τ
and a 180-degree “refocusing” pulse. After a second delay τ, the
transverse magnetization transverseM is refocused as an NMR spin-echo. ΔB0
2–92
effects are canceled at the center of the echo, where the dephasing from
ΔB0 is momentarily rephased.
If the induced signal is measured at time TE after a 90º RF pulse,
then equation (2.34) can be rewritten as
( )02exp /transverse transverse EM M T T= − (2.35)
where TE=n2τ and n is the number of 180-degree pulses. T2 is the true
NMR transverse relaxation time or spin-spin relaxation time. In Eq.(2.35), *
2T is substituted with T2 upon the correction of ΔB0 effects.
T2 can be measured by repeating the spin-echo experiment with
different TE, then fitting the resulting transverseM versus TE curve to
Eq.(2.35). Another more accurate method, and one that is used in NMR
logging, is to employ a multiple-spin-echo sequence, where the NMR
signal is refocused multiple times as shown in Figure 2.47. This method
removes unwanted molecular diffusion effects. In NMR logging, 400 or
more echos are used in the measurements. The exponential curve, which
measures the amplitudes of the decaying echos is the fundamental well
log measurement and is used to compute the T2 spectrum.
2-93
Figure 2.47. Carr-Purcell-Meiboom-Gill spin echo pulse sequence.
Longitudinal relaxation is a process that restores the longitudinal
magnetization to its equilibrium state after an RF pulse is turned off.
Immediately after a 90º pulse, the net magnetization vector lies in the
transverse plane. Therefore, longitudinal magnetization is zero. However,
as time passes, the longitudinal magnetization will approach the
equilibrium value. This buildup of the longitudinal magnetization is also
exponential in time and can be expressed as
( ) ( )0 11 exp /zM t M t T= − −⎡ ⎤⎣ ⎦ (2.36)
where M0 is the equilibrium value of the longitudinal magnetization, Mz(t)
is the longitudinal magnetization at any given time t after a 90º RF
pulse, and T1 is the NMR longitudinal relaxation time or spin-lattice
relaxation time that characterizes the rate of buildup. For any given
system, T1 is greater than or equal to T2. (T1 and T2 of 1H nuclei in water
2–94
in a test tube is about 3 seconds. In a porous medium, the T1 and T2 are
significantly lower than in a test tube and are of the order of
milliseconds). The longitudinal direction or direction of B0, is
conventionally assigned the z direction, i.e., B0=Bz.
In an NMR experiment, sufficient time must elapse between
successive 90º RF pulses to allow M to achieve equilibrium. This time,
known as the repetition time TR, depends on the sample’s T1. In general,
TR should be greater than 5T1. Substituting TR into Eq.(2.36) gives
( ) ( )0 11 exp /z R RM T M T T= − −⎡ ⎤⎣ ⎦ (2.37)
where Mz(TR) is the longitudinal magnetization at the repetition time TR
after a 90º RF pulse is turned off. Since another 90º RF pulse is applied
immediately at time TR, the 0transverseM for this subsequent 90º RF pulse
should be equal to Mz(TR)) of Eq.(2.37). Thus,
( )00 11 exp /transverse RM M T T= − −⎡ ⎤⎣ ⎦ (2.38)
Substituting Eq.(2.38) into Eq.(2.35) gives the transverse magnetization
at the time of measurement as
( ) ( )0 1 21 exp / exp /transverse R EM M T T T T= − − −⎡ ⎤⎣ ⎦ (2.39)
As Eq.(2.38) implies, T1 can be measured by repeating an NMR
experiment with several different TR values, then fitting the resulting
transverseM versus TR to Eq.(2.38).
Molecular Diffusion Effect
2-95
Random thermal motion will affect the NMR signal as spins move
from one part of the sample to another and experience a different
magnetic field strength because of static field inhomogeneities. Variation
in the frequency and phase of these mobile spins introduces a phase
incoherence that causes a reduction in the signal amplitude. This effect
can be expressed by Eq.(2.40) (Hahn, 1950):
32 2
0 2exp exp3 2
E Etransverse transverse
R
T TD GM MT
γ⎡ ⎤⎛ ⎞ ⎛ ⎞= − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎢ ⎥⎝ ⎠ ⎣ ⎦
(2.40)
where D is the diffusion coefficient and G (∝ΔB0) is the magnetic field
gradient.
NMR Signal and Corresponding T2 Spectrum
The exponential decay curve shown in Figure 2.47 suggests that all
the spins relax at the same T2 relaxation time. This would be applicable
to the spins in a bulk fluid such as the protons in water in a test tube.
However, when the fluids are confined in the pore space of a porous
medium, the protons near the pore wall will relax faster (shorter T2) than
those in the center of the pores. Further, the protons in small pores will
relax faster than those in large pores. As a result, the NMR decay signal
will contain a spectrum of T2 relaxation times. The signal can be
decomposed into its T2 spectrum by using a multi-exponential model as
( ) ( )210
exp /j m
j jj
M tf t T
M
=
=
= −∑ (2.41)
2–96
where fj is proportional to the population of protons which have a
relaxation time of T2j. The inversion problem is to find a set if T2
amplitudes, fj, from a set of measurements of echos, gi. Application of
Eq.(2.41) to each echo gives the following system of linear equations in fj
( ) ( )210
exp / for 1,2,...,j m
ii j i j i
j
M tg f t T i n
Mε
=
=
= = − + =∑ (2.42)
where gi is the amplitude of the echo measured at time ti, εi is the error
caused by noise in the data and n is the total number of echos. A least
square fit is used to determine the fjs that minimize the sum
( )2
22
1 1 1
exp /n m m
j i j i ji j j
f t T g fα= = =
⎛ ⎞− − +⎜ ⎟
⎝ ⎠∑ ∑ ∑ (2.43)
where αi is a regularization constant or smoothing parameter.
Figure 2.48 shows an example log measurement for n = 500, TE =
1.2 milliseconds, with 12 measurements averaged together. Note the
noisy data. The magnitude of the NMR signal at t =0 (M0) is proportional
to the number of hydrogen nuclei in the measurement volume. This
number can be calibrated into the total NMR porosity. Figure 2.49 shows
the corresponding T2 spectrum obtained from the data of Figure 2.48.
2-97
Time (ms)
Figure 2.48. Carr-Purcell-Meiboom-Gill spin echo measurements.
T2 (ms)
Figure 2.49. T2 spectrum of the data of Figure 2.49.
2–98
Based on laboratory measurements in sandstone cores, the NMR
spectrum can be divided into several segments to represent the fluid
distribution in the pore space as shown in Figure 2.50. In this figure, the
area under the spectrum is equal to the total NMR porosity.
Figure 2.50. T2 spectrum and fluid distribution in the pore space.
The nomenclature for Figure 2.50 is as follows. The Free Fluid
Index (FFI) is the percent of the bulk volume occupied by movable fluids
(water + hydrocarbon). In Figure 2.50, this is shown as the fraction of the
2-99
bulk volume for T2 greater than 30 milliseconds. The Free Fluid Index is
also known as Bulk Volume Movable (BVM). The Bulk Volume
Irreducible (BVI) is the percent of the bulk volume containing irreducible
water. The irreducible water consists of two components: (1) Clay Bound
Water (CBW) (T2 from 0.3 to 3 milliseconds), which is the water bound to
the clay minerals and (2) Bulk Volume Capillary (BVC) (T2 from 3 to 30
milliseconds), which is the water trapped by capillary forces. The Bulk
Volume Water (BVW) (T2 from 0.3 to 300 milliseconds) is the percent of
the bulk volume occupied by water (moveable, capillary bound water and
clay bound water). These bulk volumes are related to porosity and
saturations as follows:
ptotal
b
VV
φ = (2.44)
ii
p
VSV
= (2.45)
p CBW BVC BVMV V V V= + + (2.46)
CBW BCV BVMtotal
b b
V V VV V
φ += + (2.47)
CBW BCV
b
V VBVIV+
= (2.48)
BVM
b
VBVMV
= (2.49)
total BVI BVMφ = + (2.50)
2–100
total MPHIφ = (2.51)
totalBVM BVI MPHI BVI FFIφ= − = − = (2.52)
w pww total
b b
S VVBVW SV V
φ= = = (2.53)
BVC BVMe
p
V VV
φ += (2.54)
BVC BVMe
total b
V VV
φφ
+= (2.55)
etotal
BVC BMVφφ+
= (2.56)
CBW BVC bwirr
p b total
V V V BVISV V φ
⎛ ⎞⎛ ⎞+= =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(2.57)
total wirrBVI Sφ= (2.58)
( )1total wirrFFI Sφ= − (2.59)
Figure 2.51 shows an NMR log in which the pore fluids in the
second track have been divided into free fluid (above 30 ms), capillary
bound fluid (3 to 30 ms), small pore bound fluid (0.28 to 3 ms), and very
small pore bound fluid (0.2 to 0.28 ms). The two small pore bound fluids
(0.2 to 3 ms) correspond to clay bound fluid. Also shown in the third
track is the T2 spectrum, with a T2 cutoff of 30 ms for free fluid. Note the
two streaks with free fluid at depths of X163 m and X189 m, where the
2-101
T2 is greater than 30 ms. This example is from a predominantly shaly
formation where the differences in the bound fluids are more apparent.
Pore Size Distribution
NMR relaxation measurements have been shown to be a sensitive
probe in the study of the microscopic structure of porous media. The
connection between NMR measurements and pore size is based on the
strong effect that the rock surface has on promoting proton relaxation.
The NMR T1 relaxation behavior of a fluid confined within a pore is
sensitive to both the pore geometry and size, and thus yields much
useful information when related to the pore-size distribution via an
appropriate mathematical model. The simplest and most common model
is the “two-fraction fast-exchange” model (Senturia and Robinson, 1970;
Howard et al., 1990), which assumes that there are two magnetically
distinct phases within the pore: a bulk phase with relaxation
characteristic of the bulk fluid and a surface phase with much faster
relaxation. Assuming that diffusion of the fluid is much faster than the
relaxation process, the observed relaxation rate is given by a single
average T1 value as
1 1 1
1 1
bulk surface
ST T V T
λ= + (2.60)
where S/V is the pore surface area to volume ratio and λ is the thickness
of the surface monolayer. The T1 technique basically determines the
surface/volume ratio as a characteristic pore-size parameter. Usually,
T1bulk>>T1surface, so Eq.(2.60) can be simplified as
2–102
Figure 2.51. Example NMR log showing fluid distribution in the pore
space.
2-103
11
1 ST V
ρ ⎛ ⎞= ⎜ ⎟⎝ ⎠
(2.61)
where ρ1 = λ/T1surface is the NMR surface relaxivity, which is a measure of
the ability of the surface to cause relaxation of proton magnetization. The
surface relaxivity ρ1 has dimensions of length/time. Eq.(2.61) shows that
T1 responds to pore size. Small pores (large S/V ratio) exhibit small
values for T1, and the converse is true for large pores.
Since transverse relaxation time T2 is closely related to T1, it is
expected that a similar relationship exists between the distribution of T2
and the pore-size distribution. However, in this case, apart from the bulk
relaxation process, which can often be neglected, the T2 relaxation is also
controlled by a surface relaxation mechanism as well as the diffusion
effect because of the magnetic field gradient. The relaxation rate equation
is of the form
( )2
2 2 2
1 1 13bulk surface
S B DT T V T
λ τγ= + + ∇ (2.62)
where the first two terms on the right side correspond to similar
expressions in Eq.(2.60), while the last term accounts for spin dephasing
because of restricted diffusion in a magnetic field gradient ∇B. τ is the
pulse spacing (pulse-to-echo delay), γ is the gyromagnetic ratio for
protons, and D is the diffusion coefficient. Usually, the first and the last
2–104
terms on the right side of Eq.(2.62)are small and can be neglected. Thus,
22
1 ST V
ρ ⎛ ⎞= ⎜ ⎟⎝ ⎠
(2.63)
where ρ2 = λ/T2surface is the NMR T2 surface relaxivity. Eq.(2.63) shows
that T2 also responds to pore-size. Thus, in a water bearing zone without
the complications of hydrocarbons, the T2 spectrum such as shown in
Figure 2.49 could be viewed as a pore-size distribution.
It should be emphasized that what is measured by NMR relaxation
is the distribution of volume-to-surface ratio, (V/S). Since (V/S) has the
dimension of length, then its distribution could be viewed as a “pore-size
distribution.” The ratio (V/S) is not a pore size or a pore diameter. Its
magnitude is markedly affected by the pore shape. The ratio will be a
maximum for a spherical pore (a sphere has the smallest surface area for
a given volume) and will decrease for other pore shapes. If all pores are
geometrically similar, then the T2 spectrum could be viewed qualitatively
as a pore size distribution.
Figure 2.52 shows a clastic sequence with shales overlying a
sandstone in a water zone. The volumetric calculations in the first track
indicate an upward decreasing clay in the sandstone interval. Geological
analysis suggests that this corresponds to a coarsening-upward
sequence. While this assertion is based on inference, the T2 spectrum in
the third track shows an upward increase in the relaxation times, a trend
2-105
which can be explained by an increase in the pore size and, therefore, an
increase in the grain size.
Figure 2.52. Example NMR log showing pore size distribution in a water-bearing sandstone section.
2–106
Carbonates often have a variety of pore types such as moldic,
intercrystalline and interparticle pores. The relationship between
relaxation times and pore size distribution is therefore more complex in
carbonates than in sandstones. Further, the surface relaxivity of
carbonates is less than that of sandstones, typically 1.7μm/s for
carbonates compared to 5μm/s for sandstones. As a result, the protons
align and relax faster in sandstones than in carbonates. Therefore, the
cut off T2 for free fluid in carbonates is higher than in sandstones,
typically on the order of 100 ms.
Estimation of Permeability from NMR Relaxation Times
The fact that T1 and T2 NMR relaxation times can be used to
estimate permeability stems from the fact that these relaxation times can
be correlated with pore size distribution in water-bearing zones.
Permeability is proportional to the square of some characteristic length of
the porous medium and as such would be proportional to 21T or 2
2T .
Various empirical equations have been proposed for estimating the
absolute permeability of a porous medium. Wyllie and Rose (1950)
suggested an empirical permeability equation of the form
x ywirrk C Sφ −= (2.64)
where C is an empirical constant, φ is the porosity, Swirr is the irreducible
water saturation and x and y are numerical exponents. Timur developed
a permeability equation in the spirit of Eq.(2.64) of the form
4.5
4210wirr
kSφ
= (2.65)
2-107
where the porosity and irreducible water saturation are in fractions and
the permeability is in millidarcy. One version of the Timur/Coates
equation that is widely used to estimate permeability from NMR logs is
given by
4 2
NMR FFIkC BVI
φ⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(2.66)
where C=10 or can be determined from laboratory measurements on
cores. In Eq.(2.66) the porosity, FFI and BVI are in porosity units (p.u.)
and the permeability is in millidarcy. BVI and FFI are related to the
irreducible water saturation by Eqs.(2.58) and (2.59).
Another permeability equation based on laboratory measurements
at Schlumber Doll Research (SDR) is given by
4 22gmk a Tφ= (2.67)
where T2gm is the logarithmic (geometric) mean of the T2 spectrum
defined as
2
12
1
logm
j jj
gm m
jj
f TT
f
=
=
=∑
∑ (2.68)
Other equations proposed by Kenyon et al. are
4 21 1NMRk C Tφ= (2.69)
and
2–108
4 22 2NMRk C Tφ= (2.70)
Figure 2.53 shows a comparison of the permeability from NMR
measurements using Eq.(2.66) with those from core analysis. The
agreement between the two sets of data is good. Also shown is the
comparison of the NMR porosity measurements with those from core
analysis. The agreement also is good.
Figure 2.53. A comparison of NMR-derived permeability and porosity with measurements from core analysis (from Dunn et al., 2002).
2-109
2.7.7 NMR Imaging of Laboratory Cores
Nuclear magnetic resonance imaging (NMRI) can be used to map
the spatial distribution of NMR observables in core analysis. Thus,
porosity distribution and fluid saturation distribution or solvent
concentration distributions can be imaged in the laboratory.
In order to create an image, the NMR experiments must be
modified to spatially encode the NMR signals. Figure 3.54 is a typical
2DFT (2-dimensional Fourier Transform) spin-echo NMR imaging
sequence showing (a) radiofrequency pulses, (b) slice-selection gradient,
(c) frequency-encoding gradient, (d) phase-encoding gradient, and (e)
NMR signal (echo). The experiment is repeated with multiple phase-
encoding gradient amplitudes.
The Effect of Magnetic Field Gradients
For most NMR imaging applications, the B0 field must be made to
vary in a linear fashion with distance. A magnetic field gradient or simply
gradient refers to the spatial variation of the strength of the B0 field.
Magnetic field gradient is a key factor in NMR imaging. A magnetic field
gradient causes the transverse magnetization to precess at a frequency
that is proportional to position along the gradient axis as follows:
( )0r rB rGω γ= + (2.71)
where ωr is the Larmor frequency at position r, Gr is the gradient, and r
is the position along the gradient axis. Since we can measure ωr and we
know B0, Gr and γ, the position of the resonating nucleus can be
determined. Gr can be applied concurrently with slice selection,
frequency encoding or independently of other events (phase encoding).
2–110
The magnitude of the gradient, its direction (i.e., along what axis), and
timing need to be controlled.
180o
90o
a. RF
b. ssG
c. ROG
d. PEG
Timing
RT p1 pw tro
ET
Echo
e. NMR
tpe
Figure 2.54. Simplified timing diagram for a 2DFT spin-echo NMR imaging sequence
2-111
Slice-Selective Excitation
In most NMR imaging applications, it is desirable to generate a
single slice or multislice images. The initial step in generating such an
image is the localization of the RF excitation to a region of space. This is
accomplished by the use of frequency-selective excitation in conjunction
with a slice-selection gradient, Gss, on an axis perpendicular to the
chosen slice plane. A frequency-selective RF pulse has two parts
associated with it, a central frequency and a bandwidth of frequencies
Δωss defined by the shape of the pulse envelope. When such a pulse is
broadcast in the presence of the slice-selection gradient, a narrow region
of the object will achieve the resonance condition and will absorb the RF
energy. The central frequency of the pulse determines the particular
location that is excited by the pulse when the slice-selection gradient is
present. Different slice positions are achieved by changing the central
frequency. The slice thickness is determined by the bandwidth of
frequencies Δωss incorporated into the RF pulse and is given by
1ss ss sG dω γΔ = (2.72)
where ds1 is the slice thickness. Typically, Δωss is fixed so that the slice
thickness is changed by changing the amplitude of Gss. Thinner slices
require larger Gss. Once Gss is determined, the central frequency ωr is
calculated using Eq.(2.71) to bring the desired location into resonance.
Multislice imaging uses the same Gss for each slice but a unique RF
pulse during excitation. Each RF pulse has the same bandwidth but a
different central frequency, thereby exciting a different region of the
object.
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Frequency Encoding
The next task in the imaging process is encoding the image
information within the excited slice. The image information is actually
the amplitude of the NMR signal arising from the various locations in the
slice. Two distinct processes are used for encoding the two dimensions:
frequency encoding and phase encoding.
The frequency encoding provides one of the two visual dimensions
of the image. The NMR signal is always detected in the presence of a
frequency encoding gradient in an imaging pulse sequence. After slice-
selective excitation, the frequency encoding gradient, also known as the
readout gradient, GRO, is applied perpendicularly to the slice direction.
Under the influence of this new gradient field, the nuclei within the slice
will precess at different frequencies depending on their positions in the
readout gradient’s direction, in accordance with Eq.(2.71). Each of these
frequencies will be superimposed in the echo. The echo signal is detected
in the presence of GRO and digitized at a chosen sampling interval for
later Fourier transformation. The magnitude of GRO and the frequencies
that are detected enable the positions of the nuclei to be determined.
Two user-selectable parameters determine the image resolution in
the frequency encoding, or the readout, direction: the field of view (FOV)
in the readout direction and the number of readout data points in the
matrix, NRO. The pixel size in the readout direction is given by
( )RO
RORO
FOVN
Δ = (2.73)
and
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( )2
RORO
RO RO ro
NBWFOVG G tγ γ
= = (2.74)
where BW is the receiver bandwidth, γ is the gyromagnetic ratio and tro is
the duration of the frequency encoding gradient.
Phase Encoding
In order to produce a two-dimensional image of the slice, one can
cause a systematic variation in phase that would encode the spatial
information along the one remaining axis of the image plane. This is
accomplished by the use of a phase encoding gradient, GPE. GPE is
perpendicular to both Gss and GRO and is the only gradient that changes
amplitude during the data acquisition loop of a standard two-
dimensional imaging sequence. The NMR imaging information is
obtained by repeating the slice excitation and signal detection many
times (typically 128 or 256 times), each with a different amplitude of GPE
applied before detection. The resulting signals are stored separately for
subsequent processing.
Separate Fourier transformation of each of these data sets yields a
set of projections onto the readout axis. Specifically, the second Fourier
transformation converts signal amplitude at each readout frequency from
a function of GPE to a function of frequency.
The image resolution in the phase encoding direction depends on
two user-selectable parameters, the FOV in the phase encoding direction
and the number of phase encoding steps in the matrix, NPE. The pixel
size in the phase encoding direction is given by
( )PE
PEPE
FOVN
Δ = (2.75)
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and
( ) PEPE
PE pe
NFOVG tπ
γ= (2.76)
where γ is the gyromagnetic ratio and tpe is the duration of the phase
encoding gradient.
Image Reconstruction
Two types of matrices are used in NMR imaging: raw data and
image data. The raw data matrix is a grid of complex points with the
frequency encoding direction displayed in the horizontal direction and
the phase encoding direction displayed in the vertical direction. All image
information is contained within the raw data matrix.
The image data matrix is obtained by a two-dimensional Fourier
transformation of the raw data matrix. The first Fourier transformation of
each row of the raw data matrix yields a set of modulated projections of
the slice onto the frequency encoding axis. The second Fourier
transformation of each column of the temporary data matrix converts the
signal magnitude at each readout frequency from a function of GPE to a
function of frequency, resulting in the image. The image matrix is a
spatial map of the nuclei signal intensity. While the Fourier
transformation contains information regarding both the magnitude and
the phase of the measured signals, the phase information is often
discarded so that the normal image matrix contains only the magnitude
information. The image matrix is usually displayed as a rectangular
image with readout in one direction and phase encoding in the other
direction.
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Figure 2.55 shows a graphical representation of how an image is
spatially encoded during 2DFT imaging. In this figure, the frequency
encoding direction is horizontal and the phase encoding direction is
vertical. A set of NMR signals is acquired using the same frequency
encoding gradient but different values of the phase encoding gradient.
Each of these NMR signals is Fourier transformed to provide a frequency
spectrum of each phase encoding step, which constitutes a one-
dimensional projection in the frequency encoding direction. Each column
of the data from the first Fourier Transform projection images is Fourier
transformed again to determine the spatial projection in the vertical
image plane.
Three-Dimensional NMR Imaging
Three-dimensional volume imaging technique is, in essence, a
double phase encoding technique. The slice-selective excitation is
replaced with another phase encoding process along that axis. Each RF
pulse excites the entire imaging volume instead of just one slice. The
second phase encoding is applied to partition or subdivide the volume
into individual layers. The number of layers is determined by the number
of phase encode steps. For example, if the number of the second phase
encoding gradient steps is changed from 32 to 128, 3D-FFT of the 3D
data set then yields 32 to 128 image layers in that direction.
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Figure 2.55. Graphical demonstration of 2DFT NMR image data acquisition and reconstruction.
The advantage of the 3D-imaging technique is that very thin
contiguous layer images can be obtained with minimal interslice
crosstalk. Also, the signal-to-noise ratio is greater than for a comparable
2D-sequential imaging method.
The disadvantage of the technique is the time required. The total
scan time for 3D volume imaging is much longer than for 2D slice or
multislice imaging. In practice, 3D volume imaging is not widely used.
Instead, 2D-multislice imaging is often used as a replacement for 3D
volume imaging.
Signal-to-Noise Ratio and Image Contrast
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Voxel size is a very important factor in increasing the signal-to-
noise ratio (SNR). Voxel size is defined by
( ) ( )
1RO PE
sRO PE
FOV FOVv d
N N⎛ ⎞⎛ ⎞
Δ = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(2.77)
If the voxel volume is large, then there are more spins in each voxel to
contribute to the signal, thus increasing the SNR. But large voxels imply
a low spatial resolution. The converse is also true. Small voxels imply a
low SNR but a high spatial resolution. Therefore, the user-controlled
parameters, slice thickness ds1, FOV, NRO and NPE, are important
parameters that affect the signal-to-noise ratio and the image resolution.
The echo time TE, repetition time TR, longitudinal relaxation time T1
and transverse relaxation time T2 are the important factors that
determine image contrast as defined in Eq.(2.39). Remembering that T2
relaxation describes the rate of decay of the NMR signal in the transverse
plane, a long TE would yield different signal intensity from objects
possessing different T2 values. The long T2 object will contribute more
signal, causing it to appear hyperintense in the NMR image relative to
the short T2 object. This is termed a “T2-weighted image.” Similarly, since
T1 relaxation describes the rate of recovery of the longitudinal
magnetization, a short TR would yield different signal intensity from
objects possessing different T1 values. The object exhibiting a short T1
value will contribute more signal, causing it to appear hyperintense in
the NMR image relative to the long T1 object. This is termed a “T1-
weighted image.”
2–118
In summary, TE controls T2-weighting and TR controls T1-
weighting of an image. Short T2 objects are dark on T2-weighted images,
but short T1 objects are bright on T1-weighted images.
Example NMR Images of Laboratory Cores
Figures 2.56 and 2.57 show NMR-derived porosity images of a
brine saturated sandstone core. Figure 2.56 shows the images at six
cross sections along the core whereas Figure 2.57 shows the images of
four, longitudinal vertical slices. The layered nature of the core is clearly
apparent from these images. Although the images are not calibrated with
numerical porosity values, nevertheless, the first three slices in Figure
2.56 indicate that the core appears to be less porous at the top than
elsewhere. The core is 10 cm long, 5 cm in diameter, and has a
permeability of 97 md and an average porosity of 15.9 %.
Figure 2.58 shows NMR-derived images of the solvent
concentration for a first-contact miscible displacement conducted in the
core of Figures 2.56 and 2.57. In this experiment, a more viscous
deuterium oxide (D2O) was used to displace the less viscous brine from
2-119
Figure 2.56. NMR images of transverse slices of a brine saturated sandstone core.
the core at a favorable mobility ratio of 0.84. D2O is practically devoid of
protons, so the injection of D2O reduces the NMR signal in the voxels
2–120
invaded by the D2O. The reduced NMR signals can be used to image the
progress of the displacement as shown in Figure 2.58. Because the
mobility ratio is favorable, the displacement pattern is controlled entirely
by the permeability variations in the core. Note the chanelling of the
injected solvent due to permeability variation in the core. Clearly, the top
of the core is less permeable than the rest of the core. This correlates well
with the low porosity of the top of the core shown in Figure 2.56. Figure
2.59 shows one-dimensional solvent concentration profiles for the same
displacement. The information is quantitative and can be used to
calibrate a numerical model of the displacement.
Figure 2.57. NMR images of longitudinal, vertical slices of a brine saturated sandstone core.
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Figure 2.58. Solvent concentration images of a first-contact miscible displacement in the sandstone core of Figures 2.55 and 2.56.
2–122
Figure 2.59. Solvent concentration profiles of the first-contact miscible displacement of Figure 2.57.
Figure 2.60 shows NMR-derived permeability images of another
layered sandstone core. The permeability for each voxel of the image was
calculated from T1 distributions using Eq.(2.69). Figure 2.61 shows a
comparison of NMR-derived permeability with flow-derived permeability
for five core samples. The agreement between the two sets of permeability
values is reasonable for four of the five measurements.
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Figure 2.60. Permeability images of a layered sandstone core.
2–124
Figure 2.61. A comparison of NMR-derived permeability with flow-derived
permeability of core samples.
2.7.8 A Comparison of Various Porosity Measurements for Shaly Sand
Figure 2.62 shows the porosities measured by the various logging
tools along with the porosity from core analysis. This figure shows that
the total porosity measured by a neutron tool is larger than those
measured by the density tool, the NMR tool or the sonic tool. The total
porosity from the sonic tool is less than that of the density tool because
the density tool measures the porosity in isolated pore whereas the sonic
2-125
tool does not. If isolated pores are present, the effective porosity
measured in core analysis will be less than the effective permeability
measured by NMR. Thus, Figure 2.62 could be helpful in reconciling
differences in the porosity measurements from the various methods.
Figure 2.62. A comparison of porosity measurements by various methods for a shaly sand.
2.8 RESERVE ESTIMATION PROJECT
The objective of this project is to compute the recoverable oil
reserve and the anticipated undiscounted net cash flow for a new oilfield
discovery using Monte Carlo Simulation. The petrophysical parameters
that go into the reserve estimation are uncertain and as such should be
2–126
treated as random variables with distinct probability distributions. The
outcome of the simulation will be a range of reserve and net cash flow
estimates with their associated probabilities and uncertainties. This
project gives an example practical application of the petrophysical
properties of porosity and water saturation, which are the main subjects
of this chapter. The project also introduces the important subject of risk
analysis of petroleum development, which should be of interest to all
petroleum engineers.
2.8.1 Reserve Estimation
Based on volumetric considerations (Figure 2.63), the recoverable
oil reserve is given by
( )7758 1 wr f
o
Ah SN R
Bφ −
= (2.78)
where
Nr = recoverable oil reserve (stock tank barrels, STB)
A = area of the reservoir (acres)
h = net pay thickness (feet)
φ = porosity (fraction)
Sw = average water saturation (fraction)
Rf = recovery factor (fraction)
Bo = oil formation volume factor (reservoir barrels/stock tank barrel, RB/STB)
7758 = conversion constant (barrels/acre-foot)
2-127
Figure 2.63. Reservoir volume.
To account for the uncertainties in the variables on the right side
of Eq.(2.78) various experts within the company have been requested to
provide their best estimates of these variables based on their professional
judgments. These estimates are shown in Table 2.8 along with their
assumed probability distributions.
Table 2.8 Reservoir parameter estimates from experts.
Property Minimum (x1) Most Likely (x2) Maximum (x3) Probability
Distributio
n
A (acres) 2500 6000 9000 Triangular
h (ft) 200 300 500 Triangular
φ 0.15 0.25 0.35 Triangular
Bo (RB/STB) 1.20 1.30 1.35 Triangular
RF 0.20 0.40 Uniform
Oil Price ($/STB) 25 30 40 Triangular
2–128
Based on preliminary evaluation, the water saturation has been
determined to correlate with porosity as shown in Eq.(2.79).
0.325 0.500wS φ= − (2.79)
2.8.2 Economic Evaluation
In order to assess the profitability of the proposed development, it
will be necessary to perform a detailed year-by-year discounted cash flow
projection for the field. However, for the purpose of this project, a
preliminary undiscounted net cash flow analysis will be sufficient. It is
assumed that the petroleum fiscal regime applicable to this field is a
royalty/tax system.
The undiscounted net cash flow is given by
Net Cash Flow (NCF) = (Gross Revenue - Royalty - Costs)(1 - Tax Rate) (2.80)
Gross Revenue = Reserve x Price (2.81)
For this project, assume the following:
Royalty = 12.5% of Gross Revenue
Costs (CAPEX + OPEX) = 38% of Gross Revenue
(CAPEX = capital expenditures, OPEX = operating expenses)
Tax Rate = 40%
2-129
2.8.3 Simulation Procedure
You should draw a random sample for each variable of interest
from its probability distribution and compute the recoverable reserve and
the net cash flow for each iteration of the simulation. Each variable can
be sampled using a random number generator. A different random
number should be used to sample each variable because the variables
are assumed to be uncorrelated, except for the water saturation, which is
correlated to porosity. Normally, you should perform enough simulation
iterations so that the means and standard deviations of the reserve and
the net cash flow are approximately constant (i.e., they are no longer
sensitive to the number of iterations). This could require several
thousand iterations. For this exercise, you should perform at least 5,000
iterations.
2.8.4 Sampling Procedure
Presented herein is the procedure for sampling from a triangular
distribution and a uniform distribution using a random number
generator. Figure 2.64 shows the probability density function (pdf) for a
triangular distribution. The first step is to compute the cumulative
distribution function (F) as a function of x. Two cases are examined.
Case 1. x1≤ x ≤x2
For this case, the probability density function is given by
( )( )( )
11
3 1 2 1
2( )
x xf x
x x x x−
=− −
(2.82)
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Figure 2.64. Probability density function for a triangular distribution.
The cumulative distribution function (F), which is the probability that x
is less than or equal to a prescribed value, is obtained by integrating
Eq.(2.82) to obtain
( )
( )( )1
21
13 1 2 1
( )x
x
x xF f x dx
x x x x−
= =− −∫ (2.83)
Solving Eq.(2.83) for x gives
( )( )1 3 1 2 1x x F x x x x= + − − (2.84)
It turns out that F is uniformly distributed between 0 and 1, just like the
random number generator in spreadsheets or other computer software.
Therefore, to sample from the first part of a triangular distribution,
generate a uniformly distributed random number (Rn) between 0 and 1
and substitute it for F in Eq.(2.84) to obtain the sample value for x as
( )( )1 3 1 2 1x x Rn x x x x= + − − (2.85)
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Case 2. x2≤ x ≤x3
For this case, the probability density function is given by
( )
( )( )3
23 1 3 2
2( )
x xf x
x x x x−
=− −
(2.86)
The cumulative distribution function (F) is obtained by integrating
Eqs.(2.82) and (2.86) to obtain
( )( )( )
2
1 2
23
1 23 1 3 2
( ) ( ) 1x x
x x
x xF f x dx f x dx
x x x x−
= + = −− −∫ ∫ (2.87)
Solving Eq.(2.87) for x gives
( )( )( )3 3 1 3 21x x F x x x x= − − − − (2.88)
To sample from the second part of the triangular distribution, substitute
the random number for F in Eq.(2.88) to obtain
( )( )( )3 3 1 3 21x x Rn x x x x= − − − − (2.89)
For each iteration, it is necessary to test the random number to
determine if Eq.(2.85) or Eq.(2.89) should be used to calculate x. Such a
test is straightforward. For x = x2, Eq. (2.85) gives the critical value of Rn
as
( )( )
2 1
3 1
x xRn
x x−
=−
(2.90)
2–132
Eq.(2.85) should be used to calculate x if
( )( )
2 1
3 1
x xRn
x x−
≤−
(2.91)
Otherwise, Eq.(2.89) should be used to calculate x.
Figure 2.65 shows a graphical demonstration of Monte Carlo
sampling using the triangular distribution for net pay thickness of Table
2.8. The upper part of the figure shows the probability density function
whereas the lower part shows the cumulative distribution function. Also
shown in the lower part of the figure is the sampled value of 380 ft of net
pay for a random number of 0.760. The data used to construct Figure
2.65 are shown in Table 2.9.
Figure 2.66 shows the probability density function for a uniform
distribution. For this case, the probability density function is given by
( )3 1
1( )f xx x
=−
(2.92)
The cumulative distribution function (F) is obtained by integrating
Eq.(2.92) to obtain
( )( )1
1
3 1
( )x
x
x xF f x dx
x x−
= =−∫ (2.93)
Solving Eq.(2.93) for x gives
( )1 3 1x x F x x= + − (2.94)
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0.000
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0 100 200 300 400 500 600
x (ft)
f(x)
F
0.0000.1000.2000.3000.4000.5000.6000.7000.8000.9001.000
0 100 200 300 400 500 600
x (ft)
F
Figure 2.65. Graphical demonstration of Monte Carlo sampling
Table 2.9: Data for Monte Carlo Sampling of Net Pay Thickness
x F f(x)
200 0.000 0.0000
0
210 0.003 0.0006
7
220 0.013 0.0013
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3
230 0.030 0.0020
0
240 0.053 0.0026
7
250 0.083 0.0033
3
260 0.120 0.0040
0
270 0.163 0.0046
7
280 0.213 0.0053
3
290 0.270 0.0060
0
300 0.333 0.0066
7
310 0.398 0.0063
3
320 0.460 0.0060
0
330 0.518 0.0056
7
340 0.573 0.0053
3
350 0.625 0.0050
0
360 0.673 0.0046
7
370 0.718 0.0043
3
380 0.760 0.0040
0
390 0.798 0.0036
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7
400 0.833 0.0033
3
410 0.865 0.0030
0
420 0.893 0.0026
7
430 0.918 0.0023
3
440 0.940 0.0020
0
450 0.958 0.0016
7
460 0.973 0.0013
3
470 0.985 0.0010
0
480 0.993 0.0006
7
490 0.998 0.0003
3
500 1.000 0.0000
0
2–136
Figure 2.66. Probability density function for a uniform distribution
To sample from a uniform distribution, substitute the random number
for F in Eq.(2.94) to obtain
( )1 3 1x x Rn x x= + − (2.95)
The sampling procedure outlined above is the basic Monte Carlo
Sampling procedure. It is not a very efficient sampling technique because
there is no guarantee that all parts of the distribution will be sampled
equally. A more efficient and sophisticated stratified sampling procedure,
known as the Latin Hypercube Sampling, is available. This sampling
procedure ensures that all parts of the distribution are sampled equally
and will result in a faster convergence of the simulation to the final
results than the traditional Monte Carlo Sampling Method.
A brief description of the Latin Hypercube Sampling is as follows.
The cumulative distribution function, 0<F<1, for each variable is divided
into n subintervals of equal probability and numbered 0, 1, 2, 3, …, n-1.
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The objective is to sample from each subinterval during the first n
iterations. However, the subinterval to be sampled is selected randomly.
For example, if n = 100, the subintervals will be numbered 0, 1, 2, 3, …,
99. Each subinterval can be selected randomly by generating a random
integer from 0 to 99. The subintervals could be picked at random in
advance and stored in a one-dimensional array of 100 elements. Let’s
look at the array for the Area, A(100), as an example. The contents of this
array after the random selections might look like this:
A(1) = 45
A(2) = 2
A(3) = 7
A(4) = 92
A(5) = 23
A(6) = 0
…
A(100) = 8
Make sure every subinterval is represented in the array. A similar array
is generated with a new set of random numbers for each of the other
variables. During the first simulation iteration, the area will be sampled
from subinterval 45. To do so, generate a random number (r) between 0
and 1, say 0.263. The random number used to sample subinterval 45 is
computed as
(1) 45 0.263 0.45263100 100 100 100A rRn = + = + = (2.96)
2–138
In the second iteration, the area will be sampled from subinterval 2 by
generating another random number (r) between 0 and 1, say 0.950 and
computing
(2) 2 0.950 0.02950100 100 100 100A rRn = + = + = (2.97)
At the end of the first 100 iterations, each of the subintervals has been
sampled once. A similar sampling procedure is applied to the other
variables.
To perform the next 100 iterations of the simulation, the sequence
in which the subintervals are sampled is picked randomly. For example,
for the second 100 iterations, the array A may look like this:
A(1) = 3
A(2) = 86
A(3) = 50
A(4) = 14
A(5) = 6
A(6) = 99
…
A(100) = 72
The sampling for the second 100 iterations then proceeds in the same
manner as for the first 100 iterations. Thus, the entire simulation is
performed in increments of 100 iterations until the planned total number
of iterations is achieved. Figure 2.67 shows qualitatively the effect of
2-139
stratification in Latin Hypercube sampling from a normal distribution for
n = 20. The bands get progressively wider towards the tails of the
distribution as the probability density drops off.
Figure 2.67. Effect of stratification in Latin Hypercube sampling.
2.8.5 Simulation Output
Based on your simulation results, calculate the following for the
Reserve (Nr) and Net Cash Flow (NCF) estimates:
1. Minimum 2. Maximum 3. Mean 4. Standard deviation 5. Skewness 6. Kurtosis
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7. Mode (most likely value) 8. P90 value (There is a 90% probability that the variable of interest is
equal to or greater than the P90 value – Low Value) 9. P50 value (median) 10. P10 value (There is a 10% probability that the variable of interest is equal to or greater than the P10 value – High Value)
Note: Based on the above definitions, P10 > P90.
Plot the following graphs:
1. Reserve histograms (Frequency vs N, and Frequency vs natural log of N)
2. Net Cash Flow (NCF) histograms (Frequency vs NCF, and Frequency vs natural log of NCF)
3. Cumulative Distribution Function (F) for the reserve (F vs N) 4. Expectation Curve for the reserve ( (1– F) vs N on the same graph
as F vs N) 5. Cumulative Distribution Function (F) for the net cash flow (F vs
NCF) 6. Expectation Curve for the net cash flow ( (1– F) vs NCF on the
same graph as F vs NCF) 7. Histograms for A, h, φ, Sw, Bo, RF and Oil Price based on your
sampling to see how close they are to their theoretical input probability density functions .
Suggestion:
You can easily perform this simulation in Excel with VBA or by
writing a high level computer program in Fortran, C++ or Matlab. The
choice is yours. Please do not perform the simulation with a commercial
risk analysis software such as @Risk or Crystal Ball because I want you
to experience the simulation first hand rather than using the commercial
software as a black box.
Note that a similar simulation approach can be used to compute
the recoverable gas reserve in the case of a gas reservoir. In this case, the
appropriate equations are as follows:
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( )43560 1 wr
g
Ah SG RF
Bφ −
= (2.98)
0.02827gZTBP
= (2.99)
where
Gr = recoverable gas reserve (standard cubic feet, scf)
Bg = gas formation volume factor (reservoir cubic feet/scf)
Z = gas compressibility factor (dimensionless)
T = absolute reservoir temperature (ºRankine)
P = reservoir pressure (psia)
43560 = conversion constant (ft2/acre)
2.9 PORE VOLUME COMPRESSIBILITY
Reservoir rock at in situ conditions is subjected to overburden
stress, whereas at the surface, recovered core has been relieved of the
overburden stress. It is not usual to perform routine porosity
measurements under stress approaching reservoir conditions. Because
of this, laboratory-measured porosities are generally expected to be
higher than in situ values. Pore volume compressibility can be used to
correct laboratory-measured porosity to an in situ value and for other
reservoir engineering calculations.
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In general, the isothermal coefficient of compressibility of a
material is defined as
1
T
VcV P
∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠ (2.100)
where the negative sign convention is used to ensure that c is a positive
number because when a material is compressed, its volume decreases
thereby making T
VP
∂⎛ ⎞⎜ ⎟∂⎝ ⎠
a negative quantity. Pore volume compressibility is
defined without the negative sign as
1 1pf
Tp T
Vc
V P Pφ
φ∂⎛ ⎞ ∂⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠
(2.101)
where P is the pore pressure (internal pressure) rather than the
overburden pressure (external pressure). When the pore pressure
declines during production, the overburden pressure causes the reservoir
to compact. Thus, a decrease in pore pressure causes a decrease in the
pore volume or porosity thereby making T
VP
∂⎛ ⎞⎜ ⎟∂⎝ ⎠
or TP
φ∂⎛ ⎞⎜ ⎟∂⎝ ⎠
positive.
For many years, the petroleum industry has relied on Hall’s (1953)
correlation, shown in Figure 2.68, for estimating pore volume
compressibility. This correlation was developed from measurements on
seven consolidated limestone and five consolidated sandstone samples, a
rather small dataset for a universal correlation.
Compressibilities, however, are highly affected by reservoir type
and overburden conditions. Newman (1973) has presented a more
comprehensive pore volume compressibility data based on 256 samples.
2-143
Of the 256 samples, 197 were sandstones from 29 sandstone reservoirs,
and 59 were limestones from 11 limestone reservoirs. Figure 2.69 shows
a correlation of the pore volume compressibilities with initial sample
porosities for all 256 samples. Also shown is Hall's correlation. Viewed
collectively, the data show little or no correlation between pore volume
compressibility and initial sample porosity. The data show considerable
disagreement with Hall's correlation.
Figure 2.68. Pore volume compressibility versus porosity (Hall, 1953).
2–144
Figure 2.69. Pore volume compressibility at 75% lithostatic pressure versus initial sample porosity for sandstone and limestone samples
(Newman, 1973).
Newman divided his data into four qualitative classifications in an
effort to improve the correlation between pore volume compressibility and
initial porosity. He classified the samples as (1) consolidated limestone,
(2) consolidated sandstone, (3) poorly consolidated (friable) sandstone,
and (4) unconsolidated sandstone. Poorly consolidated or friable
sandstone could be broken by hand whereas consolidated sandstone
2-145
could not. Unconsolidated sandstone would collapse under its own
weight if it is not protected in a sleeve.
Figure 2.70 shows the correlation for consolidated limestones
along with the correlations from Hall and van der Knaap (1959).
Qualitatively, all three dataset show a general decrease in pore volume
compressibility with an increase in porosity. However, there is
considerable scatter in the data.
Figure 2.70. Pore volume compressibility at 75% lithostatic pressure versus initial sample porosity for limestone samples (Newman, 1973).
2–146
Figure 2.71 shows the correlation for consolidated sandstones
along with Hall's correlation. Qualitatively, the data show a general
decrease in pore volume compressibility with an increase in porosity with
some scatter. Newman's data decrease more rapidly with an increase in
porosity than Hall's data.
Figure 2.71. Pore volume compressibility at 75% lithostatic pressure
versus initial sample porosity for consolidated sandstone samples (Newman, 1973).
2-147
Figure 2.72 shows the correlation for poorly consolidated or friable
sandstones along with Hall's correlation. There appears to be no
correlation between pore volume compressibility and porosity for this
class of sandstones.
Figure 2.72. Pore volume compressibility at 75% lithostatic pressure
versus initial sample porosity for poorly consolidated sandstone samples (Newman, 1973).
2–148
Figure 2.73 shows the correlation for unconsolidated sandstones
along with Hall's correlation. For this class of sandstones, the pore
volume compressibility increases with an increase in porosity, in contrast
to the trend for consolidated samples.
Figure 2.73. Pore volume compressibility at 75% lithostatic pressure versus initial sample porosity for unconsolidated sandstone samples
(Newman, 1973).
2-149
Newman's data clearly show that there is no universal correlation
for pore volume compressibility and porosity. For detailed analysis, pore
volume compressibility should be measured in the laboratory on core
samples of interest. Figure 2.74 shows a schematic diagram of a typical
device for measuring pore volume compressibility. These measurements
are fairly difficult to make, and must include calibrations for the
compressibilities of the fluids and the equipment.
Figure 2.74. Typical apparatus for measuring pore volume
compressibility (Hall, 1953).
Pore volume compressibility is not always constant. For example,
in abnormally pressured reservoirs, it is common to find that pore
volume compressibility at high reservoir pressures is much larger than
2–150
the pore volume compressibility at low pressures. This is due to
compaction effects as the pore pressure declines.
Pore volume compaction can lead to undesirable operational
problems. A famous case of this is the subsidence of Ekofisk field in the
Norwegian sector of North Sea. The seafloor subsided about 3 meters
from 1971 to 1984 due to reservoir depletion. Continuous monitoring
indicated subsidence of 33 to 38 cm per year. By 1997, the cumulative
subsidence was 7.8 meters. In 1998, the operator undertook expensive
measures to jack up their platforms by 6 meters. On the other hand,
pore volume compaction provides valuable reservoir drive energy to
increase production from the field.
Total system compressibility is a parameter of importance in
pressure transient testing and reservoir material balance analysis,
especially above the bubble point pressure of the oil. Total system
compressibility is defined as the combined compressibility of the pore
volume and all the fluids saturating the medium:
t o o w w g g fc S c S c S c c= + + + (2.102)
NOMENCLATURE
amf = activity coefficient of mud filtrate
aw = activity coefficient of formation water
A = atomic weight
Bg = gas formation volume factor
Bo = oil formation volume factor
B0 = magnetic field strength
2-151
BVI = bound fluid index
c = isothermal coefficient of compressibility
cf = isothermal coefficient of compressibility of pore volume
cg = isothermal coefficient of compressibility of gas
co = isothermal coefficient of compressibility of oil
cw = isothermal coefficient of compressibility of water
dh = borehole diameter
di = diameter of invaded zone
ds1 = slice thickness
dxo = diameter of flushed zone
D = diffusion coefficient
Ec = electrochemical potential
Ek = electrokinetic potential
Ekmc = electrokinetic potential of mud cake
F = formation resistivity factor
FFI = free fluid index
FOV = field of view
G = gradient
GPE = phase encode gradient
GRO = readout gradient
Gss = slice selection gradient
Gr = recoverable gas reserve
h = net pay thickness
I = formation resistivity index
k = permeability
m = cementation factor
M = net magnetization
Mz = net magnetization in the z direction
n = saturation exponent
Nr = recoverable oil reserve
MNR = nuclear magnetic resonance
2–152
P = pressure
RF = radiofrequency
Rf = recovery factor
Ri = resistivity of invaded zone
Rmf = resistivity of mud filtrate
Rn = random number
Ro = resistivity of water saturated zone
Rs = resistivity of surrounding formation
Rt = true formation resistivity
Rw = resistivity of formation water
Rxo = resistivity of flushed zone
S = surface area of pore
Sh = hydrocarbon saturation
Sg = gas saturation
So = oil saturation
Soi = initial oil saturation
Sor = residual oil saturation
Sxo = water saturation of flushed zone
Sw = water saturation
Swirr = irreducible water saturation
t = time
tpe = duration of phase encoding gradient pulse
tro = duration of frequency encoding gradient pulse
T = temperature
T1 = longitudinal relaxation time or spin-lattice relaxation time
T2 = transverse relaxation time or spin-spin relaxation time
T2gm = geometric mean time *
2T = apparent transverse relaxation time spin-spin relaxation time
T2bulk = transverse relaxation time of bulk fluid
T2surface = transverse relaxation time of surface fluid
TE = echo time
2-153
TR = repetition time
v = compressional wave velocity
V = volume of pore
Vp = pore volume
Vb = bulk volume
Vs = grain volume
Z = atomic number
φ = porosity
φe = effective porosity
φtotal = total porosity
ρ1 = NMR T1 surface relaxivity
ρ2 = NMR T2 surface relaxivity
ρa = apparent density
ρb = bulk density
ρe = electron density
ρf = fluid density
ρm = rock matrix density
ΔB0 = magnetic field inhomogeneity
ΔPE = pixel size in the phase encoding direction
ΔRO = pixel size in the readout direction
Δv = voxel size
Δt = interval transit time
Δtf = fluid interval transit time
Δtm = rock matrix interval transit time
Δtsh = shale interval transit time
∇B = magnetic field gradient
α = regularization constant or smoothing parameter
λ = thickness of surface monolayer
ε(tj) = noise
τ = interpulse spacing time
2–154
ωo = Lamor frequency
ωr = Lamor frequency at position r
γ = gyromagnetic ratio
Ψair = x-ray attenuation coefficient of air (= 0)
Ψbrine = x-ray attenuation coefficient of brine
Ψdry = x-ray attenuation coefficient of dry sample
Ψm = x-ray attenuation coefficient of rock matrix
Ψwet = x-ray attenuation coefficient of brine saturated sample
REFERENCES AND SUGGESTED READINGS
Amyx, J.W., Bass, D.M., Jr. and Whiting, R.L. : Petroleum Reservoir Engineering, McGraw-Hill Book Company, New York, 1960.
Anderson, G. : Coring and Core Analysis Handbook, Petroleum Publishing Company, Tulsa, Oklahoma, 1975.
Archer, J.S. and Wall, C.G. : Petroleum Engineering, Graham & Trotman, London, England, 1986.
Archie, G.E. :“The Electrical Resistivity Log as an Aid in Determining Some Reservoir Characteristics,” Trans. AIME (1942) 146, 54-62.
Baker Atlas Publication: Introduction to Wireline Log Analysis, 1992.
Bassiouni, Z. : Theory, Measurement, and Interpretation of Well Logs, SPE Textbook Series, Vol. 4, Society of Petroleum Engineers, Richardson, TX, 1994.
Beard, D.C. and Weyl, P.K. : “Influence of Texture on Porosity and Permeability of Unconsolidated Sand,” AAPG Bull. (Feb. 1973) 57, 349-369.
Bradley, H.B. (Editor-in-Chief) : Petroleum Engineering Handbook, SPE, Richardson, Texas, 1987.
Calhoun, J.C. : Fundamentals of Reservoir Engineering, University of Oklahoma Press, Norman, Oklahoma, 1982.
2-155
Carman, P.C. : "Fluid Flow Through A Granular Bed,” Trans. Inst. Chem. Eng. London (1937) 15, 150-156.
Carman, P.C. : "Determination of the Specific Surface of Powders,” J. Soc. Chem. Indus (1938) 57, 225-234.
Carpenter, C.B. and Spencer, G.B. : “Measurements of the Compressibility of Consolidated Oil-Bearing Sandstones,” RI 3540, USBM (Oct. 1940).
Coats, G.R., Xiao, L. and Prammer, M.G. : NMR Logging: Principles and Applications, Halliburton Energy Services, Houston, 1999.
Collins, R.E. : Flow of Fluids Through Porous Materials, Research & Engineering Consultants Inc., 1990.
Core Laboratories Inc. : Special Core Analysis, 1976.
Corey, A.T. : Mechanics of Heterogeneous Fluids in Porous Media, Water Resources Publications, Fort Collins, Colorado, 1977.
Cosse, R. : Basics of Reservoir Engineering, Editions Technip, Paris, 1993.
Dobrynin, V.M. : "The Effect of Overburden Pressure on Some Properties of Sandstones,” Soc. Pet. Eng. J. (Dec. 1962) 360-366.
Dunn, K.J., Bergman, D.J. and Latorraca, G.A.: Nuclear Magnetic Resonance Petrophysical and Logging Applications, Handbook of Geophysical Exploration, Seismic Exploration Vol. 32, Pergamon, New York, 2002.
Fatt, I. : “Pore Volume Compressibilities of Sandstone Reservoir Rocks,” Trans., AIME (1958) 213, 362-364.
Fraser, H.J. and Graton, L.C. : "Systematic Packing of Spheres - With Particular Relation to Porosity and Permeability,” J. Geol., Vol. 43, No. 8, Nov. - Dec., 1935, 789-909.
Geertsma, J. : “The Effect of Fluid Pressure Decline on Volumetric Changes of a Porous Medium,” Trans. AIME (1957) 210, 331-340.
Hall, H.N. : “Compressibility of Reservoir Rocks,” Trans., AIME (1953) 198, 309-311.
Haughey, D.P. and Beveridge, G.S.G. :”Structural Properties of Packed Beds - A Review,” Can. J. Chem. Eng., Vol. 47 (April 1969) 130-149.
Jennings, H.J. : “How to Handle and Process Soft and Unconsolidated Cores,” World Oil (June 1965) 116-119.
Jorden, J.R. and Campbell, F.L. : Well Logging I - Rock Properties, Borehole Environment, Mud and Temperature Logging, Monograph
2–156
Vol. 9, Society of Petroleum Engineers of AIME , Richardson, Texas, 1984, Ch. 2.
Keelan, D.K. : "A Critical Review of Core Analysis Techniques” The Jour. Can. Pet. Tech. (April-June 1972) 42-55.
Kleinberg, I. : “Probing Oil Wells with NMR,” The Industrial Physicist, (June/July 1996) 18-21.
Li, Ping: Nuclear Magnetic Resonance Imaging of Fluid Displacements in Porous Media, PhD Dissertation, The University of Texas at Austin, Austin, Texas, August 1997.
Mann, R.L. and Fatt, I. : “Effect of Pore Fluids on the Elastic Properties of Sandstones,” Geophysics (1960) 25, 433-444.
Mayer-Gurr, A. : Petroleum Engineering, John Wiley & Sons, New York, 1976.
Monicard, R.P. : Properties of Reservoir Rocks, Gulf Publishing Company, Houston, TX, 1980.
Neasham, J.W.: "The Morphology of Dispersed Clay in Sandstone Reservoirs and Its Effect on Sandstone Shaliness, Pore Space and Fluid Flow Properties," SPE 6858, Presented at the 52nd Annual Technical Conference and Exhibition of the Society of Petroleum Engineers, Denver, Oct. 9-12, 1977.
Newman, G.H. : "Pore-Volume Compressibility of Consolidated, Friable, and Unconsolidated Reservoir Rocks Under Hydrostatic Loading,” J. Pet. Tech. (Feb. 1973) 129-134.
Peters, E.J. and Afzal, N. : “Characterization of Heterogeneities in Permeable Media with Computed Tomography Imaging,” Journal of Petroleum Science and Engineering, 7, No. 3/4, (May 1992) 283-296.
Peters, E.J. and Hardham, W.D. : “Visualization of Fluid Displacements in Porous Media Using Computed Tomography Imaging,” Journal of Petroleum Science and Engineering, 4, No. 2, (May 1990) 155-168.
Pirson, S.J. : Oil Reservoir Engineering, McGraw-Hill Book Company, Inc., New York, 1958.
Ruth, D. and Pohjoisrinne, T. : "The Precision of Grain Volume Porosimeters” The Log Analyst (Nov.-Dec. 1993) 29-36.
Schlumberger Publication: Log Interpretation Principles/Applications, Schlumberger Educational Series, 1987.
2-157
Tamari, S. : "Optimum Design of the Constant-Volume Gas Pycnometer for Determining the Volume of Solid Particles," Meas. Sci. Tech.,Vol.15, 2004, 549-558.
Tiab, D. and Donaldson, E.C. : Petrophysics, Second Edition, Elsevier, New York, 2004.
van der Knaap, W. : "Nonlinear Behavior of Elastic Porous Media,” Trans., AIME (1959) 216, 179-187.
Vose, D. : Quantitative Risk Analysis: A Guide to Monte Carlo Simulation Modeling, John Wiley and Sons, New York, 1998.
Winsauer, W.O., Shearin, H.M., Jr., Masson, P.H. and Williams, M. : "Resistivity of Brine-Saturated Sands in Relation to Pore Geometry," AAPG Bull., Vol. 36, No. 2 (Feb. 1952) 253-277.
Wyllie, M.R.J. and Rose, W.D. : "Some Theoretical Considerations Related to the Quantitative Evaluation of the Physical Characteristics of Reservoir Rock from Electrical Log Data,” Trans., AIME (1950) 189, 105-118.
Wyllie, M.R.J. and Spangler, M.B. : "Application of Electrical Resistivity Measurements to Problems of Fluid Flow in Porous Media,” AAPG Bull., Vol. 36, No. 2 (Feb. 1952) 359-403.
Wyllie, M.R.J., Gregory, A.R. and Gardner, G.H.F. : "Elastic Wave Velocities in Heterogeneous and Porous Media” Geophysics, Vol. 21, No. 1 (1956) 41-70.
3-1
CHAPTER 3
PERMEABILITY
3.1 DEFINITION
Permeability gives an indication of the porous medium’s ability to
transmit fluids (i.e., permit fluid flow). It is defined through Darcy’s law. For a
horizontal system, Darcy’s law for single phase flow in differential form is
kA dPqdxμ
= − (3.1)
For steady-state linear flow of a single phase liquid in a horizontal medium as
shown in Figure 3.1, Darcy’s Law may be integrated to give
( )1 2P PkA kA PqL Lμ μ− Δ
= = (3.2)
Figure 3.1. Linear flow geometry.
3-2
Eq(3.2) can be solved for the absolute permeability of the porous medium as
q LkA P
μ=
Δ (3.3)
The absolute permeability is the permeability of the medium when it is
saturated 100% by a single phase, non-reactive liquid. It is a property of the
porous medium and is independent of the fluid properties.
Eq.(3.2) can be written as
LP qkAμ⎛ ⎞Δ = ⎜ ⎟
⎝ ⎠ (3.4)
Eq.(3.4) is useful for processing laboratory data to determine the absolute
permeability of a sample using a non-reactive liquid. If a steady state liquid
flow experiment is performed at several rates, Eq.(3.4) shows that a graph of
ΔP versus q should be linear with a slope, m, given by
LmkAμ
= (3.5)
from which the absolute permeability can be calculated as
LkmAμ
= (3.6)
Table 3.1 shows the data from such an experiment performed on an
unconsolidated sandpack using Dow Corning mineral oil. The fluid and
sandpack properties were as follows:
μ = 105.363 cp
L = 115.6 cm
d = 4.961 cm
φ = 37.80%
3-3
Table 3.1: Pressure Drop and Flow Rate Data for Steady State Flow Experiment in a Sandpack (from Peters, 1979).
q
(cc/s)
ΔP
(atm)
0 0
0.0014 0.0476
0.0556 1.9284
0.0889 3.0573
0.1333 4.5439
0.2222 7.5303
0.3111 10.465
Figure 3.2 shows the graph of ΔP versus q from the experiment. The data plot
as a straight line through the origin, thereby verifying the validity of Darcy's
law for the experiment. The slope of the line is m = 33.658. The permeability
is calculated with Eq.(3.6) as
( )( )
( )2
105.363 115.618.72
4.96133.6582
kπ
= =⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
darcys.
If the graph of Figure 3.2 were nonlinear, then Darcy's law would not be
valid for the experiment. This situation would occur if the liquid reacted with
the porous medium. An example of such a reaction would occur if the
experiment were performed on a sandstone containing a significant amount of
montmorrillonite-type clays using fresh water. Can you sketch how ΔP would
vary with q in such an experiment?
3-4
Figure 3.2. Graph of ΔP verus q for steady state flow of a non-reactive liquid
through an unconsolidated sandpack.
Table 3.2 compares Darcy units with oilfield units. Using the
information in the table, Eq.(3.2) can be written in oilfield units as
0.001127kA PqB Lμ
Δ= (3.7)
or
1887.2
kA PqB Lμ
Δ= (3.8)
For radial flow into a wellbore as shown in Figure 3.3, Darcy’s law may
be expressed in radial coordinates in oilfield units as
3-5
( )0.00708
ln
e w
e
w
kh P Pq
rBr
μ
−=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.9)
or
( )1141.2
ln
e w
e
w
kh P Pq
rBr
μ
−=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.10)
Table 3.2 Comparison of Darcy and Oilfield Units
Variable Darcy Unit Oilfield Unit Pressure atm psia Time second day Flow rate cm3/s STB/D Wellbore radius cm ft Well drainage radius cm ft Porosity fraction fraction Permeability darcy millidarcy Pay thickness cm ft Fluid viscosity cp cp Compressibility atm-1 psi-1
Figure 3.3. Radial flow into a wellbore.
3-6
3.2 DIMENSIONS AND UNITS OF PERMEABILITY
Eq.(3.2) can be rearranged to determine the dimensions of permeability
as follows:
[ ] [ ][ ][ ][ ][ ]q L
kA p
μ=
Δ (3.11)
where the square brackets refer to the dimensions of the enclosed variable.
Using mass (M), length (L) and time (T) as the fundamental dimensions, the
dimensions of the variables on the right side of Eq.(3.11) are
[ ]3Lq
T=
[ ] MLT
μ =
[ ]L L=
[ ] 2A L=
[ ] 2
MpLT
Δ =
Thus,
[ ]( )
( )
3
22
L M LT LT
kML
LT
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=⎛ ⎞⎜ ⎟⎝ ⎠
or
[ ] 2k L= (3.12)
Eq.(3.12) shows that the permeability of a porous medium has the dimensions
of length squared. This means that permeability is proportional to the square
of some characteristic dimension of the porous medium.
3-7
Based on Eq.(3.12), a rational unit for permeability would be foot
squared or centimeter squared. Both units were found to be too large for use
with porous media. Therefore, the petroleum industry adopted the darcy as
the unit of permeability. It can be shown that
9 2 13 2 11 21 9.869 10 9.869 10 1.062 10darcy x cm x m x ft− − −= = = (3.13)
3.3 LABORATORY DETERMINATION OF PERMEABILITY
Permeability is an empirical parameter, which must be determined by
measuring all the parameters in Darcy’s law. Core permeabilities are usually
measured in the laboratory using dry gas (air, nitrogen or helium) as the
flowing fluid to minimize rock-fluid reaction. In this case, another form of
Darcy’s law for steady-state flow of an ideal gas must be used, recognizing
that the volumetric flow rate of a gas varies with pressure. For an ideal gas at
a constant temperature, Boyle's law can be written for a fixed mass of gas at
two conditions as
sc scqP q P= (3.14)
where qsc is the gas volumetric flow rate at a reference pressure Psc.
Substituting Eq.(3.14) into (3.1) and rearranging gives
21
2g
scsc
k A dPqP dxμ
⎛ ⎞= − ⎜ ⎟
⎝ ⎠ (3.15)
where kg is the permeability to gas of the medium. Eq.(3.15) can be integrated
to give the working equation for a gas permeameter as
2 2
1 2
2g
scsc
k A P PqP Lμ
⎛ ⎞−= ⎜ ⎟
⎝ ⎠ (3.16)
which can be used to calculate the permeability to gas as
( )2 2
1 2
2 sc scg
q LPkA P P
μ=
− (3.17)
3-8
The permeability to gas determined from Eq.(3.17) at the low pressures
typically used in the laboratory measurement is usually higher than the
absolute permeability of the porous medium because of an electro-kinetic
phenomenon known as the Klinkenberg (1941) effect. Klinkenberg effect is
due to the fact that at low mean pressures, the mean free path of the gas
molecules is about the same size as the pores in the rock. This gives rise to
gas slippage at the wall of the pores. As a result, the permeability to gas at
low pressure is higher than the absolute permeability of the porous medium.
Klinkenberg found that the permeability to gas, kg, is related to the absolute
permeability of the medium, kL, by the equation
1g Lbk kP
⎛ ⎞= +⎜ ⎟⎝ ⎠
(3.18)
where P is the mean gas pressure given by
1 2
2P PP +
= (3.19)
and b is a constant, which depends on the gas used in the measurement. The
absolute permeability (liquid permeability) of the medium can be determined
in the laboratory by measuring permeabilities to gas, kg, at different average
core pressures. A graph of kg versus 1P
yields a straight line with an
intercept equal to kL and a slope equal to kLb as shown in Figure 3.4.
Figue 3.5 shows the Klinkenberg correction plots for a core for gas
permeabilities measured with hydrogen, nitrogen and carbon dioxide. It
should be noted that the degree of gas slippage varies with the nature of the
gas. Hydrogen with the smallest molecules has the most gas slippage and
carbon dioxide with the largest molecules has the least gas slippage. All three
straight lines extrapolate to the absolute permeability of the core at infinite
mean pressure. The Klinkenberg effect is a laboratory scale phenomenon. It
is usually not important at the high mean pressures of a petroleum reservoir.
3-9
Figure 3.4. Klinkenberg permeability correction.
Figure 3.5. Permeability of a core sample to hydrogen, nitrogen and carbon dioxide. Absolute permeability of the core to isooctane = 2.55 md (from
Klinkenberg, 1941).
3-10
Figure 3.6 shows a schematic diagram of a gas permeameter. In this
figure, the gas volumetric flow rate, qsc, is measured at the outlet pressure, P2.
Therefore, in this case, P2 is the reference pressure, Psc.
Figure 3.6. Schematic diagram of a gas permeameter.
In general, permeability is an anisotropic property of a porous medium,
that is, it is directional. Routine core analyses are usually made on core
plugs drilled horizontally from a core. It is sometimes possible to specify that
plugs be cut along bedding planes. Plugs are sometimes cut vertically if it is
desired to obtain vertical as well as horizontal permeabilities. These plugs are
shown in Figure 3.7.
3-11
Figure 3.7. Possible core plug samples.
Because of interactions of fluids with reservoir rocks, absolute
permeabilities to different fluids may not always be the same. The
permeability to brine, for example, is often somewhat less than the
Klinkenberg-corrected gas permeability. For this reason, it is usually
beneficial to obtain at least a few brine permeability measurements, especially
if waterflooding is anticipated. These measurements are time-consuming, and
3-12
thus are more costly. Another permeability that is sometimes referenced is
the permeability to oil at irreducible water saturation (kor). This permeability
is known as the effective permeability to oil at irreducible water saturation. Its
value is typically around 80% to 98% of the absolute permeability depending
on the quality of the porous medium. Table 3.3 presents typical values for
clean, unconsolidated sandpacks.
Table 3.3: Petrophysical Data of Sandpacks (from Peters, 1979)
Sample L d φ k kor kor/k Swirr
No (cm) (cm) (%) (darcy) (darcy) (%)
1 22.9 4.81 38.98 16.43 15.82 0.96 11.20
2 23.6 4.84 36.37 18.36 17.99 0.98 8.66
3 112.8 4.96 38.38 21.89 18.05 0.82 9.18
4 23.7 4.81 38.94 18.93 18.33 0.97 11.34
5 110.5 4.83 38.91 20.52 18.50 0.90 10.83
6 110.0 4.81 37.61 21.91 19.40 0.89 10.48
7 23.6 4.84 37.49 14.19 11.03 0.78 15.26
8 116.1 4.96 34.66 18.28 15.57 0.85 8.32
9 113.0 4.96 39.62 22.99 19.90 0.87 8.61
10 115.9 4.96 35.86 19.22 15.49 0.81 12.53
11 22.8 4.84 37.52 16.23 15.19 0.94 11.20
12 110.4 4.81 37.46 20.85 18.48 0.89 10.00
13 115.9 4.97 37.80 18.54 15.53 0.84 9.29
14 110.0 4.81 38.11 22.50 18.62 0.83 8.91
15 112.8 4.97 35.48 22.72 20.90 0.92 9.92
16 23.4 4.84 34.61 14.51 9.58 0.66 11.03
17 23.6 4.84 38.60 20.87 18.49 0.89 10.15
18 23.7 4.81 35.44 12.11 11.94 0.99 12.02
3-13
Example 3.1
The permeability of a clean, dry core plug is to be determined. The core is
cylindrical with a diameter of 24 mm and a length of 32 mm. The core was
installed in a gas permeameter and air was flowed through it at an average
rate of 100 cm3 in 2 minutes 20 seconds, measured at atmospheric pressure
(Figure 3.8). The pressure differential across the sample was kept constant at
12 cm of mercury. The upstream gauge pressure (at the inlet of the core) was
76 cm of mercury. The gas viscosity at the test temperature was 0.01808 cp.
The barometric pressure was 76 cm of mercury.
Figure 3.8. Gas permeameter for Example 3.1.
1. Calculate the permeability to air at the test conditions.
2. Does the permeability calculated in part 1 represent the true absolute permeability of this core? If yes, why? If no, why not?
3-14
Solution to Example 3.1
d = 24 mm = 2.4 cm
L = 32 mm = 3.2 cm
qsc = 100cm3/(2 minutes 20 seconds) = (100/140) cm3/s
P1 = 76cm of Hg gauge = 76+76 = 152 cm of Hg absolute
= 152/76 = 2 atm absolute
P2 = (P1 - 12) = 152 - 12 = 140 cm of Hg absolute
= 140/76 = 1.842 atm absolute
Psc = 1 atm
μg = 0.0808 cp
Substituting into Eq.(3.17) gives
( )( )( )( )( )( ) ( )2 2 2
2 100 /140 0.01808 1 3.20.0301 darcy = 30.1 md
2.4 / 2 2 1.842gkπ
= =−
No. The calculated permeability to gas of 30.1 md is not the true absolute
permeability of the core because of Klinkenberg effect. It is larger than the
absolute permeability of the core.
3.4 FIELD DETERMINATION OF PERMEABILITY
Permeability can be determined in the field by use of pressure transient
tests, essentially measuring the permeability of the in-situ reservoir rock on a
large scale. A transient pressure test consists of changing the flow rate of a
well and then recording the bottomhole pressure response as a function of
time. The pressure data can be analyzed to obtain formation permeability and
other reservoir and well parameters. Oil permeability at irreducible water
saturation is normally used for comparison with the permeability from most
pressure transient tests that involve only oil flow.
3-15
3.4.1 Diffusivity Equation for Slightly Compressible Liquid
The partial differential equation that describes the transient pressure
response of a reservoir undergoing single phase flow is obtained by combining
the law of mass conservation, Darcy’s law and an equation of state for the
fluid. The mass conservation equation, also known as the continuity equation,
is
( ) ( ). 0vt
φρρ
∂∇ + =
∂ (3.20)
Darcy's law is given by
kv Pμ
= − ∇ (3.21)
The equation of state for a slightly compressible liquid such as oil or water is
given by
( )oc P Poeρ ρ −= (3.22)
where ρo is the density at a reference pressure Po. Eqs.(3.20) through (3.22)
can be combined to give
2 1tc P PPk t t
φμα
∂ ∂∇ = =
∂ ∂ (3.23)
where α is a constant and is given by
t
kc
αφμ
= (3.24)
Eq. 3.23 is known as the diffusivity equation in the petroleum engineering
community. Others may recognize it as the diffusion equation or the heat
conduction equation. Thus, Eq.(3.23) is a standard partial differential
equation of mathematical physics. It is a second order, parabolic, linear
partial differential equation. The solutions of this equation for specific initial
3-16
and boundary conditions are used for transient pressure analysis and for
natural water influx calculations.
The constant, α, is known as the diffusivity constant and can be written
as
t t
khk Tc hc S
μα
φμ φ
⎛ ⎞⎜ ⎟⎝ ⎠= = = (3.25)
where T is the transmissibility of the reservoir given by
khTμ
= (3.26)
and S is the storativity of the reservoir given by
tS hcφ= (3.27)
Thus, the diffusivity constant is the ratio of the transmissibility to the
storativity (storage capacity) of the reservoir.
The mathematical model (initial-boundary value problem) for transient
pressure analysis is as follows:
2
2
1 tcP P Pr r r k t
φμ∂ ∂ ∂+ =
∂ ∂ ∂ (3.28)
( )P r,0 iP= (3.29)
0
lim2
sf
r
qPrr kh
μπ→
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠ (3.30)
( )lim , irP r t P
→∞= (3.31)
Eq.(3.28) is Eq.(3.23) for one-dimensional radial flow. Eq.(3.29) is the initial
condition, which specifies that before production commenced, the pressure in
the reservoir was constant and equal to the initial reservoir pressure, Pi.
3-17
Eq.(3.30) is the internal boundary condition at the wellbore, which specifies a
constant instantaneous sandface rate upon commencement of production.
This instantaneous constant sandface boundary condition is unrealistic
because we expect that when a well is put on production, the sandface rate
should increase from zero to the final constant rate over a finite time.
Eq.(3.29) also specifies a vanishingly small wellbore radius thereby treating
the well as a line source. These two specifications, though unrealistic, have
been made for mathematical expediency because with them, we can obtain a
simple approximate solution to the partial differential equation. Eq.(3.31) is
the external boundary condition, which specifies that far away from the
wellbore, the reservoir is undisturbed and as such, the reservoir pressure will
remain at the initial pressure, Pi. This boundary condition is the infinite
acting reservoir boundary condition.
Eqs.(3.28) through (3.31) constitute the initial-boundary value problem
that describes the transient pressure response of a reservoir that has been
put on instantaneous constant sandface rate production. It is useful to put
these equations in dimensionless forms. This will reduce the number of
variables and also will facilitate the use of existing solutions from other fields
for pressure transient analysis. For example, there is a wealth of solutions of
linear, second order, parabolic partial differential equation in the heat
conduction literature. Casting our model in dimensionless variables allows
the immediate use of these existing solutions from other disciplines for our
purpose.
In dimensionless forms, Eqs.(3.28) through (3.31) become
2
2
1D D D
D D D D
P P Pr r r t
∂ ∂ ∂+ =
∂ ∂ ∂ (3.32)
( )P r ,0 0D D = (3.33)
3-18
0
lim 1D
DDr
D
Prr→
⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠
(3.34)
( )lim , 0D
D D DrP r t
→∞= (3.35)
where the dimensionless variables are defined as
( ),
2
iD
sf
P P r tP
qkhμ
π
−=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.36)
Dw
rrr
= (3.37)
and
2Dr w
kttc rφμ
= (3.38)
In oilfield units, the dimensionless pressure and dimensionless time are given
by
( ),
141.2
iD
P P r tP
q Bkhμ
−=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.39)
and
2
0.0002637D
r w
kttc rφμ
= (3.40)
In Eq.(3.39), q is surface rate and qB is the sandface rate.
Eqs.(3.32) through (3.35) can be solved, for example, by Laplace
transformation. It can be shown that the solution is of the form
( ) 1 ( )2DP x Ei x= − − (3.41)
3-19
where
2
4D
D
rxt
= (3.42)
and
( )y
x
eEi x dyy
−∞− = −∫ (3.43)
Ei(-x) is the exponential integral function, which is tabulated in mathematical
handbooks. Table 3.4 gives the Ei function for various values of the argument,
x. The solution given in Eq.(3.41) can be written as
( )21,
2 4D
D D DD
rP r t Eit
⎛ ⎞= − −⎜ ⎟
⎝ ⎠ (3.44)
at any dimensionless radius and time, and
( ) 1 11,2 4D D
D
P t Eit
⎛ ⎞= − −⎜ ⎟
⎝ ⎠ (3.45)
at the wellbore, where rD = 1. In terms of actual pressure, radius and time in
oilfield units, Eq.(3.44) and (3.45) become
( )2948141.2 1,
2t
ic rq BP r t P Ei
kh ktφμμ ⎡ ⎤⎛ ⎞
= − − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
(3.46)
and
( ) ( )2948141.2 1,
2t w
wf w ic rq BP t P r t P Ei
kh ktφμμ ⎡ ⎤⎛ ⎞
= = − − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
(3.47)
3.4.2 Pressure Drawdown Equation
For 0.01x ≤ , the exponential integral function can be accurately
approximated as
3-20
( ) ln 0.5772Ei x x− + (3.48)
This approximation is usually valid at the wellbore in all well testing
situations. Using the approximation, the dimensionless solution at the
wellbore, Eq.(3.45), can be written as
( ) ( )11, ln 0.80907 for 252D D D DP t t t= + ≥ (3.49)
Eq.(3.49) can be written in terms of log to base 10 as
( ) 21, 1.1513 log log 3.23D Dt w
kP t tc rφμ
⎡ ⎤⎛ ⎞= + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (3.50)
Eq.(3.50) gives the flowing bottomhole pressure as
( ) ( ) 2
162.6, log log 3.23wf w it w
q B kP t P r t P tkh c r
μφμ
⎡ ⎤⎛ ⎞= = − + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (3.51)
Eq.(3.51) provides the theoretical basis for the semilog analysis of drawdown
pressure data. It shows that if the well is produced instantaneously at a
constant sandface rate, then a graph of Pwf versus logt will be linear with a
negative slope given by
162.6q Bmkh
μ= − (3.52)
and a pressure intercept at logt = 0 given by
int 2log 3.23it w
kP P mc rφμ
⎡ ⎤⎛ ⎞= + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (3.53)
The formation permeability can be calculated from the slope of the semilog
line as
162.6q Bkmh
μ= − (3.54)
3-21
Table 3.4: Exponential Integral Function
3-22
and the initial pressure can be calculated from the pressure intercept as
int 2log 3.23it w
kP P mc rφμ
⎡ ⎤⎛ ⎞= − −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (3.55)
3.4.3 Pressure Buildup Equation
We wish to solve the diffusivity equation for the case of a well, which
has been producing at a constant instantaneous rate q for a time and then
shut in at time tp as shown in Figure 3.9C. Since the diffusivity equation is a
linear partial differential equation, superposition principle can be used to
construct this solution from the constant rate solution that we already have.
Figure 3.9 shows a graphical construction of how two constant rates can be
used to generate the rate schedule for a pressure buildup test. If the constant
production rate A is added to the constant injection rate B, the outcome will
be the rate in C, which simulates the rate for a pressure buildup test for a
well that had produced for time tp before it was shut in for the buildup test.
Figure 3.9. Rate schedule for pressure buildup test.
3-23
Using Eq.(3.49), the dimensionless pressure for the buildup test can easily be
derived by applying superposition principle as follows:
DC DA DBP P P= + (3.56)
( )1 ln 0.809072DA p D
P t t⎡ ⎤= + Δ +⎣ ⎦ (3.57)
[ ]1 ln 0.809072DB DP t= − Δ + (3.58)
Substituting Eqs.(3.57) and (3.58) into Eq.(3.56) gives the dimensionless
pressure for the buildup test as
1 ln2
pDBU DC
t tP P
t+ Δ⎛ ⎞
= = ⎜ ⎟Δ⎝ ⎠ (3.59)
Eq.(3.59) gives the following pressure buildup equation
( ) 162.6 log pws i
t tq BP t Pkh t
μ + Δ⎛ ⎞Δ = − ⎜ ⎟Δ⎝ ⎠
(3.60)
Eq.(3.60) is known as the Horner pressure buildup equation for an infinite
acting reservoir. It shows that if the sandface rate (qB) can be reduced to zero
instantaneously as shown in Figure 3.8C, then a graph of the buildup
bottomhole pressure, Pws(Δt), versus Horner time, pt tt
+ Δ⎛ ⎞⎜ ⎟Δ⎝ ⎠
, will be linear with
a negative slope given by Eq.(3.54) and a pressure intercept equal to Pi. The
formation permeability can be calculated from the slope of the Horner line and
the initial reservoir pressure can be determined by extrapolating the Horner
line to an infinite shut in time. For a bounded reservoir (in contrast to an
infinite acting reservoir), Horner pressure buildup equation takes the form
( ) * 162.6 log pws
t tq BP t Pkh t
μ + Δ⎛ ⎞Δ = − ⎜ ⎟Δ⎝ ⎠
(3.61)
3-24
where the extrapolated pressure, P*, is less than the initial reservoir pressure,
Pi.
3.4.4 Diagnostic Plots
The drawdown equation, Eq.(3.51), was derived using an internal
boundary condition in which the sandface rate increased instantaneously
from zero to a constant value. The buildup equation, Eq.(3.60), was derived
using an internal boundary condition in which the sandface rate decreased
from a constant value to zero instantaneously. Given that the wellbore
contains compressible fluids, when the well is opened for production, the
surface rate may be constant but the sandface rate will increase from zero to
a constant value over a definite period of time as shown in Figure 3.10.
Similarly, when the well is shut in at the wellhead for a pressure buildup test,
the surface rate is zero but the sandface rate will decline from its constant
value to zero over a finite period of time as shown in Figure 3.9. This
phenomenon, whereby the sandface rate lags behind the surface rate, is
known as wellbore unloading during drawdown and afterflow during buildup.
The term wellbore storage is normally used to describe the phenomenon either
during drawdown or during buildup.
Figure 3.10. Idealized and actual sandface rates during drawdown and buildup.
3-25
The consequence of neglecting the effect of wellbore storage in our
mathematical model is that the very early time pressure data will deviate from
the semilog line of Eq.(3.51) for the drawdown test or from the Horner semilog
line of Eq.(3.60) for the buildup test as shown in Figures 3.11 and 3.12. It is
therefore necessary to perform preliminary diagnostic plots of the pressure
data to identify those data that should be fitted to the semilog lines if any. The
function normally used to diagnose the presence of wellbore storage in the
pressure data is the welltest derivative function defined as
'
lnD D
D DD D
dP dPP td t dt
= = (3.62)
Figure 3.11. Semilog plot for pressure drawdown test showing deviation from semilog line due to wellbore storage effect.
3-26
Figure 3.12. Horner plot for pressure buildiup test showing deviation from
semilog line due to wellbore storage effect.
The welltest derivative function is defined in the manner shown in Eq.(3.62) to
take advantage of the fact that the data that fit the welltest model for a
drawdown test fall on the semilog line of Eq.(3.42). Along this semilog line, the
dimensionless welltest derivative function will be a constant as shown below:
' 1 a constantln 2
D DD D
D D
dP dPP td t dt
= = = = (3.63)
In actual variables, the welltest derivative function becomes
' 70.6 a constantln
d P d P q BP td t dt kh
μΔ Δ ⎛ ⎞Δ = = = =⎜ ⎟⎝ ⎠
(3.64)
Thus, after the effect of wellbore storage has subsided, the welltest derivative
function will become constant along the semilog line. It is only the pressure
data that have a constant welltest derivative (making allowance for
fluctuations due to noise) that should be fitted to the drawdown semilog line.
3-27
The diagnostic plots consist of log-log plots of ΔP versus t and ΔP' versus t
superimposed on the same graph. Figure 3.13 shows typical diagnostic plots
for a pressure drawdown test affected by wellbore storage.
Figure 3.13. Diagnostic plots for pressure drawdown data affected by wellbore storage.
The diagnostic plots can be divided into three time segments. In the first
time segment, ΔP and ΔP' are equal and have a unit slope on the log-log scale.
The data in this time segment are dominated 100% by wellbore storage. They
cannot be used to estimate reservoir properties. They can only be used to
estimate wellbore storage coefficient. In the second time segment, ΔP and ΔP'
separate from each other. The derivative function has a characteristic hump
indicative of wellbore storage. The magnitude of the hump is a measure of the
severity of the wellbore storage problem (and skin damage). The pressure data
3-28
in this time segment are still affected by wellbore storage but to a lesser
degree than in the first time segment. The data in this segment can be
analyzed by type curve matching to estimate approximate formation
properties. In the third time segment, the welltest derivative becomes constant
in accordance with Eq.(3.64). Data in this time segment are no longer affected
by wellbore storage and will plot on the correct semilog line of Eq.(3.51). Note
that the constant value of the welltest derivative function in this time segment
is equal to 70.6 q Bkhμ⎛ ⎞
⎜ ⎟⎝ ⎠
as shown in Eq.(3.64). This fact can be used to estimate
the formation permeability from the derivative function. If the test ended
during the second time segment, which happens in many tests, such a test
cannot be analyzed with a semilog plot.
Figure 3.14 shows typical diagnostic plots for a pressure buildup test
affected by wellbore storage. In general, the diagnostic plots are similar to
those of a drawdown test. The only difference is that in the third time
segment, when the pressure data are no longer affected by wellbore storage,
the welltest derivative function for a buildup may not be constant but may be
distorted as shown in Figure 3.14. This distortion is caused by the fact that
an approximate ΔP is used to calculate the derivative function for a buildup
test instead of the true ΔP, which is not accessible.
The welltest derivative function can be calculated with the following
central difference approximation:
( ) ( )
( ) ( )
1 11 1
1 1'
1 1i
i i i ii i i i
i i i ii i
t i i i i
P P P Pt t t tt t t td PP t t
dt t t t t
− ++ −
− +
− +
⎡ ⎤⎛ ⎞ ⎛ ⎞Δ − Δ Δ − Δ− + −⎢ ⎥⎜ ⎟ ⎜ ⎟− −Δ⎛ ⎞ ⎝ ⎠ ⎝ ⎠⎢ ⎥Δ = =⎜ ⎟ ⎢ ⎥− + −⎝ ⎠
⎢ ⎥⎢ ⎥⎣ ⎦
(3.65)
3-29
Figure 3.14. Diagnostic plots for pressure buildup data affected by wellbore storage.
Figure 3.15 shows the pressures used to define ΔP for drawdown and buildup
tests. For drawdown test, ΔP is defined as
( )i wfP P P tΔ = − (3.66)
For buildup test, ΔP should be defined as
( ) ( )ws wf pP P t P t tΔ = Δ − + Δ (3.67)
However, because Pwf(tp+Δt) is not available, ΔP for buildup test is
approximated by
3-30
( ) ( )ws wf pP P t P tΔ = Δ − (3.68)
This approximation could distort the welltest derivative function after the
effect of wellbore storage has subsided. The derivative function for a pressure
buildup test can be calculated with Eq.(3.65) using actual shut in time, Δt, or
Horner time, pt tt
+ Δ⎛ ⎞⎜ ⎟Δ⎝ ⎠
, or effective shut in time, Δte, defined as
( )( )p
ep
t tt
t tΔ
Δ =Δ +
(3.69)
Figure 3.15. Pressures used to define ΔP for drawdown and buildup.
3.4.5 Skin Factor
During drilling operations, the near wellbore permeability could be
reduced by formation damage caused by mud filtrate invasion. Also, if the well
is treated by an acid job, the near wellbore permeability could be enhanced
and the well is said to be stimulated. Thus, for a damaged or stimulated well,
3-31
there is a region of altered permeability near the wellbore as shown in Figure
3.16.
The additional pressure change near the wellbore caused by the region
of altered permeability can be incorporated into the welltest model through
the concept of a skin factor. The skin factor, S, is a dimensionless pressure
change at the wellbore given by
2
skinPSq
khμ
π
Δ=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.70)
in Darcy units or
141.2
skinPSq Bkhμ
Δ=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.71)
in oilfield units where ΔPskin is defined as
skin wfundamaged wfdamagedP P PΔ = − (3.72)
for a damaged well or
skin wfundamaged wfstimulatedP P PΔ = − (3.73)
for a stimulated well. It can be seen from Figure 3.16 that ΔPskin and hence, S,
is positive for a damaged well and negative for a stimulated well. It can be
shown that the permeability of the altered zone, ks, is related to the true
formation permeability, k, by the equation
1 ln s
s w
rkSk r
⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ (3.74)
Eq.(3.74) contains two unknowns, ks and rs, an as such is not as useful for
estimating the skin factor as it appears to be.
3-32
Figure 3.16. Schematic diagram showing skin.
Skin factor can be incorporated into the welltest model as
( ) ( )11, ln 0.80907 +S for 252D D D DP t t t= + ≥ (3.75)
Eq.(3.75) can be written in terms of log to base 10 as
3-33
( ) 21, 1.1513 log log 3.23 0.87D Dt w
kP t t Sc rφμ
⎡ ⎤⎛ ⎞= + − +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (3.76)
Eq.(3.76) gives the flowing bottomhole pressure in the presence of a skin as
( ) ( ) 2
162.6, log log 3.23 0.87wf w it w
q B kP t P r t P t Skh c r
μφμ
⎡ ⎤⎛ ⎞= = − + − +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (3.77)
Eq.(3.77) can be solved for the skin factor from the drawdown semilog line as
( )
2
11.1513 log 3.23
162.6wf i
t w
P hr P kSq B c r
khμ φμ
⎡ ⎤⎢ ⎥− ⎛ ⎞⎢ ⎥= − +⎜ ⎟⎛ ⎞⎢ ⎥⎝ ⎠−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(3.78)
where Pwf(1hr) is read on the semilog straight line or its extrapolation as
shown in Figure 3.10. For a buildup test, the skin factor is given by
( ) ( )
2
11.1513 log 3.23
162.6wf p ws
t w
P t P hr kSq B c r
khμ φμ
⎡ ⎤⎢ ⎥− ⎛ ⎞⎢ ⎥= − +⎜ ⎟⎛ ⎞⎢ ⎥⎝ ⎠−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(3.79)
where Pws(1hr) is read on the Horner straight line or its extrapolation as
shown in Figure 3.12. Thus, the skin factor, S, can be determined from a
drawdown or a buildup test. Note that the presence of a skin factor does not
affect the shape of the welltest derivative function and as such, it can still be
used to perform its diagnostic function.
3.4.6 Homogeneous Reservoir Model with Wellbore Storage and Skin
The welltest model can be reformulated to include wellbore storage and
skin. The resulting initial-boundary value problem to be solved consists of the
following equations in Darcy units
3-34
2
2
1 tcP P Pr r r k t
φμ∂ ∂ ∂+ =
∂ ∂ ∂ (3.28)
( )P r,0 iP= (3.29)
( )lim , irP r t P
→∞= (3.31)
2
w
wf
r r
dPkh Pr C qr dt
πμ =
∂⎛ ⎞ − =⎜ ⎟∂⎝ ⎠ (3.80)
( ),w
wfr r
PP P r t S rr =
⎡ ∂Δ ⎤⎛ ⎞Δ = Δ − ⎜ ⎟⎢ ⎥∂⎝ ⎠⎣ ⎦ (3.81)
The previous single internal boundary condition, Eq.(3.30), has now
been replaced by two internal boundary conditions, Eqs.(3.80) and (3.81).
Eq.(3.80) states that the constant surface rate, q, consists of the sum of the
sandface rate, given by the first term on the left side of the equation, and the
wellbore storage rate, given by the second term on the left side of the
equation. The constant, C, is the wellbore storage coefficient defined as
( )( )
i
i wf
V V tVCP P P t
−Δ= =
Δ − (3.82)
where
C = wellbore storage coefficient, reservoir cm3/atm
ΔV = change in wellbore fluid volume, reservoir cm3
ΔP = change in bottomhole pressure, atm
Vi = initial wellbore fluid volume before unloading, reservoir cm3
V(t) = wellbore fluid volume during unloading, reservoir cm3
Pi = initial wellbore pressure before unloading, atm
Pwf(t) = wellbore pressure during unloading, atm
3-35
The wellbore storage coefficient, C, is a measure of the severity of the wellbore
storage problem and can be determined from the unit slope line of the first
time segment on the diagnostic plots. The unit slope line in oilfield units is
given by
24qBP t
C⎛ ⎞Δ = ⎜ ⎟⎝ ⎠
(3.83)
from which C can be calculated in RB/psi as
unit slope line24
qB tCP
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟Δ⎝ ⎠⎝ ⎠ (3.84)
Eq.(3.81) states that the pressure change at the wellbore, (Pi-Pwf),
consists of the sum of the pressure change for the undamaged or
unstimulated well, given by the first term on the right side of the equation,
and ΔPskin, given by the second term on the right side of the equation.
The initial-boundary value problem can be put in dimensionless form as
2
2
1D D D
D D D D
P P Pr r r t
∂ ∂ ∂+ =
∂ ∂ ∂ (3.32)
( ),0 0D DP r = (3.33)
( )lim , 0D
D D DrP r t
→∞= (3.35)
1
1D
wD DD D
D D r
dP PC rdt r
=
⎛ ⎞∂− =⎜ ⎟∂⎝ ⎠
(3.85)
1D
DwD D D
D r
PP P S rr
=
⎡ ⎤⎛ ⎞∂= −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦
(3.86)
where the dimensionless wellbore storage coefficient is defined as
3-36
22Dt w
CCc hrπφ
= (3.87)
in Darcy units or
2
5.6152D
t w
CCc hrπφ
= (3.88)
in oilfield units. The larger the dimensionless wellbore storage coefficient, the
more severe is the wellbore storage problem.
The initial-boundary value problem represented by Eqs.(3.32), (3.33),
(3.35), (3.85) and (3.86) can be solved by Laplace transformation. It can be
shown that the dimensionless pressure at the wellbore in Laplace space is
given by
( )( ) ( )
( ) ( ) ( )1
3/ 2 21 1
1,o
wD
D o
K z S zK zP z
z K z z C K z S zK z
+=
⎡ ⎤+ +⎣ ⎦
(3.89)
where z is the Laplace parameter and Ko and K1 are Bessel functions of the
second kind of order 0 and 1. Eq.(3.89) can easily be inverted numerically, by
say Stehfest algorithm, to obtain the PwD function in the time domain. To
obtain the welltest derivative function, the Calculus derivative with respect to
tD in Laplace space is first obtained, inverted and multiplied by tD. The
Calculus derivative in Laplace space is given by
( ) ( )( ) ( )
( ) ( ) ( )1'
3/ 2 21 1
1, 1,o
wD wD
D o
z K z S zK zP z zP z
z K z z C K z S zK z
⎡ ⎤+⎣ ⎦= =⎡ ⎤+ +⎣ ⎦
(3.90)
Figure 3.17 shows the solutions and welltest derivative functions for
various values of CDe2S plotted on log-log scales. This family of solutions
constitutes Bourdet et al.'s wellbore storage type curves, which can be used to
analyze transient pressure tests that fit a homogeneous reservoir model with
wellbore storage and skin by type curve matching.
3-37
3.4.7 Type Curve Matching
A type curve is a graph of dimensionless pressure together with the
corresponding welltest derivative function versus dimensionless time, often on
a log-log scale. Such a graph is a convenient way to present the solution to
the diffusivity equation obtained by numerical computations. Figure 3.17 is
an example of type curves obtained by solving the diffusivity equation in the
presence of wellbore storage and skin. From the definitions of PwD, tD and CD,
BOURDET TYPE CURVES
0.1
1
10
100
0.1 1 10 100 1000 10000
tD/CD
PwD
- Pw
D'
CDe2S
1060
1050
1040
1030
1020
1015
1010
108
106
104
103
102
Figure 3.17. Bourdet et al type curves for a homogeneous reservoir with wellbore storage and skin for CDe2S from 102 to 1060.
3-38
log log log141.2wD
khP Pq Bμ
⎛ ⎞= Δ + ⎜ ⎟
⎝ ⎠ (3.91)
42.951 10log log logD
D
t x khtC Cμ
−⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ (3.92)
Eqs.(3.91) and (3.92) show that the graph of log wDP versus log D
D
tC
⎛ ⎞⎜ ⎟⎝ ⎠
is the
same as the graph of log PΔ versus log t with the axes shifted by the constants
log141.2
khq Bμ
⎛ ⎞⎜ ⎟⎝ ⎠
and 42.951 10log x kh
Cμ
−⎛ ⎞⎜ ⎟⎝ ⎠
. Thus, the ΔP versus t obtained from a
transient pressure test can be matched graphically with the appropriate type
curve to estimate the reservoir and well parameters.
The procedure for using the Bourdet et al.'s type curves to analyze a
transient pressure test is as follows:
1. Prepare a log-log plot of ΔP versus t on tracing paper using the same
scale as the type curves. Use only major grid lines on this plot. This is
the field plot.
2. Slide the field plot over the type curves both vertically and horizontally
to obtain the best match of the PD function and the corresponding
welltest derivative function. Trace the matched type curve on the field
plot to obtain a visual record of the best match.
3. Choose a convenient match point and record DMP , D
D M
tC
⎛ ⎞⎜ ⎟⎝ ⎠
, Mt and
( )2SD M
C e .
4. Calculate the formation permeability from the pressure match as
( )( )( )( )
141.2 DM
M
q B Pk
h Pμ
=Δ
(3.93)
3-39
5. Calculate the wellbore storage coefficient from the time match as
( )0.0002951
M
D
D M
kh tC
tC
μ⎛ ⎞
Δ⎜ ⎟⎝ ⎠=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.94)
6. Calculate the dimensionless wellbore storage coefficient.
7. Calculate the skin factor as
( )2
1 ln2
SD M
D
C eS
C
⎡ ⎤⎢ ⎥=⎢ ⎥⎣ ⎦
(3.95)
Figure 3.18 shows qualitatively a type match using the Bourdet et al.'s type
curves.
Figure 3.18. Example type curve match.
3-40
3.4.8 Radius of Investigation of a WellTest
An approximate equation for estimating the radius investigated by a
welltest is given by
0.03248invt
ktrcφμ
= (3.96)
where the variables are in oilfield units.
3.4.9 Field Example of a Well Test Analysis
A pressure buildup test was conducted on a new well in a new reservoir with
the results shown in Table 3.5.
Table 3.5: Pressure Buildup Data for Field Example
t tΔ ( )wsP tΔ
(hrs) (hrs) (psia) 15.33000 0.00000 3086.33 15.33417 0.00417 3090.57 15.33833 0.00833 3093.81 15.34250 0.01250 3096.55 15.34667 0.01667 3100.03 15.35083 0.02083 3103.27 15.35500 0.02500 3106.77 15.35917 0.02917 3110.01 15.36333 0.03333 3113.25 15.36750 0.03750 3116.49 15.37583 0.04583 3119.48 15.38000 0.05000 3122.48 15.38833 0.05833 3128.96 15.39667 0.06667 3135.92 15.40500 0.07500 3141.17 15.41333 0.08333 3147.64 15.42583 0.09583 3161.95 15.43833 0.10833 3170.68 15.45083 0.12083 3178.39 15.46333 0.13333 3187.12 15.47583 0.14583 3194.24 15.49250 0.16250 3205.96 15.50917 0.17917 3216.68 15.52583 0.19583 3227.89 15.54250 0.21250 3238.37 15.55917 0.22917 3249.07 15.58000 0.25000 3261.79
3-41
15.62167 0.29167 3287.21 15.66333 0.33333 3310.15 15.70500 0.37500 3334.34 15.74667 0.41667 3356.27 15.78833 0.45833 3374.98 15.83000 0.50000 3394.44 15.87167 0.54167 3413.90 15.91333 0.58333 3433.83 15.95500 0.62500 3448.05 15.99667 0.66667 3466.26 16.03833 0.70833 3481.97 16.08000 0.75000 3493.69 16.14250 0.81250 3518.63 16.20500 0.87500 3537.34 16.26750 0.93750 3553.55 16.33000 1.00000 3571.75 16.39250 1.06250 3586.23 16.45500 1.12500 3602.95 16.51750 1.18750 3617.41 16.58000 1.25000 3631.15 16.64250 1.31250 3640.86 16.70500 1.37500 3652.85 16.76750 1.43750 3664.32 16.83000 1.50000 3673.81 16.95500 1.62500 3692.27 17.08000 1.75000 3705.52 17.20500 1.87500 3719.26 17.33000 2.00000 3732.23 17.58000 2.25000 3749.71 17.70500 2.37500 3757.19 17.83000 2.50000 3763.44 18.08000 2.75000 3774.65 18.33000 3.00000 3785.11 18.58000 3.25000 3794.06 18.83000 3.50000 3799.80 19.08000 3.75000 3809.50 19.33000 4.00000 3815.97 19.58000 4.25000 3820.20 19.83000 4.50000 3821.95 20.08000 4.75000 3823.70 20.33000 5.00000 3826.45 20.58000 5.25000 3829.69 20.83000 5.50000 3832.64 21.08000 5.75000 3834.70 21.33000 6.00000 3837.19 21.58000 6.25000 3838.94 22.08000 6.75000 3838.02 22.58000 7.25000 3840.78 23.08000 7.75000 3843.01 23.58000 8.25000 3844.52
3-42
24.08000 8.75000 3846.27 24.58000 9.25000 3847.51 25.08000 9.75000 3848.52 25.58000 10.25000 3850.01 26.08000 10.75000 3850.75 26.58000 11.25000 3851.76 27.08000 11.75000 3852.50 27.58000 12.25000 3853.51 28.08000 12.75000 3854.25 28.58000 13.25000 3855.07 29.08000 13.75000 3855.50 29.83000 14.50000 3856.50 30.58000 15.25000 3857.25 31.33000 16.00000 3857.99 32.08000 16.75000 3858.74 32.83000 17.50000 3859.48 33.58000 18.25000 3859.99 34.33000 19.00000 3860.73 35.08000 19.75000 3860.99 35.83000 20.50000 3861.49 36.58000 21.25000 3862.24 37.58000 22.25000 3862.74 38.58000 23.25000 3863.22 39.58000 24.25000 3863.48 40.58000 25.25000 3863.99 41.58000 26.25000 3864.49 42.58000 27.25000 3864.73 43.83000 28.50000 3865.23 45.33000 30.00000 3865.74
Other pertinent data are as follows:
Producing time before the test = 15.33 hours
Production rate before the test = 174 STB/D
Wellbore radius = 0.29 ft
Formation thickness = 107 ft
Porosity = 25 %
Total compressibility = 4.2x10-6 psi-1
Formation volume factor = 1.06 RB/STB
Oil viscosity = 2.5 cp
3-43
a. Screen the pressure data using appropriate diagnostic plots.
b. Analyze the test data and determine the formation permeability, the
total skin factor and the initial reservoir pressure.
c. Calculate the radius of investigation of the test.
d. Simulate the buildup test assuming a homogeneous reservoir with
wellbore storage and skin. Show the best match graphically by
superimposing the simulated and the test data on the plots of P versus
t, log-log diagnostic plots and the Horner plot. What are the values of k,
S and CD , Pwf(tp) and Pi that gave the best match?
The pertinent data for the diagnostic plots and Horner plot are shown in
Table 3.6.
Table 3.6: Calculated Values of ΔP and ΔP'
t tΔ etΔ Horner ( )wsP tΔ PΔ '
eP wrt tΔ Δ 'P wrt tΔ Δ
(hrs) (hrs) (hrs) Time (psia) (psi) (psi) (psi)
15.33000 0.00000 3086.33
15.33417 0.00417 0.00417 3677.26 3090.57 4.24
15.33833 0.00833 0.00833 1841.34 3093.81 7.48 5.98 5.98
15.34250 0.01250 0.01249 1227.40 3096.55 10.22 9.33 9.32
15.34667 0.01667 0.01665 920.62 3100.03 13.70 13.46 13.45
15.35083 0.02083 0.02080 736.96 3103.27 16.94 16.88 16.85
15.35500 0.02500 0.02496 614.20 3106.77 20.44 20.24 20.20
15.35917 0.02917 0.02911 526.54 3110.01 23.68 22.73 22.69
15.36333 0.03333 0.03326 460.95 3113.25 26.92 25.98 25.93
15.36750 0.03750 0.03741 409.80 3116.49 30.16 23.96 23.91
15.37583 0.04583 0.04569 335.50 3119.48 33.15 27.55 27.46
15.38000 0.05000 0.04984 307.60 3122.48 36.15 37.07 36.95
15.38833 0.05833 0.05811 263.82 3128.96 42.63 47.21 47.03
3-44
15.39667 0.06667 0.06638 230.94 3135.92 49.59 49.03 48.82
15.40500 0.07500 0.07463 205.40 3141.17 54.84 53.03 52.76
15.41333 0.08333 0.08288 184.97 3147.64 61.31 77.43 76.99
15.42583 0.09583 0.09523 160.97 3161.95 75.62 88.83 88.32
15.43833 0.10833 0.10757 142.51 3170.68 84.35 71.73 71.24
15.45083 0.12083 0.11989 127.87 3178.39 92.06 80.09 79.46
15.46333 0.13333 0.13218 115.98 3187.12 100.79 85.25 84.53
15.47583 0.14583 0.14446 106.12 3194.24 107.91 92.29 91.40
15.49250 0.16250 0.16080 95.34 3205.96 119.63 110.52 109.37
15.50917 0.17917 0.17710 86.56 3216.68 130.35 119.27 117.89
15.52583 0.19583 0.19336 79.28 3227.89 141.56 129.06 127.44
15.54250 0.21250 0.20959 73.14 3238.37 152.04 136.87 135.00
15.55917 0.22917 0.22579 67.89 3249.07 162.74 146.06 143.92
15.58000 0.25000 0.24599 62.32 3261.79 175.46 155.10 152.61
15.62167 0.29167 0.28622 53.56 3287.21 200.88 172.44 169.27
15.66333 0.33333 0.32624 46.99 3310.15 223.82 192.65 188.52
15.70500 0.37500 0.36605 41.88 3334.34 248.01 212.55 207.52
15.74667 0.41667 0.40564 37.79 3356.27 269.94 208.64 203.21
15.78833 0.45833 0.44502 34.45 3374.98 288.65 216.24 209.94
15.83000 0.50000 0.48421 31.66 3394.44 308.11 241.12 233.50
15.87167 0.54167 0.52318 29.30 3413.90 327.57 265.11 256.05
15.91333 0.58333 0.56195 27.28 3433.83 347.50 247.95 239.07
15.95500 0.62500 0.60052 25.53 3448.05 361.72 253.29 243.21
15.99667 0.66667 0.63889 23.99 3466.26 379.93 283.06 271.37
16.03833 0.70833 0.67705 22.64 3481.97 395.64 243.76 233.17
16.08000 0.75000 0.71502 21.44 3493.69 407.36 258.62 246.28
16.14250 0.81250 0.77160 19.87 3518.63 432.30 298.44 283.73
16.20500 0.87500 0.82775 18.52 3537.34 451.01 258.25 244.44
16.26750 0.93750 0.88347 17.35 3553.55 467.22 273.98 258.07
16.33000 1.00000 0.93876 16.33 3571.75 485.42 278.26 261.44
16.39250 1.06250 0.99363 15.43 3586.23 499.90 283.74 265.20
16.45500 1.12500 1.04809 14.63 3602.95 516.62 301.05 280.62
16.51750 1.18750 1.10213 13.91 3617.41 531.08 288.60 267.90
16.58000 1.25000 1.15576 13.26 3631.15 544.82 253.30 234.50
16.64250 1.31250 1.20899 12.68 3640.86 554.53 247.56 227.85
3-45
16.70500 1.37500 1.26182 12.15 3652.85 566.52 281.16 258.06
16.76750 1.43750 1.31426 11.66 3664.32 577.99 263.46 241.04
16.83000 1.50000 1.36631 11.22 3673.81 587.48 247.74 225.68
16.95500 1.62500 1.46926 10.43 3692.27 605.94 227.42 206.12
17.08000 1.75000 1.57070 9.76 3705.52 619.19 210.56 188.93
17.20500 1.87500 1.67066 9.18 3719.26 632.93 224.74 200.33
17.33000 2.00000 1.76919 8.67 3732.23 645.90 208.37 184.96
17.58000 2.25000 1.96203 7.81 3749.71 663.38 162.84 142.20
17.70500 2.37500 2.05641 7.45 3757.19 670.86 150.46 130.44
17.83000 2.50000 2.14947 7.13 3763.44 677.11 140.26 120.70
18.08000 2.75000 2.33172 6.57 3774.65 688.32 140.46 119.19
18.33000 3.00000 2.50900 6.11 3785.11 698.78 138.98 116.46
18.58000 3.25000 2.68151 5.72 3794.06 707.73 115.07 95.49
18.83000 3.50000 2.84944 5.38 3799.80 713.47 133.68 108.08
19.08000 3.75000 3.01297 5.09 3809.50 723.17 150.18 121.27
19.33000 4.00000 3.17227 4.83 3815.97 729.64 107.37 85.60
19.58000 4.25000 3.32750 4.61 3820.20 733.87 64.24 50.83
19.83000 4.50000 3.47882 4.41 3821.95 735.62 40.75 31.50
20.08000 4.75000 3.62637 4.23 3823.70 737.37 56.31 42.75
20.33000 5.00000 3.77029 4.07 3826.45 740.12 79.61 59.90
20.58000 5.25000 3.91071 3.92 3829.69 743.36 87.17 65.00
20.83000 5.50000 4.04777 3.79 3832.64 746.31 74.57 55.11
21.08000 5.75000 4.18157 3.67 3834.70 748.37 72.12 52.33
21.33000 6.00000 4.31224 3.56 3837.19 750.86 70.51 50.88
21.58000 6.25000 4.43987 3.45 3838.94 752.61 34.46 25.33
22.08000 6.75000 4.68648 3.27 3838.02 751.69 19.52 12.42
22.58000 7.25000 4.92217 3.11 3840.78 754.45 53.06 36.18
23.08000 7.75000 5.14764 2.98 3843.01 756.68 43.29 28.98
23.58000 8.25000 5.36355 2.86 3844.52 758.19 41.52 26.89
24.08000 8.75000 5.57049 2.75 3846.27 759.94 40.82 26.16
24.58000 9.25000 5.76902 2.66 3847.51 761.18 33.25 20.81
25.08000 9.75000 5.95963 2.57 3848.52 762.19 40.20 24.38
25.58000 10.25000 6.14279 2.50 3850.01 763.68 37.65 22.86
26.08000 10.75000 6.31892 2.43 3850.75 764.42 32.21 18.81
26.58000 11.25000 6.48843 2.36 3851.76 765.43 33.95 19.69
3-46
27.08000 11.75000 6.65168 2.30 3852.50 766.17 36.54 20.56
27.58000 12.25000 6.80901 2.25 3853.51 767.18 38.36 21.44
28.08000 12.75000 6.96074 2.20 3854.25 767.92 36.51 19.89
28.58000 13.25000 7.10716 2.16 3855.07 768.74 30.55 16.56
29.08000 13.75000 7.24854 2.11 3855.50 769.17 27.64 14.43
29.83000 14.50000 7.45173 2.06 3856.50 770.17 32.70 16.92
30.58000 15.25000 7.64495 2.01 3857.25 770.92 30.23 15.15
31.33000 16.00000 7.82892 1.96 3857.99 771.66 32.51 15.89
32.08000 16.75000 8.00429 1.92 3858.74 772.41 34.83 16.64
32.83000 17.50000 8.17164 1.88 3859.48 773.15 30.98 14.58
33.58000 18.25000 8.33152 1.84 3859.99 773.66 33.60 15.21
34.33000 19.00000 8.48442 1.81 3860.73 774.40 27.78 12.67
35.08000 19.75000 8.63077 1.78 3860.99 774.66 23.22 10.01
35.83000 20.50000 8.77100 1.75 3861.49 775.16 40.28 17.08
36.58000 21.25000 8.90548 1.72 3862.24 775.91 39.27 16.70
37.58000 22.25000 9.07644 1.69 3862.74 776.41 26.72 10.90
38.58000 23.25000 9.23853 1.66 3863.22 776.89 21.33 8.60
39.58000 24.25000 9.39243 1.63 3863.48 777.15 24.52 9.34
40.58000 25.25000 9.53875 1.61 3863.99 777.66 33.76 12.75
41.58000 26.25000 9.67803 1.58 3864.49 778.16 25.91 9.71
42.58000 27.25000 9.81077 1.56 3864.73 778.40 23.88 8.48
43.83000 28.50000 9.96817 1.54 3865.23 778.90 30.25 10.62
45.33000 30.00000 10.14560 1.51 3865.74 779.41
The diagnostic plots depicted in Figure 3.19 show the classic response of a
homogeneous reservoir model with wellbore storage and skin. From the
welltest derivative function, it would appear that the effect of wellbore storage
had subsided at about 10 hrs, in which case, the pressure data beyond 10 hrs
might be fitted to the Horner semilog line. Notice the noise in the derivative
function. This is not surprising because numerical differentiation is a noisy
process. The wellbore storage coefficient can be calculated from the unit slope
line as
3-47
( )( )174 1.06 1 0.0096 RB/psi24 800
C ⎛ ⎞= =⎜ ⎟⎝ ⎠
The dimensionless wellbore storage coefficient is given by
( )( )( )( )( )( )
226
5.615 0.00969.09 10
2 0.25 4.20 10 107 0.29DC xxπ −
= =
Figure 3.19. Diagnostic plots for field example.
Figure 3.20 shows the Horner plot. The Horner straight line is given by
( ) 15.333880.60 78.52logwstP t
t+ Δ⎛ ⎞Δ = − ⎜ ⎟Δ⎝ ⎠
3-48
The slope of the Horner line is
78.52 psi/log cyclem = −
from which the permeability is calculated as
( )( )( )( )( )( )
162.6 174 2.5 1.068.92 md
78.52 107k = − =
−
Figure 3.20. Horner plot for field example.
From the Horner line,
( )1 3785.35wsP hr = psia
The skin factor can be calculated as
3-49
( )( )( )( )26
3086.33 3785.35 8.921.1513 log 3.2378.52 0.25 2.5 4.20 10 0.29
Sx −
⎧ ⎫⎡ ⎤−⎪ ⎪⎛ ⎞ ⎢ ⎥= − +⎨ ⎬⎜ ⎟− ⎢ ⎥⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
( )1.1513 8.90 7.61 3.23 5.21S = − + =
Thus, the well is damaged. The initial reservoir pressure is given by
* 3880.6iP P= = psia.
The radius of investigation for the drawdown portion of the test is given by
( )( )( )( )( )6
8.92 15.330.03248 234 ft
0.25 2.5 4.20 10invrx −
= =
The radius of investigation for the buildup portion of the test is given by
( )( )( )( )( )6
8.92 300.03248 330 ft
0.25 2.5 4.20 10invrx −
= =
The buildup test cannot investigate what the drawdown did not investigate.
Therefore, the radius of investigation of the buildup test test is obtained from
the drawdown portion of the test as 234 ft. The reservoir bulk volume
investigated by the test is given by
( )( )( )2 2 6 3234 107 18.406 10 ftinvV r h xπ π= = =
The estimated permeability of 8.92 md can be assigned to this volume of the
reservoir. This is in contrast to the core permeability, which is measured on a
minute portion of the reservoir.
The last step in a pressure transient analysis is to simulate the test. A
successful simulation of the test, though non-unique, tends to lend credibility
to the results of the analysis. Using the k, S and CD obtained from the semilog
analysis as preliminary estimates, the homogeneous reservoir model with
3-50
wellbore storage and skin can be used to simulate the test by a trial and error
procedure.
Figures 3.21 through 3.23 show graphical comparisons of the simulated
test and the field test. Figure 3.21 compares the pressure buildup responses;
Figure 3.22 compares the log-diagnostic plots; and Figure 3.23 compares the
Horner plots. Clearly, the agreement between the simulated test and the field
test is good. The best match was obtained with the following parameters:
9.4 k = md
6S = 29 10DC x=
2 81.46 10SDC e x=
( ) 3089.46wf pP t = psia
3880.60iP = psia
Thus, the simulation serves to fine tune the parameters obtained by the
conventional semilog analysis of the test.
Figure 3.21. A comparison of the simulated and the measured pressure buildup response for field example.
3-51
Figure 3.22. A comparison of the simulated and the measured
diagnostic plots for field example.
Figure 3.23. A comparison of the simulated and the measured Horner plots for field example.
3-52
3.4.10 Welltest Model for Dry Gas Reservoir
The diffusivity equation for real gas flow is given by
2 1tc M MMk t t
φμα
∂ ∂∇ = =
∂ ∂ (3.97)
The dependent variable in Eq.(3.97), M(P), is defined as
2( )o
P
P
PM P dPZμ
= ∫ (3.98)
where Po is a reference pressure. M(P) is known as the real gas potential or
the real gas pseudo pressure. Eq.(3.97) is similar in structure to Eq.(3.23)
that formed the basis for the welltest model for oil (and water) reservoirs. This
suggests that all the equations used to analyze transient pressure tests in an
oil reservoir can be adapted to analyze transient pressure tests in a dry gas
reservoir provided the pressure data from the gas well test are first
transformed into the real gas potential, M(P).
The dimensionless pressure for real gas flow in oilfield units is defined
as
( ) ( )1422
i wfD
sc
M P M PP
q Tkh
−=
⎛ ⎞⎜ ⎟⎝ ⎠
(3.99)
where
qsc = gas production rate, Mscf/D
T = absolute temperature of the reservoir, ºRankine (ºR = ºF+460)
The semilog drawdown line is given
( ) ( ) 2
1637 log log 3.23 0.87scwf i
t w
q T kM P M P t Skh c rφμ
⎡ ⎤⎛ ⎞= − + − +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (3.100)
3-53
The gas velocity in the reservoir can be quite high given the low viscosity of
gas. This could lead to non-darcy flow with an additional pressure drop above
that of darcy flow. The effect of non-darcy flow can be expressed as a skin
factor as shown in Section 3.12. As a result, the total skin factor for a gas well
test, S, is normally split into the sum of two skin factors S* and Dqsc, where D
is a non-darcy coefficient and S* is the part of the total skin factor that does
not include the non-darcy effect. With this modification, Eq.(3.100) becomes
( ) ( ) *2
1637 log log 3.23 0.87 0.87scwf i sc
i ti w
q T kM P M P t S Dqkh c rφμ
⎡ ⎤⎛ ⎞= − + − + +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦(3.101)
The slope of the semilog line is given by
1637 scq Tmkh
= − (3.102)
from which the formation permeability can be determined. The total skin
factor for drawdown test is given by
( ) ( )
121.1513 log 3.23
1637wf ihr
sc i ti w
M P M P kSq T c r
khφμ
⎡ ⎤⎢ ⎥− ⎛ ⎞⎢ ⎥= − +⎜ ⎟⎛ ⎞⎢ ⎥⎝ ⎠−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(3.103)
Horner's buildup equation is given by
( ) ( )* 1637 log pscws
t tq TM P M Pkh t
+ Δ⎛ ⎞= − ⎜ ⎟Δ⎝ ⎠
(3.104)
Eq.(3.104) suggests that a graph of ( )wsM P versus log pt tt
+ Δ⎛ ⎞⎜ ⎟Δ⎝ ⎠
will be linear
with a negative slope, m, given by
1637 scq Tmkh
= − (3.96)
from which the formation permeability can be determined. The total skin
3-54
factor for a buildup test is given by
( ) ( )1
* * 21.1513 log 3.231637
wf ws hr
sc t w
M P M P kSq T c r
khφμ
⎡ ⎤⎢ ⎥− ⎛ ⎞⎢ ⎥= − +⎜ ⎟⎛ ⎞⎢ ⎥⎝ ⎠−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(3.105)
where M(Pws)1hr is obtained from the semilog straight line or its extrapolation,
and *μ and *tc are obtained at the pressure intercept, P*. Diagnostic plots
should be used to screen the M(P) data for wellbore storage in the manner
described for oil welltests to determine the correct semilog lines if any.
The M(P) function for a dry gas reservoir can be computed in advance
using the gas properties. To do so requires the ability to estimate the gas Z
factor and viscosity as functions of pressure and temperature. Correlations
exist for these functions in the literature. Table 3.7 shows the gas properties
and the M(P) function for a dry gas reservoir with the following properties:
Reservoir temperature = 246 ºF + 459.67 = 705.67 ºR
Pseudocritical pressure = 665.0 psia
Pseudocritical temperature = 374.9 ºR
Apparent molecular weight of gas = 18.870 lb mass/lb mole
Apparent molecular weight of air = 28.9625 lb mass/lb mole
Density of fresh water = 62.368 lb mass/cu ft
Table 3.7: Variation of Gas Properties with Pressure
Pressure P/Z Bg ρg cg μg M(P)
(psia) Z (psia) (ft3/Mscf) (lbm/ft3) (106/psi) (cp) (psia2/cp)
200 0.986 203 98.120 0.505 5066.65 0.01393 2.8933E+06
400 0.974 411 48.428 1.024 2562.91 0.01413 1.1625E+07
600 0.962 624 31.895 1.554 1725.07 0.01437 2.6130E+07
800 0.951 841 23.656 2.095 1303.13 0.01466 4.6298E+07
1000 0.942 1062 18.736 2.646 1047.10 0.01498 7.1964E+07
3-55
1200 0.934 1285 15.477 3.203 873.69 0.01534 1.0292E+08
1400 0.927 1511 13.169 3.764 747.30 0.01573 1.3890E+08
1600 0.921 1737 11.456 4.327 650.20 0.01616 1.7962E+08
1800 0.917 1962 10.140 4.888 572.64 0.01661 2.2476E+08
2000 0.915 2186 9.102 5.446 508.85 0.01710 2.7397E+08
2200 0.914 2406 8.267 5.996 455.22 0.01760 3.2690E+08
2400 0.915 2623 7.583 6.537 409.42 0.01813 3.8321E+08
2600 0.917 2836 7.015 7.066 369.82 0.01868 4.4254E+08
2800 0.920 3043 6.538 7.582 335.30 0.01924 5.0455E+08
3000 0.925 3244 6.133 8.082 305.01 0.01982 5.6894E+08
3200 0.931 3438 5.786 8.567 278.31 0.02040 6.3540E+08
3400 0.938 3626 5.486 9.036 254.68 0.02100 7.0367E+08
3600 0.945 3808 5.225 9.488 233.71 0.02159 7.7350E+08
3800 0.954 3982 4.995 9.923 215.03 0.02220 8.4466E+08
4000 0.964 4150 4.793 10.341 198.36 0.02280 9.1697E+08
4200 0.974 4312 4.614 10.744 183.45 0.02340 9.9024E+08
4400 0.985 4467 4.454 11.130 170.07 0.02400 1.0643E+09
4600 0.997 4616 4.310 11.501 158.03 0.02460 1.1391E+09
4800 1.009 4759 4.181 11.857 147.19 0.02520 1.2144E+09
5000 1.021 4896 4.063 12.199 137.39 0.02579 1.2901E+09
5200 1.034 5028 3.957 12.528 128.52 0.02637 1.3662E+09
5400 1.048 5154 3.859 12.843 120.47 0.02695 1.4426E+09
5600 1.061 5276 3.770 13.147 113.14 0.02753 1.5192E+09
5800 1.075 5393 3.689 13.439 106.46 0.02810 1.5959E+09
6000 1.090 5506 3.613 13.719 100.37 0.02866 1.6727E+09
6200 1.104 5614 3.543 13.990 94.79 0.02921 1.7496E+09
6400 1.119 5719 3.479 14.250 89.67 0.02976 1.8265E+09
6600 1.134 5820 3.418 14.501 84.96 0.03030 1.9033E+09
6800 1.149 5917 3.362 14.743 80.63 0.03084 1.9801E+09
7000 1.165 6011 3.310 14.977 76.64 0.03137 2.0568E+09
3-56
3.5 FACTORS AFFECTING PERMEABILITY
Permeability is affected by compaction, pore size, sorting, cementation,
layering and clay swelling.
3.5.1 Compaction
Just as compaction reduces porosity, it also reduces permeability. As a
result of compaction, the permeability of rocks tends to decrease with depth of
burial.
3.5.2 Pore Size (Grain Size)
In general, for a sand, the permeability is proportional to the square of
the mean pore size. For a well sorted sand, the pore size is proportional to the
grain size. Therefore, for well sorted sands, the permeability is proportional to
the square of the grain size. Thus, a well sorted sand with a larger grain size
will have a higher permeability than a well sorted sand with a smaller grain
size (Table 3.8).
Table 3.8: Measured Porosities and Permeabilities of Artificial Sandpacks (adapted from Beard and Weyl, 1973)
3-57
To see why grain size affects the permeability of a medium, let us
compare the wetted surface area per unit bulk volume (specific surface area)
for flow through a pipe with no grains and through a porous medium with two
grain sizes. Figure 3.24 shows a cube of each medium of volume L3 meter3.
Let the porous medium consist of uniform spherical sand grains of radius r
meter and porosity φ. The wetted surface area of the pipe per unit bulk
volume (S) is
2
2 33
Surface Area 4 4 /Bulk Volume
LS m mL L
⎛ ⎞= = = ⎜ ⎟⎝ ⎠
(3.106)
Figure 3.24. Comparison of flow in a pipe with flow in a porous medium.
For the porous medium, the number of grains, N, contained in the L3
meter3 of bulk volume is given by
( )3
3
1Total Solid Volume4Volume of 1 sand grain3
LN
r
φ
π
−= =
⎛ ⎞⎜ ⎟⎝ ⎠
(3.107)
The wetted surface area is given by
3-58
( ) ( )3 32
3
1 3 1Area of 1sand grain 4
43
L LArea Nx x r
rr
φ φπ
π
− −= = =
⎛ ⎞⎜ ⎟⎝ ⎠
(3.108)
The wetted area per unit bulk volume (specific surface area) is given by
( ) ( )32 3
3
3 1 3 11 /L
S m mr L r
φ φ⎡ ⎤− −⎛ ⎞= =⎢ ⎥ ⎜ ⎟⎝ ⎠⎣ ⎦
(3.109)
Let us examine some numerical values. Let L = 1 meter and let the porous
medium consist of very fine sand with a grain diameter of 6100 10x − meter (r = 650 10x − meter) and a porosity of 28%. Figure 1.1 gives the diameters of the
grains of various clastic rocks. The specific surface area for the pipe flow is
2 34 /S m m=
The specific surface area for the porous medium is
( ) 2 36
3 1 .2843200 /
50 10S m m
x −
−= =
Thus, the specific surface area of the porous medium is 10,800 times that of
the pipe. Since the fluid velocity is zero at the wetted surface during flow (no
slip condition), it is obvious that much more energy will be required to sustain
the same volumetric flow rate through the porous medium than through the
pipe, everything else being equal. Put another way, it is obvious that the
porous medium is much less permeable than the pipe.
Let us examine the effect of grain size on the specific surface area of the
porous medium, and hence the permeability, by changing the grain size to a
fine silt with a grain diameter of 610 10x − meter (r = 65 10x − meter). In this case,
the specific surface area is
( ) 2 36
3 1 .28432000 /
5 10S m m
x −
−= =
3-59
Thus, the specific surface area of the fine silt has increased by a factor of 10
compared to the very fine sand. Using the same argument as before, it is
obvious that the silt will be less permeable than the fine sand. Therefore,
permeability decreases as the grain size decreases.
Example 3.2
An idealized petroleum reservoir consists of an unconsolidated sand of
uniform grain size. The sand grains were deposited in such a way as to give
the closest packing possible with spherical grains. This extremely well-sorted,
homogeneous and isotropic reservoir has the following dimensions:
Length = 3 miles
Width = 2 miles
Thickness = 250 feet
Grain diameter = 1/8 mm (fine sand)
Estimate the following properties for the reservoir:
a. Porosity.
b. Surface area per unit bulk volume (specific surface area).
c. Wetted surface area of the sand grains in square miles.
Notes:
1 mile = 5280 ft
1 ft = 30.48 cm
Solution to Example 3.2
L = 3 miles = (3)(5280)(30.48) = 482803.20 cm
W = 2 miles = (2)(5280)(30.48) = 321868.80 cm
h = 250 ft = (250)(30.48) = 7620.00 cm
3-60
d = 1/8 mm = 0.0125 cm (fine sand)
Rhombohedral packing with a porosity of 25.9% is the closest packing.
The specific surface area based on bulk volume is given by Eq.(3.109) as
( )( )
2 33 1 0.259355.68 /
0.0125 / 2S cm cm
−= =
The wetted surface are is given by
( )( ) 17 2Wetted Surface Area 355.68 482803.20 321868.80 7620.00 4.212 10bSxV x x x cm= = =
( ) ( )17 2
72 2 2
4.212 10Wetted Surface Area 1.626 105280 30.48 /
x cm xx cm mile
= = square miles.
It should be observed that the wetted surface area for the idealized
reservoir is enormous and would be considerably more if the grains were of
clay sized particles. This enormous surface area has implications for the
occurrence of surface phenomena in reservoir rocks.
3.5.3 Sorting
Poor sorting reduces the pore size and consequently reduces the
permeability of a medium (Table 3.8).
3.5.4 Cementation
Cementation reduces the pore size and consequently reduces the
permeability of the rock.
3.5.5 Layering
Permeability is a tensor and can therefore be different in different
directions. Because of layering in sedimentary rocks, horizontal permeabilities
in petroleum reservoirs tend to be higher than vertical permeabilities in the
absence of vertical fractures.
3-61
3.5.6 Clay swelling
Many consolidated sandstones contain clay and silt, e.g., arkoses and
graywackes. Since montmorillonite-type clays absorb fresh water and swell,
the permeabilities of such sandstones will be greatly reduced when measured
with fresh water. The addition of salts, such as sodium chloride or potassium
chloride, will in most cases, eliminate the clay swelling.
3.6 TYPICAL RESERVOIR PERMEABILITY VALUES
Reservoir permeabilities vary widely, from 0.001 md for a tight gas sand
in East Texas to 4000 md for an unconsolidated sand in the Niger Delta.
Reservoir permeabilities may be loosely described as follows:
Very low: k < 1md
Low: 1md < k < 10 md
Fair: 10 md < k < 50 md
Average: 50 md < k < 200 md
Good: 200 md < k < 500 md
Excellent: k > 500 md
Because of low gas viscosity, gas reservoirs with permeabilities less than 1 md
can still produce at economic rates if the reservoir is hydraulically fractured.
Because of the higher oil viscosities compared to gas viscosities, oil reservoirs
with permeabilities less than 1 md are unlikely to produce at economic rates
even with hydraulic fracturing.
3.7 PERMEABILITY-POROSITY CORRELATIONS
Since permeability depends on the continuity of pore space, there is not
in theory, or in fact, a unique relationship between the porosity of a rock and
its permeability. For unconsolidated sands, it is possible to establish
relationships between porosity and either some measure of apparent pore
3-62
diameter or specific surface area and permeability. However, these have very
limited applications.
In the same depositional and diagenetic environment, there is
sometimes a reasonable relationship between porosity and permeability,
although for a given porosity, permeabilities can vary widely. Figure 3.25
shows a typical permeability-porosity correlation obtained from core analysis.
The figure shows a correlation of permeability versus porosity for 500 samples
from a uniform sandstone reservoir in Pennsylvania. The permeabilities were
measured with gas and corrected for Klinkenberg effect. The porosity was
measured by saturating a well machined core with a wetting liquid. The
scatter in the data is not the result of experimental errors. Apart from
showing the general increase of permeability with an increase in porosity, the
correlation shows the wide spread, or lack of a close relationship, between
porosity and permeability. Nelson (1994) has presented a comprehensive
review of permeability - porosity correlations using data from the open
literature.
Attempts have been made to correlate permeability and porosity using
an equation of the form:
ln k a bφ= + (3.110)
where a and b are constants. This relationship is at best applicable at a local
level. Nevertheless, where it exists, it does provide one method of estimating
permeability from well logs and even possibly from drill cuttings. To test the
strength of the linear relationship between lnk and φ, let us examine the data
from sandstone cores shown in Table 3.9. Figure 3.26 shows the regression
line for the permeability - porosity correlation. The regression line is given by
ln 0.327 3.049k φ= −
with the square of the correlation coefficient (R2) equal to 0.3514. The
correlation is weak.
3-63
Figure 3.25. Permeability-porosity correlation for a sandstone reservoir (Ryder, 1948).
Table 3.9: Permeability - Porosity Data for Sandstone Cores (from Kenyon et al., 1986)
Sample φ k
No (%) (md)
1 31.2 2719
2 30.4 1382
3 24.8 3.3
4 28.5 215
5 33 151
3-64
6 31 426
7 30.3 1072
8 9.2 0.0012
9 22 32.1
10 22.8 83.2
11 20.4 4.56
12 19 311
13 18.6 7.47
14 20.9 20.5
15 21.7 52.5
16 14.8 4.17
17 17.6 10.9
18 11.3 4.5
19 11 3
20 10.1 2.8
21 10.4 3.4
22 13.7 2.1
23 12.8 1.3
24 12.2 7.3
25 11.7 4.8
26 17.2 91.9
27 16.7 190.4
28 18.6 424.9
29 18.3 270.3
30 18.3 137.3
31 16.3 28.5
32 15.6 200.5
33 17.2 36.5
34 16.4 243.2
35 18.3 337.3
36 18.1 39.9
37 14.9 64.6
38 19.6 669.5
39 18.3 540.5
40 18.8 890.4
3-65
41 17.8 429.8
42 16.1 0.011
43 18 0.52
44 16.6 0.52
45 16.1 0.087
46 17.2 12.1
47 22.5 5.97
48 19.6 1.72
49 12.7 0.063
50 15.2 0.93
51 19.1 20.3
52 14.8 4.72
53 20.5 45
54 20.5 45
55 23.9 663
56 23.8 591
57 22.9 511
58 21.6 478
59 22.2 131
60 22.3 1305
61 16.8 621
62 6.3 10.4
63 24.6 1425
64 24.3 2590
65 20 0.849
66 21.9 6
67 6.3 0.003
68 19.8 0.499
69 31.3 139
70 25.7 12.6
71 14.7 1.68
3-66
Figure 3.26. Permeability-porosity correlation for sandstone cores (from
Kenyon et al., 1986).
Figure 3.27 shows permeability - porosity correlations in which an effort
was made to classify the grain size of the sandstones. Samples containing
more than 50% of 1-2 mm grain size were classified as very coarse-grained;
0.5-1 mm, coarse-grained; 0.25-0.5 mm, medium-grained; and 0.1-0.25 mm,
fine-grained. Samples containing more than 10% of silt (<0.1 mm) were called
silty, whereas samples with clay (<0.004 mm) content greater than 7% were
classified as clayey. Both the absolute permeabilities and the effective
permeabilities to gas at irreducible water saturations were measured.
Permeabilities also were corrected for Klinkenberg effect. The permeabilities
shown in Figure 3.27 are the effective permeabilities to the nonwetting gas at
irreducible wetting phase saturation. These end point permeabilities are not
functions of the porous medium alone as in the case of the absolute
permeabilities. They are functions of the porous medium and the wetting and
nonwetting fluids. It would appear that accounting for the grain size
3-67
distribution improved the correlation of the end point effective permeabilites
with porosity. The author indicates that the correlations of the absolute
permeabilities with porosity showed the usual scatter regardless of the grain
size classification but the data were not presented.
Figure 3.27. Effect of grain size on the permeability-porosity correlations for various sandstones (from Chillingar, 1964).
The permeability - porosity correlation for unconsolidated sands is even
weaker than for consolidated sandstones. Figure 2.28 shows the correlations
based on the data of Table 3.3 for clean, unconsolidated and well sorted
sandpacks. The data are highly scattered. The R2 for the regression line for
the absolute permeability - porosity correlation is only 0.1742, which is a very
weak correlation. The R2 of 0.2305 for the correlation of the effective
3-68
permeability at irreducible water saturation versus porosity is higher than for
the absolute permeability. Thus, the effective permeability in the presence of
irreducible water saturation correlates better with porosity than the absolute
permeability of the medium. This observation is consistent with the
observation of the author of Figure 3.27.
Figure 3.28. Permeability - porosity correlations for unconsolidated
sandpacks (from Peters, 1979).
3-69
3.8 CAPILLARY TUBE MODELS OF POROUS MEDIA
3.8.1 Carman-Kozeny Equation
Because rock pore space is made up of a network of interconnected,
tortuous, flow conduits, attempts have been made to calculate permeability by
modeling pore-level flow as if through cylindrical capillary tubes. Such
models are sometimes termed “bundle-of-capillary-tubes” models and can
provide some insights into permeability - porosity relationships and the effect
of grain size on permeability. Figure 2.29 shows such a porous medium model
with n cylindrical and tortuous pores (capillary tubes). The Hagen-Poiseuille’s
law for steady flow through a single tortuous, circular capillary tube of radius
r is
( )4 41 2
8 8ie e
p pr r pqL L
π πμ μ
− Δ= = (3.111)
where
q1 = volumetric flow rate for a single capillary tube, cm3/s
r = the radius of the capillary tubes, cm
ΔP = pressure drop across the medium, dynes/cm2
μ = fluid viscosity, poise
Le = tortuous length of the capillary tube, cm
For n capillary tubes,
( )4 41 2
8 8Te e
p pn r n r pqL L
π πμ μ
− Δ= = (3.112)
where qT is the total volumetric flow rate for the porous medium. The porosity
of the medium is given by
3-70
2
p e
b T
V n r LV A L
πφ = = (3.113)
Figure 3.29. Bundle of capillary tubes model of a porous medium with tortuous pores.
where AT is the total cross-sectional area and L is the length of the porous
medium. Solving Eq.(3.113) for AT gives
2
eT
Ln rAL
πφ
⎛ ⎞= ⎜ ⎟⎝ ⎠
(3.114)
The integrated form of Darcy's law for single phase flow through the medium
is given by
TT
kA PqLμ
Δ= (3.115)
A comparison of Eq.(3.115) with Eq.(3.112) shows that the permeability of the
bundle of capillary tube model is given by
3-71
4
8 T e
n r LkA L
π ⎛ ⎞= ⎜ ⎟
⎝ ⎠ (3.116)
Substituting Eq.(3.114) into (3.116) gives the permeability for the model as
( )
2
28 /e
rkL Lφ
= (3.117)
The tortuosity of the porous medium is defined as
2
eLL
τ ⎛ ⎞= ⎜ ⎟⎝ ⎠
(3.118)
Thus, the permeability is given by
2
8rk φτ
= (3.119)
Eq.(3.119) shows once again that permeability is proportional to the square of
a characteristic dimension of the porous medium. In addition, it shows that
the permeability is also proportional to the porosity of the medium. The
tortuosity, τ, is a parameter that is greater than 1. Note that some authors
define tortuosity as (Le/L) instead of (Le/L)2 as defined here.
To enable the bundle of capillary tubes model to describe a porous
medium made of granular material, we introduce specific surface area into the
model. Let Sp be the wetted surface area of the pores per unit pore volume of
the porous medium. For cylindrical pores,
2
2 2ep
e
nrLSnr L rπ
π= = (3.120)
Substituting for r in Eq.(3.119) using Eq.(3.120) gives
22 p
kS
φτ
= (3.121)
3-72
What if the pore is not circular in cross-section? In that case, the 2 in
Eq.(3.121) can be replaced by an empirical numerical factor, ko , to give
2o p
kk S
φτ
= (3.122)
Eq.(3.122) is one version of the Carman-Kozeny equation. The factor, koτ, is
known as the Kozeny constant. Based on laboratory measurements, Carman
suggested that for granular material, the Kozeny constant, koτ, is equal to 5.
Thus, to estimate the permeability of such materials, Eq.(3.122) can be
written as
25 p
kSφ
= (3.123)
The permeability in Eq.(3.123) is in units of length squared. It can be easily
converted to darcy or millidarcy using Eq.(3.13).
Other versions of the Carman-Kozeny equation can be obtained
depending on how the specific surface area is defined. Let S be the wetted
surface area per unit bulk volume of the porous medium. Then
pS Sφ= (3.124)
Substituting Eq.(3.124) into (3.122) gives the following alternative version of
the Carman-Kozeny equation:
3
2o
kk S
φτ
= (3.125)
Eq.(3.125) shows that the permeability is proportional to the porosity cubed.
Let Ss be the wetted surface area per unit grain volume of the porous
medium. S and Ss are related by
( )1 sS Sφ= − (3.126)
3-73
Substituting Eq.(3.126) into (3.122) gives yet another version of the Carman-
Kozeny equation as
( )
3
22 1o s
kk S
φτ φ
=−
(3.127)
The three versions of the Carman-Kozeny equation, Eqs.(3.122), (3.125) and
(3.127) show that permeability is inversely proportional to the square of the
specific surface area of the grains. Since small grains have a higher specific
surface area than large grains, a medium composed of small grains will have
a lower permeability than one composed of large grains, a conclusion we
arrived at previously. All three versions of the Carman-Kozeny equation will
give the same permeability. Which one to use depends on what information is
on hand.
Two more versions of the Carman-Kozeny equation can be obtained by
introducing the concept of hydraulic radius into the model and introducing
the grain diameter in the case of granular material. The hydraulic radius, rH,
is defined as
Volume of Pore 1Wetted Surface Area of PoreH
p
rS
= = (3.128)
For cylindrical pores,
2
2 2e
He
n r L rrnrL
ππ
= = (3.129)
Substituting Eq.(3.129) into (3.119) gives
2
2Hrk φτ
= (3.130)
It is unfortunate that the hydraulic radius of a circular pore is equal to (r/2)
instead of r.
3-74
For granular material with spherical grains of diameter, Dp,
( )( )
2
3
4 / 2 64 / 23
ps
pp
DS
DD
π
π= = (3.131)
Substituting Eq.(3.131) into (3.127) gives
( )
2 3
236 1p
o
Dk
k
φ
τ φ=
− (3.132)
If ko = 2 is substituted into Eq.(3.132), then
( )
2 3
272 1pD
kφ
τ φ=
− (3.133)
Eq.(3.133) suggests that permeability is proportional to the square of the
grain size, an assertion we have encountered before.
To apply any of the Carman-Kozeny equations to estimate permeability,
the porosity, specific surface area and the Kozeny constant must be known.
Several techniques have been proposed for the determination of the specific
surface area of porous media. These include (1) a statistical method, (2)
adsorption methods, (3) the heat of wetting method and (4) a method based on
fluid flow developed by Kozeny.
Example 3.3
Estimate the permeability of the idealized petroleum reservoir of Example 3.2.
Solution to Example 3.3
Since the specific surface area with respect to bulk volume (S) is available
from Example 3.2, Eq.(3.125), which contains S is the most convenient
version of the Carman-Kozeny equation to use for the permeability
esstimation. Substituting numerical values into Eq.(3.125) gives
3-75
( )( )( ) ( )
3 8 28 2
2 9 2
0.259 2.7467 102.7467 10 2.78 9.869 10 /5 355.68
x cmk x cm darcysx cm darcy
−−
−= = = =
where the Kozeny constant of 5 has been used.
3.8.2 Tortuosity
Based on the bundle of capillary tube model, we see that tortuosity is a
geometric property of the porous medium that reflects the length of the flow
path at the pore level as the fluid flows around the grain obstacles relative to
the length of the porous medium. Therefore, the lower the porosity, the higher
the tortuosity should be. Winsauer et al. (1952) have measured the tortuosity
of sandstones along with other properties. The tortuosity was measured
electrically based on the analogy between the flow of electrical current and
fluid flow. Their results are shown in Table 3.10. They defined tortuosity as
(Le/L). Their original data have been squared to obtain the more current
definition of tortuosity.
Table 3.10: Tortuosity and Other Data from Winsauer et al., 1952.
Porosity Permeability Resistivity Tortuosity
Core φ k Factor τ
No % md F (Le/L) (Le/L)2
1 17.0 90 23.3 2.30 5.29
2 14.7 7 51.0 3.30 10.89
3 6.7 4 67.0 3.20 10.24
4 17.6 220 16.6 2.00 4.00
5 26.3 1920 8.6 1.60 2.56
6 25.6 4400 9.4 1.60 2.56
7 13.9 145 33.0 2.40 5.76
8 18.6 25 22.9 2.30 5.29
9 18.8 410 18.6 2.00 4.00
3-76
10 16.1 3 42.0 3.20 10.24
11 15.0 9 41.0 2.90 8.41
12 22.1 200 13.1 1.90 3.61
13 20.6 36 16.6 2.10 4.41
14 30.7 70 8.4 1.70 2.89
15 16.4 330 21.1 2.10 4.41
16 18.8 98 19.3 2.20 4.84
17 24.8 1560 10.8 1.80 3.24
18 19.1 36 17.2 2.30 5.29
19 29.8 1180 8.4 1.75 3.06
20 27.1 3200 11.7 2.05 4.20
21 28.2 2100 10.9 2.05 4.20
22 19.4 8 24.0 2.50 6.25
23 19.7 18 20.8 2.35 5.52
24 31.5 2200 6.9 1.50 2.25
25 19.3 19 24.4 2.70 7.29
26 27.3 88 12.4 2.20 4.84
27 25.1 370 11.6 1.97 3.88
28 15.0 115 37.3 2.90 8.41
29 18.4 130 19.0 2.00 4.00
30
31 39.5 4.7 1.37 1.88
Figure 3.30 shows the correlation between tortuosity and porosity. As
expected, there is a negative correlation between the two variables. The lower
the porosity, the longer is the length of the flow path that the fluid particles
must take to flow from one end of the porous medium to the other. The longer
the flow path, the larger the tortuosity of the medium.
Figure 3.31 shows the correlation between tortuosity and the formation
resistivity factor. As expected, there is a strong positive correlation between
the two variables. Resistivity factor represents the obstacle to the flow of
electric current in the medium whereas tortuosity represents the obstacle to
the flow of fluid in the medium. Since there is a close relationship between the
3-77
flow path for electric current and for fluid flow, a property that obstructs one
will also obstruct the other. Hence, tortuosity and formation resistivity factor
should be positively correlated.
Figure 3.30. Tortuosity-porosity correlation (from Winsauer et al., 1952).
3-78
Figure 3.31. Tortuosity-formation resistivity factor correlation (from Winsauer
et al., 1952).
An approximate relationship between tortuosity and formation
resistivity factor can be derived as follows. Consider the flow of electric
current through a core sample of porosity φ, length L and cross-sectional area
A saturated with water of resistivity Rw. The electrical resistance of the core is
given by
o oLr RA
⎛ ⎞= ⎜ ⎟⎝ ⎠
(3.134)
where Ro is the resistivity of the core fully saturated with the water. The
electrical current in the core is conducted by the water alone since the rock
minerals are nonconductors. Let us replace the core with the same volume of
water in a test tube such that the resistance of the water is equal to that of
the saturated core. Since the electrical current in the core follows a tortuous
path in flowing from one end to the other, the length of the equivalent water
3-79
circuit Le will be longer than L. Let Ae be the cross-sectional area of the
equivalent water circuit. Then
e eA L ALφ= (3.135)
The equivalent water circuit must have the same volume as the water in the
core to maintain the same salinity. The resistance of the equivalent water
circuit is given by
ew w
e
Lr RA
⎛ ⎞= ⎜ ⎟
⎝ ⎠ (3.136)
Substituting Eq.(3.135) into (3.136) gives
2e
w wLr RALφ
⎛ ⎞= ⎜ ⎟
⎝ ⎠ (3.137)
Since the equivalent water circuit has the same resistance as the saturated
core, Eqs.(3.134) and (3.137) give
2e
o wLLR R
A ALφ⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠ (3.138)
Eq.(3.138) can be rearranged as
2 1o e
w
R LR L
τφ φ
⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
(3.139)
Thus, an approximate relationship between turtuosity and the formation
resistivity factor is
Fτ φ= (3.140)
3.8.3 Calculation of Permeability from Pore Size Distribution
The bundle of capillary tubes model presented in Section 3.7.1 assumed
that the n capillary tubes had the same pore size. Here, we extend the model
to include an arbitrary pore size distribution. Given the probability density
3-80
function (pdf) for the pore size (pore diameter) distribution, we wish to
calculate the permeability of the bundle of capillary tubes model.
Consider the probability density function for pore diameter distribution
as shown in Figure 3.32. The probability density function satisfies
( )0
1f dδ δ∞
=∫ (3.141)
Let nδ be the number of pores with diameter between δ and δ+dδ and n be the
total number of pores. Then
( )n nf dδ δ δ= (3.142)
Figure 3.32. Probability density function for pore size distribution.
The cross-sectional area occupied by pores with diameter between δ and δ+dδ
is given by
3-81
( )2
4cdA nf dδπ δ δ⎛ ⎞
= ⎜ ⎟⎝ ⎠
(3.143)
The cross-sectional area occupied by all the pores is obtained by integrating
Eq.(3.143) as
( )2
2
04 4cn nRA f dπ πδ δ δ
∞= =∫ (3.144)
where 2R is a constant defined as
( )2 2
0R f dδ δ δ
∞= ∫ (3.145)
The cross-sectional area occupied by all the pores is related to the cross-
sectional area of the porous medium by the porosity as
c TA A φ= (3.146)
Substituting Eq.(3.144) into (3.146) and solving for n, one obtains
2
4 TAnR
φπ
= (3.147)
The volumetric flow rate for pores with diameter between δ and δ+dδ is
obtained from Hagen-Poiseuille's law as
( )4
128Te
Pdq nf dL
πδ δ δμ
Δ= ⎡ ⎤⎣ ⎦ (3.148)
The total volumetric flow rate for the porous medium is obtained by
integrating Eq.(3.148) as
( ) 4
0128Te
n Pq f dL
π δ δ δμ
∞Δ= ∫ (3.149)
Substituting Eq.(3.147) into (3.149) gives
3-82
( ) 42 032
TT
e
A Pq f dL R
φ δ δ δμ
∞Δ= ∫ (3.150)
A comparison of Eq.(3.150) with Darcy's law, Eq.(3.115), gives the
permeability of the bundle of capillary tube model as
( ) ( ) 4
2 032 /e
k f dL L R
φ δ δ δ∞
= ∫ (3.151)
Substituting Eq.(3.118) into (3.151) gives
( ) ( ) 4
2 032k f d
Rφ δ δ δτ
∞= ∫ (3.152)
Substituting Eq.(3.145) into (3.152) gives the final result
( )
( )( )
4
0
2
032
f dk
f d
δ δ δφτ δ δ δ
∞
∞
⎡ ⎤⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
∫∫
(3.153)
Example 3.4
The probability density function for the pore size distribution of a bundle of
capillary tubes model is given by a triangular distribution as shown in Figure
3.33. The porosity of the medium is 15%. If the minimum pore diameter (δ1) =
0 μm, the most likely pore diameter (δ2) = 8 μm and the largest pore diameter
(δ3) = 10 μm, calculate the permeability of the medium. Assume a tortuosity of
1.0.
3-83
Figure 3.33. Probability density function for pore size distribution for Example
Solution to Example 3.4
The required permeability is given by Eq.(3.153). All we need to do is to
evaluate the two integrals on the right of the equation for the triangular
probability density function. Eq.(3.153) then takes the form
( )
( ) ( )
( ) ( )
2 3
1 2
2 3
1 2
4 41 2
2 21 2
32
f d f dk
f d f d
δ δ
δ δδ δ
δ δ
δ δ δ δ δ δφτ δ δ δ δ δ δ
⎡ ⎤+⎢ ⎥= ⎢ ⎥+⎢ ⎥⎣ ⎦
∫ ∫∫ ∫
(3.154)
Eqs.(2.82) and (2.86) give
( ) ( )( )( )
11
3 1 2 1
2f
δ δδ
δ δ δ δ−
=− −
(3.155)
( ) ( )( )( )
32
3 1 3 2
2f
δ δδ
δ δ δ δ−
=− −
(3.156)
Substituting Eqs.(3.155) and (3.156) into Eq.(3.154) and performing the
integrations gives
3-84
3 13 22
1 13 2
0.15 2.241 10 2.583 10 0.2617 261.74.067 10 9.869 10 /32 1.0
x x mk m darcy mdx x m darcy
μ−
−
⎛ ⎞= = = =⎜ ⎟
⎝ ⎠
3.9 STEADY STATE FLOW THROUGH FRACTURES
To increase the productivities of certain reservoirs, it is often necessary
to hydraulically fracture the reservoirs. Consider the single vertical fracture
shown in Figure 3.34. It can be shown that Hagen-Poiseuille’s law for steady
state flow through the fracture of width w in cgs units is given by
2
12w A Pq
LμΔ
= (3.157)
where A = wh is the cross-sectional area of the fracture. A comparison of
Eq.(3.157) with Darcy’s law, Eq.(3.115), shows that the permeability of the
fracture is given by
2
2
12wk cm= (3.158)
Example 3.5
Calculate the permeability of a fracture of width 2 millimeters.
Solution to Example 3.5
The fracture permeability is given by Eq.(3.158) as
( )2 25
9 2
0.20 1 3.378 1012 9.869 10 /
cmk x x darcys
x cm darcy−= =
The permeability of the fracture is enormous and therein lies the benefit of
fracturing for the improvement of the productivity of a well.
3-85
Figure 3.34. Flow through a fracture (Flow between parallel plates).
3.10 AVERAGING PERMEABILITY DATA
Because of heterogeneity, different portions of the same reservoir may
have different permeabilities. It becomes necessary to be able to estimate the
apparent or average permeability of various permeability combinations.
For linear beds in series (Figure 3.35), the volumetric flow rate is the
same for each bed and the total pressure drop is equal to the sum of the
pressure drop across each bed. For n beds in series, the average permeability
is given by
1
1 1
n
iin n
i i
i ii i
LLk
L Lk k
=
= =
= =∑
∑ ∑ (3.159)
where L Is total length of the porous medium in the flow direction, Li is the
length of bed i and ki is the permeability of bed i.
3-86
Figure 3.35. Linear beds in series For radial beds in series (Figure 3.36), the volumetric flow rate is the
same for each bed and the total pressure drop is equal to the sum of the
pressure drop across each bed. For n beds in series, the average permeability
is given by
( )( )1
1
ln /ln /
e wn
i i
i i
r rk
r rk
−
=
=
∑ (3.160)
where re is the well's drainage radius and r0 is equal to the wellbore radius, rw
Figure 3.36. Radial beds in series
3-87
For linear beds in parallel (Figure 3.37), the total volumetric flow rate is
the sum of the flow rate of each bed and the pressure drop is the same for all
the beds. For n beds in parallel, the average permeability is given by
1
1
n
i ii
n
ii
k Ak
A
=
=
=∑
∑ (3.161)
where Ai is the area of bed i and ki is the permeability of bed i. If the beds
have the same width, then
1
1
n
i ii
n
ii
k hk
h
=
=
=∑
∑ (3.162)
Figure 3.37. Linear beds in parallel.
For radial beds in parallel (Figure 3.38), the total volumetric flow rate is
the sum of the flow rate of each bed and the pressure drop is the same for all
the beds. For n beds in parallel, the average permeability is given by
Eq.(3.162).
3-88
Figure 3.38. Radial beds in parallel.
3.11 DARCY'S LAW FOR INCLINED FLOW
For inclined flow, Darcy's Law in differential form can be written as
61.0133 10kA dP g dzq
ds x dsρ
μ⎛ ⎞= − ±⎜ ⎟⎝ ⎠
(3.163)
where
q = volumetric flow rate, cm3/s
k = absolute permeability of the porous medium, darcy
A = total area of the medium normal to the flow direction, cm2
μ = fluid viscosity, centipoise
P = absolute pressure, atm
s = distance in the direction of flow and is always positive, cm
dPds
= pressure gradient along s, atm/cm
ρ = fluid density, gm/cm3
g = acceleration of gravity, 981 cm/s2
1 atm = 1.0133x106 dynes/cm2
The sign convention for applying Darcy's law for inclined flow is as follows. If
the z direction is positive upwards as shown in Figure 3.39, then the positive
3-89
sign is used in the term in parenthesis in Eq.(3.163) and the equation
becomes
Figure 3.39. Coordinate system for applying Darcy's law for z positive upwards.
61.0133 10kA dP g dzq
ds x dsρ
μ⎛ ⎞= − +⎜ ⎟⎝ ⎠
(3.164)
On the other hand, if the z coordinate is positive downwards as shown in
Figure 3.40, then the negative sign is used in the term in parenthesis in
Eq.(3.163) to obtain
61.0133 10kA dP g dzq
ds x dsρ
μ⎛ ⎞= − −⎜ ⎟⎝ ⎠
(3.165)
3-90
Figure 3.40. Coordinate system for applying Darcy's law for z positive downwards.
dzds
in Eqs.(3.163) to (3.165) could be written as sinθ, where θ is the angle the
porous medium along s makes with the horizontal plane. Alternatively, it
could be written as cosα, where α is the angle that the porous medium makes
with the upward vertical z axis and is equal to 2π θ⎛ ⎞−⎜ ⎟
⎝ ⎠.
Eq.(3.163) can be written in oilfield units as
0.001127 0.433kA dP dzqds ds
γμ
⎛ ⎞= − ±⎜ ⎟⎝ ⎠
(3.166)
where γ is the specific gravity of the fluid and 0.433 is the pressure gradient of
fresh water in psi/ft.
Eq.(3.163) also can be written in terms of a velocity potential as
kA dqdsμΦ
= − (3.167)
3-91
where the velocity potential is defined by
61.0133 10gzPx
ρΦ = ± (3.168)
in Darcy units or
0.433P zγΦ = ± (3.169)
in oilfield units. Eq.(3.167) shows that flow will occur in the direction of
decreasing velocity potential. Note that for inclined systems, flow does not
necessarily occur in the direction of decreasing pressure. For example, for a
static system in which the porous medium is oriented vertically, the pressure
at the top of the medium is less than at the bottom and yet there is no flow.
Eq.(3.156) may also be written as
6
6
1.0133 101.0133 10s
q k g d x Pv zA x ds g
ρμ ρ
⎛ ⎞= = − ±⎜ ⎟
⎝ ⎠ (3.170)
where vs is known as the Darcy velocity or superficial velocity. An interstitial
velocity can be defined as
si
vvφ
= (3.171)
Eq.(3.170) can be further written in terms of the hydraulic conductivity, K,
and hydraulic or piezometric head, h, as
sq dhv KA ds
= = − (3.172)
where
61.0133 10k gK
xρ
μ= (3.173)
and
3-92
Ph zgρ
= ± (3.174)
Eq.(3.172) is the form of Darcy's law normally used in the groundwater
community. Note that hydraulic head has dimension of length and the
hydraulic conductivity has dimension of length/time. Possible units of
hydraulic conductivity are cm/s, m/s, m/d, US gal/day ft2 and US gal/min
ft2. Useful conversion factors are as follows:
2 5 21 gal/day ft 4.75 10 cm/s 4.08 10 /US x x m d− −= = (3.175)
41 9.613 10 cm/s (for water at 20 )darcy x C−= (3.176)
2 21 1.4156 10 gal/min ft (for water at 20 )darcy x US C−= (3.177)
Table 3.11 shows typical hydraulic conductivity values for various rock types
and soils. Figure 3.41 shows the range of hydraulic conductivity and
permeability for various rock types and soils. Clearly, rocks and soils have a
wide range of permeability and hydraulic conductivity values.
Sometimes, it is easier to solve single phase flow problems in terms of
hydraulic head (piezometric head) rather than in terms of pressure. This is
especially true for inclined or vertical flow in which gravity plays a role.
Figure 3.42 shows the definition of the hydraulic head (piezometric head) for a
point O in an inclined porous medium located above a datum level at which z
is zero. The hydraulic head is given by
h zψ= + (3.178)
3-93
Table 3.11: Typical Hydraulic Conductivity for Various Rock Types
3-94
Figure 3.41. Range of permeability and hydraulic conductivity of various rock types and soils.
where h is the hydraulic head, z is the height of O above (or below) an
arbitrary datum and ψ is the pressure head. It should be emphasized that z in
Eq.(3.178) is not a cartesian coordinate in the flow direction but rather the
elevation of the point P above or below a reference datum. In a sense, z is a
gravity head that reflects the potential energy of the fluid at O in the earth’s
gravitational field. The datum at which z = 0 is arbitrary and can be selected
anywhere in the system. If the point O is below the datum, then z is negative.
The pressure head is given by
Pg
ψρ
= (3.179)
3-95
It should be observed in Figure 3.42 that the hydraulic head at point O is
given by the elevation of a liquid manometer at point O above (or below) the
arbitrary datum.
Figure 3.42. Definition of hydraulic head h, pressure head Ψ and elevation head z for a laboratory manometer.
Consider the flow in an inclined porous medium shown in Figure 3.43.
Darcy’s law for 1D flow is given by
dhq KA h KAds
= − ∇ = − (3.180)
Examination of the manometer levels (hydraulic heads) immediately shows
the direction of flow. Flow always occurs from a high hydraulic head to a low
hydraulic head. Eq.(3.180) may be integrated to obtain
2 1
2 1
h hq KAs s
⎛ ⎞−= − ⎜ ⎟−⎝ ⎠
(3.181)
Eq.(3.174) can be rewritten as
1 2
2 1
h h hq KA KAs s L
⎛ ⎞− Δ⎛ ⎞= =⎜ ⎟ ⎜ ⎟− Δ⎝ ⎠⎝ ⎠ (3.182)
3-96
Figure 3.43. Darcy’s experiment for inclined flow.
where Δh is the positive difference in the hydraulic heads at points 1 and 2
and ΔL is the positive length of the porous medium between points 1 and 2 in
the flow direction. It should be observed that the elevations of points 1 and 2
(z1 and z2) do not appear explicitly in Eq.(3.182) although they do appear
implicitly in the hydraulic heads. However, Eq.(3.182) can be rewritten such
that z1 and z2 appear explicitly as follows. From Figure 3.43,
1 2cosL z zαΔ = − (3.183)
where α is the angle the porous medium makes with the vertical direction.
Substituting Eq.(3.183) into Eq.(3.182) gives
1 2
1 2
cosh hq KAz z
α⎛ ⎞−
= ⎜ ⎟−⎝ ⎠ (3.184)
3-97
In the special case in which the porous medium is vertical, α is zero, cosα is
1.0 and z1 – z2 = ΔL. Substituting these values into Eq.(3.184) gives
hq KAL
Δ⎛ ⎞= ⎜ ⎟Δ⎝ ⎠ (3.185)
which is identical to Eq.(3.182). Thus, Eq.(3.182) applies for all angles of
inclination. The hydraulic conductivity is related to the permeability of the
porous medium in consistent units by
k gK ρμ
= (3.186)
Figure 3.44 shows schematic diagrams of two types of liquid
permeameters normally used to measure permeability in groundwater
hydrology. Figure 3.44 (a) is a constant head permeameter whereas Figure
3.44 (b) is a falling head permeameter. Using the principles of inclined flow
outline above, the working equations for the permeameters can easily be
derived as follows. For the constant head permeameter, Eq.(3.182) gives
h k g hq KA AL L
ρμ
⎛ ⎞= = ⎜ ⎟
⎝ ⎠ (3.187)
in consistent units, where A is the cross-sectional area of the porous medium.
For the falling head permeameter, Darcy's law at time t is
hq KAL
= (3.188)
The volumetric balance for flow in the manometer gives
dhq adt
= − (3.189)
where a is the cross-sectional area of the manometer. Substituting Eq.(3.187)
into (3.186) gives the differential equation for the falling head, h, as
3-98
Figure 3.44. Schematic diagram of liquid permeameters; (a) constant head
permeameter, (b) falling head permeameter.
dh KA hdt aL
⎛ ⎞= −⎜ ⎟⎝ ⎠
(3.190)
with the initial condition
( )0 oh h= (3.191)
Eq.(3.190) along with (3.191) can easily be solved to obtain the working
equation for the falling head permeameter as
3-99
lno
h KA k gAt th aL aL
ρμ
⎛ ⎞ ⎛ ⎞⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
(3.192)
in consistent units.
3.12 VALIDITY OF DARCY'S LAW
Darcy's law is valid for slow, laminar flow in porous media without
chemical reaction. Figure 3.45 shows the Fanning friction factor versus
Reynolds number for single phase flow in a porous medium. For flow through
porous media, Reynolds number is defined as
Re pv Dρμ
= (3.193)
where
v = Darcy velocity, cm/s
Dp = mean grain diameter of the granular porous medium, cm
ρ = fluid density, gm/cm3
μ = fluid viscosity, poise
Figure 3.45. Fanning friction factor factor for flow in porous media (Rose,
1945)
3-100
For laminar flow, in which Darcy's law is valid, the friction factor versus
Reynolds number is given by the line shown in Figure 3.45 having the
equation
1000Re
f = (3.194)
It can seen in the figure that the experimental data begin to deviate from the
line at a Reynolds number of about 1.0. Thus, Darcy's law is valid for flow in
porous media for Reynolds number up to about 1. At Reynolds numbers
greater than 1, Darcy's law is no longer valid. Therefore, flow at Reynolds
number greater than 1 can be characterized as non-darcy flow.
Based on analogy with flow in pipes, the flow regimes for non-darcy flow
can be classified as shown in Figure 3.46 as transition flow for Reynolds
number in the range 1 to 100, and turbulent flow for Reynolds number
greater than 100.
Figure 3.46. Classification of flow in porous media.
3-101
3.13 NON-DARCY FLOW
Darcy’s law is inadequate for modeling high velocity flow in porous
media such as high velocity gas flow near the wellbore in a natural gas
reservoir in which the Reynolds number is greater than 1. For this non-darcy
flow, Forchheimer has proposed a flow equation of the form
2dP v vdx k
μ βρ− = + (3.195)
where the variables are in Darcy units. The first term on the right-side of
Eq.(3.195) gives the viscous component of the pressure gradient (Darcy flow)
whereas the second term gives the inertial-turbulent component of the
pressure gradient, which constitutes the non-darcy flow effect. Because the
non-darcy flow effect is most severe near the wellbore, this additional effect is
usually incorporated into the gas flow equation as a rate-dependent skin
factor. The velocity coefficient (β) is a function of the permeability and porosity
of the porous medium and is independent of rate. Figure 3.47 shows a
correlation of the velocity coefficient with permeability from core analysis.
Consider radial steady state flow of a real gas from the reservoir into the
well. For this case, Eq.(3.195) can be written as
2dP v vdr k
μ βρ= + (3.196)
Let us multiply Eq.(3.196) by (2p/μZ) and integrate to get
22 2 2e e e
wf w w
P r r
P r r
P P v PdP dr v drZ Z k Z
μ βρμ μ μ
= +∫ ∫ ∫ (3.197)
The gas density in Eq.(3.197) is at reservoir conditions. We would like to
evaluate the density at standard conditions. The density at reservoir
conditions and at standard conditions are related by the gas formation
volume factor as
3-102
Figure 3.47. Velocity coefficient in Forchheimer equation (Firoozabadi and Katz, 1979.
sc
gBρρ = (3.198)
where Bg is the gas formation volume factor and ρsc is the gas density at
standard conditions. For real gas flow, the gas formation volume factor is
given by
scg
sc
P ZTBT P
= (3.199)
The Darcy velocity at radius r is given by
3-103
2
qvrhπ
= (3.200)
The equation of state for real gas flow is given by
sc sc
sc
P qPqZT Z T
= (3.201)
Substituting Eqs.(3.198) to (3.201) into (3.197) gives
2
2 22
e wf e e
o o w w
P P r rsc sc sc sc sc scP P r r
sc sc
q P T q P T kqP P dr drdP dPZ Z khT r khT h r
βρμ μ π π πμ
⎛ ⎞⎛ ⎞− = + ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠∫ ∫ ∫ ∫ (3.202)
where Po is an arbitrary base pressure. In Eq.(3.202), the left–hand side of
Eq.(3.197) has been split into two integrals for convenience. Integration of
Eq.(3.202) gives
( ) ( ) 1 1ln2
sc sc e sc sc sc sce w
sc w sc w e
q P T r q P T kqM P M PkhT r khT h r r
βρπ π πμ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞− = + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (3.203)
where M(P) is the real gas potential or real gas pseudo pressure. At field scale,
(1/re) is normally negligibly small compared to (1/rw). With this
approximation, Eq.(3.203) becomes
( ) ( ) 1ln2
sc sc e sc sc sc sce wf
sc w sc w
q P T r q P T kqM P M PkhT r khT h r
βρπ π πμ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞− = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (3.204)
Eq.(3.204) can be written in dimensionless form as
ln eD sc
w
rP Dqr
⎛ ⎞= +⎜ ⎟
⎝ ⎠ (3.205)
where the dimensionless pressure in terms of M(P) is defined as
( ) ( )e wf
Dsc sc
sc
M P M PP q P T
khTπ
−= (3.206)
3-104
and a non-darcy flow coefficient is defined as
2
sc
w
kDhr
βρπμ
= (3.207)
Eq.(3.205) can be modified to include the mechanical skin factor S* to give
*ln eD sc
w
rP S Dqr
⎛ ⎞= + +⎜ ⎟
⎝ ⎠ (3.208)
Eq.(3.208) can be rewritten in the form of an inflow performance relationship
in Darcy units as
( ) ( )
*ln
e wfscsc
sc esc
w
M P M PkhTqP T r S Dq
r
π⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(3.209)
In oilfield units, Eq.(3.209) becomes
( ) ( )
*1422ln
e wfsc
esc
w
M P M PkhqT r S Dq
r
⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(3.210)
It should be noted that the skin factor normally determined from well
test analysis is the total skin factor given by
*scS S Dq= + (3.211)
Therefore, the total skin factor determined from a gas well test should be
investigated further to determine the contribution of the non-darcy flow
component. The non-darcy flow component of the total skin factor cannot be
eliminated by well stimulation. It can be estimated by conducting drawdown
tests at two different flow rates.
3-105
It can be shown that the additional potential drop due to non-darcy flow
occurs mostly near the wellbore. This additional potential drop at any radius r
can be obtained from Eq.(3.203) as
( ) 1 12
sc sc sc sc
sc w
q P T kqM rkhT h r r
βρπ πμ
⎛ ⎞ ⎛ ⎞⎛ ⎞Δ = −⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠ (3.212)
At r =re,
( ) 12
sc sc sc sce
sc w
q P T kqM rkhT h r
βρπ πμ
⎛ ⎞ ⎛ ⎞⎛ ⎞Δ = ⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠ (3.213)
where (1/re) is negligibly small compared to (1/rw). Thus,
( )( ) 1 w
e
rM rM r r
Δ= −
Δ (3.214)
Table 3.12 shows the variation of Δm(r)/Δm(re) with r as calculated from
Eq.(3.214). It can be observed that 90% of the total non-darcy flow potential
drop occurs within r = 10rw. Thus, if rw is 0.3 ft, then 90% of the total non-
darcy flow potential drop occurs within 3 ft of the well. This is the justification
for treating the non-darcy flow effect as a flow-dependent skin factor.
Table 3.12: Non-Darcy Potential Drop
w
rr
( )( )
e
M rM r
ΔΔ
1 0 2 0.50 5 0.80 10 0.90
3-106
3.14 DARCY’S LAW FOR ANISOTROPIC POROUS MEDIA
3.14.1 Definition of Homogeneity and Anisotropy
A medium is homogeneous with respect to a certain property if that
property is independent of position within the medium. Otherwise, the
medium is heterogeneous. A medium is isotropic with respect to a certain
property if that property is independent of direction within the medium. If at a
point in the medium, a property of the medium varies with direction, the
medium is said to be anisotropic with respect to that property. Figure 3.48
shows the principal values of permeability anisotropy at two locations (x1, z1)
and (x2, z2) in a vertical cross-section of a porous medium. Although arrows
Figure 3.48. Diagrams showing heterogeneity and anisotropy (Freeze and Cherry, 1979).
3-107
and arrow lengths have been used to indicate the directions and magnitudes
of the principal values of the permeability anisotropy, they should not be
misinterpreted as vectors. Permeability is a tensor, not a vector. The special
form of the permeability tensor shown in the figure is
( )0
,0x
z
kk x z
k⎡ ⎤
= ⎢ ⎥⎣ ⎦
(3.215)
and not
( ), x
z
kk x z
k⎡ ⎤
= ⎢ ⎥⎣ ⎦
(3.216)
3.14.2 Darcy's Law for Homogeneous and Anisotropic Medium
For a homogeneous and anisotropic porous medium, Darcy’s Law is
given by
.kvμ
= − ∇Φ (3.217)
Eq.(3.217) can be expanded in 3D Cartesian coordinates as
1x xx xy xz
y yx yy yz
z zx zy zz
xv k k kv k k k
yv k k k
z
μ
⎡ ⎤∂Φ⎢ ⎥∂⎡ ⎤ ⎢ ⎥⎡ ⎤∂Φ⎢ ⎥ ⎢ ⎥⎢ ⎥ = − ⎢ ⎥ ⎢ ⎥⎢ ⎥ ∂⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ∂Φ⎢ ⎥
⎢ ⎥∂⎣ ⎦
(3.218)
The permeability tensor given by
( ), ,xx xy xz
yx yy yz
zx zy zz
k k kk x y z k k k
k k k
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
(3.219)
3-108
is a second rank tensor with nine elements. The tensor is symmetric and as
such kxy = kyx, kxx = kzx and kyz = kzy. The corresponding equations for 2D flow
are
1x xx xy
y yx yy
v k k xv k k
yμ
∂Φ⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤ ∂⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ∂Φ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥∂⎣ ⎦
(3.220)
and
( ), xx xy
yx yy
k kk x y
k k⎡ ⎤
= ⎢ ⎥⎣ ⎦
(3.221)
Darcy's law can also be written in terms of hydraulic conductivity and
hydraulic head as
.v K h= − ∇ (3.222)
where K is the hydraulic conductivity tensor. All the properties of the
permeability tensor are applicable to the hydraulic conductivity tensor.
An important property of a symmetric tensor is that by a suitable
change of axes, it is possible to transform the full tensor into another tensor
consisting of only diagonal terms with the off-diagonal terms being zero. This
is a considerable simplification. The axes that permit this transformation of
the tensor to diagonal form are called the principal axes of the permeability
anisotropy of the porous medium. These principal axes contain the maximum
and minimum values of the directional permeabilities which are the diagonal
terms of the transformed tensor. When viewed in the coordinates of the
principal axes of anisotropy, the permeability (hydraulic conductivity) tensor
in 3D becomes
3-109
( )0 0
, 0 00 0
u
v
z
kk u v k
k
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
(3.223)
where u, v, w are the coordinates in the directions of the principal axes of the
anisotropy. When viewed in the coordinates of the principal axes of the
anisotropy, Darcy's law for homogeneous and anisotropic media in 3D
simplifies to
1
u
u
v v
w
w
kuv
v kv
vk
w
μ
∂Φ⎡ ⎤⎢ ⎥∂⎡ ⎤ ⎢ ⎥
∂Φ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥∂⎢ ⎥ ⎢ ⎥⎣ ⎦ ∂Φ⎢ ⎥
⎢ ⎥∂⎣ ⎦
(3.224)
In this special coordinate system, the Darcy velocity vector and the velocity
potential gradient vector are collinear and the equation can be written as
1u v w u v wv i v j v w k i k j k w
u v uμ∂Φ ∂Φ ∂Φ⎛ ⎞+ + = − + +⎜ ⎟∂ ∂ ∂⎝ ⎠
(3.225)
where i, j and w are unit vectors. The corresponding equations for 2D flow are
( )0
,0u
v
kk u v
k⎡ ⎤
= ⎢ ⎥⎣ ⎦
(3.226)
1 ux
yv
kv uv
kv
μ
∂Φ⎡ ⎤⎢ ⎥⎡ ⎤ ∂= − ⎢ ⎥⎢ ⎥ ∂Φ⎢ ⎥⎣ ⎦⎢ ⎥∂⎣ ⎦
(3.227)
and
1u v u vv i v j k i k j
u vμ∂Φ ∂Φ⎛ ⎞+ = − +⎜ ⎟∂ ∂⎝ ⎠
(3.228)
3-110
Example 3.5
Steady state flow of a single phase liquid occurs in a horizontal, 2D,
homogeneous and anisotropic reservoir, which has a permeability tensor
given by
k(x,y)= 200 –50–50 750
md
The pressure in the reservoir has been measured at three observations wells
whose coordinates are given in Table 3.13. The liquid viscosity is 1 cp. Do the
negative entries in the permeability tensor bother you?
Table 3.13: Data from Observation Wells
Observation
Well
x – Coordinate
(ft)
y – Coordinate
(ft)
Pressure
(psia)
1 0 0 2000
2 0 800 1800
3 800 0 1500
1. Calculate the magnitude (in ft/day) and the direction of the Darcy
velocity for the flow. Give the direction in terms of the angle the Darcy
velocity vector makes with the positive x- axis.
2. Sketch two vectors with one pointing in the flow direction and the other
pointing in the direction of the pressure gradient. What is the angle
between the two vectors in degrees?
3. Sketch the flow field and the pressure gradient field.
Solution to Example 3.5
Figure 3.49 shows the positions of the wells and their bottomhole pressures.
3-111
Figure 3.49. Positions of the observation wells and their bottomhole pressures.
We apply Eq.(3.217) in oilfield units to obtain
( )( )0.001127 5.615 .kv Pμ
= − ∇ (3.229)
For the 2D flow, Eq.(3.229) can be expanded to give
( )( )0.001127 5.615x xx xy
y yx yy
Pv k k x
Pv k ky
μ
∂⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤ ∂⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ∂⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥∂⎣ ⎦
(3.230)
From the given data,
1500 2000 5800 8
Px
∂ − ⎛ ⎞= = −⎜ ⎟∂ ⎝ ⎠ psi/ft
1800 2000 1800 4
Py
∂ − ⎛ ⎞= = −⎜ ⎟∂ ⎝ ⎠ psi/ft
3-112
5 1 0.625 0.2508 4
P i j i j∇ = − − = − −
( ) ( )2 20.625 0.250 0.6731P∇ = − + − = psi/ft
Let the pressure gradient make an angle θ with the negative x-axis. Then
( ) ( ) ( ) ( )( )( )2 20.625 0.250 . 0.625 0.250 1 cosi j i θ− − − = − + −
2 2
0.625cos 0.92850.625 0.250
θ = =+
21.80θ =
The components of the Darcy velocity are given by
( )( )5
200 500.001127 5.615 850 750 11
4
x
y
vv
⎡ ⎤−⎢ ⎥−⎡ ⎤ ⎡ ⎤= − ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎢ ⎥⎣ ⎦ −⎢ ⎥⎣ ⎦
( )( ) ( ) ( )5 10.001127 5.615 200 50 0.71208 4xv ⎡ ⎤⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
ft/D
( )( ) ( ) ( )5 10.001127 5.615 50 750 0.98888 4yv ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
ft/D
The Darcy velocity vector is
0.7120 0.9888v i j= +
( ) ( )2 20.7120 0.9888 1.2185v = + = ft/D
Let the Darcy velocity make an angle α with the positive x-axis.
0.9888tan 1.38880.7120
α = =
3-113
54.24α =
Figure 3.50 shows the magnitudes and the directions of the darcy velocity and
the pressure gradient vectors. The figure shows that the Darcy velocity and
pressure gradient vectors are not collinear for flow in the anisotropic porous
medium.
Figure 3.50. Directions and magnitudes of Darcy velocity and pressure gradient vectors for flow in an anisotropic reservoir.
The angle between the Darcy velocity vector and the pressure gradient vector
is 147.56º. This angle can also be calculated directly from vector calculus as
. . cosv P v P β∇ = ∇
3-114
where β is the angle between the two vectors. Thus,
( ) ( )( )( )
0.7120 0.9888 . 0.625 0.25 0.6922cos 0.84401.2185 0.6731 0.8202
i j i jβ
+ − −= = − = −
147.56β =
Figure 3.51 shows sketches of the flow field and the pressure gradient field.
Figure 3.51. Flow and pressure gradient fields for Example 3.5.
3.14.3 Transformation of Permeability Tensor from one Coordinate System to Another
The objective is to derive the equations for transforming the
permeability tensor (hydraulic conductivity tensor) for an anisotropic reservoir
from one coordinate system to another coordinate system that makes and
angle θ with the first coordinate system, where θ is considered positive for
3-115
anticlockwise rotation. This can be achieved by first establishing the
geometric relationship between the two coordinate systems and then
transforming Darcy’s Law from the first coordinate system to the second
coordinate system. Such a transformation can be used to (1) determine the
principal values of the permeability anisotropy, (2) determine the principal
axes of the permeability anisotropy and (3) examine the flow field in any
coordinate system.
Figure 3.52 shows the relationship between the two coordinate systems.
It should be emphasized that the (u,v) coordinates shown in Figure 3.52 are
not the principal axes of the permeability anisotropy. The two coordinate
systems are related by the following equations:
cos sinx u vθ θ= − (3.231)
sin cosy u vθ θ= + (3.232)
Eqs.(3.231) and (3.232) can be written in matrix form as
cos sinsin cos
x uy v
θ θθ θ
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (3.233)
Using the inverse of the rotation matrix of Eq.(3.233), the relationship
between the two coordinate systems also can be written as
cos sinsin cos
u xv y
θ θθ θ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(3.234)
Darcy’s law in the xy coordinate system is given by
3-116
Figure 3.52. Relationship between the two coordinate systems.
1x xx xy
y yx yy
v k k xv k k
yμ
∂Φ⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤ ∂⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ∂Φ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥∂⎣ ⎦
(3.220)
The Darcy velocity in the uv coordinate system is related to the Darcy velocity
in the xy coordinate system by Eq.(3.234) as
3-117
cos sinsin cos
xu
yv
vvvv
θ θθ θ
⎡ ⎤⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦ ⎣ ⎦
(3.235)
Substituting Eq.(3.220) into (3.235) gives
cos sin1sin cos
xx xyu
yx yyv
k kv xk kv
y
θ θθ θμ
∂Φ⎡ ⎤⎢ ⎥⎡ ⎤⎡ ⎤ ⎡ ⎤ ∂⎢ ⎥= − ⎢ ⎥⎢ ⎥ ⎢ ⎥ ∂Φ− ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦⎢ ⎥∂⎣ ⎦
(3.236)
The flow potential in the xy coordinate system is related to the flow potential
in the uv coordinate system by Eq.(2.233) as
cos sinsin cos
x u
y v
θ θθ θ
∂Φ⎡ ⎤ ∂Φ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎡ ⎤∂ ∂⎢ ⎥ = ⎢ ⎥⎢ ⎥∂Φ ∂Φ⎢ ⎥ ⎣ ⎦ ⎢ ⎥⎢ ⎥ ⎢ ⎥∂ ∂⎣ ⎦⎣ ⎦
(3.237)
Substituting Eq.(3.237) into (3.236) gives
cos sin cos sin1sin cos sin cos
xx xyu
yx yyv
k kv uk kv
v
θ θ θ θθ θ θ θμ
∂Φ⎡ ⎤⎢ ⎥−⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ∂= − ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ∂Φ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎢ ⎥⎣ ⎦⎢ ⎥∂⎣ ⎦
(3.238)
Eq.(3.238) is Darcy’s law in the uv coordinate system and is of the form
1u uu uv
v vu vv
v k k uv k k
vμ
∂Φ⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤ ∂= − ⎢ ⎥⎢ ⎥ ⎢ ⎥ ∂Φ⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥∂⎣ ⎦
(3.239)
A comparison of Eqs.(3.238) and (3.239) shows that the permeability tensor in
the uv coordinate system is related to the tensor in the xy coordinate system
as follows:
cos sin cos sinsin cos sin cos
xx xyuu uv
yx yyvu vv
k kk kk kk k
θ θ θ θθ θ θ θ
−⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
(3.240)
3-118
Carrying out the matrix multiplications on the right side of Eq.(3.240) and
equating the result to the left side matrix gives
2 2cos sin cos sin cos sinuu xx yx xy yyk k k k kθ θ θ θ θ θ= + + + (3.241)
2 2sin cos sin cos sin cosuv xx yx xy yyk k k k kθ θ θ θ θ θ= − − + + (3.242)
2 2sin cos cos sin sin cosvu xx yx xy yyk k k k kθ θ θ θ θ θ= − + − + (3.243)
2 2sin sin cos sin cos cosvv xx yx xy yyk k k k kθ θ θ θ θ θ= − − + (3.244)
From trigonometry,
2 12cos (1 cos 2 )θ θ= + (3.245)
2 12sin (1 cos 2 )θ θ= − (3.246)
12sin cos sin 2θ θ θ= (3.247)
Substituting Eqs.(3.245) to (3.247) into Eq.(3.241) gives
1 1 1 12 2 2 2(1 cos 2 ) sin 2 sin 2 (1 cos 2 )uu xx yx xy yyk k k k kθ θ θ θ= + + + + − (3.248)
Since kxy = kyx, Eq.(3.248) can be simplified to
cos 2 sin 22 2
xx yy xx yyuu xy
k k k kk kθ θ
+ −= + + (3.249)
A comparison of Eqs.(3.242) and (3.243) shows that kuv = kvu. Substituting
Eqs. Eqs.(3.245) to (3.247) into Eq.(3.242) gives
1 1 1 12 2 2 2sin 2 (1 cos 2 ) (1 cos 2 ) sin 2uv xx yx xy yyk k k k kθ θ θ θ= − − − + + + (3.250)
….Since kxy = kyx and kuv = kvu, Eq.(3.250) can be simplified to
3-119
sin 2 cos 22
xx yyuv vu xy
k kk k kθ θ
−⎛ ⎞= = − +⎜ ⎟
⎝ ⎠ (3.251)
Substituting Eqs.(3.245) to (3.247) into Eq.(3.244) gives
1 1 1 12 2 2 2(1 sin 2 ) sin 2 sin 2 (1 cos 2 )vv xx yx xy yyk k k k kθ θ θ θ= − − − + + (3.252)
which can be simplified to
cos 2 sin 22 2
xx yy xx yyvv xy
k k k kk kθ θ
+ −⎛ ⎞= − −⎜ ⎟
⎝ ⎠ (3.253)
One of the principal axes of the permeability anisotropy is obtained from
0uv vuk k= = (3.254)
Substituting Eq.(3.254) into Eq.(3.251) gives
sin 2 cos 2 02
xx yyxy
k kkθ θ
−⎛ ⎞− + =⎜ ⎟
⎝ ⎠ (3.255)
Eq.(3.255) yields
2sin 2tan 2
cos 2xy
xx yy
kk k
θθθ
= =−
(3.256)
or
112
2tan xy
xx yy
kk k
θ −⎛ ⎞
= ⎜ ⎟⎜ ⎟−⎝ ⎠ (3.257)
The other principal axis of the permeability anisotropy is given by (θ +90)
degrees.
Example 3.6
For the flow field of Example 3.5,
1. Calculate the principal axes of the permeability anisotropy.
3-120
2. Calculate the principal values of the permeability anisotropy. Show the
transformed permeability tensor for the principal coordinate system.
3. Calculate the components of the Darcy velocity vector in the principal
directions of the permeability anisotropy.
Solution to Example 3.6
One of the principal axes of the permeability anisotropy is given by Eq.(3.257)
as
( ) ( ) ( )1 1 12 50tan 2 tan tan 0.1818 10.3048
200 750θ− − −−⎛ ⎞
= = =⎜ ⎟−⎝ ⎠
10.3048 5.152
θ = =
The principal values of the permeability anisotropy are given by Eqs.(3.249)
and (3.253) as
( ) ( )200 750 200 750 cos 10.3048 50sin 10.30482 2uk + −
= + −
475 270.5642 8.9443 195.49uk md= − − =
( ) ( )200 750 200 750 cos 10.3048 50sin 10.30482 2vk + −⎛ ⎞= − +⎜ ⎟
⎝ ⎠
475 270.5642 8.9443 754.51vk md= + + =
The transformed permeability tensor is given by
( )0 195.5 0
,0 0 754.5u
v
kk u v md
k⎡ ⎤ ⎡ ⎤
= =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
, where ku and kv are the principal values of
the permeability anisotropy. Figure 3.53 shows that the coordinate system
along the principal axes of the permeability anisotropy (u,v) makes an angle of
+5.15º with the x,y coordinate system.
3-121
Figure 3.53. Relationship between the principal axes of the permeability anisotropy and the (x,y) coordinates.
Also shown in Figure 3.53 is the Darcy velocity vector in relation to the x,y
and the u,v coordinate systems, where the u,v coordinates are the principal
axes of the permeability anisotropy.
The components of the Darcy velocity in the u,v coordinates can be
calculated as follows.
1.2185cos(54.24 5.15 ) 1.2185cos 49.09 0.7980uv = − = = ft/D
( )1.2185sin 54.24 5.15 1.2185sin 49.09 0.9209vv = − = = ft/D
The components of the Darcy velocity in the u,v coordinates also can be
calculated from Eq.(3.234) as
3-122
0.7120 0.7980cos5.15 sin 5.15 cos5.15 sin 5.150.7120 0.9888
0.9888 0.9209sin 5.15 cos5.15 sin 5.15 cos5.15u
v
vv
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
0.7980uv = ft/D
0.9209vv = ft/D
These are the results obtained previously. It should be noted that because the
flow is invariant with a rotation of the axes, the magnitude of the Darcy
velocity (1.2185 ft/D) remains the same in the new coordinate system. The
direction of the Darcy velocity also remains the same. Only the components of
the Darcy velocity change in the new coordinate system.
3.14.4 Alternative Calculation of the Principal Values and the Principal Axes of the Permeability Anisotropy
From linear algebra, the principal values of the permeability tensor are
given by the eigenvalues of the permeability tensor whereas the principal axes
of the permeability anisotropy are given by the eigenvectors of the
permeability tensor. This can be demonstrated by reworking Example 3.6 as
an eigenvalue problem.
Example 3.7
Rework Example 3.6 by calculating the eigenvalues and eigenvectors of the
permeability tensor.
Solution to Example 3.7
Let the eigenvalues of the permeability tensor be λ. The characteristic
equation is given by
200 50det 0
50 750λ
λ− −⎡ ⎤
=⎢ ⎥− −⎣ ⎦
( )( ) ( )( )200 750 50 50 0λ λ− − − − − =
3-123
2 950 147500 0λ λ− + =
( )( )( )2950 950 4 1 147500 950 559.01702 2
λ± − ±
= =
1 195.5mdλ =
2 754.5mdλ =
These principal values are the same as obtained in Example 3.6. For
1 195.5λ = , the eigenvector is obtained by solving the following simultaneous
equations:
200 195.5 50 050 750 195.5 0
xy
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
4.5085 50 050 554.5085 0
xy
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
4.5085 50 0x y− =
50 554.5085 0x y− + =
11.09021
xy
y⎡ ⎤ ⎡ ⎤
=⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
The eigenvector is
11.09021
u ⎡ ⎤= ⎢ ⎥
⎣ ⎦
This eigenvector makes an angle θ with the positive x-axis given by
1tan 0.090211.0902
θ = =
5.15θ =
3-124
For 2 754.5λ = , the eigenvector is obtained by solving the following
simultaneous equations:
200 754.5 50 050 750 754.5 0
xy
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
554.5 50 050 4.5 0
xy
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
554.5 50 0x y− − =
50 4.5 0x y− − =
0.09021
xy
y−⎡ ⎤ ⎡ ⎤
=⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
The eigenvector is
0.09021
v−⎡ ⎤
= ⎢ ⎥⎣ ⎦
We can easily show that the two eigenvectors are orthogonal.
( )( ). 11.0902 0.0902 1 0u v = − − =
The eigenvectors point in the directions of the principal axes of the
permeability anisotropy. These are the same results obtained in Example 3.6.
3.14.5 Directional Permeability
For an isotropic medium, the Darcy velocity vector and the velocity
potential gradient vector are collinear as shown in Figure 3.54. Because the
two vectors lie along the same line with an angle of 180º between them, we
ordinarily do not specify whether the permeability is in the direction of flow or
in the direction of the velocity potential gradient. For this case, the
"directional" permeability in the direction of flow (kdf) is equal to the
3-125
"directional" permeability in the direction of the velocity potential gradient
(kdp) and is calculated in Darcy units as
df dp
vk k
μ= =
∇Φ (3.258)
Figure 3.54. Collinear relationship between the direction of flow and the direction of the velocity potential gradient for an isotropic medium.
For an anisotropic medium, in general, the Darcy velocity vector and
the velocity potential gradient vector are not collinear as shown in Figure
3.55. The concept of directional permeability seeks to extend the definition of
permeability of an isotropic medium expressed in Eq.(3.258) to an anisotropic
medium. Accordingly, the directional permeability in the direction of flow is
defined in Darcy units by the equation
cosdf
vk
μα
=∇Φ
(3.259)
where cosα∇Φ is the component of the velocity potential gradient vector that
is collinear with the Darcy velocity. Similarly, the directional permeability in
the direction of the velocity potential gradient is defined in Darcy units by the
equation
3-126
cos
dp
vk
μ α=
∇Φ (3.260)
where cosv α is the component of the Darcy velocity vector that is collinear
with the velocity potential gradient vector.
Figure 3.55. Relationship between the direction of flow and the direction of the velocity potential gradient in an anisotropic porous medium.
Example 3.8
Calculate the directional permeabilities for the flow field of Example 3.5.
Solution to Example 3.8
The directional permeability in the direction of flow is calculated with
Eq.(3.259) in oilfield units as
( )( )( ) ( )
(0.001127(5.615) cos
1 1.2185 338.96 md
(0.001127(5.615) 0.6731 cos 180 147.56
df
vk
μα
=∇Φ
= =−
3-127
The directional permeability in the direction of the velocity potential gradient
is calculated with Eq.(3.260) in oilfield units as
( )( ) ( )( )
cos(0.001127(5.615)
1 1.2185 cos 180 147.56 241.46 md
(0.001127(5.615) 0.6731
dp
vk
μ α=
∇Φ
−= =
We now derive the relationship between the directional permeabilities
and the principal values of the permeability anisotropy. Let kx and ky be the
principal values of the permeability anisotropy where the x,y coordinates are
the principal axes of the anisotropy in a 2D reservoir. Let kdf be the directional
permeability in the direction of flow (in the direction of the Darcy velocity
vector) that makes an angle θ with the positive x-axis as shown in Figure
3.56. The permeability tensor for this medium in the x,y coordinate system is
( )0
,0x
y
kk x y
k⎡ ⎤
= ⎢ ⎥⎣ ⎦
(3.261)
Figure 3.56. Darcy velocity vector in the flow direction that makes an angle θ with the positive x axis.
3-128
Darcy’s law can be written in the x, y and s directions as
xx
kvxμ
∂Φ= −
∂ (3.262)
yy
kv
yμ∂Φ
= −∂
(3.263)
dfs
kv
sμ∂Φ
= −∂
(3.264)
where s is the flow direction. Figure 3.56 shows the relationships between the
Darcy velocity in the flow direction vs and its x and y components vx and vy.
From Figure 3.56, we find
cosx sv v θ= (3.265)
siny sv v θ= (3.266)
Eqs. (3.362) and (3.263) give the potential gradients in the x and y directions
as
cossx
vx k
μ θ∂Φ= −
∂ (3.267)
sinsy
vy k
μ θ∂Φ= −
∂ (3.268)
Because Φ = Φ(x,y), from Calculus,
dx dys x ds x ds
∂Φ ∂Φ ∂Φ= +
∂ ∂ ∂ (3.269)
From Figure 3.56,
cosdxds
θ= (3.270)
3-129
sindyds
θ= (3.271)
Substituting Eqs.(3.267) to (3.271) into Eq.(3.264) and rearranging gives the
directional permeability in the direction of flow as
2 21 cos sin
df x yk k kθ θ
= + (3.272)
Given the principal values of the permeability anisotropy (kx, ky) and the
principal coordinate system (x,y), Eq.(3.272) can be used to calculate the
directional permeability in the flow direction that makes an angle θ with the
positive x-axis.
Eq.(3.272) can be put in rectangular coordinates by choosing a vector r
in the s direction (flow direction) and noting that
cosx r θ= (3.273)
siny r θ= (3.274)
2
22cos x
rθ = (3.275)
2
22sin y
rθ = (3.276)
Substituting Eqs.(3.275) and (3.276) into (3.272) gives
2 2 2
df x y
r x yk k k
= + (3.277)
If we make the magnitude of the vector dfr k= , then Eq.(3.277) becomes
2 2
1x y
x yk k
= + (3.278)
3-130
or
( ) ( )
2 2
2 21x y
x y
k k= + (3.279)
Eq.(3.279) is the cannonical equation of an ellipse with major semiaxes
xk and yk . Thus, by constructing the permeability ellipse as shown in
Figure 3.57, the directional permeability can easily be determined graphically
for any given flow direction.
Figure 3.57. Permeability ellipse for calculating the directional permeability in the direction of flow.
For flow 3D, the directional permeability in the direction of flow will be given
by the permeability ellipsoid
( ) ( ) ( )
2 2 2
2 2 21x y z
x y z
k k k= + + (3.280)
with semiaxes xk , yk and zk .
3-131
Let the velocity potential gradient vector make an angle α with the
positive x-axis. It can be shown that the directional permeability in the
direction of the gradient vector is given by
2 21 cos sin
11 1xdp ykk k
α α= +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
(3.281)
The permeability ellipse in 2D is
2 2
11 1x y
x y
k k
= +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
(3.282)
or
2 2
2 211 1
x y
x y
k k
= +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(3.283)
with semiaxes ( )1/ xk and ( )1/ yk . The permeability ellipsoid in 3D is
2 2 2
2 2 2111 1
zx y
x y y
kk k
= + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
(3.284)
with semiaxes ( )1/ xk , ( )1/ xk and ( )1/ zk .
In an anisotropic medium, the directional permeability in the direction
of the velocity potential gradient is distinct from the directional permeability
in the direction of flow. Thus, there are two distinct definitions of directional
permeability for an anisotropic medium. It should be noted that principal
values of the permeability anisotropy (kx, ky and kz) are the true permeabilities
in the principal directions (x, y and z). They are also the directional
3-132
permeabilities in the direction of flow and the direction of the potential
gradient because in the principal coordinate system, the Darcy velocity and
the velocity potential gradient are collinear.
Figures 3.58 shows a polar plot of permeability data from Scheidegger
(1954) for permeabilities measured around a full-sized core. The measured
permeability data appear to fit a complicated curve. However, when the same
data are plotted as k and 1k
as shown in Figures 3.59 and 3.60, they fit on
ellipses. This observation verifies the tensorial nature of permeability. These
data can be used to identify the principal axes and the principal values of the
permeability anisotropy.
Figure 3.58. Permeability data measured around a cylindrical core (Data from Scheidegger, 1954).
3-133
Figure 3.59. Permeability ellipse for permeability in the direction of flow for the data of Figure 3.58.
3-134
Figure 3.60. Permeability ellipse for permeability in the direction of the
velocity potential gradient for the data of Figure 3.58.
3-135
Example 3.9
Calculate and plot the permeability ellipses for the flow field of Example 3.5.
Solution to Example 3.9
The permeability ellipses are given by Eqs.(3.279) for the flow direction and
(3.283) for the potential gradient direction, where x and y are the principal
axes of the permeability anisotropy. These equations can be adapted for our
purpose as follows:
( ) ( ) ( ) ( )2 2 2 2
2 2 2 21195.5 754.5u v
u v u v
k k= + = +
and
2 2 2 2
2 2 2 211 11 1
195.5 754.5u v
u v u v
k k
= + = +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
These ellipses are plotted in Figures 3.61 and 3.62.
3-136
Figure 3.61. Permeability ellipse for permeability in the direction of flow for Example 3.9.
3-137
Figure 3.62. Permeability ellipse for permeability in the direction of the
velocity potential gradient for Example 3.9.
3.14.6 Measurement of Transverse Permeability of a Cylindrical Core
If the transverse permeabilities around a full-sized core could be
measured, then the permeability anisotropy of the core could be determined.
Figure 3.63 shows a schematic diagram of an apparatus that can be used to
make the measurements. After a permeability measurement, the sleeve is
rotated to a new angle and the measurement is repeated. Using conformal
mapping, Collins (1961) has derived the following equation for calculating the
transverse permeability from a gas flow experiment (in Darcy units):
( )1 1
2 1
Pqk GbPL PP
μ α=⎛ ⎞+ Δ⎜ ⎟⎝ ⎠
(3.285)
3-138
Figure 3.63. Apparatus for measuring transverse permeability of a full-sized cylindrical core.
where P1 is the inlet pressure, q1 is the gas volumetric flow rate measured at
P1, P is the mean pressure 1 2
2P P+⎛ ⎞
⎜ ⎟⎝ ⎠
, PΔ is the pressure drop between the
inlet and the outlet ( )1 2P P− , b is the Klinkenberg parameter, L is the height of
the core, G(α) is a geometric correction factor to account for the complex flow
3-139
geometry of the gas from inlet to outlet and α is equal to 2θ⎛ ⎞
⎜ ⎟⎝ ⎠
. Figure 3.64
shows the plot of G(α) versus α.
Figure 3.64. Geometric factor for transverse permeability calculation (from Collins, 1961).
By making such measurements around the core, a polar plot of the
permeability can be made as shown in Figure 3.65. Such a plot can be used
to determine the principal values and the principal directions of the
permeability anisotropy. The permeability tensor of the medium can then be
constructed from this information for any coordinate system.
3-140
Figure 3.65. Polar plot of transverse permeabilities around a core (from Scheidegger, 1954).
3.15 EXAMPLE APPLICATIONS OF PERMEABILITY
Permeability is used in numerical reservoir simulations to predict
reservoir performance. Permeability (absolute and relative permeability) must
be assigned to every one of the approximately one million grid blocks in a full-
field simulation, often with a full tensor representation. The purpose of this
3-141
section is to present simple examples of how permeability could be used to
predict reservoir performance. We examine the productivity indices of a
horizontal well and a hydraulically fractured well with infinite fracture
conductivity. The analysis shows that a horizontal well and a fractured well
are similar in the manner in which they affect reservoir performance.
3.15.1 Productivity of a Horizontal Well Introduction
Horizontal wells are increasingly being drilled for petroleum recovery.
One of the major advantages of a horizontal well over a vertical well is the
larger contact area between a horizontal well and the reservoir compared to a
vertical well in the same reservoir. This larger areal contact can significantly
enhance the productivity of a horizontal well compared to a vertical well
draining the same reservoir volume. However, the productivity of a horizontal
well can be affected significantly by the permeability anisotropy of the
reservoir.
The objectives of this section are: (1) to provide a simple formula for
estimating the productivity index of a horizontal well for single phase flow in
terms of the reservoir permeability and other relevant factors, (2) to compare
the productivities of a horizontal well and a vertical well in the same reservoir
to determine if a horizontal well is superior to a vertical well under the
particular circumstance, and (3) to examine how the permeability anisotropy
of the reservoir affects the productivity of a horizontal well.
Homogeneous and Isotropic Reservoirs
The steady state productivity index (PI) for a horizontal well in a
homogeneous and isotropic reservoir is given by Giger et al. (1984) as
3-142
2
2
1 12
ln ln2
2
H D
e
w
e
kLBPI F
LrL h
h rLr
πμ
π
⎛ ⎞⎜ ⎟⎝ ⎠=
⎡ ⎤⎛ ⎞⎢ ⎥+ − ⎜ ⎟⎢ ⎥ ⎛ ⎞⎝ ⎠ +⎢ ⎥ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
(3.286)
where
PIH = productivity index of a horizontal well, surface rate/unit pressure drop
FD = units conversion constant
k = effective permeability to oil
L = total length of the horizontal well
μ = oil viscosity
B = oil formation volume factor, reservoir volume/surface volume
h = formation thickness
rw = wellbore radius
re = well drainage radius
Figure 3.66 shows the flow geometry. The horizontal well of length L is located
half-way between the top and the bottom of the reservoir of thickness h. L is
assumed to be small relative to 2re. The units conversion constant, FD, is
given in Table 3.14.
3-143
Figure 3.66: Flow geometry for steady state productivity index equation.
Table 3.14: Units and Dimensions
Dimension
Darcy Units
SI Units
Oilfield Units
Field Metric Units
Distance L cm m ft m Area L2 cm2 m2 ft2 m2
Pressure ML-1T-2 atm Pa psi MPa Permeability L2 darcy m2 millidarcy
(md) μm2
Fluid Viscosity
ML-1T-1 cp Pa.s cp mPa.s
Liquid Flow Rate
L3T-1 cm3/s m3/s B/D m3/D
FD 1 1 0.001127 86.4
The steady state productivity index of a vertical well can be obtained by
integration of Darcy’s Law as
3-144
2
lnV D
e
w
khBPI F
rr
πμ
⎛ ⎞⎜ ⎟⎝ ⎠=⎛ ⎞⎜ ⎟⎝ ⎠
(3.287)
Therefore, the ratio of the productivity index of a horizontal well to a vertical
well draining the same reservoir volume is obtained from Eqs.(3.286) and
(3.287) as
2
ln
1 12
ln ln2
2
e
wHD
V
e
w
e
rrPI F
PI LrL h
h rLr
π
⎛ ⎞⎜ ⎟⎝ ⎠=
⎡ ⎤⎛ ⎞⎢ ⎥+ − ⎜ ⎟⎢ ⎥ ⎛ ⎞⎝ ⎠ +⎢ ⎥ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
(3.288)
Note that because the right side of Eq.(3.288) involves ratios of lengths,
Eq.(3.288) is valid in any system of units.
Example 3.10
Given the following data for an isotropic reservoir, calculate (a) the
productivity index for a horizontal well, (b) the productivity index for a vertical
well, and (c) the ratio of the productivity indices of the horizontal and the
vertical wells.
Length of horizontal well (L) = 500 m
Formation thickness (h) = 40 m
Drainage radius (re) = 1000 m
Wellbore radius (rw) = 0.12 m
Horizontal permeability (kH) = 400 md
Vertical permeability (kV) = 400 md
3-145
Oil viscosity (μo) = 5 cp
Oil formation volume factor (B)= 1.2 RB/STB
Solution to Example 3.10
In oilfield units, FD = 0.001127. Substituting the numerical values into
Eq.(3.286) gives the productivity index for the horizontal well as
2
400 500 3.280825 1.20.001127 26.02 STB/D/psi
5001 12 1000500 40ln ln
50040 2 0.122 1000
H
x xxPI
xx
x
π
π
⎛ ⎞⎜ ⎟⎝ ⎠= =
⎡ ⎤⎛ ⎞⎢ ⎥+ − ⎜ ⎟ ⎛ ⎞⎝ ⎠⎢ ⎥ + ⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
Substituting the numerical values into Eq.(3.287) gives the productivity index
for the vertical well as
400 40 3.280825 1.20.001127 6.86 STB/D/psi1000ln0.12
V
x xxPI
π ⎛ ⎞⎜ ⎟⎝ ⎠= =
⎛ ⎞⎜ ⎟⎝ ⎠
The ratio of the horizontal to vertical well productivity indices is
26.02 3.86.86
H
V
PIPI
= =
Clearly, in this isotropic reservoir, the productivity of the horizontal well is
much higher than that of the vertical well.
Homogeneous and Anisotropic Reservoirs
Many petroleum reservoirs are anisotropic and have different
permeabilities in the different directions. For example, in a layered reservoir,
the vertical permeability is usually much less than the horizontal
permeability. Also, a reservoir that is intersected by a large number of vertical
3-146
fractures will have a higher vertical permeability than the horizontal
permeability. Permeability anisotropy can have a significant effect on the
productivity of a horizontal well. This effect is examined in this section.
Consider an anisotropic reservoir with horizontal and vertical
permeabilities kH and kV, respectively. These are the principal values of the
permeability anisotropy and x and z are the principal axes of the anisotropy.
The diffusivity equation for steady state single phase flow for the anisotropic
medium is given by
2 2 0H VP Pk kx z
∂ ∂+ =
∂ ∂ (3.289)
By a suitable change of coordinates, Eq.(3.289) can be transformed into a
form that describes the flow in an equivalent isotropic medium having a
permeability *k . The required transformation equations are as follows:
*H Vk k k= (3.290)
*
H
kX xk
= (3.291)
and
*
V
kZ zk
= (3.292)
where X and Z are the transformed spatial coordinates. With these
transformations, Eq.(3.289) becomes
*2 2 0P Pk
X Z∂ ∂⎛ ⎞+ =⎜ ⎟∂ ∂⎝ ⎠
(3.293)
Since *k is nonzero, Eq.(3.293) yields
3-147
2 2 0P PX Z∂ ∂⎛ ⎞+ =⎜ ⎟∂ ∂⎝ ⎠
(3.294)
which is the diffusivity equation for an equivalent isotropic medium with a
permeability of *k . Eqs.(3.290) to (3.292) can be used in conjunction with
Eq.(3.286) to calculate the productivity index of a horizontal well in an
anisotropic reservoir as shown in the following example.
Example 3.11
Given the following data for an anisotropic reservoir, calculate (a) the
productivity index for a horizontal well, (b) the productivity index for a vertical
well, and (c) the ratio of the productivity indices of the horizontal and the
vertical wells.
Length of horizontal well (L) = 500 m
Formation thickness (h) = 40 m
Drainage radius (re) = 1000 m
Wellbore radius (rw) = 0.12 m
Horizontal permeability (kH) = 400 md
Vertical permeability (kV) = 25 md
Oil viscosity (μo) = 5 cp
Oil formation volume factor (B) = 1.2 RB/STB
Solution to Example 3.11
From Eq.(3.290),
* 400 25 100 mdk x= =
From Eq.(3.291),
* 100500 250 m400
L = =
3-148
* 1001000 500 m400er = =
From Eq.(3.292),
* 10040 80 m25
h = =
Substituting these numerical values into Eq.(3.286) gives the productivity
index for the horizontal well in the anisotropic reservoir as
2
100 250 3.280825 1.20.001127 8.71 STB/D/psi
2501 12 500250 80ln ln
25080 2 0.122 500
H
x xxPI
xx
x
π
π
⎛ ⎞⎜ ⎟⎝ ⎠= =
⎡ ⎤⎛ ⎞⎢ ⎥+ − ⎜ ⎟ ⎛ ⎞⎝ ⎠⎢ ⎥ + ⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
The productivity index for the vertical well in the anisotropic reservoir is given
by Eq.(3.287) with k = kH. Thus, for the vertical well,
400 40 3.280825 1.20.001127 6.86 STB/D/psi1000ln0.12
V
x xxPI
π ⎛ ⎞⎜ ⎟⎝ ⎠= =
⎛ ⎞⎜ ⎟⎝ ⎠
The ratio of the horizontal to the vertical well productivity indices in the
anisotropic medium is
8.71 1.276.86
H
V
PIPI
= =
Thus, the productivity advantage of a horizontal well is reduced considerably
in an anisotropic reservoir in which the vertical permeability is significantly
lower than the horizontal permeability.
3-149
In a naturally fractured reservoir, if the azimuth of the horizontal is
such that it intersects the vertical fractures, then the productivity of the well
will enhanced as shown in the following example.
Example 3.12
A horizontal well has intersected a large number of vertical fractures as
shown in Figure 3.67 such that the vertical permeability is larger than the
horizontal permeability. Given the following data for this anisotropic reservoir,
calculate (a) the productivity index for a horizontal well, (b) the productivity
index for a vertical well, and (c) the ratio of the productivity indices of the
horizontal and the vertical wells.
Length of horizontal well (L) = 500 m
Formation thickness (h) = 40 m
Drainage radius (re) = 1000 m
Wellbore radius (rw) = 0.12 m
Horizontal permeability (kH) = 400 md
Vertical permeability (kV) = 800 md (Due to the vertical fractures)
Oil viscosity (μo) = 5 cp
Oil formation volume factor (B) = 1.2 RB/STB
3-150
Figure 3.67. A horizontal well that intersects vertical fractures.
Solution to Example 3.12
From Eq.(3.290),
* 400 800 565.69 mdk x= =
From Eq.(3.291),
* 565.69500 594.60 m400
L = =
* 565.691000 1189.21 m400er = =
From Eq.(3.292),
* 565.6940 33.64 m800
h = =
Substituting these numerical values into Eq.(3.286) gives the productivity
index for the horizontal well in the fractured anisotropic reservoir as
3-151
2
565.69 594.60 3.280825 1.20.001127 32.34 STB/D/psi
594.601 12 1189.21594.60 33.64ln ln25033.64 2 0.12
2 1189.21
H
x xxPI
xx
x
π
π
⎛ ⎞⎜ ⎟⎝ ⎠= =
⎡ ⎤⎛ ⎞⎢ ⎥+ − ⎜ ⎟ ⎛ ⎞⎝ ⎠⎢ ⎥ + ⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
The productivity index for the vertical well in the anisotropic reservoir is given
by Eq.(3.286) with k = kH. Thus, for the vertical well,
400 40 3.280825 1.20.001127 6.86 STB/D/psi1000ln0.12
V
x xxPI
π ⎛ ⎞⎜ ⎟⎝ ⎠= =
⎛ ⎞⎜ ⎟⎝ ⎠
The ratio of the horizontal to the vertical well productivity indices in the
anisotropic medium with vertical fractures is
32.34 4.716.86
H
V
PIPI
= =
Clearly, the advantage of a horizontal well that intersects vertical fractures in
an anisotropic reservoir compared to a vertical well is apparent from this
example. Note, however, that if the horizontal well was wrongly drilled parallel
to the fractures instead of normal to the fractures as shown in Figure 3.67,
the productivity of the horizontal well will not be enhanced by the fractures.
The horizontal well placement relative to the fracture azimuth is therefore
important to take advantage of the fractures.
The change in coordinates given by Eqs.(3.290) to (3.292) also
transforms the circular wellbore into an elliptical wellbore with the same cross
sectional area as the circular wellbore. This change in the shape of the
wellbore has been neglected in the calculations in Examples 3.11 and 3.12
because its effect is negligibly small.
3-152
There are more elaborate equations for calculating the productivity
index of a horizontal well in the literature. These more complicated equations
and their associated assumptions may be found in the references.
3.14.2 Productivity of a Vertically Fractured Well
The productivity index of a vertically fractured well with an infinite
fracture conductivity in a homogeneous and isotropic reservoir as shown in
Figure 3.68 is given by Karcher et al. (1986) as
2
2
1 12
ln
2
H
F D
e
e
k LBPI F
LrL
h Lr
πμ
⎛ ⎞⎜ ⎟⎝ ⎠=
⎡ ⎤⎛ ⎞⎢ ⎥+ − ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥
⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
(3.295)
where L is the total length of the fracture. The assumptions underlying the
above equation are:
• Fracture height is equal to the reservoir pay thickness
• The fracture is plane and is centered on the well
• The well drainage radius is large compared to the fracture length
• The fracture has infinite conductivity
The fracture conductivity is defined as
C fF k w= (3.296)
where kf is the fracture permeability and w is the fracture width. A
dimensioless fracture conductivity is defined as
3-153
fCD
H f
k wF
k L= (3.297)
where Lf is the fracture half-length 2L⎛ ⎞
⎜ ⎟⎝ ⎠
. A hydraulic fracture behaves as if it
has an infinite conductivity for
300CDF ≥ (3.298)
Figure 3.68. Vertically fractured well.
3-154
If 12 e
Lr
⎛ ⎞⎜ ⎟⎝ ⎠
, then Eq.(3.295) simplifies to
2
4ln
H
V De
k hBPI FrL
πμ
⎛ ⎞⎜ ⎟⎝ ⎠=⎛ ⎞⎜ ⎟⎝ ⎠
(3.299)
which is the productivity index of a vertical well with an equivalent wellbore
radius given by
4wLr = (3.300)
The ratio of the productivity index of the fracture well to the unfractured well
is given by
2
ln
1 12
ln
2
e
wFD
V
e
e
rrPI F
PI LrL
h Lr
⎛ ⎞⎜ ⎟⎝ ⎠=
⎡ ⎤⎛ ⎞⎢ ⎥+ − ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥
⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
(3.301)
Eq.(3.301) can be used to demonstrate the increased productivity of a
fractured well compared to the unfractured well.
A comparison of Eqs.(3.295) and (3.286) shows that the productivity
index of a vertical fracture with an infinite conductivity is similar to but
somewhat higher than the productivity index of a horizontal well. The
horizontal well has a positive pseudo skin represented by the term ln2 w
hrπ
⎛ ⎞⎜ ⎟⎝ ⎠
in
Eq.(3.286) compared to the fractured well. The pseudoskin is caused by the
convergence of flow into the horizontal wellbore whereas the flow into the
fracture occurs over the entire fracture face with no convergence of flow. The
3-155
similarity in the productivity index equations for a fracture and a horizontal
well indicates that fracturing a vertical well is an alternative to drilling a
horizontal well.
NOMENCLATURE
A = cross sectional area in the flow direction
Ai = Area of bed i
AT = total area
b = Klinkenberg empirical constant dependent on rock and gas
d = core diameter
B = oil formation volume factor
Bg = gas formation volume factor
cf = compressibility of porous medium
ct = total compressibility
C = wellbore storage coefficient
CD = dimensionless wellbore storage coefficient
D = non-Darcy coefficient
Dp = grain diameter, mean grain diameter
Ei = exponential integral function
f(δ) = probability density function for pore diameter for a bundle of capillary tubes model
F = formation resisitivity factor
FC = fracture conductivity
FCD= dimensionless fracture conductivity
g = gravitational acceleration
G(α) = geometric factor for calculating transverse permeability of a cylindrical porous medium
h = pay thickness, hydraulic head
hi = thickness of bed i
K = hydraulic conductivity
k = absolute permeability of the medium
kdf = directional permeability in the direction of flow
3-156
kdp = directional permeability in the direction of the velocity potential gradient
ki = permeability of bed i
k = permeability tensor, average permeability
kij = individual entries in the permeability tensor
kg = measured gas permeability
kL = permeability to liquid (absolute permeability of medium)
kor = permeability to nonwetting phase at irreducible wetting phase saturation
koτ = Kozeny constant
kx, ky, kz = principal values of permeability anisotropy where xyz is the principal coordinate system
ku, kv, kw = principal values of permeability anisotropy where uvw is the principal coordinate system
L = length
Li = length of bed i
Le = tortuous length of flow path
ln = natural logarithm (log to base e)
log = log to base 10
M = real gas potential or real gas pseudo pressure
P = pressure
PD = dimensionless pressure
'DP = dimensionless welltest derivative function
Pi = initial reservoir pressure
P* = extrapolated pressure from a Horner plot
Psc = reference pressure
Pe = pressure at external radius
Pw = pressure at the wellbore
PwD= dimensionless wellbore pressure
wDP = dimensionless wellbore pressure in Laplace space
'wDP = Calculus derivative of dimensionless wellbore pressure in Laplace
space
3-157
Pwf = flowing bottomhole pressure
Pws = shut-in bottomhole pressure
P1 = inlet pressure
P2 = outlet pressure
P = mean pressure 1 2
2P P+⎛ ⎞
⎜ ⎟⎝ ⎠
dPds
= pressure gradient
PI = productivity index
PIH = productivity index of a horizontal well
PIV = productivity index of a vertical well
q = volumetric flowrate
qsf = sandface rate
qsc = volumetric rate at a reference pressure, Psc
qT = total volumetric flow rate
r = radial distance from the wellbore
rD = dimensionless radius
re = external drainage radius
rH = hydraulic radius
ri = radius of bed i
rinv = radius of investigation of a welltest
ro = resistance of a porous medium fully saturated with water
rw = resistance of water
rw = wellbore radius
Re = Reynolds Number
Ro = resistivity of a porous medium fully saturated with water
Rw = resistivity of water
s = cartesian coordinate in the flow direction.
S = skin factor
S = storativity
S = surface area per unit bulk volume (specific surface area)
3-158
Sp = surface area per unit pore volume (specific surface area)
Ss = surface area per unit grain volume (specific surface area)
Swirr = irreducible wetting phase saturation
S* = mechanical skin facor
t = time
tD = dimensionless time
tp = producing time before shut-in
T = absolute temperature
T = transmissibility
Tsc = reference temperature
v = flux vector, Darcy velocity vector
vx, vy, vz = components of Darcy velocity in the xyz Cartesian coordinate system.
vu, vv, vw = components of Darcy velocity in the uvw Cartesian coordinate system.
w = fracture width
z = height above (or below) an arbitrary datum
z = Cartesian coordinate in the vertical direction
Z = gas Z-factor
β = non-darcy velocity coefficient
ρ = fluid density.
μ = viscosity
φ = porosity, fraction
τ = tortuosity
γ = liquid specific gravity
Φ = velocity flow potential function
Ψ = pressure head
∇ = gradient operator
ΔP = pressure change
ΔP' = welltest derivative function
Δt = shut-in time
Δte = effective shut-in time
3-159
Δh = positive difference in the hydraulic heads at points 1 and 2.
ΔL = positive length of the porous medium between points 1 and 2 in the flow direction.
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De Boodt, M.F. and Kirkham, D. : "Anisotropy and Measurement of Air Permeability of Soil Clods," Soil Sci., Vol. 76 (1953) 127-313.
Dobrynin, V.M. : "The Effect of Overburden Pressure on Some Properties of Sandstones,” Soc. Pet. Eng. J. (Dec. 1962) 360-366.
Dullien, F.A.L. : Porous Media - Fluid Transport and Pore Structure, Academic Press, New York, 1979.
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Freeze, R.A. and Cherry, J.A. : Goundwater, Prentice-Hall, Englewood Cliffs, 1979.
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Gilman, J.R. and Jargon, J.R. : "Evaluating Horizontal Versus Vertical Well Performance," World Oil (April, 1992) 67-72.
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Goode, P.A. and Kuchuk, F.J. : "Inflow Performance for Horizontal Wells," SPE Reservoir Engineering (August, 1991) 319-323.
Hutta, J.J. and Griffiths, : "Directional Permeability of Sandstones: A Test Technique," Producers Monthly, Nov. 26, Vol.34 and Oct.24, Vol. 31, 1955.
Jennings, H.J. : “How to Handle and Process Soft and Unconsolidated Cores,” World Oil (June 1965) 116-119.
3-161
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Joshi, S. D. : "Production Forecasting Methods for Horizontal Wells," paper SPE 17580, presented at the SPE International Meeting on Petroleum Engineering, Tianjin, China, November 1-4, 1988.
Joshi, S. D. : Horizontal Well Technology, PennWell Publishing Co., Tulsa, 1991.
Karcher, B.J., Giger, F.M. and Combe, J. : "Some Practical Formulas to Predict Horizontal Well Behavior," paper SPE 15430, presented at the 61st SPE Annual Technical Meeting,New Orleans, LA, October 5-8, 1986.
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Klinkenberg, L.J. : “The Permeability of Porous Media to Liquids and Gases,” Drilling and Production Practices, American Petroleum Institute (1941) 200.
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Tulsa, Tulsa, Oklahoma, 1988.
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4-1
CHAPTER 4
HETEROGENEITY
4.1 INTRODUCTION
Petroleum reservoir rocks are, in general, heterogeneous and very
often anisotropic in nature. This means that the petrophysical properties
of interest such as permeability, porosity, fluid saturation, lithology and
others vary in space both vertically and laterally. Evidence of
heterogeneity was seen in the results of the core analysis presented in
Table 2.2. It was found that the permeability, porosity and fluid
saturations varied from top to bottom within the same reservoir.
Variability of rock and fluid properties is a reality that must be dealt with
in petroleum resource assessment and reservoir performance prediction.
Suppose it has been decided to model reservoir performance
through numerical simulation. In the construction of the simulation
model, every one of the nearly one million grid blocks requires the
assignment of porosity, absolute permeability, relative permeability,
capillary pressure and various fluid properties. How can one generate the
values of the rock and fluid properties to be used in the simulation
model? One may assume that the reservoir is homogeneous and use the
same value of each property in every grid block. Such a simulation model
4–2
ignores the geology of the reservoir and will consequently yield
misleading and optimistic results. One may resort to populating the grid
blocks with property values from a random number generator. This
model also ignores the geology of the reservoir and the general
observation that data from nearby locations tend to be similar whereas
data from locations that are far apart tend to be dissimilar. In other
words, petrophysical data tend to be correlated. Of course, some of these
properties can be estimated from core analysis and well logs. However,
because wells sample only a very minute volume of the reservoir, the
properties in the bulk of the reservoir remain unknown and must be
estimated by other means. There is therefore a need to devise a rational
way to estimate the petrophysical properties for the reservoir simulation
model.
The estimation of reservoir properties at locations for which no
measurements have been made can be accomplished with a new type of
statistics known as geostatistics (compare with geochemistry and
geophysics). Geostatistics is the practical applications of the Theory of
Regionalized Variables developed by Georges Matheron in Fontainebleau.
The main difference between ordinary statistics and geostatistics is that
ordinary statistics is typically based on random, independent and
uncorrelated data whereas geostatistics is based on random and spatially
correlated data.
The basic concepts of geostatistics are presented in this chapter.
Before estimating the values of the properties of the heterogeneous
reservoir, one must first characterize the heterogeneity from the samples
of the property that have been collected from core analysis, well logs and
other sources. After characterizing the nature of the heterogeneity, an
4-3
estimation is undertaken that honors both the measured data (samples)
and the structure of the heterogeneity observed from the samples.
4.2 MEASURES OF CENTRAL TENDENCY AND VARIABILITY (HETEROGENEITY)
4.2.1 Measures of central tendency Mean
The mean (arithmetic mean) is the best known measure of central
tendency. Given N sample data 1 2 3, , ,..., NΦ Φ Φ Φ , the mean (arithmetic
mean) of the sample data is defined as
1
1 i N
iiN
=
=
Φ = Φ∑ (4.1)
Geometric Mean
The geometric mean is a better measure of central tendency for
data from a log normal distribution than the arithmetic mean. The
geometric mean is defined as
1
1ln lni N
geom iiN
=
=
Φ = Φ∑ (4.2)
Taking antilog in Eq.(4.2) gives the geometric mean as
1 2 3...Ngeom NΦ = Φ Φ Φ Φ (4.3)
Median
The median of a sample data set is found by determining that
value of iΦ which has equal number of values above it and below it. To
determine the median, the data are arranged in ascending (or
4–4
descending) order. If the number of data N is odd, then the median is
determined from the sorted data as
( 1) / 2 for N oddmedian N +Φ = Φ (4.4)
If N is even, then the mean of the two most central values is used to
calculate the median as
( )( )/ 2 / 2 11 for N even2median N N +Φ = Φ + Φ (4.5)
Mode
The mode of a probability distribution ( )p Φ is the value of Φ at
which the probability distribution is a maximum. The mode is useful
primarily when there is a single sharp maximum, in which case the mode
estimates the central value. Sometimes, a distribution is bimodal, with
two relative maxima. In such a case, one may wish to know the two
modes individually. For a bimodal distribution, the mean and median are
not very useful as measures of central tendency.
4.2.2 Measures of Variability (Heterogeneity or Spread) Variance
The most useful measure of variability around the central value is
the variance defined for sample data as
( ) ( )2
21
1
1...1
i N
N ii
Var sN
=
=
Φ Φ = = Φ − Φ− ∑ (4.6)
The standard deviation is the positive square root of the variance and is
given by
4-5
( ) ( )1 1... ...N Ns VarΦ Φ = Φ Φ (4.7)
Dykstra-Parsons Coefficient of Variation
A measure of permeability variability that is widely used in the
petroleum industry is the Dykstra-Parsons (1950) coefficient of variation,
V. This coefficient of variation is determined based on the assumption
that permeability data are drawn from a log normal distribution. The
calculation of the Dykstra-Parsons coefficient of permeability variation
involves plotting the frequency distribution of the permeability data on a
log-normal probability graph paper. This is done by arranging the
permeability values in descending order and then calculating for each
permeability, the percent of the samples with permeabilities greater than
or equal to that value. Table 4.1 shows an example calculation. Note
that to avoid values of 0 or 100%, which are not present on the
probability scale, the percent greater than or equal to value is normalized
by N+1, where N is the number of samples.
The data are plotted on a log-normal probability graph paper as
shown in Figure 4.1. Normally, such a plot gives a straight line, at least
when the central portion is used. The midpoint of the permeability
distribution ( )% 50≥ is the median permeability. The Dykstra-Parsons
coefficient of permeability variation, V, is defined as
50 84.1
50
k kVk−
= (4.8)
where k50 is the permeability value at ( )% 50≥ , which is the log mean
permeability and k84.1 is the permeability value at ( )% 84.1≥ , which is one
4–6
standard deviation from the mean. For the data of Table 4.1 and Figure
4.1,
N+1 = 20
k50 = 17.8 md
k84.1 = 6.1 md
50 84.1
50
17.8 6.1 0.6617.8
k kVk− −
= = =
Table 4.1. Calculation of Dykstra-Parsons Coefficient of Variation (V)
k (md)
No of Samples >= k % >= k
100 1 5 65 2 10 50 3 15 32 4 20 29 5 25 27 27 7 35 23 23 9 45 18 10 50 17 17 12 60 16 13 65 8 14 70 7 7 16 80 6 17 85 4 18 90 1 19 95
4-7
Figure 4.1. Log normal permeability distribution.
Dykstra-Parsons coefficient of variation is a dimensionless number
that ranges from 0 to 1. A homogeneous reservoir has a coefficient of
permeability variation that approaches 0 whereas an extremely
heterogeneous reservoir has a coefficient of permeability variation that
approaches 1. Petroleum reservoirs typically have Dykstra-Parsons
coefficients of permeability variation between 0.5 and 0.9. Figure 4.2
shows theoretical log normal permeability distributions and their
corresponding Dykstra-Parsons coefficients.
4–8
Figure 4.2. Theoretical log normal permeability distributions and their corresponding Dykstra-Parsons coefficients (Carlson, 2003).
Lorenz Coefficient
Another measure of heterogeneity used in the petroleum industry
is the Lorenz coefficient . The Lorenz coefficient of variation is obtained
by plotting a graph of cumulative kh versus cumulative φh, sometimes
called a flow capacity plot. Table 4.2 shows an example calculation
whereas Figure 4.3 shows the plot of cumulative kh versus cumulative
φh for determining the Lorenz coefficient. The Lorenz coefficient is
defined from Figure 4.2 as
area ABCALorenz Coefficientarea ADCA
= (4.9)
4-9
where ABCA is the cross-hatched area in the figure and ADCA is the
triangle below the cross-hatched area. From Figure 4.2, the Lorenz
coefficient of the data of Table 4.2 is about 0.65.
The Lorenz coefficient of variation also varies from 0 to 1.
Unfortunately, the Lorenz coefficient is not a unique measure of reservoir
heterogeneity. Several different permeability distributions can give the
same value of Lorenz coefficient. For log-normal permeability
distribtutions, the Lorenz coefficient is very similar to the Dykstra-
Parsons coefficient of permeability variation.
Table 4.2 . Calculation of Lorenz Coefficient of Variation
h k φ kh Σkh Σkh/Sumk
h
φh Σφh Σφh/Sumφh
(ft) (md) (fraction) (md-ft) (md-ft) (ft) (ft)
0.000 0.000
8.1 4388 0.22 35542.
8
35542.
8
0.399 1.782 1.782 0.062
2.0 2640 0.22 5280.0 40822.
8
0.459 0.440 2.222 0.078
3.7 2543 0.21 9409.1 50231.
9
0.564 0.777 2.999 0.105
5.0 1579 0.20 7895.0 58126.
9
0.653 1.000 3.999 0.140
4.0 930 0.20 3720.0 61846.
9
0.695 0.800 4.799 0.168
9.6 662 0.19 6355.2 68202.
1
0.766 1.824 6.623 0.232
6.4 441 0.20 2822.4 71024.
5
0.798 1.280 7.903 0.277
2.7 402 0.16 1085.4 72109.
9
0.810 0.432 8.335 0.292
4–10
5.6 401 0.20 2245.6 74355.
5
0.836 1.120 9.455 0.331
8.0 378 0.18 3024.0 77379.
5
0.870 1.440 10.895 0.381
4.0 267 0.21 1068.0 78447.
5
0.882 0.840 11.735 0.411
15.1 250 0.21 3775.0 82222.
5
0.924 3.171 14.906 0.522
5.9 249 0.17 1469.1 83691.
6
0.940 1.003 15.909 0.557
2.8 232 0.22 649.6 84341.
2
0.948 0.616 16.525 0.578
7.4 200 0.17 1480.0 85821.
2
0.964 1.258 17.783 0.622
9.2 136 0.20 1251.2 87072.
4
0.978 1.840 19.623 0.687
7.6 98 0.19 744.8 87817.
2
0.987 1.444 21.067 0.737
10.1 47 0.19 474.7 88291.
9
0.992 1.919 22.986 0.804
9.6 30 0.18 288.0 88579.
9
0.995 1.728 24.714 0.865
3.5 28 0.17 98.0 88677.
9
0.997 0.595 25.309 0.886
6.6 16 0.16 105.6 88783.
5
0.998 1.056 26.365 0.923
15.8 13 0.14 205.4 88988.
9
1.000 2.212 28.577 1.000
Sumkh = 88988.
9
Sumφh = 28.577
4-11
Figure 4.3. Flow capacity distribution.
While the Dykstra-Parsons and Lorenz coefficients give
quantitative measures of the permeability variation, they provide no
information on the spatial relationship between the permeability values.
It is well known that permeability and other petrophysical properties of
reservoir rocks are not randomly distributed but are spatially correlated.
There is a need for other measures of heterogeneity that take into
account the spatial correlation of the data.
4.3 MEASURES OF SPATIAL CONTINUITY
Figure 4.4 shows the spatial distributions of a petrophysical
property, Φ , measured at equally spaced coordinates in two linear
reservoirs A and B. Which of the two reservoirs is more heterogeneous
with respect to the property Φ ? Most professionals will say that reservoir
4–12
A is more heterogeneous than reservoir B. However, a careful
examination of the numerical values of the property shows them to be
the same in both reservoirs. For both reservoirs, the mean of Φ is 5.0,
the variance is 6.67 and the standard deviation is 2.58. Thus, by the
usual measure of ordinary statistics, the degree of heterogeneity of the
two reservoirs is the same. Yet, there is something about the two
reservoirs that leads one to conclude that A is more heterogeneous than
B. It is the spatial arrangement of the values of the property relative to
each that leads one to conclude that A is more heterogeneous than B. In
reservoir A, the property appears to be randomly distributed in space
whereas in reservoir B, it is distributed in an orderly and continuous
fashion. Thus, to fully characterize heterogeneity, the spatial correlation
structure of the property must be taken into account.
Three related functions are normally used to characterize the
spatial continuity of the data from a heterogeneous reservoir. These are
(1) the variogram (semi-variogram), (2) the covariance function and (3)
the correlation coefficient function.
4-13
Figure 4.4. Spatial distribution of a petrophysical property in two linear reservoirs A and B.
4–14
4.3.1. Variogram Definition
The variogram is a function obtained by plotting the semivariance
of the differences between the properties at two locations separated by a
distance h versus h. The variogram is defined as
( ) ( ) ( ) ( ) ( ) 2
2 2 h
x x hVar x x hh
Nγ
Φ − Φ +⎡ ⎤⎣ ⎦Φ − Φ +⎡ ⎤⎣ ⎦= =∑
(4.10)
where
γ = semivariance
h = lag distance
( )xΦ = value of property at location x
( )x hΦ + = value of property at location x+h
Nh = number of data pairs separated by the distance h.
Examination of Eq.(4.10) shows that each numerical value of γ is the
variance of ( ) ( )x x hΦ − Φ +⎡ ⎤⎣ ⎦ divided by 2, where the mean of
( ) ( )x x hΦ − Φ +⎡ ⎤⎣ ⎦ is normally assumed to be zero. Thus, each numerical
value of γ is the semivariance of ( ) ( )x x hΦ − Φ +⎡ ⎤⎣ ⎦ for a lag distance h and
the function γ (h) is the variogram or semivariogram.
Figure 4.5 shows an ideal variogram. It starts at zero and increases
with increasing lag distance until a certain distance is reached at which
it levels off and becomes constant. The lag distance at which the
variogram levels off (a in the figure) is defined as the correlation length,
4-15
or the range of influence, and the value of the variogram at this point is
called the sill. The sill is the semivariance of the entire data set. Thus,
Figure 4.5. Typical variogram
hidden in the variogram are the variance and standard deviation of the
data set, the usual measures of heterogeneity of ordinary statistics. If the
correlation length is zero, the spatial distribution of the property is fully
random. With increasing correlation length, the range of influence of one
value on its neighbors increases up to the correlation length. At lag
distances beyond the correlation length, the data are no longer
correlated.
The variogram and correlation length are directional quantities
(anisotropy again), and in general, will be different in different directions.
4–16
Sometimes, the variogram has a discontinuity at the origin as shown in
Figure 4.5. The value of the variogram at the discontinuity (Co in the
figure) is known as the nugget effect, a term that originates from the
mining industry which was the first industry to widely apply geostatistics
to estimate ore grade.
It should be emphasized that not all variograms have a nugget
effect, a sill or even a correlation length. Variograms come in different
shapes depending on the underlying geological structure of the
heterogeneity. Figure 4.5 was presented as one possible variogram shape
to introduce the general features of the function and the associated
nomenclature.
In order to demonstrate that the variogram does capture the
spatial continuity of the data from a heterogeneous reservoir, let us
compare the variograms for reservoirs A and B as shown in Figure 4.6.
The variogram for reservoir A is cyclical with a constant average value of
about 5.5 for all lag distances. The constant average value can be viewed
as a pure nugget effect, which indicates that the values of the property
are randomly distributed in space as is apparent from Figure 4.4. By
contrast, the variogram for reservoir B is continuous for all lag distances,
an indication of the orderly and continuous spatial distribution of the
property as evident in Figure 4.4. Therefore, the variogram (and later the
covariance and correlation functions) does capture the spatial continuity
of the data.
4-17
Figure 4.6. Comparison of the variograms for reservoirs A and B of Figure 4.4.
How to Calculate the Variogram
Let us consider the simple case of a linear reservoir in which the
property Φ is measured at 10 equally spaced locations with a distance of
Δx between each datum. Let the values of the property at locations 1
through 10 be 1 2 3 10, , ,...,Φ Φ Φ Φ . To compute the semivariance at each lag
distance, one can generate a table of ( )xΦ and ( )x hΦ + for each lag
distance as shown in Table 4.3. Using the entries in the table, the
semivariances can easily be calculated with Eq.(4.10). In fact, the
semivariance for h = 0 can be obtained by inspection as γ(0) = 0. The
4–18
semivariance at h = 0 is always equal to 0 even if the variogram has a
nugget effect.
Table 4.3. Table of Φ(x) and Φ(x+h) for Computing Semivariances of
Sample Data
h = 0 h = Δx h = 2Δx h = 3Δx h = 4Δx h = 5Δx
Nh = 10 Nh = 9 Nh = 8 Nh = 7 Nh = 6 Nh = 5
Φ(x
) Φ (x+h)
Φ(x
) Φ (x+h)
Φ(x
) Φ (x+h)
Φ(x
) Φ (x+h) Φ(x) Φ (x+h) Φ(x) Φ (x+h)
Φ1 Φ1 Φ1 Φ2 Φ1 Φ3 Φ1 Φ4 Φ1 Φ5 Φ1 Φ6
Φ2 Φ2 Φ2 Φ3 Φ2 Φ4 Φ2 Φ5 Φ2 Φ6 Φ2 Φ7
Φ3 Φ3 Φ3 Φ4 Φ3 Φ5 Φ3 Φ6 Φ3 Φ7 Φ3 Φ8
Φ4 Φ4 Φ4 Φ5 Φ4 Φ6 Φ4 Φ7 Φ4 Φ8 Φ4 Φ9
Φ5 Φ5 Φ5 Φ6 Φ5 Φ7 Φ5 Φ8 Φ5 Φ9 Φ5 Φ10
Φ6 Φ6 Φ6 Φ7 Φ6 Φ8 Φ6 Φ9 Φ6 Φ10
Φ7 Φ7 Φ7 Φ8 Φ7 Φ9 Φ7 Φ10
Φ8 Φ8 Φ8 Φ9 Φ8 Φ10
Φ9 Φ9 Φ9 Φ10
Φ10 Φ10
The following observations can be made about the calculation of
semivariances and the variogram.
1. The number of data pairs, Nh, decreases as h increases. Beyond h
= NΔx/2, the reduction in Nh causes the variogram to fluctuate
excessively and unmeaningfully. Therefore, the variogram is
typically truncated beyond h = NΔx /2.
2. The variogram is a non-negative function.
4-19
3. The variogram can be computed even for irregularly spaced data.
In this case, a different computational strategy must be used. Here
is one possible algorithm. Consider the non-uniformly distributed
data at ten locations in a linear reservoir as shown in Figure 4.7. If
there are N data points, there will be NC2 data pairs in the set
where NC2 is the combination of N things taken 2 at a time and is
given by
( )
( )2
1!2 !2! 2N
N NNCN
−= =
− (4.11)
For 10 data points, there will be 45 data pairs. The first task is to
compute and store the lag distances, hij, and the corresponding
values of ( )2
2i jΦ − Φ
for all the data pairs. The algorithm is as
follows:
1. Create two one-dimensional arrays of size equal to the
number of data pairs. Let these be H(M) and B(M) where M is
the number of data pairs as determined from Eq.(4.11).
2. Visit location 1, compute and store sequentially, the
following data pairs:
( )
( )
( )
22 1
2 1
23 1
3 1
210 1
10 1
,2
,2
...
,2
x x
x x
x x
Φ − Φ−
Φ − Φ−
Φ − Φ−
4–20
Figure 4.7. Irregularly spaced data.
where the first number is stored in array H and the second
number is stored in array B.
3. Eliminate 1Φ from the data set to prevent duplication.
4. Move to location 2 and continue to compute and store the
following data in the arrays
4-21
( )
( )
( )
23 2
3 2
24 2
4 2
210 2
10 2
,2
,2
...
,2
x x
x x
x x
Φ − Φ−
Φ − Φ−
Φ − Φ−
5. Eliminate 2Φ from the data set to avoid duplication.
6. Continue to compute and fill the arrays by visiting each
location in the manner described above until the last data
pair is added to the array. The last data pair for our example
will be
( )210 9
10 9 ,2
x xΦ − Φ
−
7. Perform a scatter plot of B(i) versus H(i), where array H now
contains the lag distances, hij, that were computed as
i jx x− . Such a scatter plot is shown in Figure 4.8.
8. Divide the data in the scatter diagram into bins as shown in
Figure 4.8. For each bin, compute and plot
( )21 1 versus
2i j
ijn n
hn n
Φ − Φ∑ ∑
where n is the number of data points in the bin, which can
be different for each bin. In constructing the bins, there
should be enough data points in each bin to prevent the
4–22
variogram from fluctuating excessively. Several bin sizes
may be tried to determine the optimum bin sizes.
9. The plot in step 8 is the experimental variogram as shown in
Figure 4.8.
Figure 4.8. Scatter plot for computing experimental variogram.
4-23
4. If there is an underlying trend in the data, the trend should be
subtracted from the data before computing the variogram. For
example, the following data set from a linear reservoir C has an
underlying trend or drift of the form 1trend xΦ = + :
3 6 9 12 15 15 14 13 12
This data set is presented in Figure 4.9. This trend or drift should
be subtracted from the data before the variogram is computed.
Figure 4.10 compares the variograms for reservoir C with and
without the underlying trend or drift. The variogram with the drift
is an ever increasing function of h because of the underlying trend
and will never reach the sill, whereas after removing the drift, the
resulting variogram is lower and reaches the sill at h = 4 km. This
is the true experimental variogram for reservoir C. In this example,
after subtracting the drift, the resulting distribution is the same as
in reservoir B. Thus, the true variogram for reservoir C is the same
as for reservoir B.
5. If data are missing from some locations, those locations should be
skipped over. Resist the temptation to interpolate and fill in the
missing data. Table 4.4 shows a data set in which data are
missing from location 7. Calculate the semivariance of the
permeability (NOT the natural log of permeability) at lag distances
of 1 foot and 2 feet. Start your calculations from the top of the
reservoir and work your way down.
4–24
Figure 4.9. Spatial distribution of a petrophysical property in a linear reservoir C with an underlying trend or drift.
Figure 4.10. Variograms for reservoir C with and without the underlying trend or drift.
4-25
Table 4.4. Porosity and Permeability Distributions with Missing Data.
Relativ
e
Depth
Porosit
y
Permeabilit
y
(ft) (%) (md)
1 20 93
2 16.3 18
3 9.7 8.4
4 16.2 21
5 14.9 10
6 12.7 1.7
7
8 5.5 25
9 5.8 17
10 6.5 4.8
11 4.7 22
12 7.3 5.6
6. If the data are distributed in 2D or 3D, the variogram can still be
computed using the basic method outlined for 1D data. In this
case, the variogram should be computed in several directions to
reveal any anisotropy that may be present. Figure 4.11 shows the
distribution of a petrophysical property in a 2D reservoir. In this
case, variograms should be computed in the following directions:
N-S, E-W, NE-SW and NW-SE.
4–26
Figure 4.11. Distribution of a petrophysical property in a two-dimensional reservoir.
7. Unless you are already familiar with the variogram, you may not
see what it has to do with spatial correlation of the data. We can
see that the entries in Table 4.3 have a lot to do with spatial
correlation by plotting the scatter diagrams of Φ(x+h) versus Φ(x)
for each value of h. For h = 0, Φ(x+h) and Φ(x) are perfectly
correlated and the data will follow the 45º line on the scatter plot
as shown in Figure 4.12. As h increases, the cloud of data points
scattered about the 45º line increases, indicating less and less
correlation between Φ(x+h) and Φ(x) as shown in Figures 4.13 and
4.14.
4-27
Figure 4.12. Scatter plot of Φ(x+h) versus Φ(x) for h = 0 km for reservoir B.
Figure 4.13. Scatter plot of Φ(x+h) versus Φ(x) for h = 1 km for reservoir B.
4–28
Figure 4.14. Scatter plot of Φ(x+h) versus Φ(x) for h = 3 km for reservoir B.
Physical Meaning of the Variogram
The scatter plot of Φ(x+h) versus Φ(x) at a fixed h can be used to
derive the physical meaning of the variogram. Consider the scatter plot
shown in Figure 4.15. The distance d from a datum point to the 45º line
is given by
( ) ( ) ( ) ( )cos 45
2
x h xd x h x
Φ + − Φ⎡ ⎤⎣ ⎦= Φ + − Φ =⎡ ⎤⎣ ⎦ (4.12)
( ) ( ) ( ) ( ) 222 2cos 45
2x h x
d x h xΦ + − Φ⎡ ⎤⎣ ⎦= Φ + − Φ =⎡ ⎤⎣ ⎦ (4.13)
From statistics, the expectation of d2 is given by
4-29
( )( ) ( ){ } ( ) ( )
( )
22
2
2 2 h
x h xE x h xE d h
Nγ
Φ + − Φ⎡ ⎤Φ + − Φ⎡ ⎤ ⎣ ⎦⎣ ⎦= = =
∑ (4.14)
Thus, the variogram is the mean of d2 about the 45º line at each lag
distance as a function of the lag distance h.
Figure 4.15. The h-scatter plot.
Variogram Models
The variogram is a means to an end not an end in itself. The
variogram is used to quantify the correlation structure of the variable of
interest for the purpose of estimation and conditional simulation. The
variogram of the sample data is known as the experimental variogram.
After computing the experimental variogram, a smooth theoretical
4–30
variogram model is usually fitted to the experimental variogram and the
model is then used for estimation. As will be shown later, the estimation
process involves the solution of a set of linear simultaneous algebraic
equations, whose coefficients are derived from the variogram. Unless the
variogram is well behaved, the simultaneous equations may not have a
solution. Hence, the need to fit a smooth and well behaved theoretical
model to the rough experimental variogram for the purpose of estimation.
Popular variogram models include (1) the spherical model, (2) the
exponential model, (3) the guassian model, (4) the linear model, (5) the
generalized linear model, (6) the nugget effect model, and (7) the cardinal
sine model (also known as the hole effect model).
1. The Spherical Model
The spherical model is given by
( )
( )
3
3
3 1 for 2 2
for
o
o
h hh C C h a
a a
h C C h a
γ
γ
⎛ ⎞= + ⎜ − ⎟ <
⎜ ⎟⎝ ⎠
= + ≥
(4.15)
where a is the correlation length or range, Co is the nugget effect if
present and C is the sill minus Co.
2. The Exponential Model
The exponential model is given by
( ) 1 ha
oh C C eγ⎛ ⎞
−⎜ ⎟⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟= + −⎜ ⎟⎝ ⎠
(4.16)
4-31
Figure 4.16 compares the spherical and the exponential models
with the same range and sill whereas Figure 4.17 compares the
two models with the same initial slope and sill. Note that there is
no nugget effect in these figures.
Figure 4.16. A comparison of the spherical and exponential models with the same range and sill.
4–32
Figure 4.17. A comparison of the spherical and exponential models with the same initial slope and sill.
3. The Gaussian Model
The Gaussian model is given by
( )2
2
1 ha
oh C C eγ
⎛ ⎞⎜ ⎟−⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟= + −⎜ ⎟⎜ ⎟⎝ ⎠
(4.17)
Figure 4.18 shows an example of a Gaussian model with Co = 0, C
= 10 and a = 4 units.
4-33
Figure 4.18. Example Gaussian model.
4. The Linear Model
The linear model is given by
( ) oh C m hγ = + (4.18)
where m is the slope. This model does not have a sill.
5. The Generalized Linear Model
The generalized linear model is given by
( ) 0 2oh C m h forαγ α= + < ≤ (4.19)
4–34
where m is a constant and α is an exponent between 0 and 2.
When α = 1, the model degenerates to the linear model. This model
does not have a sill. Figure 4.19 compares the linear model to the
generalized linear model.
Figure 4.19. The linear model and the generalized linear model.
5. The Nugget Effect Model
The nugget effect model is given by
( )( )0 0
for 0oh C h
γ
γ
=
= > (4.20)
This model gives the variogram of a property that has a random
spatial distribution. It is basically a spherical model with a very
4-35
small range of influence. Figure 4.20 shows an example nugget
effect model.
Figure 4.20. The nugget effect model.
6. The Cardinal Sine Model (Hole Effect Model)
The cardinal sine model or hole effect model is given by
( ) ( )sin /1
/o
h ah C C
h aγ
⎡ ⎤= + −⎢ ⎥
⎣ ⎦ (4.21)
Figure 4.21 shows an example cardinal sine model with Co = 0, C =
10 and a = 1 unit. The sinusoidal nature of the variogram is an
indication of the periodic nature of the underlying heterogeneity.
4–36
Figure 4.21. The cardinal sine model (hole effect model).
Fitting a Theoretical Variogram Model to an Experimental Variogram
Every experimental variogram can be fitted with a theoretical
variogram model. If one model does not fit the experimental variogram,
several models can be combined or nested to fit the experimental
variogram. The only restriction is that each of the combined models must
be applied to all the lag distances. It is not permissible to apply a model
up to a certain lag distance and then switch to a different model for the
remaining lag distances. The procedure for fitting a theoretical model to
an experimental variogram is a trial and error procedure.
To demonstrate the procedure for fitting a theoretical model to an
experimental variogram, let us compute the experimental variogram for
the permeability data from the results of the core analysis of Table 2.2
and then fit a theoretical model to it. To keep the magnitude of the
4-37
variogram manageable, we compute the variogram for the natural log of
the permeability rather than the permeability itself. Such a
transformation is frequently done in geostatistics to make the data more
manageable. The transformation affects the magnitude of the variogram
but not its shape. It is not unusual to transform the data, compute the
variogram, fit a theoretical model to the experimental variogram, use the
model to perform estimation and then transform the estimated data back
to its original units. It should be noted that the transformation in this
example does not imply that the permeability data is log normally
distributed. The transformation has been done purely for convenience.
The results of the calculations are summarized in Table 4.5.
Table 4.5. Variograms for Permeability Data of Table 2.2.
1 2 3 4 5 6 7 8
Uncorrecte
d
Correcte
d
γ(h) γ(h)
Depth X-
Coord
k lnk lnk h (ft) With Drift Without Drift
4807.
5
0 2.5 0.916 -3.812 0 0.000 0.000
4808.
5
1 59 4.078 -0.688 1 0.038 0.362
4809.
5
2 221 5.398 0.596 2 0.049 0.488
4810.
5
3 211 5.352 0.512 3 0.051 0.468
4811.
5
4 275 5.617 0.740 4 0.057 0.618
4812.
5
5 384 5.951 1.037 5 0.057 0.591
4813. 6 108 4.682 -0.269 6 0.050 0.545
4–38
5
4814.
5
7 147 4.990 0.002 7 0.054 0.574
4815.
5
8 290 5.670 0.644 8 0.061 0.630
4816.
5
9 170 5.136 0.073 9 0.060 0.623
4817.
5
10 278 5.628 0.528 10 0.065 0.632
4818.
5
11 238 5.472 0.335 11 0.065 0.641
4819.
5
12 167 5.118 -0.057 12 0.068 0.728
4820.
5
13 304 5.717 0.505 13 0.075 0.780
4821.
5
14 98 4.585 -0.664 14 0.064 0.605
4822.
5
15 191 5.252 -0.034 15 0.074 0.684
4823.
5
16 266 5.583 0.260 16 0.077 0.628
4824.
5
17 40 3.689 -1.672 17 0.057 0.485
4825.
5
18 260 5.561 0.163 18 0.084 0.632
4826.
5
19 179 5.187 -0.248 19 0.086 0.642
4827.
5
20 312 5.743 0.271 20 0.097 0.720
4828.
5
21 272 5.606 0.097 21 0.098 0.655
4829.
5
22 395 5.979 0.432 22 0.105 0.769
4830. 23 405 6.004 0.420 23 0.115 0.787
4-39
5
4831.
5
24 275 5.617 -0.004 24 0.109 0.701
4832.
5
25 852 6.748 1.089 25 0.135 0.946
4833.
5
26 610 6.413 0.718 26 0.136 0.830
4834.
5
27 406 6.006 0.274 27 0.132 0.880
4835.
5
28 535 6.282 0.513 28 0.149 0.965
4836.
5
29 663 6.497 0.690 29 0.164 1.170
4837.
5
30 597 6.392 0.548
4838.
5
31 434 6.073 0.192
4839.
5
32 339 5.826 -0.092
4840.
5
33 216 5.375 -0.580
4841.
5
34 332 5.805 -0.188
4842.
5
35 295 5.687 -0.343
4843.
5
36 882 6.782 0.715
4844.
5
37 600 6.397 0.292
4845.
5
38 407 6.009 -0.133
4847.
5
40 479 6.172 -0.044
4–40
4848.
5
41 0
4849.
5
42 139 4.934 -1.356
4850.
5
43 135 4.905 -1.422
4851.
5
44 0
Figure 4.22 shows the experimental variogram for the natural log of
permeability, lnk. The graph is the plot of the data in columns 6 and 7 of
Table 4.5. The ever increasing nature of the variogram indicates an
underlying trend or drift in the permeability data. Before addressing the
problem of the underlying trend, let us fit a theoretical model to the
experimental variogram to demonstrate the procedure.
Figure 4.22. Experimental variogram for natural log of permeability for core analysis data.
4-41
Figure 4.23 shows the fitted theoretical model from trial and error.
The nested model is
( )3
23
3 10.027 0.025 0.0001012 3 2 3
h hh hγ
⎛ ⎞= + ⎜ − ⎟ +
⎜ ⎟⎝ ⎠
The nested model is of the form
( ) h Spherical Model Generalized Linear Modelγ = +
Figure 4.23. Theoretical variogram model fit to the experimental
variogram of core analysis data.
4–42
Let us now deal with the underlying trend in the permeability data.
Figure 4.24 shows a scatter plot of the log permeability data versus
relative depth together with the regression line. Clearly, there is an
underlying trend or drift in the permeability data. The trend line is
ln 0.0372 4.7281k x= +
where x is the relative depth. This trend was subtracted from the lnk
data to obtain the corrected lnk data shown in column 5 of Table 4.5.
Figure 4.24. Permeability data showing underlying trend or drift.
4-43
Figure 4.25 shows the scatter plot for the corrected permeability data
together with the regression line. Clearly, the underlying trend has been
removed.
Figure 4.26 shows the experimental variogram and the theoretical
model fit after the drift has been removed. The experimental variogram is
the graph of the data in columns 6 and 8 of Table 4.5. We see that after
removing the underlying trend, the experimental variogram can be fitted
with a spherical model of the form
( )3
3
3 10.31 0.352 9 2 9
h hhγ
⎛ ⎞= + ⎜ − ⎟
⎜ ⎟⎝ ⎠
.
Figure 4.25. Permeability data after the drift has been removed.
4–44
The true variogram of the natural log of permeability has a nugget effect
of 0.31, a sill of 0.66 and a correlation length of 9 ft.
Variogram Anisotropy
Variograms computed in different directions can show anisotropy.
Figure 4.27 shows two such anisotropies. In Figure 4.27a, the sills of the
variograms in the two directions are the same but the correlations
lengths are different. In Figure 4.27b, the slopes of the variograms in the
two directions are different. We can compute the correlation lengths or
the slopes of the variograms in different directions and plot them as
Figure 4.26. Variogram of natural log of permeability after removing the
underlying trend.
4-45
shown in Figure 4.28. If the plot turns out to be an ellipse as shown in
the figure, then a simple coordinate transformation can be used to
compute an equivalent isotropic variogram. If γ1(h) is the variogram in
direction 1, then the equivalent isotropic variogram is given by
( )2
2 211 1 2
2
ah h ha
γ γ⎛ ⎞⎛ ⎞⎜ ⎟= + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(4.22)
for the case of the correlation length anisotropy and
( )2
2 211 1 2
2
slopeh h hslope
γ γ⎛ ⎞⎛ ⎞⎜ ⎟= + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(4.23)
for the case of the slope anisotropy.
Figure 4.27. Geometric anisotropy.
4–46
Figure 4.28. Ellipses showing variogram anisotropy.
Figure 4.28b shows the importance of computing the variograms in at
least four directions in a 2D data set to reveal possible anisotropy. If the
variograms were computed only in the horizontal and vertical directions,
the anisotropy would be missed because, in the case shown in the figure,
the principal axes of the anisotropy make an angle of 45º with the
vertical and horizontal axes. As a result, the correlation lengths in the
vertical and horizontal directions are the same and would not reveal the
presence of the anisotropy.
In sedimentary rocks, because of layering, the sill of the variogram
in the vertical direction is usually different from that in the horizontal
direction because the degree of heterogeneity normal to the layers is
higher than along the layers. In this case, the variogram can be split into
two components, an isotropic component given by ( )2 2 21 2 3o h h hγ + + and a
vertical component given by ( )3v hγ . The overall variogram is then given
by
( ) ( ) ( )2 2 21 2 3 3o vh h h h hγ γ γ= + + + (4.24)
4-47
Example Experimental Variograms
Presented in this section are the experimental variograms for the
sandpack and Berea sandstone cores whose porosity distributions were
presented in Figures 2.7 to 2.10. Experimental variograms can be
computed for any property that varies in space. Here, we present the
variograms for the CT numbers of the dry scans of the sandpack and the
Berea sandstone. The CT number is proportional to the bulk density of
the sample and is similar to the measurements from a density log. Each
3D image data set consisted of 128x128x50 voxels (volume elements) for
a total of 812,200 data points. For each sample, variograms were
computed in three orthogonal directions (Z, Y and X) as shown in Figure
4.29.
Figure 4.29. Orthogonal directions in which variograms were computed for CT data.
4–48
Figure 4.30 shows CT images for a vertical slice and a typical
cross-sectional slice of the sandpack. The packing method introduced
radial and longitudinal heterogeneities into the sandpack. Figures 4.31
and 4.32 show the experimental variograms for the sandpack in the
transverse and longitudinal directions. Figure 4.31 shows that the
vertical and horizontal variograms are essentially the same, confirming
the radial symmetry observed in the CT images of the sandpack. The
variogram in the longitudinal direction (Figure 4.32) has a wavy or
sinusoidal characteristic. This is caused by the longitudinal
heterogeneity introduced into the sandpack by the packing method. This
characteristic is known as a trend surface.
4-49
Figure 4.30. Vertical and cross-sectional slices of CT image of dry
sandpack (Peters and Afzal, 1992).
Figure 4.31. Experimental variograms for sandpack in the transeverse
directions (Y, Z) (Peters and Afzal, 1992).
4–50
Figure 4.32. Experimental variogram for sandpack in the longitudinal
direction (X) (Peters and Afzal, 1992).
Figure 4.33 shows CT images for a vertical slice and a typical
cross-sectional slice of the Berea sandstone. Figures 4.34 and 4.35 show
the experimental variograms for the Berea sandstone in the transverse
and longitudinal directions. Figure 4.34 shows a wavy characteristic in
the vertical direction (Y-direction). This is caused by the layering in that
direction. As expected, the magnitudes of the variograms show that the
medium is more heterogeneous in the Y-direction (across the layers) than
in the Z-direction (along the layers). Figure 4.35 shows that the
correlation length in the longitudinal direction (X-direction) is about 40
cm.
Clearly, the variograms are able to characterize the spatial
structure of the heterogeneity in the property of interest, in this case, the
x-ray absorption coefficient of a sandpack and a Berea sandstone.
4-51
4–52
Figure 4.33. Vertical and cross-sectional slices of CT image of dry Berea sandstone (Peters and Afzal, 1992).
Figure 4.34. Experimental variograms for Berea sandstone in the transeverse directions (Y, Z) (Peters and Afzal, 1992).
4-53
Figure 4.35. Experimental variogram for Berea sandstone in the longitudinal direction (X) (Peters and Afzal, 1992).
4.3.2. Covariance (Autocovariance) Function Definition
For sample data, the covariance function at a lag distance h is
defined as
( )( ) ( ) ( ) ( )
1h
x x x h x hC h
N
⎡Φ − Φ ⎤ ⎡Φ + − Φ + ⎤⎣ ⎦ ⎣ ⎦=
−
∑ (4.25)
The following observations can be made about the covariance function:
1. The covariance function gives the strength of the linear
relationship between Φ(x) and Φ (x+h).
2. The covariance function can be positive or negative. Recall that the
variogram function was non-negative.
4–54
3. Because the covariance function depends on the units of
measurement of Φ (x), it is not always possible to tell from the
magnitude of the covariance if the linear correlation is strong or
weak. A better measure of the strength of the linear correlation is
a dimensionless covariance function known as correlation
coefficient to be discussed next.
4. The table of data used to calculate the variogram can also be used
to calculate the covariance function. The algorithm for computing
the covariance function for unequally spaced data is similar to that
for computing the variogram. In this case, three one-dimensional
arrays should be used to store hij, iΦ and jΦ as the locations are
visited.
5. It will be shown that for a stationary random field, the variogram
and the covariance function are related by the equation
( ) ( ) ( )0h C C hγ = − (4.26)
where C(0) is the covariance at a zero lag distance and is equal the
variance of the data, s2. The fact that C(0) is the variance of the
data set is apparent from the entries in Table 4.3 for h = 0. The
variance of the data is given by
( )( ) ( ) ( ) ( ) ( ) ( ) 2
201 1
x x x x x xC s
N N
⎡Φ − Φ ⎤ ⎡Φ − Φ ⎤ ⎡Φ − Φ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦= = =
− −
∑ ∑ (4.27)
Thus, like the variogram, the covariance function also has hidden
in it the usual measure of variability by ordinary statistics.
4-55
Figure 4.36 shows a comparison of the covariance functions for
reservoirs A and B. The covariance function for reservoir A is periodic
indicating the random spatial distribution of the property in reservoir A
whereas the covariance function for reservoir B is smooth and orderly,
indicating the continuous nature of the spatial distribution of the
property in reservoir B. These are the same features observed in the
variograms. In fact, a comparison of Figures 4.36 and 4.6 shows that the
covariance function is roughly the variogram turned up-side-down about
the h-axis in the spirit of Eq.(4.26). This observation is more apparent in
Figures 4.37 and 4.38, which compare the variogram and the covariance
function for each reservoir. Thus, like the variogram, the covariance
function captures the correlation structure of the spatial distribution of
the property of interest and can be used for the purpose of estimation
instead of the variogram.
Figure 4.36. A comparison of the covariance functions for reservoirs A and B.
4–56
Figure 4.37. A comparison of the variogram and the covariance function for reservoir A.
4-57
Figure 4.38. A comparison of the variogram and the covariance function for reservoir B.
Physical Meaning of Covariance Function
The covariance function measures the covariation of Φ(x) and
Φ(x+h) about their respective means. Figure 4.39 shows the scatter plot
for the covariation of ( ) ( )x h x h⎡Φ + − Φ + ⎤⎣ ⎦ versus ( ) ( )x x⎡Φ − Φ ⎤⎣ ⎦ at a fixed
lag distance h. The sum of the shaded areas divided by (Nh-1) is the
covariance function at that lag distance. Because the areas can be
positive or negative depending on which quadrant the scatter plot data
fall in, the covariance can be positive or negative. If the scatter plots are
concentrated along the axis AB in Figure 4.40, then all the areas will be
positive and the covariance will show a strong positive correlation. If the
scatter plots are concentrated along the axis CD, then all the areas are
negative and the covariance will show a strong negative correlation.
4–58
Figure 4.39. Scatter plot of covariation at a fixed lag distance.
4-59
Figure 4.40. Patterns of covariation at a fixed lag distance.
Figure 4.41 shows the scatter plot of the covariation for reservoir A at a
lag distance of 1 unit. The data are concentrated in the quadrants in
which all the areas are negative. Therefore, the covariance function will
be negative at this lag distance as can be confirmed in Figure 4.36 or
4.37.
4–60
Figure 4.41. Covariation for reservoir A at a lag distance of 1 unit.
A cross covariance function can be computed between two different
variables such as permeability and porosity. In that case, the cross
covariance is defined as
( )( )( )
,1
i ik kC k
N
φ φφ
− −=
−
∑ (4.28)
4.3.3. Correlation Coefficient Function (Autocorrelation Function) Definition
The correlation coefficient function is the dimensionless version of
the covariance function and is defined as
4-61
( ) ( )( ) ( )x x h
C hh
s sρ
Φ Φ +
= (4.29)
where ( )xsΦ is the standard deviation of ( )xΦ and ( )x hsΦ + is the standard
deviation of ( )x hΦ + .
The following observations can be made about the correlation
coefficient function.
1. Like the covariance function, the correlation coefficient function
gives the strength of the linear relationship between Φ(x) and
Φ(x+h).
2. The correlation coefficient function is dimensionless and can vary
in value from –1 to +1 as may be seen in Figure 4.42, which shows
the correlation coefficient function for reservoir A.
Figure 4.42. Correlation coefficient function for reservoir A.
4–62
3. The magnitude of the correlation coefficient function is a measure
of the strength of the linear relationship between Φ(x) and Φ (x+h).
4. The sign of the correlation coefficient indicates the nature of the
correlation. If low Φ (x) are paired with low Φ (x+h) or high Φ (x) are
paired with high Φ (x+h), the correlation coefficient will be positive.
If low Φ (x) are paired with high Φ (x+h) or high Φ (x) are paired
with low Φ (x+h), the correlation coefficient will be negative.
5. If Φ (x) and Φ (x+h) are independent, then the correlation
coefficient (and the covariance) will be zero.
6. A zero correlation coefficient (or covariance) does not necessarily
mean that Φ (x) and Φ (x+h) are independent. They could be related
as a quadratic or other nonlinear function. Remember that the
correlation coefficient is a test for a linear relationship not a
quadratic relationship.
A cross correlation function between two variables such as
permeability and porosity can be computed as
( ) ( ),,
k
C kk
s sφ
φρ φ = (4.30)
4.4 PROBABILITY DISTRIBUTIONS
There are several probability distributions in statistics, but only
the normal and the log normal distributions are presented in this section
because they are the two distributions most commonly observed in
petrophysical properties. Based on core analyses and field
4-63
measurements, it has been observed that porosity tends to follow a
normal distribution (see Figures 2.9 and 2.10) whereas permeability
tends to follow a log normal distribution.
4.4.1 Normal (Gaussian) Distribution
The probability density function for a normal distribution, also
commonly known as the Gaussian, is given by
( ) ( )2
2
1 exp for 22
xP x x
μσσ π
⎡ ⎤−= − − ∞ ≤ ≤ ∞⎢ ⎥
⎢ ⎥⎣ ⎦ (4.31)
where x is a random variable and μ is the mean, σ2 is the variance and σ
is the standard deviation of the distribution. The distribution is
characterized by two parameters, the mean (μ) and the standard
deviation (σ). Figure 4.43 shows a normal distribution with μ = 5 and
standard σ = 4. The distribution is centered on the mean and is
symmetric about the mean. For this distribution, all the measures of the
central tendency of the distribution (mean, median and mode) have the
same numerical value. Note that the distribution extends from −∞ to +∞ .
Using the transformation
xz μσ−
= (4.32)
all normal distributions can be can standardized as
( )21 exp for
22zP z z
π⎛ ⎞
= − − ∞ ≤ ≤ +∞⎜ ⎟⎝ ⎠
(4.33)
4–64
where z is a random variable with μ = 0 and σ = 1. Figure 4.44 shows a
standard normal distribution with the horizontal axis calibrated in units
Figure 4.43. Normal (Gaussian) distribution with μ = 5 and σ = 4.
of standard deviation. Also shown are the areas under the curve for
certain ranges of the horizontal axis. For example, about 68% of the data
are contained within one standard deviation below and above the mean.
Thus, for a normal distribution, the following are true
( ) 0.6826P z σ≤ = (4.34)
( )1.96 0.9500P z σ≤ = (4.35)
( )2 0.9544P z σ≤ = (4.36)
4-65
( )3 0.9973P z σ≤ = (4.37)
( ) 0.1588P z σ−∞ ≤ ≤ − = (4.38)
( ) 0.8413P z σ−∞ ≤ ≤ + = (4.39)
Figure 4.44. Standard normal distribution with baseline in standard deviation units.
The cumulative distribution function for a normal distribution is
defined as
( ) ( )2
2
1 exp22
x xF x dx
μσσ π −∞
⎡ ⎤−= −⎢ ⎥
⎢ ⎥⎣ ⎦∫ (4.40)
4–66
Figure 4.45 shows the cumulative distribution plot on linear graph
paper. When potted on a special probability graph paper, the cumulative
distribution function plots as a straight line as shown in Figure 4.46.
Figure 4.45. Cumulative normal distribution, mean μ and standard deviation σ.
The cumulative distribution function is difficult to compute because the
integration called for in Eq.(4.40) cannot be performed analytically. Only
numerical integration is possible. It is not particularly convenient to
integrate any function numerically from to +−∞ ∞ . To overcome this
difficulty, we undertake a transformation that allows the integration of
Eq.(4.40) using the complementary error function, which is tabulated in
mathematical handbooks.
4-67
Figure 4.46. Cumulative normal distribution plotted on probability paper.
4–68
Let
2
xu μσ
−= (4.41)
2
dxduσ
= (4.42)
Substituting Eqs.(4.41) and (4.42) into (4.40) gives the cumulative
distribution function as
( ) 21 u uF u e duπ
−
−∞= ∫ (4.43)
The function 2ue
π
−
is symmetric about u = 0. Therefore,
2 21 11
u u u
ue du e du
π π∞− −
−∞= −∫ ∫ (4.44)
Substituting Eq.(4.44) into (4.43) and rearranging gives
( ) 21 222
u
uF u e du
π∞ −⎛ ⎞= −⎜ ⎟
⎝ ⎠∫ (4.45)
Eq.(4.45) can be written as
( ) ( )1 22
F u erfc u= −⎡ ⎤⎣ ⎦ (4.46)
where
4-69
( ) 22 u
uerfc u e du
π∞ −= ∫ (4.47)
The function, erfc(u), is the complementary error function, which arises
frequently in the solutions of certain partial differential equations of
mathematical physics such as the diffusivity equation, heat conduction
equation, diffusion equation and convection-dispersion equation. The
error function, erf(u), is defined as
( ) 2
0
2 u uerf u e duπ
−= ∫ (4.48)
Thus,
( ) ( )1erf u erfc u= − (4.49)
Table 4.6 gives the error and the complementary error functions whereas
Figure 4.48 shows graphs of the two functions. Using the information in
Table 4.6, the cumulative distribution function F(x) can be computed
with Eq.(4.46) instead of Eq.(4.40) and plotted against x. Figure 4.48
shows the F(x) obtained from such a computation for the normal
distribution of Figure 4.43.
Table 4.6. Error and Complementary Error Functions
x erf(x) erfc(x)
0.00 0 1
0.05 0.05637
2
0.94362
8
0.10 0.11246
3
0.88753
7
Useful
relationships
0.15 0.16799 0.83200
4–70
6 4
0.20 0.22270
3
0.77729
7
erfc(-x) = 1 + erf(x)
0.25 0.27632
6
0.72367
4
0.30 0.32862
7
0.67137
3
erf(-x) = -erf(x)
0.35 0.37938
2
0.62061
8
0.40 0.42839
2
0.57160
8
erfc(x) = 1 - erf(x)
0.45 0.47548
2
0.52451
8
0.50 0.52050
0
0.47950
0
0.55 0.56332
3
0.43667
7
0.60 0.60385
6
0.39614
4
0.65 0.64202
9
0.35797
1
0.70 0.67780
1
0.32219
9
0.75 0.71115
6
0.28884
4
0.80 0.74210
1
0.25789
9
0.85 0.77066
8
0.22933
2
0.90 0.79690
8
0.20309
2
0.95 0.82089
1
0.17910
9
1.00 0.84270 0.15729
4-71
1 9
1.10 0.88020
5
0.11979
5
1.20 0.91031
4
0.08968
6
1.30 0.93400
8
0.06599
2
1.40 0.95228
5
0.04771
5
1.50 0.96610
5
0.03389
5
1.60 0.97634
8
0.02365
2
1.70 0.98379
0
0.01621
0
1.80 0.98909
1
0.01090
9
1.90 0.99279
0
0.00721
0
2.00 0.99532
2
0.00467
8
2.10 0.99702
1
0.00297
9
2.20 0.99813
7
0.00186
3
2.30 0.99885
7
0.00114
3
2.40 0.99931
1
0.00068
9
2.50 0.99959
3
0.00040
7
2.60 0.99976
4
0.00023
6
2.70 0.99986 0.00013
4–72
6 4
2.80 0.99992
5
0.00007
5
2.90 0.99995
9
0.00004
1
3.00 0.99997
8
0.00002
2
Figure 4.47. Graphs of error and complementary error functions.
4-73
Figure 4.48. Cumulative distribution function calculated with the complementary error function for the normal distribution of Figure 4.43.
Generation of Random Deviates from a Normal (Gaussian) Distribution
Sometimes, we wish to generate sample data from a normal
distribution. Two methods are presented for generating such data. The
first method is based on the Central Limit Theorem (CLT) and the second
method is the Box Muller method.
The Central Limit Theorem shows that the mean of a group of
numbers drawn randomly from any distribution tends to a normal
(Gaussian) distribution as the number of means increases. Thus, if we
calculate many times the sums of N variates drawn from a uniform
distribution, we should expect the sums to fall into a truncated normal
(Gaussian) distribution bounded by 0 and N, with a mean value of N/2. If
we generate N values of a random number ri from a uniform distribution
( )0 1ir≤ ≤ and calculate
4–74
1 2
N
ii
Nz r=
= −∑ (4.50)
the random variable z will be drawn from an approximately Gaussian
distribution with a mean μ and a standard deviation σ given by
0μ = (4.51)
12Nσ = (4.52)
The choice of N = 12 is particularly convenient because z will be a normal
(Gaussian) distribution with a mean of 0 and a standard deviation of 1.
Such a random variate can be used to draw a random sample, x, from a
normal (Gaussian) distribution of mean μ and standard deviation σ as
x zμ σ= + (4.53)
Eqs.(4.51) and (4.52) can be proved as follows. For a uniform
distribution of r between a and b, the mean or expectation of the
distribution is given by
[ ]2
b
a
r a bE r drb a
μ += = =
−∫ (4.54)
The variance or second moment is given by
( ) ( )( ) [ ]( )222 2Var r E r E r E r E rσ ⎡ ⎤ ⎡ ⎤= = − = −⎣ ⎦⎣ ⎦ (4.55)
But
4-75
2 2 2
2
3b
a
r a ab bE r drb a
+ +⎡ ⎤ = =⎣ ⎦ −∫ (4.56)
Substituting Eqs.(4.54) and (4.56) into (4.55) gives
( ) ( ) ( )2 22 22
3 4 12a b b aa ab bVar rσ
+ −+ += = − = (4.57)
The mean of the variable z is given by
[ ] [ ] [ ] [ ]1 21
...2 2
i N
i Ni
N NE z E r E r E r E r Eμ=
=
⎡ ⎤ ⎡ ⎤= = − = + + + −⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦∑ (4.58)
For the case of a = 0 and b = 1, Eq.(4.54) gives E[r]=1/2. Thus,
substituting this into Eq.(5.58) gives
[ ] 1 1 1... 02 2 2 2 2 2
N N NE zμ = = + + + − = − = (4.59)
The variance of z is given by
( )21 2 ...
2NNVar z Var r r rσ ⎡ ⎤= = + + + −⎢ ⎥⎣ ⎦
(4.60)
Since r1, r2, …, rN are independent, then
( ) ( ) ( ) ( )21 2 ...
2NNVar z Var r Var r Var r Varσ ⎛ ⎞= = + + + − ⎜ ⎟
⎝ ⎠ (4.61)
Substituting Eq.(4.57) with a = 0 and b = 1 into Eq.(4.61) gives
( )2 1 1 1... 012 12 12 12
NVar zσ = = + + + − = (4.62)
4–76
The standard deviation for z is
12Nσ = (4.63)
Eqs.(4.59) and (4.63) are the same as Eqs.(4.51) and (4.52).
The method of Box and Muller for generating normal deviates from
a uniform random number generator is as follows. Select two random
numbers r1 and r2 from a uniform distribution ( )0 1ir≤ ≤ . Calculate
( )1 1 22 ln cos 2z r rπ= − (4.64)
( )2 1 22 ln sin 2z r rπ= − (4.65)
It can be shown that z1 and z2 are random deviates from a normal
(Gaussian) distribution with a mean of 0 and a standard deviation of 1.
The deviates can be used to draw two random variates (x1 and x2) with
mean μ and standard deviation σ as
1 1x zμ σ= + (4.66)
2 2x zμ σ= + (4.67)
4.4.2 Log Normal Distribution
The probability density function for a log normal distribution is
given by
4-77
( ) ( )2
2
ln1 1 exp 22x
P xx
μσσ π
⎡ ⎤−= −⎢ ⎥
⎢ ⎥⎣ ⎦ (4.68)
where x is a random variable and μ is the mean, σ2 is the variance and σ
is the standard deviation of lnx. The mean, τ, and the variance, ω2, of the
log normal distribution are related to the parameters of lnx as
2
exp2
στ μ⎛ ⎞
= +⎜ ⎟⎝ ⎠
(4.69)
( ) ( )22 21 exp 2eσω μ σ= − + (4.70)
It should be noted that x has a log normal distribution whereas lnx has a
normal distribution.
Figure 4.49 shows a log normal distribution. The distribution is
positively skewed with a long tail towards the high values of x. Note that
x is always positive. Because of the high values in the tail, the mean is
larger than the median, which in turn is larger than the mode or
geometric mean. Note that in this case, the geometric mean is a better
measure of central tendency than the mean or the median.
4–78
Figure 4.49. Log normal distribution.
Figure 4.50 shows the cumulative distribution function for a log
normal distribution plotted on a log normal probability graph paper. On
this scale, the log normal distribution plots as a straight line. Such a plot
was used in Figure 4.1 to determine the Dykstra-Parsons coefficient of
permeability variation on the assumption the data were drawn from a log
normal distribution.
4-79
Figure 4.50. Cumulative log normal distribution plotted on log normal probability paper.
4.5 ESTIMATION
4.5.1 Introduction
Estimation is one of the major applications of geostatistics, the
other being conditional simulation. The objective is to estimate the
4–80
variable of interest at a location xo for which no datum has been
measured, using the data measured at other locations xi distributed in
space. Suppose you are given some petrophysical data from a linear
reservoir as shown in Figure 4.51 and requested to estimate the values at
locations 3, 5 and 8 at which no measurements were made. What would
you do? Most professionals would construct the diagram shown
Figure 4.51. Sample data from a linear reservoir.
in Figure 4.52 to estimate * *3 5and Φ Φ but would be at a loss about how to
estimate *8 Φ , where the asterisk is used in the symbols to distinguish
the estimated from the measured values. In fact, some would claim that
there was not enough information to estimate *8Φ because no data were
measured beyond location 8.
4-81
Figure 4.52. Estimation of unmeasured values by linear interpolation.
Let us write the equation for the estimator *3Φ . From linear
interpolation
( )* 3 23 2 4 2
4 2
x xx x
⎛ ⎞−Φ = Φ + Φ − Φ⎜ ⎟−⎝ ⎠
(4.71)
Let
3 2 234
4 2 42
x x hx x h
λ⎛ ⎞−
= =⎜ ⎟−⎝ ⎠ (4.72)
where hij are lag distances. Substituting Eq.(4.72) into (4.71) and
rearranging gives the estimator as
4–82
*3 2 2 4 4λ λΦ = Φ + Φ (4.73)
where
2 41λ λ= − (4.74)
or
2 4 1λ λ+ = (4.75)
We can easily generalize the estimator of Eq.(4.73) and the constraint of
Eq.(4.75) as follows:
*
1
i N
o i ii
λ=
=
Φ = Φ∑ (4.76)
subject to the constraint
1
1N
ii
λ=
=∑ (4.77)
In Eq.(4.76), the subscript o has been used to indicate the location at
which an estimate is to be made and N could be the entire available data
set or a subset of the available data depending on the structure of the
heterogeneity. Let us now use Eqs.(4.76) and (4.77) to estimate * * *3 5 8, and Φ Φ Φ . For *
3Φ , λ2 = λ4 = 0.5 and
( )( ) ( )( )*3 0.50 30 0.50 50 40Φ = + =
Similarly
4-83
( )( ) ( )( )*5 0.50 50 0.50 20 35Φ = + =
To estimate *8Φ , we proceed as follows. The weights, λi, in Eq.(4.76)
should be assigned to each measured value used in the estimation
depending on how close the measured value is to the point xo at which
the estimate is to be made. Unfortunately, there is no mathematical
expression for closeness. We can measure how far away each sample is
from the point of estimation but we cannot measure how close it is to the
point of estimation. Let us measure the lag distances of the measured
data from the point of estimation as
2
4
6
642
= 12
o
o
o
x x xx x xx x xTotal x
− = Δ− = Δ
− = ΔΔ
Based on the above lag distances, it is clear that 6Φ should have the most
weight, followed by 4Φ and 2Φ . We can adopt several different strategies
to accomplish this. Here is one strategy. Let us make the weights
proportional to the inverse lag distances as follows:
2
4
6
0.16676
0.25004
0.50002 = 0.9167Total
αλ α
αλ α
αλ α
α
= =
= =
= =
where α is to be chosen to satisfy the constraint of Eq.(4.77). In this case,
α = 1.0909. Thus,
4–84
( )( )
( )( )
( )( )
2
4
6
0.1667 0.1667 1.0909 0.18196
0.2500 0.2500 1.0909 0.27274
0.5000 0.5000 1.0909 0.54552 = 1.0000Total
αλ α
αλ α
αλ α
= = = =
= = = =
= = = =
Therefore,
( )( ) ( )( ) ( )( )*8 0.1819 30 0.2727 50 0.5455 20 30Φ = + + =
It should be noted that in estimating * *3 5and Φ Φ , only a subset of the
measured data was used whereas in estimating *8Φ , all the measured
data were used. However, there was no good reason to use the subset of
the measured data to estimate * *3 5and Φ Φ . All the measured data could
have been used to estimate * *3 5and Φ Φ in the manner similar to the
estimation of *8Φ .
Having completed the estimation, the next question is what is the
reliability of the estimates? What are the error bounds on the estimates?
Most professionals would give up at this point. This is where geostatistics
can help. Before presenting the geostatistical estimation technique
known as ordinary kriging, it is worthwhile to mention the major
limitation of the estimation undertaken above. It is that the structure of
the heterogeneity as represented by either the variogram, the covariance
function or the correlation coefficient function has not be taken into
account in the estimation.
Geostatistics uses a probabilistic framework to make estimations
and in so doing provides estimates that honor the correlation structure of
4-85
the heterogeneity. It also provides quantitative assessment of the
reliability of the estimates. Let us demonstrate a simple geostatistical
estimation using a probabilistic framework. Suppose we wish to estimate
the missing datum in Figure 4.53. Proceeding in the spirit of Eqs.(4.76)
and (4.47), let us use the first "ring" of data around the missing datum
(or rather first "square" of data) to obtain an estimate as
*1 2 3 4 5 6 7 819 18 17 15 18 15 15 18o λ λ λ λ λ λ λ λΦ = + + + + + + + (4.78)
The error of the estimation, which is unknown, is given by
8
*1 2 3 4 5 6 7 8
119 18 17 15 18 15 15 18
i
o o o o ii
e λ λ λ λ λ λ λ λ λ=
=
= Φ − Φ = + + + + + + + − Φ ∑ (4.79)
where oΦ is the unknown missing datum. Notice that the constraint of
Eq.(4.47) has been used in Eq.(4.79). The error can be expanded as
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
*1 2 3 4
5 6 7 8
19 18 17 15
18 15 15 18
o o o o o o o
o o o o
e λ λ λ λ
λ λ λ λ
= Φ − Φ = − Φ + − Φ + − Φ + − Φ
+ − Φ + − Φ + − Φ + − Φ (4.80)
4–86
Figure 4.53. Missing data from a 2D reservoir.
To simplify the calculations, let us choose λ2 = λ3 = λ4 = λ5 = λ6 = λ7 = λ8 =
0 and λ1 = 1. This is not the optimum choice of the weights and will not
lead to the best estimate. However, because the weights sum to 1, we
have not violated the estimation model given by Eqs.(4.46) and (4.47) and
as such the estimation is legitimate. With this simplification, our
estimate is
* 19oΦ = (4.81)
and the error is now given by
4-87
( )* 19o o o oe = Φ − Φ = − Φ (4.82)
Although the error given by Eq.(4.82) is unknown, there is enough
information in the available data for us to determine its statistics. If we
accept the premise that the error is a function of the lag distance, 2xΔ ,
then we can look at the field of data in the SW-NE direction to find
similar errors and from them determine the statistics of our error. There
are in fact 47 such errors in the data set. They are tabulated in column 4
of Table 4.7.
Table 4.7. Errors in the SW-NE Direction of the Data Set at a Lag
Distance of 2xΔ .
1 2 3 4 5
Error
#
Φ(x) Φ(x+h) Φ(x+h - Φ(x) [Φ(x+h)-Φ(x)]2
1 23 22 -1 1
2 22 20 -2 4
3 20 19 -1 1
4 21 17 -4 16
5 17 17 0 0
6 17 14 -3 9
7 19 15 -4 16
8 15 18 3 9
9 18 20 2 4
10 20 16 -4 16
11 18 18 0 0
12 19 14 -5 25
13 14 19 5 25
14 17 16 -1 1
15 16 15 -1 1
16 15 18 3 9
4–88
17 18 18 0 0
18 18 23 5 25
19 23 16 -7 49
20 15 14 -1 1
21 14 10 -4 16
22 10 15 5 25
23 15 20 5 25
24 20 25 5 25
25 25 21 -4 16
26 21 14 -7 49
27 13 10 -3 9
28 10 16 6 36
29 16 18 2 4
30 18 20 2 4
31 20 20 0 0
32 20 17 -3 9
33 11 13 2 4
34 13 14 1 1
35 14 23 9 81
36 23 18 -5 25
37 18 19 1 1
38 10 13 3 9
39 13 18 5 25
40 18 22 4 16
41 22 13 -9 81
42 17 15 -2 4
43 15 20 5 25
44 20 20 0 0
45 16 14 -2 4
46 14 18 4 16
47 15 17 2 4
Mean 0.128
Variance 15.766 15.447
4-89
Std Dev 3.971 3.930
Semivarianc
e
7.723
As shown at the bottom of Table 4.7, the mean, variance and standard
deviation of the error are
2
0.12815.766
3.791
μσσ
=
==
(4.83)
If we assume that the error is normally distributed, then we are 95%
confident that the error of our estimation is given by
1.96 1.96oeμ σ μ σ− ≤ ≤ + (4.84)
Thus,
( ) ( ) ( )0.128 1.96 3.791 19 0.128 1.96 3.791o− ≤ − Φ ≤ + (4.85)
( )7.272 19 7.558o− ≤ − Φ ≤ (4.86)
or
11.442 26.272o≤ Φ ≤ (4.87)
Given our knowledge of the missing datum, is the statement of Eq.(4.87)
true? You bet. Of course, it would have been nice if the 95% confidence
interval was narrower than that given by Eq.(4.87). It would have been
narrower if we had chosen the weights to minimize the variance of the
4–90
error of Eq.(4.80) instead of choosing them arbitrarily as we did to
simplify the calculations. Figure 4.54 shows the histogram of the errors
of Table 4.7, which indicates that the assumption of a normal
distribution of the errors in the calculation of the 95% confidence
interval is reasonable.
Figure 4.54. Histogram of errors in the SW-NE direction at h = 2xΔ .
It should be noted that the data in columns 2 to 5 of Table 4.7 are
the same data needed to calculate the semivariance at a lag distance of
( )2h x= Δ in the SW-NE direction. From the calculations, we have
( )2
2 7.7232
xσ γ= Δ = (4.88)
4-91
Thus, geostatistics gives the mean, variance and standard deviation of
the error as
2
0.00015.447
3.930
μσσ
=
==
(4.89)
These statistics of the error are very similar to those given in Eq.(4.83).
Applying Eq.(4.89) gives the following 95% confidence interval:
( ) ( ) ( )0.000 1.96 3.930 19 0.000 1.96 3.930o− ≤ − Φ ≤ + (4.90)
( )7.703 19 7.703o− ≤ − Φ ≤ (4.91)
or
11.297 26.703o≤ Φ ≤ (4.92)
The result given in Eq.(4.92) is essentially the same as that given by
Eq.(4.87). Therefore, in geostatistics, the variogram (or the covariance
function) is used to calculate the error variance. Having now
demonstrated a simple geostatistical estimation that gave both the
estimate and the confidence limits for the estimate, we are ready to
formally derive ordinary kriging equations for geostatistical estimation.
4.5.2 Ordinary Kriging Equations
The objective is to estimate an unknown variable at the location x0 at
which no measurement has been made using the measured data at
locations xi as shown in Figure 4.55. In geostatistical estimation, the
variable of interest is treated as a stationary random function with a
4–92
normal probability distribution. If the measured data do not exhibit a
normal distribution, they must first be transformed into a normal
distribution before proceeding further. Such a transformation can always
be done by using the cumulative distribution function of the data and
the cumulative distribution function of a standard normal distribution
with a mean of zero and a standard deviation of 1.0 as shown
schematically in Figure 4.56. After the estimation, the estimated value is
transformed back to its original distribution by using the same two
cumulative distribution functions in reverse order.
Figure 4.55. Schematic showing the locations sample data and the location at which an estimation is to be made.
Figure 4.57 shows two random functions ( )AZ x and ( )BZ x together
with their probability density functions at four locations, x1, …, x4. The
mean of random function ( )AZ x is constant throughout the field. As a
result, ( )AZ x is described as a stationary random function. By contrast,
4-93
the mean of random function ( )BZ x is not constant throughout the field.
In fact, it increases as x increases. As a result, random function ( )BZ x is
described as a non-stationary random function.
Figure 4.56. Transformation of sample data to a standard normal distribution.
The kriging estimate is calculated as the weighted average of the
measured data as in Eq.(4.76). The challenge is to determine the weights
λi in such a way as to obtain the best estimate in some sense. The
criteria used in ordinary kriging to obtain the best estimate are (1) the
estimate should be unbiased and (2) the estimation error should have
4–94
minimum variance. Thus, ordinary kriging is a Best Linear Unbiased
Estimator (BLUE).
Figure 4.57. Stationary ( ( )AZ x ) and non-stationary ( ( )BZ x ) random functions.
Derivation in Terms of the Covariance Function
The estimated value at the unsampled location xo is given by
4-95
*
1( ) ( )
N
o i ii
Z x Z xλ=
= ∑ (4.93)
where Z*(xo) is the estimated value at location xo, Z(xi) are the measured
data, Z(x) is the assumed random function model, λi are the weights to
be determined and xi are the locations where the variable of interest has
been measured (i.e., the data). Note that in this model, the Z(xi) are just
samples drawn from the random function Z(x). Moreover, Z(x) is a
stationary random function with a constant mean, m, which is
independent of the locations xo and xi. Also, Eq.(4.93) is valid in 1D, 2D
or 3D.
Let the true but unknown value at xo be Z(xo). The estimation error
is given by
*( ) ( ) ( )o o oe x Z x Z x= − (4.94)
This error is a random variable with a probability distribution. In order to
obtain an unbiased estimate, on average, the mean estimation error
must be zero. Thus, for an unbiased estimate, the expectation of the
error must be zero:
[ ] *( ) ( ) ( ) 0o o oE e x E Z x Z x⎡ ⎤= − =⎣ ⎦ (4.95)
What is an unbiased or biased estimator? Figure 4.58 shows three
estimators for a petrophysical property whose value is know to μ.
Estimator 1 is unbiased but not very precise. Estimator 2 is unbiased
4–96
and more precise than estimator 1. Estimator 3 is biased and more
precise than estimator 1.
Figure 4.58. Biased and unbiased estimators.
Substituting Eq.(4.93) into Eq.(4.94) gives the condition for an
unbiased estimate as
1
( ) ( ) 0N
i i oi
E Z x Z xλ=
⎡ ⎤− =⎢ ⎥
⎣ ⎦∑ (4.96)
Changing the order of the summation and the expectation gives
4-97
[ ] [ ]1
( ) ( ) 0N
i i oi
E Z x E Z xλ=
− =∑ (4.97)
But the expectations of the random variables, Z(xi) and Z(xo) are equal
and is given by the mean of Z(x), m, which is a constant for the
stationary random function model used in the estimation. Eq.(4.97) can
now be written as
1 1
1 0N N
i ii i
m m mλ λ= =
⎛ ⎞− = − =⎜ ⎟
⎝ ⎠∑ ∑ (4.98)
The condition for an unbiased estimate is therefore given by
1
1N
ii
λ=
=∑ (4.99)
which is the same as Eq.(4.77).
The error variance is given by
( )22 *( ) ( )e o oE Z x Z xσ ⎡ ⎤= −⎢ ⎥⎣ ⎦ (4.100)
Substituting Eq.(4.93) into Eq.(4.100) gives the error variance as
2
2
1
( ) ( )N
e i i oi
E Z x Z xσ λ=
⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦∑ (4.101)
Let us add and subtract the mean, m, from the inner bracket in
Eq.(4.101) and make use of Eq.(4.99) (∑λi = 1) to obtain
4–98
( )2
2
1
( ) ( ( ) )N
e i i oi
E Z x m Z x mσ λ=
⎡ ⎤⎛ ⎞= − − −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦∑ (4.102)
Expanding the right side of Eq.(4.102) gives the estimation variance as
( ) ( )2
22
1 1
( ) ( ) 2 ( ( ) )( ( ) )N N
e i i o i i oi i
E Z x m Z x m Z x m Z x mσ λ λ= =
⎡ ⎤⎛ ⎞= − + − − − −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦∑ ∑ (4.103)
Algebraically, the square of the simple summation in Eq.(4.103) can be
rewritten in terms of a double summation to obtain
( ) ( )( )22
1 1 1
( ( ) )( ( ) ) ( ) 2 ( ) ( )N N N
e i j i j o i i oi j i
E Z x m Z x m Z x m Z x m Z x mσ λ λ λ= = =
⎡ ⎤= − − + − − − −⎢ ⎥
⎣ ⎦∑ ∑ ∑
…………….(4.104)
Changing the order of the summation and the expectation in Eq.(4.104)
gives
( )
( )( )
22
1 1
1
( ( ) )( ( ) ) ( )
2 ( ) ( )
N N
e i j i j oi j
N
i i oi
E Z x m Z x m E Z x m
E Z x m Z x m
σ λ λ
λ
= =
=
⎡ ⎤⎡ ⎤= − − + −⎣ ⎦ ⎣ ⎦
⎡ ⎤− − −⎣ ⎦
∑ ∑
∑ (4.105)
The expectations in Eq.(4.105) can be expressed in terms of the
covariance function as follows:
2
1 1 1
( ) (0) 2 ( )N N N
e i j ij i ioi j i
C h C C hσ λ λ λ= = =
= + −∑ ∑ ∑ (4.106)
4-99
where C(hij) is the covariance function, hij is the lag distance between
locations i and j, and hio is the lag distance between locations o and i.
Recall that the location of the point at which an estimate is to be made is
o and the locations of the data to be used in the estimation are at points
i. Eq.(4.106) can be rearranged as
2
1 1 1
(0) ( ) 2 ( )N N N
e i j ij i ioi j i
C C h C hσ λ λ λ= = =
= + −∑ ∑ ∑ (4.107)
But C(0), the covariance function at a lag distance of 0, is the same as
the variance of the data, σ2. Therefore, Eq.(4.107) can be written in the
following final form:
2 2
1 1 1( ) 2 ( )
N N N
e i j ij i ioi j i
C h C hσ σ λ λ λ= = =
= + −∑ ∑ ∑ (4.108)
The estimation problem now boils down to determining the values
of λi, which will minimize the estimation variance given by Eq.(4.108)
subject to the unbiasedness constraint given by Eq.(4.99). This is a
classical problem of optimization, which can be solved by the method of
Lagrange multipliers.
Derivation in Terms of the Variogram
Eq.(4.108) can be derived in terms of the variogram instead of the
covariance function. To do so, we first derive the relationship between
the variogram and the covariance function for a stationary random
function, Z(x). By definition, the variogram is given by
( )212( ) ( ) ( )h E Z x Z x hγ ⎡ ⎤= − +⎣ ⎦ (4.109)
4–100
Adding and subtracting the mean of the random function from the inner
bracket of Eq.(4.109) gives
( ) ( )( )212( ) ( ) ( )h E Z x m Z x h mγ ⎡ ⎤= − − + −⎢ ⎥⎣ ⎦
(4.110)
Expanding the right side of Eq.(4.110) gives
( ) ( ) ( )( )2 212( ) ( ) ( ) 2 ( ) ( )h E Z x m Z x h m Z x m Z x h mγ ⎡ ⎤= − + + − − − + −⎣ ⎦ (4.111)
Eq.(4.111) can be rewritten as
( )( ) ( )( )( )( )
1 12 2
22
( ) ( ) ( ) ( ) ( )
( ) ( )
h E Z x m Z x m E Z x h m Z x h m
E Z x h m Z x m
γ ⎡ ⎤ ⎡ ⎤= − − + + − + −⎣ ⎦ ⎣ ⎦⎡ ⎤− + − −⎣ ⎦
(4.112)
The expectations in Eq.(4.112) can be expressed in terms of the
covariance function as follows:
1 12 2( ) (0) (0) ( )h C C C hγ = + − (4.113)
Finally, Eq.(4.113) yields the required relationship as
( ) (0) ( )h C C hγ = − (4.114)
which is the same as Eq.(4.26). For large lag distances, Eq.(4.114)
becomes
( ) (0) ( )C Cγ ∞ = − ∞ (4.115)
For a stationary random function, γ(∞) is the sill of the variogram, C(∞) is
0 because there is no more correlation beyond the correlation length and
4-101
C(0) is the variance of the data (σ2) as previously stated. Substituting
these facts into Eq.(4.115) gives
2( ) (0) sill of the variogramCγ σ∞ = = = (4.116)
Figure 4.59 compares the variogram and the covariance function for a
stationary random function.
Figure 4.59. A comparison of the variogram and the covariance function for a stationary random function.
We are now ready to rewrite the error variance equation, Eq.(4.108), in
terms of the variogram. From Eq.(4.114), we have
( ) (0) ( )ij ijC h C hγ= − (4.117)
( ) (0) ( )io ioC h C hγ= − (4.118)
4–102
Substituting Eqs.(4.117) and (4.118) into Eq.(4.108) gives the error
variance as
[ ]2
1 1 1(0) (0) ( ) 2 (0) ( )
N N N
e i j ij i ioi j i
C C h C hσ λ λ γ λ γ= = =
⎡ ⎤= + − − −⎣ ⎦∑ ∑ ∑ (4.119)
Eq.(4.119) can be rearranged as
2
1 1 1 1 1 1(0) (0) ( ) 2 (0) 2 ( )
N N N N N N
e i j i j ij i i ioi j i j i i
C C h C hσ λ λ λ λ γ λ λ γ= = = = = =
= + − − +∑ ∑ ∑ ∑ ∑ ∑ (4.120)
Eq.(4.120) can be simplified to
2
1 1 1
(0) (0) ( ) 2 (0) 2 ( )N N N
e i j ij i ioi j i
C C h C hσ λ λ γ λ γ= = =
= + − − +∑ ∑ ∑ (4.121)
Further simplification of Eq.(4.121) gives the error variance in terms of
the variogram in final form as
2
1 1 1
( ) 2 ( )N N N
e i j ij i ioi j i
h hσ λ λ γ λ γ= = =
= − +∑ ∑ ∑ (4.122)
Solution of the Kriging Equations in terms of the Covariance Function
The problem to be solved is to choose the weights (λi) to minimize
the error variance given by Eq.(4.108) subject to the constraint given by
Eq.(4.99). The problem statement is as follows:
2 2
1 1 1
( ) ( ) 2 ( )N N N
e i i j ij i ioi j i
Minimize C h C hσ λ σ λ λ λ= = =
= + −∑ ∑ ∑ (4.123)
4-103
1
to 1 0N
ii
Subject λ=
⎛ ⎞− =⎜ ⎟
⎝ ⎠∑ (4.124)
This problem can be solved by the method of Lagrange multipliers. The
solution steps are as follows:
1. Convert all constraints into equality constraints.
2. Multiply each equality constraint by a new variable μi, where μi is
the Lagrange multiplier for the ith constraint.
3. Add (or subtract) the resulting constraint equations to the original
objective function to obtain the Lagrangian function, L. This step
relaxes the constraint and converts the constrained optimization to
an unconstrained optimization.
4. Differentiate the Lagrangian function and equate to zero to
determine the stationary points, which constitute the required
solution.
Applications of steps 1 to 3 to the problem at hand will result in the
following Lagrangian function:
2
1 1 1 1
( , ) ( ) 2 ( ) 2 1N N N N
i i j ij i io ii j i i
L C h C hλ μ σ λ λ λ μ λ= = = =
⎛ ⎞= + − + −⎜ ⎟
⎝ ⎠∑ ∑ ∑ ∑ (4.125)
Consider the case for N = 2. The Lagrangian can be expanded to obtain
2 21 2 1 11 1 2 12 2 1 21
22 22 1 10 2 20
1 2
( , , ) ( ) ( ) ( )
( ) 2 ( ) 2 ( ) 2 2 2
L C h C h C h
C h C h C h
λ λ μ σ λ λ λ λ λ
λ λ λλ μ λ μ μ
= + + +
+ − −+ + −
(4.126)
Differentiating Eq.(4.126) and equating to zero will give the following
linear simultaneous equations:
4–104
1 11 2 12 2 21 101
2 ( ) ( ) ( ) 2 ( ) 2 0L C h C h C h C hλ λ λ μλ
∂= + + − + =
∂ (4.127)
1 12 1 21 2 22 202
( ) ( ) 2 ( ) 2 ( ) 2 0L C h C h C h C hλ λ λ μλ
∂= + + − + =
∂ (4.128)
1 22 2 2 0L λ λμ
∂= + − =
∂ (4.129)
Eqs.(4.127) to (4.129) can be rewritten in matrix form as
11 12 21 1 10
12 21 22 2 20
2 ( ) ( ) ( ) 2 2 ( )( ) ( ) 2 ( ) 2 2 ( )
2 2 0 2
C h C h C h C hC h C h C h C h
λλμ
+⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.130)
Because C(h12) = C(h21), Eq.(4.130) can be rewritten as
11 12 1 10
21 22 2 20
2 ( ) 2 ( ) 2 2 ( )2 ( ) 2 ( ) 2 2 ( )
2 2 0 2
C h C h C hC h C h C h
λλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.131)
Upon dividing by 2, Eq.(4.131) simplifies to
11 12 1 10
21 22 2 20
( ) ( ) 1 ( )( ) ( ) 1 ( )1 1 0 1
C h C h C hC h C h C h
λλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.132)
Eq.(4.132) can be solved by standard techniques such as matrix
inversion or Gaussian elimination.
4-105
After solving Eq.(4.132), λ1, λ2 and μ can be substituted into
Eq.(4.123) to calculate the minimum estimation error variance. Let us
derive an algebraic expression for the minimum error variance.
Expanding Eq.(4.123) for N = 2 gives the error variance as
2 2 21 2 1 11 1 2 12 2 1 21
22 22 1 10 2 20
( , , ) ( ) ( ) ( )
( ) 2 ( ) 2 ( )
e C h C h C h
C h C h C h
σ λ λ μ σ λ λ λ λ λ
λ λ λ
= + + +
+ − − (4.133)
Eq.(4.133) can be rearranged as
2 21 2 1 1 11 2 12 2 1 21 2 22
1 10 2 20
( , , ) ( ) ( ) ( ) ( )
2 ( ) 2 ( )
e C h C h C h C h
C h C h
σ λ λ μ σ λ λ λ λ λ λ
λ λ
⎡ ⎤ ⎡ ⎤= + + + +⎣ ⎦ ⎣ ⎦− − (4.134)
From Eq.(4.132), it can be seen that
1 11 2 12 10( ) ( ) ( )
C h C h C hλ λ μ+ = − (4.135)
and
2 21 2 22 20( ) ( ) ( )
C h C h C hλ λ μ+ = − (4.136)
Substituting Eqs.(4.135) and (4.136) into Eq.(4.134) gives the minimum
estimation error variance as
( ) ( )2 2
min 1 10 2 20
1 10 2 20
( ) ( ) 2 ( ) 2 ( )
e C h C hC h C h
σ σ λ μ λ μ
λ λ
= + − + −
− − (4.137)
4–106
which simplifies to
22 2min
1
( )
e i ioi
C hσ σ μ λ=
= − − ∑ (4.138)
Eq.(4.138) can be used to calculate the minimum error variance directly
instead of substituting λ1, λ2 and μ into Eq.(4.123) to calculate the
minimum estimation error variance. Eq.(4.132) and (4.138) can easily be
generalized to any value of N. For example, for N = 4, Eqs.(4.132) and
(4.138) will become
11 12 13 14 1 10
21 22 23 24 2 20
31 32 33 34 3 30
41 42 43 44 4 40
( ) ( ) ( ) ( ) 1 ( )( ) ( ) ( ) ( ) 1 ( )( ) ( ) ( ) ( ) 1 ( )( ) ( ) ( ) ( ) 1 ( )1 1 1 1 0 1
C h C h C h C h C hC h C h C h C h C hC h C h C h C h C hC h C h C h C h C h
λλλλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.139)
and
42 2min
1
( )
e i ioi
C hσ σ μ λ=
= − − ∑ (4.140)
For any value of N, Eq.(4.132) can be generalized to the following N+1
linear simultaneous equations:
1
1
( ) ( ) for i = 1, 2, ..., N
1
N
j ij ioj
N
jj
C h C hλ μ
λ
=
=
+ =
=
∑
∑ (4.141)
4-107
The minimum estimation error variance can be generalized to
2 2min
1
( )
N
e i ioi
C hσ σ μ λ=
= − − ∑ (4.142)
The following observations can be made about the ordinary kriging
model in terms of the covariance function.
1. In order to set up the simultaneous equations to be solved, one
must first compute the lag distance matrix, hij. For N = 4, the
lag distance matrix will look like this:
11 12 13 14 12 13 14
21 22 23 24 21 23 24
31 32 33 34 31 32 34
41 42 43 44 41 42 43
00
00
ij
h h h h h h hh h h h h h h
hh h h h h h hh h h h h h h
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤ = =⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
(4.143)
2. Except for the one zero at the bottom corner, the diagonal
entries of the matrix equation to be solved are given by
211 22( ) ( ) ... ( ) (0)NNC h C h C h C σ= = = = (4.144)
3. All the off diagonal entries of the matrix equation are less than
the diagonal entries. Thus, the matrix is diagonally dominant.
This is a desirable structure for solving the system of linear
simultaneous equations.
4. The matrix to be inverted is full. Therefore, a lot of calculations
could be involved, depending on the value of N.
5. The solution λ1, λ2, …, λN and μ depends only on the spatial
coordinates of the data and not on the values of the data.
4–108
6. The estimation error variance depends only on the spatial
coordinates of the data and not on the values of the data.
7. Kriging is an exact estimator and will return the measured
value if it is applied to the location for which datum was
measured.
Solution of the Kriging Equations in terms of the Variogram
In this case, the problem statement is
2
1 1 1 ( ) ( ) 2 ( )
N N N
e i i j ij i ioi j i
Minimize h hσ λ λ λ γ λ γ= = =
= − +∑ ∑ ∑ (4.145)
1
to 1 0N
ii
Subject λ=
⎛ ⎞− =⎜ ⎟
⎝ ⎠∑ (4.146)
The Lagrangian is given by
1 1 1 1
( , ) ( ) 2 ( ) 2 1N N N n
i i j ij i io ii j i i
L h hλ μ λ λ γ λ γ μ λ= = = =
⎛ ⎞= − + + −⎜ ⎟
⎝ ⎠∑ ∑ ∑ ∑ (4.147)
For the case of N = 2, the Lagrangian can be expanded to obtain
21 2 1 11 1 2 12 2 1 21
22 22 1 10 2 20
1 2
( , , ) ( ) ( ) ( )
( ) 2 ( ) 2 ( ) 2 2 2
L h h h
h h h
λ λ μ λ γ λ λ γ λ λ γ
λ γ λ γ λ γλ μ λ μ μ
= − − −
− + ++ + −
(4.148)
Differentiating the Lagrangian and equating to zero will give the following
linear simultaneous equations:
4-109
1 11 2 12 2 21 101
2 ( ) ( ) ( ) 2 ( ) 2 0L h h h hλ γ λ γ λ γ γ μλ
∂= − − − + + =
∂ (4.149)
1 12 1 21 2 22 202
( ) ( ) 2 ( ) 2 ( ) 2 0L h h h hλ γ λ γ λ γ γ μλ
∂= − − − + + =
∂ (4.150)
1 22 2 2 0L λ λμ
∂= + − =
∂ (4.151)
Eqs.(4.149) to (4.151) can be rewritten in matrix form as
[ ]
[ ]11 12 21 1 10
12 21 22 2 20
2 ( ) ( ) ( ) 2 2 ( )( ) ( ) 2 ( ) 2 2 ( )
2 2 0 2
h h h hh h h h
γ γ γ λ γγ γ γ λ γ
μ
⎡ ⎤− − + −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− + − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
(4.152)
Upon dividing by 2 and noting that γ(h12) = γ(h21), Eq.(4.152) can be
rewritten as
11 12 1 10
21 22 2 20
( ) ( ) 1 ( )( ) ( ) 1 ( )1 1 0 1
h h hh h h
γ γ λ γγ γ λ γ
μ
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.153)
After solving Eq.(4.153), λ1, λ2 and μ can be substituted into
Eq.(4.145) to calculate the minimum estimation error variance. Let us
derive an algebraic expression for the minimum error variance.
Expanding Eq.(4.145) for N = 2 gives the error variance as
2 21 2 1 11 1 2 12 2 1 21
22 22 1 10 2 20
( , , ) ( ) ( ) ( )
( ) 2 ( ) 2 ( )
e h h h
h h h
σ λ λ μ λ γ λ λ γ λ λ γ
λ γ λ γ λ γ
= − − −
− + + (4.154)
4–110
Eq.(4.154) can be rearranged as
21 2 1 1 11 2 12 2 1 21 2 22
1 10 2 20
( , , ) ( ) ( ) ( ) ( )
2 ( ) 2 ( )
e h h h h
h h
σ λ λ μ λ λ γ λ γ λ λ γ λ γ
λ γ λ γ
⎡ ⎤ ⎡ ⎤= − + − +⎣ ⎦ ⎣ ⎦+ + (4.155)
From Eq.(4.153), it can be seen that
1 11 2 12 10( ) ( ) ( )
h h hλ γ λ γ γ μ+ = + (4.156)
and
2 21 2 22 20( ) ( ) ( )
h h hλ γ λ γ γ μ+ = + (4.157)
Substituting Eqs.(4.156) and (4.157) into Eq.(4.155) gives the minimum
estimation error variance as
( ) ( )2
min 1 10 2 20
1 10 2 20
( ) ( ) 2 ( ) 2 ( )
e h hh h
σ λ γ μ λ γ μλ γ λ γ
= − + − +
+ + (4.158)
which simplifies to
22min
1
( )
e i ioi
hσ μ λ γ=
= − + ∑ (4.159)
Eqs.(4.153) and (4.159) can easily be generalized to any value of N. For
example, for N = 4, Eqs.(4.153) and (4.159) will become
4-111
11 12 13 14 1 10
21 22 23 24 2 20
31 32 33 34 3 30
41 42 43 44 4 40
( ) ( ) ( ) ( ) 1 ( )( ) ( ) ( ) ( ) 1 ( )( ) ( ) ( ) ( ) 1 ( )( ) ( ) ( ) ( ) 1 ( )1 1 1 1 0 1
h h h h hh h h h hh h h h hh h h h h
γ γ γ γ λ γγ γ γ γ λ γγ γ γ γ λ γγ γ γ γ λ γ
μ
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=−⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.160)
and
42min
1
( )
e i ioi
hσ μ λ γ=
= − + ∑ (4.161)
For any value of N, Eq.(4.153) can be generalized to the following N+1
linear simultaneous equations:
1
1
( ) ( ) for i = 1, 2, ..., N
1
N
j ij ioj
N
jj
h hλ γ μ γ
λ
=
=
− =
=
∑
∑ (4.162)
The minimum estimation error variance can be generalized to
2min
1
( )
N
e i ioi
hσ μ λ γ=
= − + ∑ (4.163)
The following observations can be made about the kriging model in
terms of the variogam.
1. The diagonal entries of the matrix equation to be solved are given
by
11 22( ) ( ) ... ( ) 0NNh h hγ γ γ= = = (4.164)
4–112
All the diagonal entries of the matrix equation are zero. This is an
undesirable structure for the solution of the system of linear
simultaneous equations. This is why the formulation and solution
of the estimation problem in terms of the covariance function is
usually preferred over the formulation in terms of the variogram.
2. Most of the entries in the matrix equation are numbers computed
from the variogram. It is essential that these numbers be
consistent and well behaved for the system of equations to have a
solution. Inconsistent numbers from an experimental variogram
could lead to a system of equations without a solution. This is why
a well behaved theoretical variogram model is usually fitted to the
experimental variogram and the theoretical variogram model is
then used instead of the experimental variogram for the estimation
calculations.
Example 4.1
The porosities at locations 1 and 4 in a linear reservoir have been
measured as shown Figure 4.60. The locations are evenly spaced 10
meters apart. The variogram for the porosity distribution in this reservoir
is shown in Figure 4.61. We are required to estimate the porosity at
location 3 at which no measurement was made.
4-113
Figure 4.60. Measured porosity values and their locations.
1. Calculate the best estimate of the porosity at location 3 based on
the available information and determine the 95% confidence
interval assuming a normal distribution.
2. Assuming a stationary random function model, carefully sketch on
Figure 4.60 the covariance function for the porosity distribution for
this reservoir showing important features of your sketch.
4–114
Figure 4.61. Variogram for Example 4.1.
3. Show that kriging is an exact interpolator. An exact interpolator
will return the measured value if it is applied at a location
containing a measured value. In other words, if kriging is applied
to estimate the porosity at location 1, it should return a value of
10% with a minimum error variance of 0.
Solution to Example 4.1
The best estimate of the porosity at location 3 can be obtained using
ordinary kriging. The kriging equations to be solved can be written by
inpection in terms of the covariance function as
11 12 1 10
21 22 2 20
11
1 1 0 1
C C CC C C
λλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.165)
4-115
Let us adopt the following subscripts: Location 1 = 1, location 3 = 0 and
location 4 = 3. The matrix of lag distances for location 3 is (subscript 0) is
11 12 10
21 22 20
0 30 2030 0 10
h h hmeters
h h h⎡ ⎤ ⎡ ⎤
=⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
(4.166)
For a stationary random function, the variogram and the covariance
function are related by Eq.(4.114). Therefore, the covariances needed in
Eq.(4.165) can easily be computed from the variogram. Thus, the matrix
equation to be solved is
1
2
5 2 1 32 5 1 41 1 0 1
λλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.167)
Eq.(4.167) can easily be solved by elimination to obtain
1
2
1323
0
λ
λ
μ
=
=
=
The estimated porosity is given by
( ) ( )* *0 3 1 4
1 2 1 210 25 20%3 3 3 3
φ φ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = Φ + Φ = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
The minimum error variance is computed from Eq.(4.140) as
22 2min
1
1 2 4( ) 5 0 3 43 3 3
e i ioi
C h x xσ σ μ λ=
⎛ ⎞= − − = − − + =⎜ ⎟⎝ ⎠
∑
4–116
The minimum standard deviation is
min4 1.15473
eσ = =
The 95% confidence interval is given by
( )( ) ( ) ( )( )30 1.96 1.1547 20 0 1.96 1.1547φ+ ≤ − ≤ +
( )32.26 20 2.26φ≤ − ≤
or
317.74 22.26 or 20% 2.26%φ≤ ≤ ±
The problem could also be solved in terms of the variogram. In this case,
the corresponding equations are
11 12 1 10
21 22 2 20
11
1 1 0 1
γ γ λ γγ γ λ γ
μ
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.168)
1
2
0 3 1 23 0 1 11 1 0 1
λλμ
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.169)
The solution of Eq.(4.169) gives
4-117
1
2
1323
0
λ
λ
μ
=
=
=
The estimated porosity is given by
( ) ( )* *0 3 1 4
1 2 1 210 25 20%3 3 3 3
φ φ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = Φ + Φ = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
The minimum error variance is obtained from Eq.(4.159) as
22min
1
1 2 4( ) 0 2 13 3 3
e i ioi
h x xσ μ λ γ=
⎛ ⎞= − + = − + + =⎜ ⎟⎝ ⎠
∑
The minimum standard deviation is
min4 1.15473
eσ = =
The 95% confidence interval is given by
( )( ) ( ) ( )( )30 1.96 1.1547 20 0 1.96 1.1547φ+ ≤ − ≤ +
( )32.26 20 2.26φ≤ − ≤
or
317.74 22.26 or 20% 2.26%φ≤ ≤ ±
4–118
Thus, as expected, the kriging equations based on the covariance
function and the variogram give the same results. The sketch of the
covariance function on Figure 4.61 is left as an exercise for the reader.
To demonstrate that kriging is an exact interpolator, we solve the
kriging equation at location 1, which contains a sample datum. We
should recover the sample value of 10% with an error variance of 0. The
matrix of lag distances is
11 12 10
21 22 20
0 30 030 0 30
h h hmeters
h h h⎡ ⎤ ⎡ ⎤
=⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
(4.170)
Substituting these numbers into Eq.(4.165) gives the matrix equation to
be solved as
1
2
5 2 1 52 5 1 21 1 0 1
λλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.171)
Eq.(4.171) can be solved to obtain
1
2
100
λλμ
===
( ) ( ) ( )( ) ( )( )* *0 1 1 41 0 1 10 0 25 10%φ φ= = Φ + Φ = + =
The minimum error variance is computed from Eq.(4.140) as
( )2
2 2min
1
( ) 5 0 1 5 0 2 0
e i ioi
C h x xσ σ μ λ=
= − − = − − + =∑
4-119
Therefore, kriging is an exact interpolator.
Example 4.2
Estimate the petrophysical property shown in Figure 4.51 at locations 1,
3, 5, 7 and 8 at which no measurements were made. The correlation
structure of the heterogeneous property is given by
( ) 0.3100 hC h e−= (4.172)
Solution to Example 4.2
To begin the calculations, we must determine the order in which the
estimates will be made. This order is determined by a random drawing.
Using a random number generator for integer values from 1 to 8, it has
been determined that the order for the estimations is 8, 1, 5, 7 and 3.
After estimating *8Φ , this value is treated as a known sample and is used
along with the measured data for the subsequent estimations. Thus, the
number of equations to be solved increases as the estimation progresses.
We begin the calculations by visiting location (node) 8. In order to
generate a compact matrix equation, we renumber the known values as
shown in Figure 4.62. Next, we construct the lag distance matrix for
location 8 as shown in Table 4.8. The entries in Table 4.8 should be read
as follows. The indices 1, 2, 3 represent the locations of the renumbered
known values. The index 0 represents the location at which an estimate
is to be made. The other entries are lag distances. For example, h10 is the
lag distance (6Δx) from the point of estimation to the renumbered sample
1, h20 is the lag distance (4Δx) from the point of estimation to the
renumbered sample 2 and h30 is the lag distance (2Δx) from the point of
estimation to the renumbered sample 3. These are the lag distances
needed to construct the right hand side vector of the system of equations
4–120
to be solved. Therefore, they have been transferred into the last column
of the table under the heading hi0. The other entries in the table should
be self explanatory. For example, h11 is the lag distance between the
renumbered sample 1 and itself, which is zero.
Figure 4.62. Renumbered sample values for the estimation at location 8.
Table 4.8. Lag Distance Matrix hij for Location 8
1 2 3
0 6Δx 4Δx 2Δx hi0
1 0Δx 2Δx 4Δx 6Δx
2 2Δx 0Δx 2Δx 4Δx
3 4Δx 2Δx 0Δx 2Δx
The matrix equation to be solved is
4-121
11 12 13 1 10
21 22 23 2 20
31 32 33 3 30
111
1 1 1 0 1
C C C CC C C CC C C C
λλλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.173)
In this problem, Δx = 1 unit. Using the matrix of lag distances in Table
4.8 and Eq.(4.172), the entries of the covariance matrix of Eq.(4.173) can
easily be computed. For example, C12 is given by
( ) ( ) 0.3 212 12 2 100 54.8812C h C e−= = =
The results of the other calculations are shown in Eq.(4.174).
1
2
3
100.0000 54.8812 30.1194 1 16.529954.8812 100.0000 54.8812 1 30.119430.1194 54.8812 100.0000 1 54.8812
1 1 1 0 1
λλλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.174)
Eq.(4.174) can be solved by any standard method of solving a system of
linear simultaneous equations, for example, by matrix inversion. The
inverse matrix for Eq.(4.174) is
1
0.0117 0.0090 0.0026 0.40800.0090 0.0181 0.0090 0.18410.0026 0.0090 0.0117 0.4080
0.4080 0.1841 0.4080 63.1862
ijC−
− −⎡ ⎤⎢ ⎥− −⎢ ⎥=⎢ ⎥− −⎢ ⎥−⎣ ⎦
(4.175)
Multiplying the right hand side of Eq.(4.174) into the inverse matrix of
Eq.(4.175) gives the solution vector as
4–122
1
2
3
0.18410.08300.732928.5088
λλλμ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
(4.176)
The kriged value is calculated as
( )( ) ( )( ) ( )( )* *0 8 0.1841 30 0.0830 50 0.7329 20 24.3321Φ = Φ = + + =
The minimum error variance (estimation variance) is computed from
Eq.(4.140) as
( )
( )
32 2min
1
( ) 100 28.5089
0.1841 16.5299 0.0830 30.1194 0.7329 54.8812 82.7434
e i ioi
C h
x x x
σ σ μ λ=
= − − = − −
− + + =
∑
The minimum standard deviation (estimation standard deviation) is
min 82.7435 9.0963
eσ = =
The 95% confidence interval is given by
( )( )*8 24.3321 1.96 9.0963 24.33 17.83Φ = ± = ±
The estimated value at location 8 is then added to the sample data set in
preparation for the estimation at location 1 as shown in Figure 4.63.
4-123
Figure 4.63. Sample data set for the estimation at location 1.
Next, we visit location 1 and construct the lag distance matrix shown in
Table 4.9.
Table 4.9. Lag Distance Matrix hij for Location 1
1 2 3 4
0 1Δx 3Δx 5Δx 7Δx hi0
1 0Δx 2Δx 4Δx 6Δx 1Δx
2 2Δx 0Δx 2Δx 4Δx 3Δx
3 4Δx 2Δx 0Δx 2Δx 5Δx
4 6Δx 4Δx 2Δx 0Δx 7Δx
4–124
The matrix equation to be solved is
11 12 13 14 1 10
21 22 23 24 2 20
31 32 33 34 3 30
41 42 43 44 4 40
1111
1 1 1 1 0 1
C C C C CC C C C CC C C C CC C C C C
λλλλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.177)
Using the matrix of lag distances in Table 4.9 and Eq.(4.170), Eq.(4.177)
becomes
1
2
3
4
100.0000 54.8812 30.1194 16.5299 1 74.081854.8812 100.0000 54.8812 30.1194 1 40.657030.1194 54.8812 100.0000 54.8812 1 22.313016.5299 30.1194 54.8812 100.0000 1 12.2456
1 1 1 1 0 1
λλλλμ
⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣
⎤⎥⎥
⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎦
(4.178)
The inverse matrix for Eq.(4.178) is
1
0.0121 0.0089 0.0010 0.0022 0.34450.0089 0.0182 0.0083 0.0010 0.15550.0010 0.0083 0.0182 0.0089 0.15550.0022 0.0010 0.0089 0.0121 0.3445
0.3445 0.1555 0.1555 0.3445 53.3636
ijC−
− − −⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥= − − −⎢ ⎥− − −⎢ ⎥⎢ ⎥−⎣ ⎦
(4.179)
Multiplying the right hand side of Eq.(4.178) into the inverse matrix of
Eq.(4.179) gives the solution vector as
1
2
3
4
0.83010.04030.04030.089313.8309
λλλλμ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
(4.180)
4-125
The kriged value is calculated as
( )( ) ( )( ) ( )( ) ( )( )* *0 1 0.8301 30 0.0403 50 0.0403 20 0.0893 24.3321 29.8968Φ = Φ = + + + =
The estimation variance is computed from Eq.(4.140) as
( )
( )
32 2min
1
( ) 100 13.8309
0.8301 74.0819 0.0403 40.6570 0.0403 22.3130 0.0893 12.2456 48.7035
e i ioi
C h
x x x x
σ σ μ λ=
= − − = − −
− + + + =
∑
The estimation standard deviation is
min 48.7035 6.9788
eσ = =
The 95% confidence interval is given by
( )( )*8 29.8968 1.96 6.9788 29.99 13.68Φ = ± = ±
The estimated value at location 1 is then added to the sample data set in
preparation for the estimation at location 5 as shown in Figure 4.64.
Next, we visit location 5 and construct the lag distance matrix shown in
Table 4.10.
4–126
Figure 4.64. Sample data set for the estimation at location 5.
Table 4.10. Lag Distance Matrix hij for Location 5
1 2 3 4 5
0 3Δx 1Δx 1Δx 3Δx 4Δx hi0
1 0Δx 2Δx 4Δx 6Δx 1Δx 3Δx
2 2Δx 0Δx 2Δx 4Δx 3Δx 1Δx
3 4Δx 2Δx 0Δx 2Δx 5Δx 1Δx
4 6Δx 4Δx 2Δx 0Δx 7Δx 3Δx
5 1Δx 3Δx 5Δx 7Δx 0Δx 4Δx
4-127
The matrix equation to be solved is
11 12 13 14 15 1 10
21 22 23 24 25 2 20
31 32 33 34 35 3 30
41 42 43 44 45 4 40
51 52 53 54 55 5 50
11111
1 1 1 1 1 0 1
C C C C C CC C C C C CC C C C C CC C C C C CC C C C C C
λλλλλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.181)
Using the matrix of lag distances in Table 4.10 and Eq.(4.170), Eq.(4.181)
becomes
100.0000 54.8812 30.1194 16.5299 74.0818 154.8812 100.0000 54.8812 30.1194 40.6570 130.1194 54.8812 100.0000 54.8812 22.3130 116.5299 30.1194 54.8812 100.0000 12.2456 174.0818 40.6570 22.3130 12.2456 100.0000 1
1 1 1 1 1 0
⎡⎢⎢⎢⎢⎢⎢⎢⎣
1
2
3
4
5
40.657074.081874.081840.657030.1194
1
λλλλλμ
⎤ ⎡ ⎤ ⎡ ⎤⎥ ⎢ ⎥ ⎢ ⎥⎥ ⎢ ⎥ ⎢ ⎥⎥ ⎢ ⎥ ⎢ ⎥
=⎥ ⎢ ⎥ ⎢ ⎥⎥ ⎢ ⎥ ⎢ ⎥⎥ ⎢ ⎥ ⎢ ⎥⎥ ⎢ ⎥ ⎢ ⎥⎦ ⎣ ⎦ ⎣ ⎦
(4.182)
The inverse matrix for Eq.(4.182) is
1
0.0262 0.0082 0.0003 0.0007 0.0170 0.10880.0082 0.0182 0.0083 0.0009 0.0008 0.14400.0003 0.0083 0.0182 0.0088 0.0008 0.14400.0007 0.0009 0.0088 0.0122 0.0018 0.31920.0170 0.0008 0.0008 0.0018 0.0205
ijC−
− − − −− − − −− − − −
=− − − −− − − − 0.28400.1088 0.1440 0.1440 0.3192 0.2840 49.4359
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
−⎣ ⎦
(4.183)
Multiplying the right hand side of Eq.(4.182) into the inverse matrix of
Eq.(4.183) gives the solution vector as
4–128
1
2
3
4
5
0.00470.48460.48460.01380.01232.1441
λλλλλμ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
−⎣ ⎦ ⎣ ⎦
(4.184)
The kriged value is calculated as
( )( ) ( )( ) ( )( ) ( )( )( )( )
* *0 5 0.0047 30 0.4846 50 0.4846 20 0.0138 24.3321
0.0123 29.8968 34.7659
Φ = Φ = + + +
+ =
The estimator variance is computed from Eq.(4.140) as
( )
32 2min
1( ) 100
0.0047 40.6570 0.4846 74.0818 0.4846 74.0818 2.1441 29.2243
0.0138 40.6570 0.0123 30.1194
e i ioi
C h
x x xx x
σ σ μ λ=
= − − =
+ +⎛ ⎞− − − =⎜ ⎟+ +⎝ ⎠
∑
The estimator standard deviation is
min 29.2243 5.4059
eσ = =
The 95% confidence interval is given by
( )( )*8 34.7659 1.96 5.4059 34.77 10.60Φ = ± = ±
The estimated value at location 5 is then added to the sample data set in
preparation for the estimation at location 7 as shown in Figure 4.65.
4-129
Figure 4.65. Sample data set for the estimation at location 7.
Next, we visit location 7 and construct the lag distance matrix shown in
Table 4.11.
Table 4.11. Lag Distance Matrix hij for Location 7
1 2 3 4 5 6
0 5Δx 3Δx 1Δx 1Δx 6Δx 2Δx hi0
1 0Δx 2Δx 4Δx 6Δx 1Δx 3Δx 5Δx
2 2Δx 0Δx 2Δx 4Δx 3Δx 1Δx 3Δx
3 4Δx 2Δx 0Δx 2Δx 5Δx 1Δx 1Δx
4 6Δx 4Δx 2Δx 0Δx 7Δx 3Δx 1Δx
5 1Δx 3Δx 5Δx 7Δx 0Δx 4Δx 6Δx
6 2Δx 1Δx 1Δx 3Δx 4Δx 0Δx 2Δx
4–130
The matrix equation to be solved is
11 12 13 14 15 16 1 10
21 22 23 24 25 26 2 20
31 32 33 34 35 36 3 30
41 42 43 44 45 46 4 40
51 52 53 54 55 56 5 50
61 62 63 64 65 66 6 60
111111
1 1 1 1 1 1 0 1
C C C C C C CC C C C C C CC C C C C C CC C C C C C CC C C C C C CC C C C C C C
λλλλλλμ
⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣
⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎦
(4.185)
Using the matrix of lag distances in Table 4.11 and Eq.(4.170), Eq.(4.185)
becomes
100.0000 54.8812 30.1194 16.5299 74.0818 40.6570 154.8812 100.0000 54.8812 30.1194 40.6570 74.0818 130.1194 54.8812 100.0000 54.8812 22.3130 74.0818 116.5299 30.1194 54.8812 100.0000 12.2456 40.6570 174.0818 40.6570 22.3130 12.
1
2
3
4
5
6
22.313040.657074.081874.0818
2456 100.0000 30.1194 1 16.529954.8812 74.0818 74.0818 40.6570 30.1194 100.0000 1 54.8812
1 1 1 1 1 1 0 1
λλλλλλμ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ……... (4.186)
The inverse matrix for Eq.(4.186) is
1
0.0322 0.0023 0.0073 0.0006 0.0207 0.0159 0.09930.0023 0.0239 0.0008 0.0009 0.0044 0.0155 0.1348
0.0073 0.0008 0.0279 0.0087 0.0055 0.0202 0.13200.0006 0.0009 0.0087 0.0122 0.0019 0.0002 0.31910.0207
ijC−
− − − −− − − − −
− − − −= − − − − −
− −0.0044 0.0055 0.0019 0.0228 0.0097 0.28980.0159 0.0155 0.0202 0.0002 0.0097 0.0420 0.0251
0.0993 0.1348 0.1320 0.3191 0.2898 0.0251 49.4209
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥− −⎢ ⎥− − − −⎢ ⎥
⎢ ⎥−⎣ ⎦
(4.187)
4-131
Multiplying the right hand side of Eq.(4.186) into the inverse matrix of
Eq.(4.187) gives the solution vector as
1
2
3
4
5
6
0.00470.00580.48410.49210.01230.00092.1422
λλλλλλμ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
(4.188)
The kriged value is calculated as
( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( )
* *0 7 0.0047 30 0.0058 50 0.4841 20 0.4921 24.3321
0.0123 29.8968 0.0009 34.7659 22.4886
Φ = Φ = + + +
+ + =
The estimation variance is computed from Eq.(4.140) as
( )
32 2min
1( ) 100
0.0047 22.3130 0.0058 40.6570 0.4841 74.0818 2.1422 29.2242
0.4921 74.0818 0.0123 16.5299 0.0009 54.8812
e i ioi
C h
x x xx x x
σ σ μ λ=
= − − =
+ +⎛ ⎞− − − =⎜ ⎟+ + +⎝ ⎠
∑
The estimation standard deviation is
min 29.2242 5.4059
eσ = =
The 95% confidence interval is given by
( )( )*7 22.4886 1.96 5.4059 22.4886 10.60Φ = ± = ±
4–132
The estimated value at location 7 is then added to the sample data set in
preparation for the estimation at location 3 as shown in Figure 4.66.
Figure 4.66. Sample data set for the estimation at location 3.
Next, we visit location 3 and construct the lag distance matrix shown in
Table 4.12.
Table 4.12. Lag Distance Matrix hij for Location 3
1 2 3 4 5 6 7
0 1Δx 1Δx 3Δx 5Δx 2Δx 2Δx 4Δx hi0
1 0Δx 2Δx 4Δx 6Δx 1Δx 3Δx 3Δx 1Δx
2 2Δx 0Δx 2Δx 4Δx 3Δx 1Δx 1Δx 1Δx
3 4Δx 2Δx 0Δx 2Δx 5Δx 1Δx 1Δx 3Δx
4 6Δx 4Δx 2Δx 0Δx 7Δx 3Δx 3Δx 5Δx
5 1Δx 3Δx 5Δx 7Δx 0Δx 4Δx 4Δx 2Δx
6 2Δx 1Δx 1Δx 3Δx 4Δx 0Δx 0Δx 2Δx
7 2Δx 1Δx 1Δx 3Δx 4Δx 0Δx 0Δx 4Δx
4-133
The matrix equation to be solved is
11 12 13 14 15 16 17 1
21 22 23 24 25 26 27 2
31 32 33 34 35 36 37 3
41 42 43 44 45 46 47 4
51 52 53 54 55 56 57 5
61 62 63 64 65 66 67 6
71 72 73 74 75 76 77
1111111
1 1 1 1 1 1 1 0
C C C C C C CC C C C C C CC C C C C C CC C C C C C CC C C C C C CC C C C C C CC C C C C C C
λλλλλλ
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
10
20
30
40
50
60
7 70
1
CCCCCCCλ
μ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
(4.189)
Using the matrix of lag distances in Table 4.12 and Eq.(4.170), Eq.(4.189)
becomes
100.0000 54.8812 30.1194 16.5299 74.0818 40.6570 22.3130 154.8812 100.0000 54.8812 30.1194 40.6570 74.0818 40.6570 130.1194 54.8812 100.0000 54.8812 22.3130 74.0718 74.0818 116.5299 30.1194 54.8812 100.0000 12.2456 40.6570 74.0
1
2
3
4818 174.0818 40.6570 22.3130 12.2456 100.0000 30.1194 16.5299 154.8812 74.0818 74.0818 40.6570 30.1194 100.0000 54.8812 122.3130 40.6570 74.0818 74.0818 16.5299 54.8812 100.0000 1
1 1 1 1 1 1 1 0
λλλλλ
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
5
6
7
74.081874.081840.657022.313054.881254.881230.1194
1
λλμ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
…… (4.190)
The inverse matrix for Eq.(4.190) is
4–134
1
0.0322 0.0023 0.0074 0.0006 0.0207 0.0159 0.0001 0.09900.0023 0.0239 0.0007 0.0008 0.0044 0.0155 0.0002 0.1343
0.0074 0.0007 0.0359 0.0006 0.0053 0.0202 0.0166 0.09650.0006 0.0008 0.0006 0.0205 0.0017
ijC−
− − − − −− − − − − −
− − − − −− − − −
=0.0001 0.0168 0.2830
0.0207 0.0044 0.0053 0.0017 0.0228 0.0097 0.0004 0.28880.0159 0.0155 0.0202 0.0001 0.0097 0.0420 0.00004 0.02500.0001 0.0002 0.0166 0.0168 0.0004 0.00004 0.0342 0.0733
0.0990 0.1343 0.0965
− −− − − − −− − − − −− − − − − −
0.2830 0.2888 0.2306 0.0733 49.2637
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
− −⎢ ⎥⎣ ⎦
(4.191)
Multiplying the right hand side of Eq.(4.190) into the inverse matrix of
Eq.(4.191) gives the solution vector as
1
2
3
4
5
6
7
0.48410.59720.11700.01510.01510.2325
0.00392.6342
λλλλλλλμ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
(4.192)
The kriged value is calculated as
( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )
* *0 3 0.4841 30 0.5972 50 0.1170 20 0.0151 24.3321
0.0151 29.8968 0.2325 34.7659 0.0039 22.4886 39.5490
Φ = Φ = + + +
+ − + =
The estimation variance is computed from Eq.(4.140) as
( )3
2 2min
1( ) 100 2.6342
0.4841 74.0818 0.5972 74.0818 0.1170 40.657029.2455
0.0.0151 22.3130 0.0151 54.8812 0.2325 54.8812 0.0039 30.1194
e i ioi
C h
x x xx x x x
σ σ μ λ=
= − − = − −
+ +⎛ ⎞− =⎜ ⎟+ + − +⎝ ⎠
∑
The estimation standard deviation is
4-135
min 29.2455 5.4079
eσ = =
The 95% confidence interval is given by
( )( )*7 39.5490 1.96 5.4079 39.55 10.60Φ = ± = ±
Table 4.13 presents a summary of the results of the estimation. Figure
4.67 shows the estimated values along with the measured values. It
should be noted that kriging gives the estimated values of *3Φ and *
5Φ that
are essentially the same as those obtained from the linear interpolation
of Figure 4.52.
Table 4.13. Results of Estimations for Example 4.2.
Location iΦ *iΦ 2
mineσ mineσ 1.96 mineσ *min1.96i eσΦ − *
min1.96i eσΦ +
1 29.90 48.72 6.98 13.68 16.22 43.58
2 30
3 39.55 29.27 5.41 10.60 28.95 50.15
4 50
5 34.77 29.27 5.41 10.60 24.17 45.37
6 20
7 22.49 29.27 5.41 10.60 11.89 33.09
8 24.33 82.81 9.10 17.84 6.49 42.17
4–136
Figure 4.67. Graph of the measured and the kriged values for Example 4.2.
Using kriging, we are now able to estimate the petrophysical
properties for all the grid blocks of our reservoir simulation model that
honor the measured data and the correlation structure of the
heterogeneous properties. Further, we are able to estimate the
uncertainty associated with the estimates at each grid block. We have
made significant progress in dealing with the heterogeneities of our
reservoir rock.
We have studied ordinary kriging. There are other types of kriging
such as simple kriging, universal kriging, block kriging and ordinary
4-137
cokriging. These other types of kriging are usually covered in a
geostastistics course.
4.6 CONDITIONAL SIMULATION
4.6.1 Introduction
Kriging gives a smooth estimate because the estimate is a weighted
average of the sample data. Such a weighted average can never be larger
than the largest sample value nor can it be smaller than the smallest
sample value. Thus, kriging eliminates local variability. If such local
variability is important, then it can be incorporated into the estimated
values using conditional simulation. Conditional simulation also is
referred to as stochastic simulation or Monte Carlo simulation is the
other major application of geostatistics. The simulation is conditioned on
the measured data.
The idea behind conditional simulation is as follows. Each of the
estimates obtained from kriging was associated with an uncertainty in
the estimated value measured by the estimation variance or the
estimation standard deviation. Thus, the kriged estimate is a random
variable with a known variance or standard deviation. If the kriged value
comes from a normal distribution, then it is possible to draw a simulated
value from this distribution that is centered on the kriged value and has
a variance that is equal to the estimation variance and a standard
deviation that is equal to the estimation standard deviation.
4.6.2 Sequential Gaussian Simulation
The objective is to perform a stochastic simulation to estimate the
values of a petrophysical property in a heterogeneous reservoir at
4–138
locations for which no sample data have been measured. The simulated
values should honor the measured sample data and the correlation
structure of the heterogeneity as revealed by the analysis of the sample
data. Further, it should retain the local variability. Here is an algorithm
for performing such a stochastic simulation known as Sequential
Gaussian Simulation (SGS).
1. Select at random a location or node not yet simulated in the grid.
2. Use ordinary kriging to compute the local estimate at the node
along with the estimation variance and the estimation standard
deviation.
3. Draw a random value from a normal distribution with a mean
equal to the kriged estimate and a standard deviation equal to the
estimation standard deviation at that node. This is the simulated
value at that node. This step can be accomplished by using either
Eq.(4.50, (4.64) or (4.65) to draw a sample from a standard normal
distribution and then applying either Eq.(4.53), (4.66) or (4.67) to
compute the simulated value as
*si i ei izσΦ = Φ + (4.193)
where siΦ is the simulated value at location i, *iΦ is the kriged value
at location i, eiσ is the standard deviation or the square root of the
estimation variance at location i and iz is a standard normal
variate with μ = 0 and σ = 1 drawn for location i. Note that upon
moving to a new location, a new iz must be drawn for that
location.
4-139
4. Include the newly simulated value in the set of conditioning data
and use the expanded data set to simulate the value at the next
location.
5. Proceed to the next location, which of course was selected
randomly at the start, and repeat the calculations to simulate the
value at this location.
6. Repeat the calculations until all grid nodes have a simulated value.
7. This completes one realization of the simulation. Other realizations
can be obtained by repeating the calculations from step 1. To do
so, the order of the simulation is first determined using a random
number generator. This order should be different from all previous
orders. The calculations will then yield a new realization, whose
simulated values will be different from the previous realizations. If
there are N nodes to be simulated, there will be N! (N factorial)
possible realizations.
Example 4.3
Simulate values at the locations 1, 3, 5, 7 and 8 of Figure 4.51 at which
no samples were taken using Sequential Gaussian Simulation.
Solution to Example 4.3.
Let us simulate one realization. The first part of the calculations was
done in Example 4.2 in which kriged estimates were computed in the
following random order: 8, 1, 5, 7 and 3. The estimation variance and
standard deviations also were computed and presented in Table 4.13. To
simulate values at the five nodes, we draw five variates from a standard
normal distribution using Eq.(4.50, (4.64) or (4.65). For example, using
4–140
Eq.(4.64) the following five variates were drawn from a standard normal
distribution:
0.8821
-1.1679
0.6419
2.8546
0.91009
Using these numbers, we can simulate the values at nodes 8, 1, 5, 7 and
3 as follows:
( )( )* *8 8 8 8 24.33 9.10 0.8821 32.36s e zσΦ = Φ + = + =
( )( )* *1 1 1 1 29.90 6.98 1.1679 21.75s e zσΦ = Φ + = + − =
( )( )* *5 5 5 5 34.77 5.41 0.6419 38.24s e zσΦ = Φ + = + =
( )( )* *7 7 7 7 22.49 5.41 2.8546 37.92s e zσΦ = Φ + = + =
( )( )* *3 3 3 3 39.55 5.41 0.9009 44.42s e zσΦ = Φ + = + =
Figure 4.68 shows the measured values, kriged values and the simulated
values.
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Figure 4.68. A comparison of the measured, kriged and simulated values for Example 4.3.
4.6.3 A Practical Application of Sequential Gaussian Simulation
A laboratory waterflood experiment was conducted to determine
the oil recovery curve for a viscous oil reservoir (μo = 100 cp). The
coreflood experiment was performed in an unconsolidated sandpack.
The task at hand is to use the laboratory test in a sandpack to forecast
the oil recovery curve for a heterogeneous reservoir. Figure 4.69 shows
the CT images of the laboratory waterflood at three times just before
water breakthrough. It is shows a fairly uniform displacement of oil by
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the water. Figure 4.70 shows the oil recovery curve from the experiment.
It should observed that even for a fairly uniform sandpack, the oil
recovery after 3 pore volumes of water injection is less than 60% of the
initial oil in place. What will be the oil recovery in a reservoir with
significant permeability heterogeneity?
Figure 4.69. CT images of a laboratory waterflood in a sandpack: (a) 0.05 pore volume injected, (b) 0.10 pore volume injected, ( c) 0.25 pore volume
injected (Gharbi and Peters, 1993)
.
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Figure 4.70. Oil recovery curve for a laboratory waterflood in a sandpack (Gharbi and Peters, 1993)
.
The problem of forecasting the oil recovery curve in the field based
on a laboratory waterflood boils down to investigating the effect of
permeability heterogeneity on the waterflood performance. To address
this question, we generated twelve synthetic reservoirs with varying
degrees of permeability heterogeneity and correlation structures and then
scaled the laboratory waterflood to the synthetic reservoirs through
numerical simulation. Figure 4.71 shows the twelve heterogeneous, 2D
4–144
permeability fields generated to cover a wide range of Dykstra-Parsons
coefficient and correlation lengths using Sequential Gaussian
Simulation. Permeability fields were generated at Dykstra-Parsons
coefficients (V) of 0.01, 0.55 and 0.87 and dimensionless correlation
lengths of 0, 0.2, 0.7 and 2.0. The correlation length was made
dimensionless by dividing the correlation length by the length of the
reservoir in that direction. A Dykstra-Parsons coefficient of 0.01
represents a nearly homogeneous medium whereas a Dyktra-Parsons
coefficient of 0.89 represents an extremely heterogeneous medium.
Petroleum reservoirs typically have Dykstra-Parsons coefficients that
range from 0.5 to 0.9. A dimensionless correlation length of zero
represents an uncorrelated or random permeability distribution; a
correlation length of 0.2 represents mild correlation; a correlation length
of 2.0 represents extremely strong correlation. Depending on the
depositional environment, petroleum reservoirs can have widely different
correlation lengths. The value of dimensionless correlation length in the
y direction (Ly) was constant at 0.2 for the permeability fields shown in
Figure 4.71.
Two observations can be made from the permeability distributions
of Figure 4.71. First, as the correlation length (Lx) increases, the
permeability distributions become more and more stratified. The
number of layers is inversely proportional to Ly. In fact, as Lx
approaches infinity, for Ly = 0.2, the permeability distribution will
consist of exactly five (1/0.2) distinct homogeneous layers. Second, with
increasing Dykstra-Parsons coefficient at a given Lx, the contrast in the
permeability values increases while their spatial arrangements remain
similar. Figure 4.72 shows the permeability histograms, which indicate
that the permeability fields are log-normally distributed in accordance
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with observations in sedimentary rocks. Figure 4.73 shows the
variograms for the twelve permeability fields, which give a visual
impression of the degree of correlation in each permeability field.
Figure 4.71. Simulated permeability distributions (Gharbi and Peters, 1993)
Figure 4.74 shows a comparison of the recovery curve for the waterflood
experiment and the numerical simulation of the experiment. The
agreement between the two recovery curves is good. The simulation was
used to determine the relative permeability curves to be used to scale the
laboratory waterflood to the synthetic heterogeneous reservoirs.
4–146
Figure 4.72. Simulated permeability histograms (Gharbi and Peters,
1993).
4-147
Figure 4.73. Simulated permeability variograms in the x-direction
(Gharbi and Peters, 1993).
4–148
Figure 4.74. A comparison of the oil recovery curves of the experiment and the numerical simulation of the experiment (Gharbi and Peters,
1993).
Figure 4.75 compares the simulated oil recovery curves for each of
the twelve heterogeneous synthetic reservoirs with that of the laboratory
waterflood experiment. The following observations can be made from
these results. If the heterogeneous reservoir is characterized by low
variability in the permeability distribution (low Dykstra-Parsons
coefficient), the waterflood response will be essentially the same as in the
laboratory sandpack regardless of the correlation structure of the
4-149
heterogeneity. This is indicated by the agreement between the simulated
and the experimental recovery curves in the first column of Fig. 4.75 (V =
0.01). If the heterogeneous reservoir is characterized by a low correlation
length in the permeability distribution (low Lx), the waterflood response
will be essentially the same as in the laboratory sandpack regardless of
the variability in the permeability distribution. This is indicated by the
agreement between the simulated and the experimental recovery curves
in the first row of Figure 4.75 (Lx = 0). If the heterogeneous reservoir is
characterized by high variability and high correlation length in the
permeability distribution, the waterflood response could be significantly
different from that of the laboratory sandpack. This is most clearly
shown by the response in the last permeability field in Figure 4.75 (Lx =
2 and V = 0.87). In this case, the waterflood effeciency is significantly
less in the heterogeneous reservoir than in the laboratory sandpack.
To investigate the reason for the significant disparity in
performance between the laboratory waterflood in a relatively
homogeneous sandpack and in certain kinds of heterogeneous reservoirs,
we examine the simulated water saturation maps. Figures 4.76 and 4.77
show the simulated water saturation maps for each of the twelve
heterogeneous reservoirs at 0.10 and 0.25 pore volume injected. We see
that the displacements in the heterogeneous media with high Dykstra-
Parsons coefficient and high correlation length are dominated by
channeling of the injected water due to the permeability stratification.
This results in significant bypassing of the oil in some layers, resulting in
a low oil recovery. These channels provide easy pathways for the water to
flow from the injection well to the producing well, essentially leaving
much of the reservoir unswept. By contrast, the displacements in the
reservoirs with low Dykstra-Parsons coefficients are characterized by
4–150
excellent sweep comparable with that observed in the CT images of the
laboratory waterflood experiment (Figure 4.69). This results in a
displacement performance that is comparable to the laboratory
waterflood experiment in the sandpack.
We conclude from this study that the performance of an enhanced
oil recovery (EOR) displacement in a heterogeneous reservoir could be
significantly lower than in a laboratory experiment depending on the
degree and structure of the heterogeneity of the reservoir. This
conclusion underscores the need for proper scaling when using the
results of laboratory coreflood experiments in relatively homogeneous
porous media to forecast the expected performance of an EOR process in
heterogeneous reservoirs. The methodology developed and presented in
this study can be used to accomplish this scaling and prevent erroneous
performance forecasts.
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Figure 4.75. A comparison of the experimental and simulated oil recovery
curves (Gharbi and Peters, 1993).
4–152
Figure 4.76. Simulated water saturation maps at 0.10 pore volume
injected (Gharbi and Peters, 1993).
Figure 4.77. Simulated water saturation maps at 0.25 pore volume
injected (Gharbi and Peters, 1993).
NOMENCLATURE
a = correlation length (range of influence)
C = covariance function
4-153
C0 = nugget effect
erf = error function
erfc = complementary error function
F = cumulative probability distribution
h = pay thickness
h = lag distance
k = absolute permeability of the medium
ln = natural logarithm (log to base e)
Lx = dimensionless correlation length in the x-direction
Ly = dimensionless correlation length in the y-direction
P = probability density function
s = standard deviation of sample data
s2 = variance of sample data
V = Dykstra-Parsons coefficient of variation
x = linear coordinate
x = random variable
z = variate from a standard normal distribution
ρ = correlation coefficient function
μ = population mean
μ = Lagrange parameter
σ = population standard deviation
σ2 = population variance
φ = porosity, fraction
τ = mean of log normal distribution
γ = variogram
λ = kriging weights
Φ = sample data
Φ = sample mean
ω = standard deviation of a log normal distribution
ω2 = variance of a log normal distribution
4–154
REFERENCES AND SUGGESTED READINGS
Armstrong, M. : Basic Linear Geostatistics, Springer-Verlag, New York, 1998.
Caers, J. : Petroleum Geostatistics, Society of Petroleum Engineers, Richardson, 2005.
Carlson, M.R. : Practical Reservoir Simulation, PennWell, Tulsa, 2003.
Chambers, R.L., Yarus, J.M. and Hird, K.B. : “Petroleum Geostatistics for Nongeostatisticians – Part 1,” The Leading Edge (May 2000) 474-479.
Chambers, R.L., Yarus, J.M. and Hird, K.B. : “Petroleum Geostatistics for Nongeostatisticians – Part 2,” The Leading Edge (June 2000) 592-599.
Clark, I. : Practical Geostatistics, Applied Science Publishers, London, 1979.
Clark, I. and Harper, W.V. : Practical Geostatistics 2000, Ecosse North America Llc, Columbus, 2000.
Clark, I. : "Does Geostatistics Work? ", Proc. 16th APCOM, Thomas J O'Neil, Ed., Society of Mining Engineers of AIME Inc, New York, 1979, 213-225.
Deutsch, C.V.: "What in the Reservoir is Geostatistics Good For ?", Jour. Cand. Pet. Tech. (April 2006) 14-20.
Deutsch, C.V.: Geostatistical Reservoir Modeling, Oxford University Press, New York, 2002.
Deutsch, C.V. and Journel, A.G. : GSLIB Geostatistical Software Library and User’s Guide, Oxford University Press, New York, 1992.
Dykstra, H. and Parsons, R.L. : “The Prediction of Oil Recovery by Waterflood,” Secondary Recovery of Oil in the United States, American Petroleum Institute (1950) 160-175.
Gharbi, R. and Peters, E.J. : “Scaling Coreflood Experiments to Heterogeneous Reservoirs,” Journal of Petroleum Science and Engineering, 10, (1993) 83-95.
Gharbi, R.: Numerical Modeling of Fluid Displacements in Porous Media Assisted by Computed Tomography Imaging, PhD Dissertation, The University of Texas at Austin, Austin, Texas, August 1993.
4-155
Gotway, C.A. and Hergert, G.W. : "Incorporating Spatial Trends and Anisotropy in Geostatistical Mapping of Soil Properties," Soil Science of America Journal, Vol. 61 (1977) 298-309.
Hohn, M.E. : Geostatistics and Petroleum Geology, Van Nostrand Reinhold, New York, 1988.
Hirsche, K., Porter-Hische, J., Mewhort, L. and Davis, R. : "The Use and Abuse of Geostatistics," The Leading Edge (March 1997) 253-260.
Holbrook, P. : Pore Pressure Through Earth Mechanical Systems, Force Balanced Petrophysics, 2001.
Isaaks, E.H. and Srivastava, R.M. : An Introduction to Applied Geostatistics, Oxford University Press, New York, 1989.
Jensen, J.L., Lake, L.W., Corbett, P.W.M. and Goggin, D.J. : Statistics for Petroleum Engineers and Geoscientists, 2nd Edition, Elsevier, New York, 2000.
Kerbs, L. : “GEO-Statistics: The Variogram,” Computer Oriented Geological Society Computer Contributions (August 1986) 2, No. 2, 54-59.
Metheron, G. : "Principles of Geostatistics," Economic Geology, Vol. 58 (1963) 1446-1266.
Peters, E.J., Afzal, N. and Gharbi, R. : “On Scaling Immiscible Displacements in Permeable Media,” Journal of Petroleum Science and Engineering, 9, (1993) 183-205.
Peters, E.J. and Gharbi, R. : “Numerical Modeling of Laboratory Corefloods,” Journal of Petroleum Science and Engineering, 9, (1993) 207-221.
Peters, E.J. and Afzal, N. : “Characterization of Heterogeneities in Permeable Media with Computed Tomography Imaging,” Journal of Petroleum Science and Engineering, 7, No. 3/4, (May 1992) 283-296.
Peters, E.J. and Hardham, W.D. : “Visualization of Fluid Displacements in Porous Media Using Computed Tomography Imaging,” Journal of Petroleum Science and Engineering, 4, No. 2, (May 1990) 155-168.
Zirschy, J.H. and Harris, D.J. : "Geostatistical Analysis of Hazardous Waste Site Data," J. of Environmental Engineering, Vol. 112 (1986) 770784.
4–156
CHAPTER 5
DISPERSION IN POROUS MEDIA
5.1 INTRODUCTION
When a miscible fluid displaces another in a porous medium, the
displacing fluid tends to mix with the displaced fluid. The result is that a
mixing or transition zone develops at the front in which the
concentration of the injected fluid decreases from one to zero.
Experiment shows that the mixing zone grows as the displacement
progresses. This mixing and spreading of the injected fluid is known as
dispersion.
Bear (1972) describes dispersion as the "macroscopic outcome of
actual movement of individual tracer particles through pores...".
Essentially, dispersion is the mixing caused by single-phase fluid
movement through a porous medium. What is "mixed" is usually called
a tracer, but can be thought of as a concentration of any chemical
component within a given phase that is transported through the system.
5-1
Dispersion has practical consequences in contaminant transport in
aquifers and in improved oil recovery from petroleum reservoirs. If a
miscible contaminant is accidentally introduced into an aquifer at a site,
dispersion will cause the contaminant to spread to a larger area as it is
being transported by groundwater flow. Even though the concentration
of the contaminant is reduced by dispersion, a much larger area of the
aquifer will become contaminated as a result of dispersion than the
original spill area. Thus, a much larger area than the original spill will
need to be cleaned up by any contaminant remediation measure.
Miscible displacement is the most efficient improved oil recovery
method. Because there is no capillary force to trap the displaced oil, it is
theoretically possible to recover 100% of the oil by miscible displacement.
However, because the injected solvent is usually more expensive than the
oil that is to be displaced, it is usually injected in small quantities as
slugs and chased by a less expensive fluid such as water or gas.
Dispersion will dilute and reduce the effectiveness of the miscible slug as
it is propagated through the reservoir. In this case, dispersion is
detrimental to the recovery process. On the other hand, dispersion
causes a solvent to mix, spread and contact the displaced fluid even after
it had been originally bypassed by the injected solvent. In this case,
dispersion improves the displacement efficiency.
Other industrial processes that involve dispersion include (1) use
of tracers such as dyes, electrolytes and radioactive isotopes to
characterize reservoir and aquifer properties, (2) development of a
transition zone between salt water and fresh water in coastal aquifers, (3)
radioactive and reclaimed sewage waste disposals into aquifers, (4) use of
reactors packed with granular material in the chemical industry, and (5)
5–2
movement of fertilizers in the soil and the leaching of salts from the soil
in agriculture.
5.2 LABORATORY FIRST-CONTACT MISCIBLE DISPLACEMENTS
Before embarking on the mathematics of dispersion in porous
media, it is instructive and illuminating to view images of miscible
displacements that show dispersion at work. The images presented in
this section are for first-contact miscible displacements. This means that
the two fluids used in the experiment are fully miscible upon first
contact. This is in contrast to developed miscibility encountered in
certain enhanced oil recovery process in which the injected fluid is not
initially miscible with the displaced fluid. However, after a certain time
has elapsed, mass transfer between the injected and the displaced fluids
causes the two fluids to become miscible. First-contact miscible
displacements are the most efficient type of displacements.
The efficiency of a miscible displacement or an immiscible
displacement for that matter, is controlled by, among other factors, the
mobility ratio, the effect of gravity, dispersion and the heterogeneity of
the porous medium. Mobility ratio is a dimensionless number that is
characteristic of a displacement. The mobility of a fluid phase in a porous
medium is defined as
phasephase
phase
kλ
μ= (5.1)
where is the effective permeability to that phase and phasek phaseμ is the
viscosity of that phase. Mobility ratio is defined as
5-3
displacing fluid
displaced fluid
Mλλ
= (5.2)
For a first-contact miscible displacement, we have single phase flow and
as a result, the effective permeability to each fluid is equal to the
absolute permeability of the porous medium. Therefore, for first-contact
miscible displacements, the mobility ratio simplifies to
displacing fluid displaced fluid
displacing fluid
displaced fluid
k
Mk
μ μμ
μ
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠= =⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
(5.3)
Thus, for a first-contact miscible displacement, the mobility ratio is the
viscosity ratio given by the viscosity of the displaced fluid divided by the
viscosity of the displacing fluid. A mobility ratio of 1 or less is favorable
to the displacement efficiency whereas a mobility ratio greater than 1 is
unfavorable to the displacement efficiency. The lower the mobility ratio,
the higher is the displacement efficiency. The higher the mobility ratio,
the lower is the displacement efficiency.
Gravity can be as a double-edged sword in displacements. Gravity
override and gravity underride or tonguing can significantly reduce the
displacement efficiency. However, by careful engineering, gravity override
can be used to enhance the displacement efficiency if the less dense
solvent is injected up dip while the denser displaced fluid is withdrawn
down dip.
Dispersion can also be as a double-edged sword in displacements.
If a miscible slug is used in a displacement process to reduce cost,
5–4
dispersion will dilute and degrade the effectiveness of the slug as the
process progresses. On the other hand, mixing caused by dispersion can
allow the injected solvent to spread and contact more of the displaced
fluid than if dispersion was absent. In this case, dispersion is favorable
to the displacement efficiency.
Permeability heterogeneity of the porous medium tends to reduce
the displacement efficiency for the most part. Because the heterogeneity
has a correlation structure, high permeabilities tend to occur next to
high permeabilities and low permeabilities tend to occur next to low
permeabilities as the porous medium is traversed from one point to
another. These permeability arrangements cause the injected fluid to
channel through the high permeability layers, thereby leaving the low
permeability layers unswept.
Figure 5.1 shows CT images of a first-contact miscible
displacement in an unconsolidated sandpack at a favorable mobility ratio
of 0.82. In the experiment, a brine containing sodium chloride was used
to displace another brine containing barium chloride, which is an x-ray
doppant. The images show the concentrations of the injected solvent in a
vertical slice of the 3D sandpack at dimensionless times of 0.13, 0.50
and 1.0 pore volume injected. Mixing caused by dispersion is clearly
visible at the displacement front. The mixing zone length, defined as the
distance between the solvent concentrations of 0.90 and 0.10, is slightly
tilted in Figure 5.1B because of a small density difference between the
injected brine and the displaced brine. The injected brine was slightly
less dense that the displaced brine. Thus, the tilt is a small gravity
override. Because of the favorable mobility ratio, the displacement is very
efficient with almost 100% displacement efficiency at 1 pore volume
injected. The displacement efficiency is slightly less than 100% because
5-5
of the small gravity override problem. This experiment shows that a
miscible displacement can be quite susceptible to gravity instability in
the form of gravity override if the injected fluid is less dense than the
displaced fluid and gravity tonguing if the injected fluid is denser than
the displaced fluid.
Figure 5.1. CT images of solvent concentration for a first-contact miscible displacement in a sandpack. A: tD = 0.13, B: tD = 0.50, C: tD = 1.0 pore
volume injected. M = 0.82, φ = 35.26%, k = 15.76 darcies, v = 0.154x10-2 cm/s, BT recovery = 96.53% (Peters and Hardham, 1990).
5–6
Figure 5.2 shows the 1D profiles of the 3D solvent concentration
distributions at the three dimensionless times. The profiles have the
classical shapes predicted by the convection-dispersion model of
dispersion to be presented later in this chapter. The profiles can be used
to characterize the dispersion phenomenon in the direction of flow.
Figure 5.2. Average solvent concentration profiles for the miscible
displacement experiment of Figure 5.1 (Peters and Hardham, 1990).
Figure 5.3 shows the images of a first-contact miscible
displacement at an unfavorable mobility ratio of 74. Here, brine with an
x-ray doppant was used to displace a mixture of glycerine and brine. The
injected brine was less dense than the displaced fluid. It can be seen that
the density difference results in a significant gravity override. Such an
override is detrimental to the displacement efficiency. Mixing caused by
dispersion is also clearly evident. The transverse or lateral dispersion is
quite orderly as one moves from the pure solvent to the pure displaced
5-7
fluid. There is also viscous instability or viscous fingering caused by the
adverse mobility ratio. At a mobility ratio greater than 1, the displacing
fluid is more mobile than the displaced fluid. As a result, at the interface
between the two fluids, the displacing fluid tends to penetrate the
displaced fluid due to the inevitable imperfections in the porous medium.
These perturbations may grow to form viscous fingers as shown in Figure
5.3B. Figure 5.4 shows the corresponding solvent concentration profiles
for this displacement. The shapes of the profiles are significantly different
from those of Figure 5.2 and cannot be predicted by the idealized
convection-dispersion model. Clearly, this displacement is less efficient
than that of Figure 5.1.
5–8
Figure 5.3. CT images of solvent concentration for a first-contact miscible
displacement in a sandpack. A: tD = 0.13, B: tD = 0.50, C: tD = 1.0 pore volume injected. M = 74, φ = 35.68%, k = 16.69 darcies, v = 0.014x10-2
cm/s, BT recovery = 31.23% (Peters and Hardham, 1990).
Figure 5.4. Average solvent concentration profiles for the miscible
displacement experiment of Figure 5.3 (Peters and Hardham, 1990).
5-9
Figure 5.5 shows the images of a first-contact miscible
displacement at an unfavorable mobility ratio of 54 with gravity override.
The images show that portions of the porous medium that were initially
bypassed by the solvent due to gravity override, were subsequently
displaced with continued solvent injection. This is a testament to the
efficiency of miscible displacements. By contrast, in an immiscible
displacement, portions of the porous medium initially bypassed for
whatever reason typically remain unswept with continued injection.
Figure 5.6 shows the corresponding solvent concentration profiles. Notice
how the disturbances in the profiles propagate in time and space. This is
characteristic of miscible displacements. Such disturbances typically
remain stationary in immiscible displacements.
Figure 5.7 shows a first-contact miscible displacement in a quarter
five-spot pattern at a mobility ratio of 1. Here, dyed water was used to
displace clear water in a thin porous medium consisting of uniform glass
beads. The images were obtained with a home-built imaging system
described by Peters and Reid (1990). As expected, the displacement is
efficient. The mixing zone length caused by dispersion is also visible. The
areal sweep efficiency can easily be measured. The areal sweep efficiency
is the ratio of the area contacted by the injected fluid and the total area
of the pattern. Note that because of the geometry of the displacement
pattern, there are dead spots between the injector and the producing
wells. In a field flood, such dead spots would be potential candidates for
infill drilling. In this experiment, the thin porous medium was oriented
horizontally. As a result, gravity effect was negligibly small.
5–10
Figure 5.5. CT images of solvent concentration for a first-contact miscible
displacement in a sandpack. M = 54, φ = 31.70%, k = 9.5 darcies, v = 0.850x10-2 cm/s, BT recovery = 28.80% (Peters and Afzal, 1992).
5-11
Figure 5.6. Average solvent concentration profiles for the miscible displacement experiment of Figure 5.5 (Peters and Afzal, 1992).
Figure 5.7. Video images of solvent concentration in a miscible
displacement in a glass bead pack. M = 1(Peters and Reid, 1990).
5–12
Figure 5.8 shows the images of another first-contact miscible
displacement in the same porous medium as in Figure 5.7 at a mobility
ratio of 100. The porous medium was oriented horizontally. There is
significant mixing due to dispersion and viscous fingering due to the
adverse mobility ratio. Clearly, this displacement is less efficient than
that of Figure 5.7.
Figure 5.8. Video images of solvent concentration in a miscible displacement in a glass bead pack. M = 100 (Peters and Reid, 1990).
5-13
Figure 5.9 shows images of a first-contact miscible displacement in
the same porous medium and at the same mobility ratio as in Figure 5.8
but with the medium oriented vertically to take advantage of the
beneficial effect of gravity override. This experiment was designed to
simulate a horizontal injection well at the top left corner and a horizontal
producing well at the bottom right corner of the medium. With this
arrangement, gravity override is beneficial to the displacement efficiency
in contrast to the displacement of Figure 5.3 in which gravity segregation
was detrimental to the displacement efficiency. It can be seen that here,
Figure 5.9. Video images of solvent concentration in a gravity-assisted miscible displacement experiment in a glass bead pack (Peters and Reid,
1990).
5–14
gravity override delays the solvent breakthrough thereby significantly
enhancing the displacement efficiency compared to the displacement of
Figure 5.8. There is viscous fingering due to the adverse mobility ratio as
well as significant mixing due to dispersion. Clearly, this displacement is
more efficient than that of Figure 5.8.
Figure 5.10 shows NMR images of a first-contact miscible
displacement in the layered sandstone core of Figure 2.57 at a favorable
mobility ratio of 0.84. One can see mixing due to dispersion. However,
the dominate effect is channeling due to the permeability heterogeneity of
the layered medium. It can be concluded from these images that
permeability stratification in a porous medium can have a significant
adverse effect on any displacement, miscible or immiscible. Figure 5.11
shows the corresponding solvent concentration profiles.
Figure 5.12 shows a miscible displacement is the same layered
sandstone core as in Figure 5.10 but at an unfavorable mobility ratio of
95. Here, the effect of viscous fingering due to the adverse mobility ratio
is superimposed on the channeling due to the heterogeneity of the core.
Notice that the channels created by the injected solvent are thinner than
in Figure 5.10 indicating a less efficient displacement than in Figure
5.10. Figure 5.13 shows the corresponding solvent concentration
profiles.
5-15
Figure 5.10. NMR images of a first-contact miscible displacement in a layered sandstone core at a mobility ratio of 0.84 (Peters and Li, 1996).
5–16
Figure 5.11. Average solvent concentration profiles for the miscible displacement of Figure 5.9 (Peters and Li, 1996).
5-17
Figure 5.12. NMR images of a first-contact miscible displacement in a layered sandstone core at a mobility ratio of 95 (Peters and Li, 1996).
5–18
Figure 5.13.Average solvent concentration profiles of a first-contact miscible displacement in a layered sandstone core at a mobility ratio of
95 (Peters and Li, 1996).
Finally, Figure 5.14 shows the effect of mobility ratio on the
efficiency of miscible displacements. The figure shows the recovery
curves for miscible displacements in the same layered sandstone core as
in Figure 5.10 for mobility ratios ranging from 0.84 to 95. Clearly,
mobility ratio has a significant impact on the efficiency and the timing of
a miscible displacement. Displacements at adverse mobility ratios suffer
from early solvent breakthroughs and require more pore volumes of
5-19
injection to attain the same displacement efficiency as favorable mobility
ratio displacements.
Figure 5.14. Recovery curves for a first-contact miscible displacement in a layered sandstone core at various mobility ratios (Peters and Li, 1996).
5.3 ORIGINS OF DISPERSION IN POROUS MEDIA
Dispersion is the net result of (a) molecular diffusion, (b) local
velocity gradients within given pores, (c) locally heterogeneous streamline
lengths and velocities, and (d) mechanical mixing in pore bodies.
5–20
Dispersion can be viewed as consisting of two components: molecular
diffusion and mechanical dispsersion.
5.3.1 Molecular Diffusion.
Molecular diffusion is a physiochemical dispersion caused by
chemical potential gradient, which is correlated to the chemical
concentration of the solute being transported. It is mixing caused by
random motions of the fluid particles due to the thermal kinetic energy of
the solute. This motion is known as Brownian motion. Molecular
diffusion is isotropic and occurs equally in all directions. Molecular
diffusion can easily be demonstrated in the laboratory. If a drop of blue
ink is carefully added to a beaker of water and allowed to sit, after
sometime, the water in the beaker will turn blue as a result of molecular
diffusion. Thus, molecular diffusion occurs whether there is flow or not.
Molecular diffusion will contribute to both longitudinal and transverse
dispersions. Molecular diffusion in a porous medium is less than it
would be in the absence of the porous medium. The solid grains hinder
diffusion just as they hindered the flow of electrical current and fluid
flow. Because of the larger molecular spacing, molecular diffusion in a
gas is much larger than in a liquid.
5.3.2 Mechanical Dispersion.
The second component of dispersion may be described as
mechanical dispersion. The origins of mechanical dispersion can be seen
in Figure 5.15. The figure shows marked fluid particles at time t and at
time t+Δt. When a fluid flows in a porous medium, its velocity
distribution within a pore is not uniform, due to boundary effects acting
in three different ways as shown in Figure 5.15. In Figure 5.15a, the no
slip condition at the pore wall creates a velocity gradient in the fluid. This
5-21
velocity gradient causes the marked fluid particles to be spread out in
the flow direction. This is longitudinal dispersion in the flow direction. In
Figure 5.15b, the variation in pore dimensions (recall k ∝ r2) causes flow
to occur faster (further) in some pores than in others. This causes the
marked fluid particles to be spread out in the direction of flow. This is
longitudinal dispersion in the direction of flow. In Figure 5.15c, the
streamlines fluctuate with respect to the mean flow direction as the fluid
particles navigate around the solid grains. This effect of the tortuosity of
the porous medium causes the fluid particles to be spread out in the
transverse direction with respect to the mean flow direction. This is
transverse dispersion, perpendicular to the direction of flow. Finally,
there is local mixing of the fluids within the pores as shown in Figure
5.16. This contributes to the mechanical dispersion.
Figure 5.15.Origins of mechanical dispersion.
5–22
Figure 5.16.Local mechanical mixing of fluid particles.
5.4 CONVECTION-DISPERSION EQUATION
5.4.1 Generalized Equation in Vector Notation
Chemical species are transported in a flowing system by two
transport mechanisms: (a) advection or convection and (b) dispersion.
Advection is mass transport due to the bulk motion of the carrying fluid
and is given by
aJ uCφ= (5.4)
where is the mass flux vector (mass/area/time) of species i due to
advection, φ is the porosity of the porous medium, u
aJ
is the interstitial
velocity vector (Darcy velocity vector/porosity) and C is the concentration
(mass/unit volume of solution) of species i in the solution. The mass flux
due to dispersion is given by
dJ Dφ C= − ∇ (5.5)
5-23
where is the mass flux vector (mass/area/time) of species i due to
dispersion and
dJ
D is the dispersion coefficient tensor. The dispersion
coefficient tensor characterizes molecular diffusion and mechanical
dispersion and is given by
dD D D= + m (5.6)
where is the molecular diffusion coefficient and dD mD is the mechanical
dispersion tensor. The continuity equation for mass transport is given by
( ) .C
Jtφ∂
0+∇ =∂
(5.7)
where t is time and is the total mass flux vector of species i due to
advection and dispersion and is given by the vector sum
J
aJ J Jd= + (5.8)
Substituting Eqs.(5.4) and (5.5) into (5.7) gives the mass transport
equation for a constant porosity medium as
( ) ( ). .C uC D Ct
∂ 0+∇ −∇ ∇ =∂
(5.9)
Eq.(5.9) is a well known equation of mathematical physics, which
is known by a variety of names such the advection-dispersion equation,
convection-dispersion equation, first-contact miscible displacement
equation, solute transport equation and mass transport equation. It is a
linear, second order, parabolic partial differential equation. It can be
5–24
used to solve a variety of transport problems. If the transport is by pure
convection with no dispersion, then Eq.(5.9) simplifies to
( ).C uCt
0∂+∇ =
∂ (5.10)
If there is no convection (static fluid), then there will be no mechanical
dispersion. In this case, transport is by molecular diffusion only and
Eq.(5.9) becomes
2 0dC D Ct
∂− ∇ =
∂ (5.11)
which is the diffusion or diffusivity equation.
5.4.2 One-Dimensional Convection-Dispersion Equation
For 1D transport in the x direction, Eq.(5.9) becomes
2
2 0x LC C Cu Dt x x
∂ ∂ ∂+ −
∂ ∂ ∂= (5.12)
where vx is the interstitial velocity in the x direction and DL is the
principal value of the dispersion coefficient in the x direction known as
the longitudinal dispersion coefficient. In this case, the flow direction is a
principal axis of the dispersion coefficient anisotropy. Eq.(5.12) can be
written in terms of Darcy velocity instead the interstitial velocity as
2
2 0xL
vC C CDt x xφ
∂ ∂ ∂+ −
∂ ∂ ∂= (5.13)
5-25
where ux is the Darcy velocity also known as the superficial velocity. It is
possible to extend the mass transport equation to include retardation of
the solute due to adsorption, chemical reaction, biological
transformations or radioactive decay. In the case of retardation due to
adsorption of the chemical species on the surface of the porous medium,
Eq.(5.12) becomes
2
2 0x L
f f
u DC C Ct R x R x
∂ ∂ ∂+ −
∂ ∂ ∂= (5.14)
where Rf is a retardation factor that accounts for adsorption. If Rf is
equal to 1, there is no adsorption whereas if Rf is greater than 1, there is
adsorption and the transport of the chemical species will be retarded.
This means that in the presence of adsorption, the concentration profiles
of the chemical species will travel at a speed that is lower than if there
was no adsorption. This fact is obvious from Eq.(5.14) in which the speed
of convection has been reduced from vx to vx/Rf. Also, retardation
reduces the effective dispersion coefficient as shown in Eq.(5.14).
5.4.3 Solution of the One-Dimensional Convection-Dispersion Equation
The initial-boundary value problem for 1D transport consists of
Eq.(5.12) together with appropriate initial and boundary conditions. For
transport in a semi infinite medium, the initial-boundary value problem
consists of the following equations:
2
2 0x LC C Cu Dt x x
∂ ∂ ∂+ −
∂ ∂ ∂= (5.12)
( ),0 iC x C= (5.15)
5–26
( )0, jC t C= (5.16)
( ), iC t C∞ = (5.17)
Eq.(5.15) is the initial condition, which specifies the initial concentration
of the injected species in the solution domain at time zero. A convenient
initial concentration in laboratory experiments is Ci equal to zero, which
means there is no injected chemical species in the initial saturating fluid.
Eq.(5.16) is the inlet boundary condition, which specifies the
concentration of the injected species at all times. A convenient inlet
boundary condition is Cj is equal to 1 at all times, although this is
difficult to arrange in an actual laboratory experiment. Eq.(5.17) is the
external boundary condition, which specifies that far away from the inlet,
the concentration of the injected species is equal to its initial value at
time zero. Such a boundary condition was used in the welltest model of
Chapter 3 for an infinite acting reservoir.
The initial-value problem can be put in dimensionless form as
follows:
2
2
1 0D D D
D D Pe D
C C Ct x N x
∂ ∂ ∂+ −
∂ ∂ ∂= (5.18)
( ),0 0DC x = (5.19)
( )0, 1D DC t = (5.20)
( ), DC t 0∞ = (5.21)
5-27
where
( ), iD
j i
C x t CC
C C−
=−
(5.22)
xD
u t qttL A Lφ
= = (5.23)
DxxL
= (5.24)
xPe
L L
u L qLND A Dφ
= = (5.25)
Eq.(5.23) gives the dimensionless time in pore volumes injected. Eq.(5.25)
defines a relevant dimensionless number for the transport known as the
Peclet Number. It is the ratio of the two transport mechanisms involved,
advection and dispersion. In the definition of Peclet Number in Eq.(5.25),
the length of the porous medium, L, has been used as a characteristic
dimension of the system. Of course, in a semi infinite medium, the length
of the porous medium would not be a convenient characteristic
dimension of the system. In general, Peclet Number is defined as
x pPe
L
u DN
D= (5.26)
where Dp is a characteristic dimension of the porous medium. A
convenient characteristic dimension could be the grain diameter in the
case in which the porous medium is composed of granular material.
5–28
Another possible characteristic dimension of a porous medium could be
kφ
in the spirit of Eq.(3.119).
Eqs.(5.12), (5.15), (5.16) and (5.17) or their dimensionless versions,
Eqs.(5.18) through (5.21), can be solved by Laplace transformation.
Ogata and Banks (1961) give the solution of Eqs.(5.12), (5.15), (5.16) and
(5.17) as
( ),2 2 2
x
L
v xDi x
L L
C x u t x u tC x t erfc e erfcD t D t
x⎡ ⎤⎛ ⎞ ⎛ ⎞−
= ++
⎢ ⎥⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ (5.27)
where erfc is the complementary error function defined by Eq.(4.47) and
reproduced here for convenience as
( ) 22 u
uerfc u e du
π∞ −= ∫ (5.28)
The error function and the complementary error function are related by
( ) ( ) 1erf u erfc u+ = (5.29)
( ) ( )1erfc u erf u− = + (5.30)
Also,
( ) ( )erf u erf u− = − (5.31)
The solution given by Eq.(5.27) can be written in dimensionless form as
5-29
( ) 1, e2
2 2
D Pex ND D D DD D D
D D
Pe Pe
x t x tC x t erfc erfct t
N N
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟
− +⎢ ⎥⎜ ⎟ ⎜= + ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
(5.32)
Ogata and Banks (1961) demonstrated that the second term in the
square bracket of the right hand sides of Eqs.(5.27) and Eq.(5.32) can be
neglected in comparison to the first term. The solution given by Eq.(5.27)
then becomes
( ),2 2
i
L
C x u tC x t erfcD t
x⎡ ⎤⎛ ⎞−
= ⎢ ⎥⎜⎜ ⎟⎟⎢ ⎥⎝ ⎠⎣ ⎦ (5.33)
A careful examination of the solution given by Eq.(5.33) shows that it is
related to the cumulative normal probability distribution with mean xv t ,
variance 2 and standard deviation LD t 2 LD t . Eq.(5.33) in dimensionless
form is
( ) 1,2
2
D DD D D
D
Pe
x tC x t erfct
N
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟
−⎢ ⎥⎜= ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(5.34)
or
( ) 1,2 2
D DD D D Pe
D
x tC x t erfc Nt
⎡ ⎤⎛ ⎞−= ⎢ ⎥⎜⎜ ⎟⎟⎢ ⎥⎝ ⎠⎣ ⎦
(5.35)
5–30
Eq.(5.34) or (5.35) also is related to the cumulative normal probability
distribution with mean Dt , variance 2 D
Pe
tN
and standard deviation 2 D
Pe
tN
.
Figure 5.17 shows the evolution of the solution predicted by
Eq.(5.34) or (5.35) for a Peclet Number of 100. It should be observed that
the solution at each dimensionless time is given by
(5.36) ( ), 1 Pr D D DC x t Cumulative Normal obability Distribution= −
with mean Dt and standard deviation 2 D
Pe
tN
. The shape of the solution is
that of an inverted S. Note also that because of dispersion, the
breakthrough time is less than 1 pore volume injected. The concentration
CD = 0.5 travels at the advection speed all times. Although it has not be
stated explicitly, the solution given by Eq.(5.34) or (5.35) is valid for an
idealized system under very restrictive conditions. The conditions are (1)
homogeneous and isotropic porous medium, (2) the injected fluid has the
same density as the displaced fluid, and (3) the injected fluid has the
same viscosity as the displaced fluid. These conditions preclude
channeling due to gross permeability heterogeneity, gravity override or
gravity tonguing and viscous fingering, all of which phenomena are not
included in the convection-dispersion equation. If these restrictions are
met, then Eq.(5.34) or (5.35) gives a good representation of the solvent
concentration profiles observed in laboratory experiments. For example,
although the experiment shown in Figures 5.1 and 5.2 does not quite
meet all the restrictions because of the small gravity override,
nevertheless Eq.(5.34) or (5.35) can be used to approximately reproduce
5-31
the laboratory measured solvent concentration profiles of Figure 5.1.
Figure 5.18 shows the result of a first attempt to reproduce the solvent
concentration profiles of Figure 5.1 using Eq.(5.34). The agreement
between laboratory measured profiles and those predicted by Eq.(5.34) is
good. Of course, the experimental profiles are distorted by the small
gravity override observed in the experiment. As a result, the experimental
profiles and the calculated profiles do not match exactly at the
displacement front. There is also evidence of retardation in the
experiment as the experimental profiles seem to lag behind the
calculated profiles.
Let us focus on the solution at a dimensionless time of 0.50 pore
volume injected shown in Figure 5.19. Marked on the figure are the
Figure 5.17. Solutions of the convection-dispersion equation for a Peclet Number of 100.
5–32
Figure 5.18. A comparison of the simulated and the measured solvent concentration profiles for the miscible displacement of Figure 5.1 and
5.2.
advection front, the dimensionless mixing zone length between CD = 0.90
and 0.10 and the dimensionless mixing zone length equal to twice the
standard deviation between CD = 0.8413 and 0.1588. It should be
observed that at tD = 0.5 pore volume injected, the advection front has
traveled exactly half the distance of the porous medium. This is as it
should be because the advection front travels at the advection speed or
at the interstitial velocity. Both the mixing zone length and the length
that represents twice the standard deviation grow in proportion to Dt ,
although this is not obvious from looking at the solution at one time. In
5-33
fact, it can be shown that the dimensionless mixing length between CD =
0.90 and CD = 0.10 is given by
3.625 DD
Pe
txN
Δ = (5.37)
or in dimensional form by
3.625 Lx D tΔ = (5.38)
Eq.(5.38) shows that if the mixing zone length can be measured as a
function of time during an experiment, then a graph of Δx versus
t could be used to determine the longitudinal dispersion coefficient.
Figure 5.19. Solution of the convection-dispersion equation for a Peclet Number of 100 at tD = 0.50 pore volume injected.
5–34
Let us replot the solutions of Figure 5.19 as CD versus z, where z is
defined as
D D
D
x tzt−
= (5.39)
A remarkable thing happens to the solutions at the various
dimensionless times. They collapse into one dimensionless curve that is
characteristic of the displacement as shown in Figure 5.20. The slope of
this curve reflects the Peclet Number or the longitudinal dispersion
coefficient. The steeper the curve, the higher is the Peclet Number or the
lower is the longitudinal dispersion coefficient. This transformation can
be used to test if the concentration profiles measured in an experiment
obey the complementary error function solution of the convection-
dispersion equation. The transformation was derived from Eq.(5.35),
which shows that CD is a function of the parameter NPe and the
independent variable D D
D
x tt− .
Figure 5.21 shows the effect of Peclet Number or longitudinal
dispersion coefficient on the solutions of the convection-dispersion
equation. At low Peclet Numbers or high longitudinal dispersion
coefficient, the mixing zone length is large whereas at high Peclet
Numbers or low longitudinal dispersion coefficient, the mixing zone
length is small.
5-35
Figure 5.20. Transformation of the solutions at a fixed Peclet Number.
5–36
Figure 5.21. The effect of Peclet Number or longitudinal dispersion coefficient on the solution of the convection-dispersion equation.
The breakthrough curve obtained by observing the concentration
at the outlet end of the porous medium (where 1Dx = ) for the case of a
finite length medium is given by
( ) 111,2 2
DD D Pe
D
tC t erfc Nt
⎡ ⎤⎛ ⎞−= ⎢ ⎥⎜⎜ ⎟⎟⎢ ⎥⎝ ⎠⎣ ⎦
(5.40)
Eq.(5.40) is the cumulative normal distribution with mean 1, variance
2
PeN and standard deviation 2
PeN. Figure 5.22 shows the breakthrough
curve given by Eq.(5.40) for a Peclet Number of 100. It is S-shaped and a
mirror image of the solvent concentration profiles. Thus, Eq.(5.40) can be
used in conjunction with the breakthrough curve measured in an
experiment to determine the Peclet Number ( )PeN and hence, the
longitudinal dispersion coefficient ( )LD .
5-37
Figure 5.22. Breakthrough curve.
Figure 5.23 shows the transformed breakthrough curve, where the
transformation equation is
1 D
D
tzt−
= (5.41)
It is similar to the solvent concentration profiles. This curve, if plotted on
a linear normal probability graph paper, will be a straight line as shown
in Figure 5.24. Such a plot can be used to fit the breakthrough curve to
the normal distribution for the purpose of determining the Peclet
Number, and hence, the longitudinal dispersion coefficient.
5–38
Figure 5.23. Transformed breakthrough curve.
5-39
Figure 5.24. Transformed breakthrough curve plotted on normal
probability scale.
5–40
In the presence of retardation due to adsorption, the solution given
by Eq.(5.34) can be modified to include the retardation factor as
( ) 1,2
2
DD
fD D D
D
f Pe
txR
C x t erfct
R N
⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟
⎢ ⎥⎜= ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
(5.42)
5.5 DISPERSION COEFFICIENT AND DISPERSIVITY
The dispersion coefficient is a second rank tensor. The dispersion
coefficient tensor in the xyz coordinate system is given by
( ), ,xx xy xz
yx yy yz
zx zy zz
D D DD x y z D D D
D D D
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
(5.43)
in 3D, and by
( ), xx xy
yx yy
D DD x y
D D⎡ ⎤
= ⎢ ⎥⎣ ⎦
(5.44)
in 2D. The dispersion coefficient tensor like the permeability or hydraulic
conductivity tensor is symmetric. Using the same transformation
equations presented in Chapter 3 for permeability and hydraulic
conductivity, the principal values of the dispersion coefficient tensor and
the principal axes of the dispersion coefficient anisotropy can be
determined. Along the principal coordinates of the anisotropy uvw, the
dispersion coefficient tensor becomes
5-41
( )0 0
, , 0 00 0
L
T
W
DD u v w D
D
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
(5.45)
in 3D, and
( )0
,0
L
T
DD u v
D⎡ ⎤
= ⎢ ⎥⎣ ⎦
(5.46)
in 2D, where DL is the longitudinal dispersion coefficient in the direction
of mean flow and DT is the transverse dispersion coefficient in the
direction perpendicular to the mean flow.
The relative magnitudes of DL and DT and the tensorial nature of
the dispersion can be demonstrated qualitatively by the following
experiment. Suppose a tracer is injected at time zero as a point source
into a homogeneous and isotropic reservoir in which there is steady flow
as shown in Figure 5.25. At times t1 and t2, the concentration of the
tracer would be as shown in Figure 5.25. The tracer has spread and the
concentration distribution has become elliptical. There is more spreading
or dispersion in the direction of mean flow than in the direction
perpendicular to it. Thus, the direction of mean flow and the direction
perpendicular to the mean flow are principal axes of the dispersion
coefficient anisotropy. The principal value of the dispersion coefficient
tensor in the direction of mean flow is DL and the one perpendicular to
the direction of mean flow is DT. In general, . L TD D≥
5–42
Figure 5.25. Variation in tracer concentration in 2D for a constant velocity flow system.
Using Eq.(5.6), the longitudinal and transverse dispersion
coefficients can be written as
L Ld LD D D m= + (5.47)
T Td TD D D m= + (5.48)
where the two terms on the right sides of Eqs.(5.46) and (5.47) represent
the contributions of molecular diffusion and mechanical dispersion to
the dispersion coefficients. Based on experimental observations,
Eqs.(5.47) and (5.48) may be written in dimensionless form as
1 2pL
o o
uDD C CD D
β⎛ ⎞
= + ⎜ ⎟⎝ ⎠
(5.49)
1 3pT
o o
uDD C CD D
β⎛ ⎞
= + ⎜ ⎟⎝ ⎠
(5.50)
where C1, C2, C3 and β are properties of the porous medium and the flow
regime, Do is the effective binary molecular diffusion coefficient between
the miscible displacing fluid and the displaced fluid, and Dp is the mean
5-43
particle diameter of the porous medium. It has been found also from
experiments that C1 is given by
11 1CFφ τ
= = (5.51)
where F is the formation resistivity factor and τ is the tortuosity of the
porous medium. Therefore, Eqs.(5.49) and (5.50) can be written as
21 pL
o o
uDD CD D
β
τ⎛ ⎞
= + ⎜ ⎟⎝ ⎠
(5.52)
31 pT
o o
uDD CD D
β
τ⎛ ⎞
= + ⎜ ⎟⎝ ⎠
(5.53)
Perkins and Johnson (1963) have shown that β is of the order of 1 to
1.25.
Figure 5.26 shows the correlations for L
o
DD
and T
o
DD
with the Peclet
Number, p
o
vDD
, for unconsolidated porous media obtained by Perkins and
Johnson (1963). Figure 5.26a shows that at Peclet Numbers less than
0.02, molecular diffusion dominates the longitudinal dispersion
coefficient and the dispersion coefficient is equal to the diffusion
coefficient in the porous medium. In this regime, the mechanical
dispersion term on the right side of Eq.(5.52) is negligible compared to
the molecular diffusion term. In this regime, the dimensionless molecular
term is about 0.67. This means that the diffusion coefficient for a tracer
in a porous medium is less than the diffusion coefficient in the same
5–44
liquid in the absence of the porous medium. This difference is caused by
the tortuosity of the porous medium. In fact, 1/0.67 or 1.5 is an estimate
of the tortuosity of unconsolidated porous media. At the transition Peclet
Numbers between 0.02 and 6, both molecular diffusion and mechanical
dispsersion contribute to the dispersion coefficient. At Peclet Numbers
above 6, the dispersion coefficient is dominated by mechanical dispersion
and the effect of molecular diffusion can be neglected.
The shape of the correlation for the dimensionless transverse
dispersion coefficient is similar to that of the dimensionless longitudinal
dispersion coefficient. However, it should be noted that the scale of the
Peclet Number in Figure 5.26b starts at 0.1 compared to 0.001 in Figure
5.26a. This means that the regime dominated by molecular diffusion
occurs over a larger range of Peclet Numbers for transverse dispersion
than for longitudinal dispersion. Thus, molecular diffusion is much more
important to transverse dispersion than to longitudinal dispersion. At
high Peclet Numbers, the longitudinal dispersion coefficient is greater
than the transverse dispersion coefficient. This is clearly evident by
plotting the two correlations for dimensionless dispersion coefficients on
the same scale as shown in Figure 5.27. Other authors have obtained
correlations for longitudinal dispersion coefficient similar to that of
Perkins and Johnston as shown in Figure 5.28.
At normal reservoir velocities, the Peclet Number is normally
greater than 6. Also, as pointed out earlier, β is approximately 1. Under
these conditions, molecular diffusion can be neglected and Eqs.(5.52)
and (5.53) can be approximated as
2p
Lo
uDD C D
D
β
o Luα⎛ ⎞
= ⎜ ⎟⎝ ⎠
≅ (5.54)
5-45
3p
To
uDD C D
D
β
o Tuα⎛ ⎞
= ⎜ ⎟⎝ ⎠
≅ (5.55)
where Lα is known as the longitudinal dispersivity and Tα is the
transverse dispersivity of the porous medium. Eqs.(5.54) and (5.55) can
be used to estimate the dispersivities if the dispersion coefficients have
been measured independently. In terms of the longitudinal dispersivity,
the Peclet Number defined in Eq.(5.25) becomes
Figure 5.26. Correlations for dimensionless longitudinal and transverse
5–46
dispersion coefficients. (a) dimensionless longitudinal dispersion coefficient, (b) transverse dispersion coefficient (Perkins and Johnston,
1963).
Figure 5.27. Correlations for dimensionless longitudinal and transverse dispersion coefficients plotted on the same scale (Perkins and Johnston,
1963).
5-47
Figure 5.28. Correlation for dimensionless longitudinal dispersion coefficient by various authors (Pfannkuch, 1963; Saffman, 1960).
PeL
LNα
= (5.56)
Dispersivity is, in general, not a function of fluid velocity, making it
a property of the porous medium. However, dispersivities are highly scale
dependent. Figures 5.29 and 5.30 show values of measured
dispersitivities as functions of the measurement scale. Note the
logarithmic scales. In general, dispersitivities measured at laboratory
scale are much smaller than those measured at field scale. Reservoir
heterogeneity is the cause of this scale effect. In field measurements, the
effect of heterogeneity is to stretch out the solvent concentration profile,
5–48
which is interpreted as a large dispersion coefficient or a large
dispersivity.
5-49
Figure 5.29. Longitudinal dispersivity versus measurement scale (Arya, 1988).
Figure 5.30. Field measured longitudinal dispersivity versus measurement scale (Gelhar, 1986).
5–50
5.6 MEASUREMENT OF DISPERSION COEFFICIENT AND DISPERSIVITY
5.6.1 Traditional Laboratory Method with Breakthrough Curve
A laboratory coreflood experiment can be used to measure the
longitudinal dispersion coefficient. Figure 5.31 shows a schematic
diagram for such a test. A tracer or solvent is used to displace a miscible
fluid from the porous medium and the solvent concentration of the
effluent at the outlet end of the core is measured as a function of time.
The figure at the upper left of Figure 5.31 shows the injected tracer or
solvent concentration as a function of time whereas the figure at the
upper right shows the concentration of the solvent in the effluent. The
injected solvent concentration is a step function in which pure solvent is
injected throughout the experiment. At the outflow end, initially, the
concentration of the injected solvent will be zero because it has not yet
broken through. After the injected solvent has arrived at the outlet end
ahead of the advection front because of dispersion, its concentration will
increase from zero to 1 over a finite period of time as shown in the
sketch. The curve of concentration versus time at the outlet end of the
core is known as the breakthrough curve and can be fitted to Eq.(5.40) to
estimate the Peclet Number and the longitudinal dispersion coefficient.
5-51
Figure 5.31. Laboratory coreflood experiment for measuring longitudinal dispersion coefficient.
In order for the solvent concentration distribution in the coreflood
to satisfy the mathematical model of Eq.(5.34) or (5.35), the restrictions
outlined previously must be implemented in the experiment. The core
should be fairly uniform, the density of the injected solvent and the
displaced fluid must be matched and the viscosity of the injected solvent
and the displaced fluid also must be matched. When these restrictions
are implemented, the breakthrough curve will approximate the
cumulative normal probability distribution. A graph of CD versus 1 D
D
tt−
on a linear probability graph paper will be a straight line as shown in
Figure 5.24. The mixing zone length between CD = 0.9 and CD = 0.1 read
from the graph is related to the Peclet Number as
0.9 0.1
1 13.625
D D
D D
Pe D DC C
t tN t t
= =
⎛ ⎞ ⎛ ⎞− −= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(5.57)
5–52
Eq.(5.57) can be rearranged to calculate the longitudinal dispersion
coefficient as
2
0.9 0.1
1 1
3.625D
D D
D DC CL
t tt t
D uL = D =
⎡ ⎤⎛ ⎞ ⎛ ⎞− −−⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
(5.58)
Alternatively, the mixing zone length corresponding to CD = 0.84 and CD
= 0.16 will be equal to twice the standard deviation of the normal
distribution. Thus,
0.84 0.16
1 122D D
D D
Pe D DC C
t tN t t
= =
⎛ ⎞ ⎛ ⎞− −= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(5.59)
Eq.(5.59) can be rearranged to calculate the longitudinal dispersion
coefficient as
2
0.84 0.16
1 18
D D
D DL
D DC C
t tuLDt t
= =
⎡ ⎤⎛ ⎞ ⎛ ⎞− −⎢ ⎥= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ (5.60)
It is much more difficult to measure transverse dispersion coefficient
experimentally. As a result, there are very few transverse dispersion
coefficient data reported in the literature beside those of Perkins and
Johnson (1963).
5.6.2 Laboratory Method of Peters et al. (1996)
Peters et al. (1996) have presented a method for measuring
longitudinal dispersion coefficient and dispersivity by imaging the
5-53
laboratory tracer test. CT or NMR imaging of the coreflood gives the
solvent concentration profiles in time and space similar to those of Figure
5.17. These profiles can be fitted to the solution of the convection-
dispersion equation, Eq.(5.34) or (5.35), by trial and error to estimate the
Peclet Number and the longitudinal dispersion coefficient. The
longitudinal dispersion coefficient so determined will be the same as that
determined from the breakthrough curve. The dispersivity is calculated
from the dispersion coefficient using Eq.(5.54). Furthermore, the growth
of the mixing zone length with time can easily be calculated from the 3D
image data such that the graph of Δx versus t can be used to calculate
the longitudinal dispersion coefficient. The advantage of this approach is
that the contribution of heterogeneity to the dispersion coefficient also
can be measured.
To demonstrate the methods, we measured the longitudinal
dispersion coefficient and the longitudinal dispsersivity of the Berea
sandstone core of Figure 4.33 and the unconsolidated sandpack of
Figure 5.30 using these methods.
A summary of the experimental conditions for the tracer test in the
Berea sandstone core is presented in Table 5.1. Figure 5.32 shows the
CT images of the tracer test in the Berea sandstone. The growth of the
mixing zone is clearly visible. The mixing zone is distorted by
permeability heterogeneity of the core. The lower half of the core is more
permeable than the upper half. This is evident in the permeability image
of the core shown in Figure 5.33. The permeability distribution of the
core was determined by a technique described by Peters and Afzal (1991).
Figure 5.34 shows a preliminary history match of the solvent
concentration profiles using Eq.(5.34). The slopes of the profiles from
5–54
Eq.(5.34) and the experiment are in good agreement indicating that the
Peclet Number is correct. However, at late dimensionless times, the
profiles obtained from Eq.(5.34) have traveled further than those from
the experiment. This is evidence of retardation caused by adsorption.
Figure 5.35 shows the history match using Eq.(5.42) with a retardation
factor, Rf, of 1.11. The agreement between the two sets of solvent
concentration profiles is good. The parameters that resulted in this
history match are , and 159PeN = 5 2600 10 /LD x cm−= s 0.379L cmα = .These are
the longitudinal dispersion coefficient and longitudinal dispersivity that
would have been obtained from a breakthrough curve. It should be noted
that the effect of the distortion in the mixing zone caused by permeability
heterogeneity is interpreted as increased dispersion coefficient and
increased dispersivity of the porous medium.
Table 5.1. Experimental Conditions for Tracer Test in Berea Sandstone
Type of Displacement First-Contact Miscible
Porous Medium Berea Sandstone
Length (cm) 60.2
Diameter (cm) 5.1
Absolute Permeability (md) 160.4
Average Porosity from CT (%) 17.3
Fluids
Displacing Fluid Distilled Water + 10% NaI
Density of Displacing Fluid (g/cm3) 1.078
Viscosity of Displacing Fluid (cp) 1.029
Displaced Fluid Distilled Water+1.4% NaCl + 10% KCl
Density of Displaced Fluid (g/cm3) 1.078
Viscosity of Displaced Fluid (cp) 1.028
Mobility Ratio (Viscosity Ratio) 1.0
Darcy Velocity (cm/s) 2.742x10-3
5-55
Interstitial Velocity (cm/s) 1.714x10-2
Figure 5.32. CT images of a tracer test in a Berea sandstone core. A: tD = 0.20; B: tD = 0.50; C: tD = 0.80 (Peters et al., 1996)
5–56
Figure 5.33. Permeability distribution for Berea sandstone core (Peters and Afzal, 1991).
Figure 5.36 shows the growth of the mixing zone length with time
for the Berea sandstone core. The mixing zone length was obtained for
the 3D CT data by calculating the distances between CD = 0.90 and CD =
0.10 for each dimensionless time. This resulted in several thousand
values of mixing zone length at each dimensionless time. The several
thousand values were then averaged to obtain one mixing zone length for
each time. Each point in Figure 5.36 is the average mixing zone length
plotted against the corresponding time. It can be seen that the mixing
zone length grows as the square root of time as predicted by Eq.(5.38).
The longitudinal dispersion coefficient and longitudinal dispersivity are
calculated from the slope of the straight line of Figure 5.36 as
5-57
5 2431 10 /LD x cm−= s and 0.272 L cmα = . It should be noted that with this
method, the dispersion coefficient and the dispersivity are less than
calculated previously because the effect of the permeability heterogeneity
was excluded from the calculations.
Figure 5.34. A comparison of the simulated and experimental solvent concentration profiles for Berea sandstone core for Rf = 0 (Peters et al.,
1996).
5–58
Figure 5.35. A comparison of the simulated and experimental solvent concentration profiles for Berea sandstone core for Rf = 1.11 (Peters et
al., 1996).
5-59
Figure 5.36. Growth of mixing zone length with time for the Berea sandstone core (Peters et al., 1996).
A summary of the experiment for the unconsolidated sandpack is
presented in Table 5.2. Figure 5.37 shows the CT images of the tracer
test in the sandpack. The growth of the mixing zone also is clearly
visible. The mixing zone is distorted by the heterogeneity of the
sandpack. It should be recalled that this sandpack contained radial
heterogeneities based on the packing method that was used in preparing
it. Figure 5.38 shows the permeability distribution for the sandpack.
Figure 5.39 shows a preliminary history match of the solvent
concentration profiles using Eq.(5.34). The slopes of the profiles from
Eq.(5.34) and the experiment are in good agreement indicating that the
Peclet Number is correct. However, at late dimensionless times, the
profiles obtained from Eq.(5.34) have traveled further than those from
the experiment. There is retardation caused by adsorption in the
sandpack. Figure 5.40 shows the history match using Eq.(5.42) with a
retardation factor of 1.04. The agreement between the two sets of solvent
concentration profiles is good. The parameters that resulted in this
history match are , and 554PeN = 5 2100 10 /LD x cm−= s 0.098 L cmα = .
Figure 5.41 shows the growth of the mixing zone length with time
for the sandpack. The longitudinal dispersion coefficient and longitudinal
dispersivity are calculated from the mixing zone length as
and 5 282 10 /LD x cm−= s 0.080 L cmα = .
Figure 5.42 shows the similarity transformations of the solvent
concentration profiles from the two tracer tests. It can be seen that there
is more dispersion in the Berea sandstone core than in the
5–60
unconsolidated sandpack. The results of the tracer tests for both porous
media are summarized in Table 5.3.
Table 5.2. Experimental Conditions for Tracer Test in Unconsolidated
Sandpack
Type of Displacement First-Contact Miscible
Porous Medium Unconsolidated Sandpack
Length (cm) 54.2
Diameter (cm) 4.8
Absolute Permeability (md) 6400
Average Porosity from CT (%) 29.7
Fluids
Displacing Fluid Distilled Water + 13% NaCl
Density of Displacing Fluid
(g/cm3) 1.089
Viscosity of Displacing Fluid (cp) 1.262
Displaced Fluid Distilled Water + 10% BaCl2
Density of Displaced Fluid (g/cm3) 1.089
Viscosity of Displaced Fluid (cp) 1.127
Mobility Ratio (Viscosity Ratio) 0.9
Darcy Velocity (cm/s) 3.037x10-3
Interstitial Velocity (cm/s) 1.023x10-2
Breakthrough Recovery (%) 95
5-61
Figure 5.37. CT images of a tracer test in unconsolidated sandpack. A: tD = 0.20; B: tD = 0.50; C: tD = 0.80 (Peters et al., 1996)
5–62
Figure 5.38. Permeability distribution for unconsolidated sandpack (Peters and Afzal, 1991).
5-63
Figure 5.39. A comparison of the simulated and experimental solvent concentration profiles for unconsolidated sandpack for Rf = 0 (Peters et
al., 1996).
5–64
Figure 5.40. A comparison of the simulated and experimental solvent concentration profiles for unconsolidated sandpack for Rf = 1.04 (Peters
et al., 1996).
5-65
Figure 5.41. Growth of mixing zone length with time for unconsolidated sandpack (Peters et al., 1996).
5–66
Figure 5.42. Similarity transformation of solvent concentration profiles (Peters et al., 1996).
Table 5.3. Summary of Results
5-67
Porous Medium Berea
Sandstone
Unconsolidated
Sandpack
Longitudinal Dispersion Coefficient
with
Heterogeneity (cm2/s) 600x10-5 100x10-5
Longitudinal Dispersivity with
Heterogeneity (cm) 0.379 0.098
Longitudinal Dispersion Coefficient
without Heterogeneity (cm2/s) 431x10-5 82x10-5
Longitudinal Dispersivity without
Heterogeneity (cm) 0.272 0.080
Retardation Factor 1.11 1.04
Peclet Number 159 554
5.6.3 Field Measurement of Dispersion Coefficient and Dispersivity
Longitudinal dispersion coefficient and dispersivity are measured
in the field by tracer tests either in a single well or between two or more
wells. Eq.(5.9) in radial coordinates can be used to calculate the
dispersion coefficient and the dispersitivity from a single-well tracer test.
Eq.(5.9) can be written in radial coordinates as
2
2 0r L rC C Cu ut r r
α∂ ∂ ∂+ −
∂ ∂ ∂= (5.61)
Gelhar and Collins (1971) give the solution of Eq.(5.61) with appropriate
initial and boundary conditions for a single-well tracer test as
5–68
( )1
1 22
163
112
2 1 1
D DiD
L D D
Di Di
t tC erfc
t tR t tα
⎡ ⎤⎢ ⎥⎢ ⎥
− −⎢ ⎥= ⎢ ⎥⎧ ⎫⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞⎪ ⎪⎛ ⎞ − − −⎨ ⎬⎢ ⎥⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦
(5.62)
where
qtRhπ φ
= (5.63)
In Eq.(5.62), Dt is the pore volume of fluid produced at various times, Dit is
the total pore volume of tracer fluid injected at the beginning of the test.
The longitudinal dispersitivity can be calculated by fitting Eq.(5.62) to the
concentration profile obtained from the tracer test. Figure 5.43 shows the
fit of Eq.(5.62) to two tracer tests in the same well in a water-bearing
aquifer. The aquifer was 8.2 meters thick, with an average hydraulic
conductivity of 1.4x10-2 cm/s and a porosity of 38%. In Test SW1, the
volume of the tracer fluid injected was such that the tracer fluid
extended to a radius of 3.13 meters from the well. In Test SW2, the
volume of tracer fluid was such that the tracer fluid extended to a radius
of 4.99 meters. Thus, the two tests had different measurement scales.
Based on the history match shown in Figure 5.42, the dispersitivity of
the aquifer from the first test was 3.0 cm and that from the second test
was 9.0 cm. The results show the dependence of the dispersivity on the
scale of the measurement. In Test SW1, the scale of measurement was
3.13 meters whereas in Test SW2, it was 4.99 meters. The two scales of
measurement resulted in two different estimates of dispersivity with the
larger measurement scale resulting in a larger dispersitivity than the
smaller measurement scale.
5-69
Figure 5.43. Comparison of tracer concentration and Eq.(5.62)for single-well injection-withdrawal test (Pickens and Grisak, 1981).
A multi-well tracer test also can be used to estimate dispersivity in
the field. However, in this case, a numerical simulator will be needed to
5–70
interpret the test. Figure 5.44 shows attempts to interpret a two-well
tracer test using a finite element numerical simulator. In the test, a slug
of tracer was injected into one well and chased by water while the tracer
concentration was measured as a function of time at the second well. A
2D homogeneous aquifer model gave a longitudinal dispersitivity of 4.0 m
whereas a 3D heterogeneous aquifer model gave a much smaller
dispersivity of 0.15 m. Of course, the history match of the breakthrough
curve for a heterogeneous aquifer does not give a unique solution to the
problem. Different configurations of the aquifer heterogeneity can result
in good history matches with widely different values of dispersivity.
Figure 5.44. A comparison of three model fits to the breakthrough data from a two-well tracer test (Huyakorn et al., 1986).
5-71
5.7 FACTORS THAT COULD AFFECT DISPERSION COEFFICIENT AND DISPERSIVITY
A poll was taken in a previous class in which the students present
were requested to name the factors they believed could possibly affect the
dispersion coefficient and the dispersivity of a porous medium. There
were no restrictions. All factors that came to mind were listed. Here are
the factors that were listed, organized into rock characteristics, fluid
properties and process characteristics.
1. Rock Characteristics
Porosity (φ)
Permeability (k)
Pore size (δ)
Pore size distribution (frequency)
Heterogeneity
Dykstra–Parsons coefficient (V)
Variogram (γ)
Correlation length (range of influence, a)
Chemical reaction
Adsorption (retardation factor, Rf)
Biological transformation
Radioactive decay
Morphology of the porous medium (Γ)
5–72
Pore structure
Cementation
Dead end pores
Tortuosity
Mean grain diameter (Dp)
Specific surface area (S)
2. Fluid Properties
Viscosity of the displaced fluid (μo)
Viscosity of the displacing fluid (μs)
Density of the displaced fluid (ρo)
Density of the displacing fluid (ρs)
Binary diffusion coefficient between the displaced and displacing fluids (Do)
3. Process (Displacement) Characteristics
Interstitial velocity field (u )
Gravitational acceleration field ( g )
After extensive discussion, it was decided that the longitudinal
dispersion coefficient is a function of some of these variables and that the
functional relationship is of the form
( )1 , , , , ,L s p o oD f D u D gμ μ ρ= Δ (5.64)
The objective is to design an experimental program to determine the
nature of the function f1. To plan the experimental program, dimensional
5-73
analysis can be used to derive the set of complete and independent
dimensionless groups that can be used in the experimental program to
determine the nature of f1. Upon performing the dimensional analysis as
presented in Appendix A, Eq.(5.64) can be written in dimensionless form
as
3
2 , ,p poL
o o s o
uD D gD fD D D s
ρμμ μ
⎛ ⎞Δ= ⎜⎜
⎝ ⎠⎟⎟ (5.65)
where f2 is a new function to be determined from experiments. In the
laboratory experiments for determining the longitudinal dispersion
coefficient, the viscosity of the injected fluid should be matched with that
of the displaced fluid. If this is done, the mobility ratio, o
s
μμ
⎛ ⎞⎜⎝ ⎠
⎟ , will be a
constant equal to 1 and its effect will be eliminated from the experiment.
Also, the density of the injected fluid should be matched with that of the
displaced fluid. In this case, the dimensionless group, 3p
o s
D gD
ρμ
⎛ ⎞Δ⎜⎜⎝ ⎠
⎟⎟ , will be a
constant equal to zero and it will not be a factor in the experiment. If
these restrictions are implemented in the experiments, then Eq.(5.65)
becomes
3pL
o o
uDD fD D
⎛ ⎞= ⎜
⎝ ⎠⎟ (5.66)
A similar analysis gives the functional relationship for the transverse
dispersion coefficient as
4pT
o o
uDD fD D
⎛ ⎞= ⎜
⎝ ⎠⎟ (5.67)
5–74
The experimental program will then consist of measuring L
o
DD
and T
o
DD
as
functions of the Peclet Number, p
o
uDD
, to determine the nature of f3 and f4.
The results of such an experimental program are shown in Figures 5.27
and 5.28.
5.8 NUMERICAL MODELING OF FIRST-CONTACT MISCIBLE DISPLACEMENT
5.8.1 Introduction
First-contact miscible displacement is sufficiently well understood
to be modeled accurately with a finite difference numerical simulator.
Such a numerical simulator can be used to interpret tracer tests or to
model field processes that can be adequately described as first-contact
miscible processes.
5.8.2 Mathematical Model of First-Contact Miscible Displacement
The mathematical model for first-contact miscible displacement in
a heterogeneous and anisotropic reservoir consists of the following mass
transport, continuity, flow and mixing equations:
( ) .( ) .( ) 0C vC D Ctφ φ∂
+∇ −∇ ∇ =∂
(5.68)
( ) .( ) 0vtφρ ρ∂
+∇ =∂
(5.69)
(kv P ρμ
= − ∇ ± )gz (5.70)
5-75
(1 )mixture s s o sC Cρ ρ ρ= + − (5.71)
4
1/ 4 1/ 4
1s smixture
s o
C Cμμ μ
−⎛ ⎞−
= +⎜⎝ ⎠
⎟ (5.72)
where
xx xy xz
yx yy yz
zx zy zz
D D DD D D D
D D D
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
(5.73)
and
xx xy xz
yx yy yz
zx zy zz
k k kk k k k
k k k
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
(5.74)
Eq.(5.68) is the convection-dispersion equation for a heterogeneous
reservoir. It is a modified version of Eq.(5.9), which was for a
homogeneous reservoir with a constant porosity. Note that the advection
term in Eq.(5.68) is in terms of Darcy velocity instead of interstitial
velocity. Eq.(5.69) is the continuity equation for flow. Eq.(5.70) is Darcy's
law for a heterogeneous and anisotropic reservoir. Eq.(5.71) is the mixing
rule for density. It specifies how to calculate the fluid density as a
function of the solvent concentration. It is a linear mixing rule, which
has been verified experimentally. Figure 5.45 compares the linear mixing
rule with experimental density measurements for mixtures of glycerol
and brine. The agreement is good. Eq.(5.72) is the mixing rule for
viscosity. It specifies how the viscosity is to be calculated as a function of
the solvent concentration. It is known as the quarter-power viscosity
5–76
mixing rule. This mixing rule also has been verified in laboratory
measurements as shown in Figure 5.46 for glycerol and brine.
Figure 5.45. Verification of linear density mixing rule for two first-contact miscible fluids: Fluid 1 is brine; Fluid 2 is glycerol and water.
5-77
Figure 5.46. Verification of quarter-power viscosity mixing rule for two
first-contact miscible fluids: Fluid 1 is brine; Fluid 2 is glycerol and water.
The mathematical model for first-contact miscible displacement,
Eqs.(5.68) through (5.74), can be solved by finite difference using
appropriate initial and boundary conditions. A possible sequence for the
calculations at one time-step is as follows. Substitute Eq.(5.70) into
(5.69) to derive a generalized diffusivity equation for the pressure field.
Solve the pressure equation by finite difference. Use the pressure field
and Eq.(5.70) to calculate the Darcy velocity field. Substitute the Darcy
velocity field into Eq.(5.68) and solve Eq.(5.68) by finite difference to
calculate the solvent concentration distribution. Of course, the
properties of the porous medium and the fluids must be specified in
advance.
5–78
5.8.3 Numerical Modeling of Laboratory Experiments
In this section, we present the results of numerical simulations of
first-contact miscible displacements experiments based on the numerical
solution of Eqs.(5.68) through (5.74). The porous media used in the
experiments and the solvent concentration distributions in time and
space were imaged by NMR. Each simulation entailed the
characterization of the porous medium by constructing its permeability
distribution using Sequential Gaussian Simulation. The porosity
distribution was obtained directly from by NMR imaging. The pertinent
dimensionless groups for the displacements are mobility ratio (M), the
Peclet Number (NPe) as defined in Eq.(5.25), the gravity or buoyancy
number (Ng), and the density number (Nρ).
The gravity number is the ratio of the gravity force to the viscous
force and is given by
( )s og
o
k gN
uρ ρμ−
= (5.75)
The higher the gravity number, the higher the potential for gravity
segregation in the experiment. If the gravity number is negative, the
gravity segregation will be in the form of gravity override, whereas if the it
is positive, the gravity segregation will be in the form of gravity tonguing.
The density number is defined as
s o
o
Nρρ ρρ−
= (5.76)
5-79
The higher the density number, the higher the potential for gravity
segregation. A negative density number correlates with gravity override
whereas a positive density number correlates with gravity tonguing.
All the simulations were performed with UTCHEM, a finite
difference chemical flooding simulator developed at The University of
Texas at Austin. A 3D Cartesian coordinate system was used to simulate
the displacements in a cylindrical core. Figure 5.47 shows how the
rectangular grids were adapted to simulate a cylindrical system. All the
simulations were performed with 40x40x20 grid blocks.
Figure 5.47. Three-dimensional simulation grids.
Experiment 1
This experiment was conducted in homogeneous Berea sandstone
core at a favorable mobility ratio of 0.84 and a Darcy velocity of 0.00014
5–80
cm/s (0.40 ft/day). Figure 5.48 shows an artist impression of the core
along with the nomenclature used in the core characterization. The core
properties were: L = 10 cm, d = 5 cm, k = 622 md and φ = 24%. Figure
5.49 shows the porosity distribution of the core obtained by NMR
imaging. Figure 5.50 shows the NMR images of the solvent concentration
distributions in time and space. The mixing zone is tilted because of
gravity tonguing. Figure 5.51 shows the solvent concentration profiles for
the experiment.
Figure 5.48. Artist impression of homogeneous Berea sandstone core.
5-81
Figure 5.49. Porosity distribution for homogeneous Berea sandstone core (Majors et al., 1997).
5–82
Figure 5.50. Solvent concentration images for Experiment 1. M = 0.84, Ng = 0.3656, Nρ = 0.0899, NPe = 45.5 (Li, 1997)
5-83
Figure 5.51. Average solvent concentration profiles for Experiment 1. M = 0.84, Ng = 0.3656, Nρ = 0.0899, NPe = 45.5 (Li, 1997)
Figure 5.52 shows the permeability field generated for the Berea
sandstone core and used as the porous medium for the numerical
simulation. The parameters of the log normal permeability field are k =
622 md, φ = 24%, ax = 5.0 cm, ay = 2.5 cm, az = 0.25 cm and V = 0.20.
The experiment was simulated using 0.0975Lα = cm and 0.0031Tα = cm.
Figure 5.53 compares the simulated and the experimental solvent
concentration distributions. The agreement between the simulation and
the experiment is good. Figure 5.54 compares the simulated and
experimental solvent concentration profiles. The agreement between the
5–84
simulation and the experiment is acceptable. Finally, Figure 5.55 shows
a head to head comparison of the simulated and experimental solvent
concentrations at the same cross-sections of the core. The data are fairly
well distributed about the 45 degree line. Therefore, the agreement
between the simulation and the experiment is reasonable.
Figure 5.52. Permeability distribution for homogeneous Berea sandstone core obtained by Sequential Gaussian Simulation. k = 622 md, φ = 24%,
ax = 5.0 cm, ay = 2.5 cm, az = 0.25 cm, V = 0.20 (Shecaira and Peters, 1998).
5-85
Figure 5.53. A comparison of the simulated and experimental solvent concentration distributions for Experiment 1 (Shecaira and Peters,
1998).
5–86
Figure 5.54. A comparison of the simulated and experimental solvent concentration profiles for Experiment 1 (Shecaira and Peters, 1998).
Figure 5.56. A comparison of the simulated and experimental solvent
concentration at the same cross-sections for Experiment 1 (Shecaira and Peters, 1998).
5-87
Experiment 2
This experiment was conducted in the same Berea sandstone core
and at the same mobility ratio as Experiment 1. However, the gravity
number was reduced from 0.3656 to 0.0180 by increasing the Darcy
velocity from 0.00014 cm/s (0.40 ft/day) to 0.00294 cm/s (8.33 ft/day)
in an effort to eliminate the gravity tonguing observed in Experiment 1.
Figures 5.57 and 5.58 show the solvent concentration images and the
solvent concentration profiles for the experiment. The images of Figure
5.57 appear to show more tonguing than in Experiment 1.
5–88
Figure 5.57. Solvent concentration images for Experiment 2. M = 0.84, Ng = 0.0180, Nρ = 0.0899, NPe = 49.5 (Li, 1997)
5-89
Figure 5.58. Average solvent concentration profiles for Experiment 2. M = 0.84, Ng = 0.0180, Nρ = 0.0899, NPe = 49.5 (Li, 1997)
Figures 5.59 to 5.61 compare the initial simulation of Experiment
2 with the experiment. The agreement between the simulation and the
experiment is poor. The simulation model is the same as for Experiment
1, except that the injection rate was increased from 0.40 ft/day to 8.33
ft/day. The images from the simulation clearly show that at the gravity
number of 0.0180, there should be no gravity tonguing. Therefore, it can
be safely concluded that the tonguing in Experiment 2 was not caused by
5–90
gravity segregation but rather by a mechanical problem with the
experiment.
Figure 5.59. A comparison of the preliminary simulated and experimental solvent concentration distributions for Experiment 2 (Shecaira and
Peters, 1998).
5-91
Figure 5.60. A comparison of the preliminary simulated and experimental solvent concentration profiles for Experiment 2 (Shecaira and Peters,
1998).
Figure 5.61. A comparison of the preliminary simulated and experimental
solvent concentration at the same cross-sections for Experiment 2 (Shecaira and Peters, 1998).
5–92
Figure 5.62 shows the end piece of the core holder through which
fluid is injected into the core. Fluid is injected as a point source. Radial
grooves are machined on the face of the end piece to assist in
distributing the injected fluid uniformly over the inlet face of the core. It
would appear that at the injection rate of 0.40 ft/day in Experiment 2,
the grooves were effective in distributing the injected fluid over the inlet
face of the core. However, when the rate was increased to 8.33 ft/day,
the grooves became ineffective in distributing the injected fluid over the
inlet face of the core. As a result, the injected fluid was distributed
unevenly over the inlet face with more fluid being injected at the center of
the core than at the periphery of the core .
Figure 5.62. Schematic diagram of end piece of core holder showing fluid injection hole and grooves.
In order to test the fluid injection hypothesis, the inlet boundary
condition of the simulation model was modified to inject more fluid at the
center than at the periphery of the core as shown in Figure 5.62. Figures
5.63 to 5.65 show comparisons of the results of the simulation with the
modified inlet boundary condition and the experiment. The agreement
between the simulation and the experiment is excellent. The hypothesis
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about the nonuniform fluid injection appears to be supported by the
simulation.
Figure 5.63. A comparison of the simulated and experimental solvent concentration distributions for Experiment 2 with a modified inlet
boundary condition (Shecaira and Peters, 1998).
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Figure 5.64. A comparison of the simulated and experimental solvent concentration profiles for Experiment 2 with a modified inlet boundary
condition (Shecaira and Peters, 1998).
Figure 5.65. A comparison of the simulated and experimental solvent
concentration at the same cross-sections for Experiment 2 with a modified inlet boundary condition (Shecaira and Peters, 1998).
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Experiment 3
This experiment was performed in the same Berea sandstone core
as Experiments 1 and 2 but at an adverse mobility ratio of 98 at the high
Darcy velocity of 0.00294 cm/s (8.33 ft/day). The combination of high
mobility ratio and high injection normally result is viscous fingering as is
evident in Figure 5.66, which shows the solvent concentration images for
this experiment. The solvent concentration profiles are shown in Figure
5.67.
Figures 5.68 to 5.70 compare the results of a preliminary
simulation with the experiment. There is no agreement between the two.
The preliminary simulation was based on a uniform permeability
distribution in the core and a uniform fluid injection at the core inlet.
The simulation was then refined by the using permeability distribution of
Figure 5.52 and the modified inlet boundary condition. More transverse
dispersion ( 0.0122Tα = cm) was included in the numerical model than in
the simulations of Experiments 1 and 2. Figures 5.71 to 5.73 compare
the results of the refined simulation with the experiment. The agreement
between the simulation and the experiment is excellent.
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Figure 5.66. Solvent concentration images for Experiment 3. M = 98, Ng = -0.0002, Nρ = -0.097, NPe = 49.5 (Li, 1997)
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Figure 5.67. Average solvent concentration images for Experiment 3. M = 98, Ng = -0.0002, Nρ = -0.097, NPe = 49.5 (Li, 1997)
5–98
Figure 5.68. A comparison of the preliminary simulated and experimental solvent concentration distributions for Experiment 3 (Shecaira and
Peters, 1998).
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Figure 5.69. A comparison of the preliminary simulated and experimental solvent concentration profiles for Experiment 3 (Shecaira and Peters,
1998).
Figure 5.70. A comparison of the preliminary simulated and experimental
solvent concentration at the same cross-sections for Experiment 3 (Shecaira and Peters, 1998).
5–100
Figure 5.71. A comparison of the simulated and experimental solvent concentration distributions for Experiment 3 after refinement of the
simulation (Shecaira and Peters, 1998).
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Figure 5.72. A comparison of the simulated and experimental solvent concentration profiles for Experiment 3 after refinement of the simulation
model (Shecaira and Peters, 1998).
Figure 5.73. A comparison of the simulated and experimental solvent
concentration at the same cross-sections for Experiment 3 after refinement of the simulation model (Shecaira and Peters, 1998).
5–102
Experiment 4
This experiment was performed in a layered Antolini sandstone
core at a favorable mobility ratio of 0.81 at a Darcy velocity of 0.00014
cm/s (0.40ft/day). Figure 5.74 shows the porosity images for sandstone
obtained by NMR. Figures 5.75 and 5.76 show the solvent concentration
images and the solvent concentration profiles for the experiment.
Using the porosity images as guidance, a three-layer permeability
distribution was generated for the sandstone core as shown in Figure
5.77. This distribution was used in the initial simulation of Experiment
4. Figures 5.78 to 5.80 compare the results of the simulation based on
the three-layer model with the experiment. The agreement between the
simulation and the experiment is fair. The permeability field was further
refined into a five-layer model as shown in Figure 5.81. Figures 5.82 to
5.84 show comparisons of the results of the simulation with the five-
layer model with the experiment. The agreement between the simulation
and the experiment is excellent.
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Figure 5.74. Porosity images of layered Antolini sandstone core of Experiment 4 (Li, 1997).
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Figure 5.75. Solvent concentration images for Experiment 4. M = 0.81, Ng = 0.0649, Nρ = 0.103, NPe = 46.8 (Li, 1997)
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Figure 5.76. Average solvent concentration profiles for Experiment 4. M = 0.81, Ng = 0.0649, Nρ = 0.103, NPe = 46.8 (Li, 1997)
5–106
Figure 5.77. A three-layer model for the Antolini sandstone core of Experiment 4.
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Figure 5.78. A comparison of the simulated and experimental solvent concentration distributions for Experiment 4 based on a three-layer
model (Shecaira and Peters, 1998).
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Figure 5.79. A comparison of the simulated and experimental solvent concentration profiles for Experiment 4 based on a three-layer model
(Shecaira and Peters, 1998).
Figure 5.80. A comparison of the simulated and experimental solvent concentration at the same cross-sections for Experiment 4 based on a
three-layer model (Shecaira and Peters, 1998).
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Figure 5.81. A Five-layer model for the Antolini sandstone core of Experiment 4.
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Figure 5.82. A comparison of the simulated and experimental solvent concentration distributions for Experiment 4 based on a five-layer model
(Shecaira and Peters, 1998).
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Figure 5.83. A comparison of the simulated and experimental solvent concentration profiles for Experiment 4 based on a five-layer model
(Shecaira and Peters, 1998).
Figure 5.84. A comparison of the simulated and experimental solvent concentration at the same cross-sections for Experiment 4 based on a
five-layer model (Shecaira and Peters, 1998).
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Experiment 5
This experiment was performed in the same core as Experiment 4
but at an adverse mobility ratio of 11 and a Darcy velocity of 0.00014
cm/s (0.40 ft/day). Figures 5.85 and 5.86 show the solvent
concentration images and the solvent concentration profiles for the
experiment. The objective of the numerical simulation of this experiment
was to predict the outcome of the experiment instead of history matching
it. Having used the five-layer model to successfully simulate Experiment
4, it was decided to use the same simulation to predict Experiment 5 by
increasing the mobility ratio from 0.81 to 11.
Figures 5.87 to 5.89 show comparisons of the predicted
performance and the experiment. Figure 5.87 shows that predicted
solvent concentration images are in qualitative agreement with the
experimental images. However, Figures 5.88 and 5.89 show that the
quantitative prediction is not as good as might be inferred from Figure
5.87. This lack of accurate quantitative prediction is not surprising
because Experiment 5 is an unstable displacement and as such is
unpredictable.
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Figure 5.85. Solvent concentration images for Experiment 5. M = 11, Ng = -0.0027, Nρ = -0.049, NPe = 46.8 (Li, 1997)
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Figure 5.86. Average solvent concentration profiles for Experiment 5. M = 11, Ng = -0.0027, Nρ = -0.049, NPe = 46.8 (Li, 1997)
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Figure 5.87. A comparison of the predicted and experimental solvent concentration distributions for Experiment 5 based on a five-layer model
(Shecaira and Peters, 1998).
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Figure 5.88. A comparison of the predicted and experimental solvent concentration profiles for Experiment 5 based on a five-layer model
(Shecaira and Peters, 1998).
Figure 5.89. A comparison of the predicted and experimental solvent concentration at the same cross-sections for Experiment 5 based on a
five-layer model (Shecaira and Peters, 1998).
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Experiment 6
The purpose of Experiment 6 and the accompanying numerical
simulation was to verify a new method for mapping the porosity and
permeability distributions in a heterogeneous core by NMR (Zuluaga et
al., 2000). A layered Antolini sandstone core was used for this
experiment. The core properties are L = 10 cm, d = 5 cm, φ = 12.8% and
k = 114 md. A technique was developed to map the 3D porosity and
permeability distributions in the core by NMR imaging. A technique was
developed to map the T1 distribution in a core. Permeability distribution
was calculated using the empirical equation of Kenyon et al. presented in
Chapter 2 as Eq.(2.69) and reproduced here for convenience as
(2.69) 41 NMRk C Tφ= 2
1
2T
The constant C1 was obtained by calibrating the NMR-derived
permeability to be equal to the permeability of the core in the direction of
flow measured by Darcy's law and found to equal to 180 μm2/s2. The
equation used to calculate the permeability of each voxel is
(5.77) 41180 NMRk φ=
Figure 5.90 shows a high resolution porosity image of the core. Dipping
layers are clearly visible in the image. Figure 5.91 shows the porosity
distribution after upscaling the high resolution porosity data
(512x512x512 voxels) to the low resolution numerical simulation grid of
40x40x40. The upscaling makes the porosity distribution somewhat
fuzzy. Figure 5.92 shows the upscaled permeability distribution. A high
permeability streak is clearly visible in the center of the core.
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After measuring the porosity and permeability distributions, a
first-contact miscible displacement was conducted in the Antolini core at
a favorable mobility ratio of 0.81 at a Darcy velocity of 0.000692 cm/s
(1.96 ft/day). In an effort to verify the validity of the porosity and
permeability distributions derived by NMR, the displacement was
simulated using the NMR-derived porosity and permeability distributions
as the input data for the core description. No attempt was made to adjust
the porosity and permeability data to history-match the experiment.
Rather the porosity and permeability distributions were used as is in
order to verify their validity.
Figure 5.90. High resolution NMR porosity image of layered Antolini sandstone core (Zuluaga, 1999).
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Figure 5.91. Upscaled porosity distribution from NMR imaging (Zuluaga et al., 2000)
Figure 5.92. Upscaled permeability distribution from NMR imaging
(Zuluaga et al., 2000)
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Figure 5.93 shows a comparison of the solvent concentration
distribution from the experiment and the simulation. The agreement
between the simulation and the experiment is excellent thereby verifying
the validity of the porosity and permeability distributions obtained by
NMR imaging. Figure 5.94 shows the head to head comparison of the
experimental and simulated solvent concentration distributions at the
same cross-sections. The agreement between the experiment and the
simulation is good. The NMR-derived porosity and permeability
distributions are verified.
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Figure 5.93. A comparison of the experimental and simulated solvent concentration distribution for Experiment 6 using NMR-derived porosity and permeability distributions. M= 0.81, Ng = 0.0160, Nρ = 0.1027, NPe =
133 (Zuluaga et al., 2000).
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Figure 5.94. A comparison of the experimental and simulated solvent concentration at the same cross-sections for Experiment 6 using NMR-
derived porosity and permeability distributions. M= 0.81, Ng = 0.0160, Nρ = 0.1027, NPe = 133 (Zuluaga et al., 2000).
NOMENCLATURE
ax = correlation length in the x direction
ay = correlation length in the y direction
az = correlation length in the z direction
A = cross-sectional area of porous medium
C = solvent concentration
CD = dimensionless solvent concentration
Ci = initial solvent concentration
Cj = injected solvent concentration
D = dispersion coefficient tensor
mD = mechanical dispersion coefficient tensor
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Dd = molecular diffusion coefficient
DL = longitudinal dispersion coefficient
Do = binary diffusion coefficient between the solvent and oil
Dp = mean grain diameter of the porous medium
erf = error function
erfc = complementary error function
F = formation resistivity factor
g = gravitational acceleration
J = total mass flux
aJ = mass flux vector due to advection
dJ = mass flux vector due to dispersion
k = permeability
L = length of porous medium
M = mobility ratio
Ng = gravity number
NPe = Peclet number
Nρ = density number
q = volumetric injection rate
Rf = retardation factor
S = specific surface area
t = time
tD = dimensionless time
u = interstitial velocity vector
v = Darcy velocity vector
V = Dykstra-Parson’s coefficient of permeability variation
xD = dimensionless distance
Lα = longitudinal dispersivity
Tα = transverse dispersivity
μo = viscosity of displaced fluid
μs = viscosity of displacing fluid
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φ = porosity
λ = fluid mobility
τ = tortuosity
ρo = density of displaced fluid
ρs = density of displacing fluid
Δx = mixing zone length
Δρ = density difference
∇ = gradient operator
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van Genuchten, M.T. and W.J. Alves: “Analytical Solutions of the One-Dimensional Convective-Dispersive Solute Transport Equation,” Tech. Bull. U.S. Dep. Agric., 1661 (1992).
van Wesenbeeck, J. and R.G. Kachanoski: “Spatial Scale Dependence of In Situ Solute Transport,” Soil Sci. Soc. Am. J., 55, No. 1, 3-7 (1991).
Wang, T.T., T.K. Kwei and H.L. Frisch: “Diffusion in Glassy Polymers, III,” J. Polymer Sci., Part A 2, 7, 2019-28 (1969).
Wheatcraft, S.W. and S.W. Tyler: “An Explanation of Scale-Dependent Dispersivity in Heterogeneous Aquifers Using Concepts of Fractal Geometry,” Water Resour. Res., 24, No. 4, 566-78 (1988).
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Zuluaga, E.: A Simulation Approach to Validate Petrophysical Data from Nuclear Magnetic Resonance Imaging, MS Thesis, University of Texas at Austin, August 1999.
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PROBLEMS
5.1. A laboratory core, which is 6 cm in diameter and 40 cm in length, has a porosity of 0.35. The core is initially saturated with water. A tracer is then injected continuously into the core at the rate of 1000 cm3/hour. The effluent data shown in Table 5.4 were recorded at the outlet end of the core. C/Co is the relative tracer concentration. Determine the dispersion coefficient of the core.
Table 5.4. Effluent Data for Tracer Test of Problem 5.1.
Time
(hour)
C/Co
0.35 0.075
0.37 0.215
0.385 0.37
0.396 0.5
0.41 0.65
0.43 0.83
0.44 0.89
0.46 0.96 5.2. Table 5.5 gives the data for a one-dimensional core tracer test.
Determine the Peclet number for the tracer test.
Table 5.5. Data for Problem 5.2.
PV
Injected
(tD)
1 D
D
tt−
CD
0.60 0.516 0.010
0.65 0.434 0.015
0.70 0.359 0.037
0.80 0.224 0.066
0.90 0.105 0.300
1.00 0.000 0.502
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1.10 -0.095 0.685
1.20 -0.183 0.820
1.30 -0.263 0.906
1.40 -0.338 0.988
1.50 -0.408 0.997 5.3. Fresh water at relative concentration C/Co = 0 is injected into a
sandpack saturated with salt water at relative concentration C/Co = 1. As the salt water is displaced the concentration measurements in Table 5.6 were made at a specific time t.
Table 5.6. Tracer Data for Problem 5.3.
Distance from Core Inlet
(cm)
C/Co
(%)
48.2 0.8
49.7 2.2
51.5 3.5
53.6 9.5
55.4 21.7
57.2 50.3
59.3 78.0
61.3 94.5
63.2 94.5
65.4 98.7
68.2 98.7
73.4 100 The average interstitial velocity of flow was 1.6 cm/minute. a. How long after the initiation of flow were these readings taken? b. Determine the dispersion coefficient and the dispersivity of the
sandpack.
5.4. The molecular diffusion coefficient of a porous medium is equal to 5x10-10 m2/s. If a tracer is placed in contact with the inlet end of the porous medium, estimate the relative tracer concentration C/Co at a distance of 5 m from the inlet after 100 years of diffusion. Comment on the effectiveness of molecular diffusion as a transport mechanism in a porous medium.
5-135
5.5. A tracer test is conducted in a relatively homogeneous cylindrical
sandpack using continuous injection of a nonreactive tracer. The injected tracer and the displaced liquid have the same density and
viscosity. The relative concentration C/Co of 0.75 was observed to arrive at the outlet end of the core after 0.9 hour from the start of injection. Other data about the test are:
Length of sandpack = 30 cm Diameter of sandpack = 10 cm
Volumetric injection rate = 1000 cm3/hour Porosity of the sandpack = 35% Hydraulic gradient = 0.1 Fluid viscosity = 1.0 cp
Fluid density = 1.0 gm/cm3
Gravitational acceleration = 981cm/s2
1 atmosphere = 1.0133x106 dynes/cm2 a. Calculate the dispersivity of the porous medium and state its
units. b. Calculate the permeability of the porous medium and state its
units. Problem 3 (15 points) A tracer test was conducted in a long core to determine the dispersivity of a porous medium using a tracer that had the same density and the same viscosity as the displaced liquid. The tracer test was imaged by CT at t = 50 minutes from the beginning of the test. The interstitial velocity for the test was 1.6 cm/minute. Figure 3 shows the 0.9 and 0.1 tracer concentration contours inside the porous medium obtained by imaging. a. Calculate the dispersivity of the porous medium and state its units. b. Is the core homogeneous? If yes, why? If no, why not? c. How would the dispersivity you calculated in part a compare with
the dispersivity obtained from the breakthrough curve in the same experiment?
5–136
Figure 3. Tracer concentration contours at t = 50 minutes.
Problem 3 (20 points)
A solution of nonabsorbing sodium iodide is injected continuously to
displace a brine solution from a core of length L, porosity φ, permeability
k and cross sectional area A at a constant volumetric rate q. The injected
and displaced fluids are miscible in all proportions and have the same
density and viscosity. Sodium iodide solution is used as a solvent in this
experiment because its high x-ray absorption coefficient permits the
solvent concentration profiles to be imaged by x-ray computed
tomography (CT scanning). Probes are installed at the inlet and outlet
5-137
ends of the core to monitor and report the solvent concentrations as
functions of time during the entire experiment.
a. Sketch the solvent concentration profile you would expect to see from the CT scanning after injecting 0.5 pore volume of the solvent. Show the inlet and outlet ends of the core on your sketch as well as the critical values of solvent concentration. Do your sketch on Figure 2.
b. Sketch the graphs of solvent concentration versus dimensionless time you expect from the probes at the inlet and outlet ends of the core. Superimpose the two graphs so that one can make a qualitative comparison of the two. Please label your graphs with inlet and outlet. Do your sketches on Figure 3.
c. Other than the absolute permeability, what transport property of the porous medium and fluids can be estimated from this experiment?
d. Predict the solvent concentration at the outlet end of the core at t = 108 minutes, given:
L = 30 cm q = 50 cm
3/hour
A = 20 cm2
φ = 0.15 DL = 400x10–5 cm
2/s
6. (20 points) A nonabsorbing tracer is pumped through a porous medium of length 30
cm at an interstitial velocity of 1x10-2 cm/s. A relative concentration of 0.42 was
measured in the effluent after 46.6 minutes from the start of the test. Determine the
dispersivity of the porous medium.
4. A tracer test is conducted in a relatively homogeneous cylindrical sandpack using
continuous injection of a nonreactive tracer. The injected tracer and the displaced
liquid have the same density and viscosity. The relative concentration C/Co of 0.75
5–138
was observed to arrive at the outlet end of the core after 0.9 hour from the start of
injection. Other data about the test are:
Length of sandpack = 30 cm
Diameter of sandpack = 10 cm
Volumetric injection rate = 1000 cm3/hour
Porosity of the sandpack = 35%
Hydraulic gradient = 0.1
Fluid viscosity = 1.0 cp
Fluid density = 1.0 gm/cm3
Gravitational acceleration = 981cm/s2
1 atmosphere = 1.0133x106 dynes/cm
2
a. Calculate the dispersivity of the porous medium. Please state the units of your
answer.
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b. Calculate the permeability of the porous medium. Please state the units of
your answer.
6. (10 points) A tracer test was conducted in a long core to determine the dispersivity of a porous medium using a tracer that had the same density and the same viscosity as the displaced liquid. The tracer test was imaged by CT at t = 50 minutes from the beginning of the test. The interstitial velocity for the test was 1.6 cm/minute. Figure 2 shows the 0.9 and 0.1 tracer concentration contours inside the porous medium obtained by imaging.
(a) Calculate the dispersivity of the porous medium and state its
units. (b) Is the core homogeneous? If yes, why? If no, why not? 2. (20 points) A nonabsorbing tracer was injected into a 30 cm long core at a
constant interstitial velocity of 36 cm/hr. The relative concentrations measured at the outlet end of the core were as follows:
Time (Hour) C/Co 0.7 0.25 0.833 0.50 0.992 0.75
Determine the dispersivity of the core.
5–140
5-141
Figure 2. Tracer concentration contours at t = 50 minutes.
CHAPTER 6
INTERFACIAL PHENOMENA AND WETTABILITY
6.1 INTRODUCTION
The pore space in a petroleum reservoir rock is usually occupied by
more than one fluid. In an oil reservoir, water and oil occupy the pore space.
In a gas reservoir, water and gas occupy the pore space. In some oil
reservoirs, at some stage of depletion, water, oil and gas may occupy the pore
space.
When more than one fluid occupies the pore space of a porous medium,
new set of problems arise. Fluid saturations must be tracked. Interfacial
forces (surface forces) between the immiscible fluids and between the fluids
and the rock surface come into play. Because the pores are of capillary
dimensions, capillarity plays a role. Interfaces separate the fluids within the
pores giving rise to differences in fluid pressure between the phases (capillary
pressure) and differences in the flow capacity (relative permeability) of the
rock and fluids. Capillarity also ensures that an immiscible displacement can
never be complete. There is always a residual saturation of the displaced fluid
that is trapped by capillarity. Further, the rock surface can show a marked
affinity for one of the fluids. Such an affinity is characterized by the concept of
wettability. Interfacial phenomena and wettability are presented in this
6-1
chapter. Capillary pressure and relative permeability are presented in
subsequent chapters.
6.2. SURFACE AND INTERFACIAL TENSIONS
6.2.1 Surface Tension
Surface tension is the contractile force that exists at the interface of a
liquid and its vapor (or air). Surface tension makes the surface of a liquid
drop act like a membrane. The force is caused by unequal molecular
attractions of the fluid particles at the surface as shown in Figure 6.1. The
force per unit length (σ = Force/Length) tending to contract the surface of a
liquid is a measure of the surface tension of the liquid. It is a property of the
liquid and is usually expressed in units of dynes/cm.
Figure 6.1: Apparent surface film caused by unequal attraction of
surface molecules of a liquid.
If the forces acting on a molecule at the surface or interface are different
from those acting on a molecule in the body of the liquid, a new interface can
6-2
only be created if work is done. It can be shown (e.g., by soap film
experiment) that the reversible work required to create a new interface is given
by
W daδ σ= − (6.1)
where δW is reversible work required to create a new area, σ is surface
(interfacial) tension and da is the increase in area. The reversible work shown
in Eq.(6.1) corresponds to a free energy quantity. It is apparent from Eq.(6.1)
that surface (interfacial) tension also can be viewed as free surface energy per
unit area with units of erg per square centimeter. Because a system at
equilibrium minimizes its free energy, liquids at equilibrium tend to minimize
their surface area. A liquid jet tends to break up into spherical drops because
a sphere has the smallest surface area per unit volume.
The surface tension of pure water at 70 °F is 72.5 dynes/cm, and at
200 °F is 60.1 dynes/cm. The surface tensions of crude oils at 70 °F range
from 24 to 38 dynes/cm. High temperatures and dissolved gas both tend to
reduce the surface tension of crude oils. Values on the order of 1 dyne/cm
may be expected at temperatures and pressures exceeding 150 °F and 3,000
psig. Table 4.1 shows the surface tensions of some selected liquids.
Factors that affect the surface tension of a liquid include pressure,
temperature and solute concentration. An increase in pressure leads to a
reduction in the surface tension of a liquid. This is because pressure exerts a
compressive force on the surface which reduces the tensile or contractile
tendency of the surface. An increase in temperature leads to a reduction in
the surface tension of a liquid. The increase in temperature causes increased
randomness at the surface which leads to an increase in the surface entropy.
It can be shown from thermodynamics that the surface entropy is given by
6-3
s
P
STσ∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠
(6.2)
where Ss is the surface entropy and T is the temperature. It is clear from
Eq.(6.2) that surface tension decreases with temperature. Figure 6.2 shows
the variation of surface tensions of hydrocarbons with temperature.
Table 4.1. Surface Tension of Pure Liquids
Liquid T (°C) σ (dynes/cm) Water 20 72.8 Water 25 72.0 Benzene 20 28.88 Benzene 25 28.22 Toluene 20 28.43 Carbon tetrachloride
20 26.9
Ethanol 20 22.39 n-Octane 20 21.8 Ethyl ether 20 17.01
The effect of solute concentration on the surface tension of a liquid
depends on the liquid and the nature of the solute. Four general cases may
be identified.
1. Liquids having fairly close values of surface tension. Generally, the
surface tension of the mixture varies approximately linearly with the
composition. For example, for a mixture of acetone and chloroform at
18° C, the surface tension increases linearly with composition (mole %
chloroform) from 22 dynes/cm for pure acetone to 27 dynes/cm for
pure chloroform.
6-4
Figure 6.2. Variation of surface tensions of hydrocarbons with temperature.
2. Liquids having widely different values of surface tension. In general,
the surface tension of a liquid is reduced substantially by addition of a
liquid of lower surface tension, but is only slightly increased by addition
of a liquid of higher surface tension. For example, addition of ethanol to
water causes a rapid reduction in the surface tension of water with
ethanol concentration. However, addition of water to benzene raises its
surface tension from 28.2 to only 29.3 dynes/cm.
3. Solutions of inorganic electrolytes. In general, the surface tension
increases with solute concentration. For example, the surface tension
6-5
of water at 20° C will be increased by addition of sodium chloride from
72.8 dynes/cm to 80 dynes/cm at a concentration of 5 moles of sodium
chloride per liter of solution.
4. Solutions of colloidal (long chain) electrolytes. In general, the surface
tension decreases with solute concentration but is followed by a region
over which the surface tension is virtually unchanged by solute
concentration. For example, the surface tension of water at 25° C will
be reduced by addition of sodium lauryl sulfate from 72 dynes/cm to 40
dynes/cm at a concentration of 0.01 moles per liter of solution. The
surface tension remains constant above this concentration.
A useful empirical relation often used to calculate surface tension is
through the concept of the parachor. The parachor for a pure substance is
defined as
14
L g
Mσρ ρ
Λ =−
(6.3)
where Λ is the parachor, M is the molecular weight of the liquid, σ is the
surface tension of the liquid in dynes/cm, ρL is the saturated liquid density in
g/cm3 and ρg is the saturated vapor density in g/cm3. Parachor has definite
values for specific atoms and structures. Parachors are predicted from the
structure of the molecules or can be calculated for pure substances and
mixtures from surface tension measurements at atmospheric pressure. The
parachors for pure substances are given in Table 6.2. Correlations for
parachors with molecular weight are shown in Figures 6.3 and 6.4. The
saturated liquid and vapor densities for various liquids are given in Figure
6.5. Equation (3.4) can be rearranged to calculate the surface tension as
6-6
4
L g
Mρ ρ
σ⎡ − ⎤⎛ ⎞
= Λ⎢ ⎥⎜⎝ ⎠
⎟⎣ ⎦
(6.4)
Table 4.2. Parachors for Computing Surface and Interfacial Tensions (Katz et
al., 1959)
Constituent Parachor Methane 77.0 Ethane 108.0 Propane 150.3 i-Butane 181.5 n-Butane 181.5 i-Pentane 225.0 n-Pentane 231.5 n-Hexane 271.0 n-Heptane 312.5 n-Octane 351.5 Hydrogen 34 (approx) Nitrogen 41 (approx)
Carbon dioxide 78
6-7
Figure 6.3. Parachors for computing interfacial tension of normal paraffin
hydrocarbons (Katz et al., 1959).
6-8
Figure 6.4. Parachors of heavy fractions for computing interfacial tension of
reservoir liquids (Firoozabadi et al., 1988).
6-9
Figure 6.5: Saturated liquid and vapor densities of various substances.
6-10
6.2.2 Interfacial Tension
Interfacial tension is the contractile force per unit length that exists at
the interface of two immiscible fluids such as oil and water (Figure 6.6). The
forces acting on the surface molecules are similar to those in the liquid-vapor
system, but the mutual attraction of unlike molecules across the interface
becomes important. The free energy required to create a fresh interface is
referred to as the excess interfacial free energy. The specific excess interfacial
free energy is dimensionally equivalent and is numerically equal to the
interfacial tension. Like surface tension, the unit of measurement of
interfacial tension is dynes/cm (or ergs/cm2).
Consider butanol (C4H9OH) with a surface tension of 24 dynes/cm in
contact with water (H2O) with a surface tension of 72 dynes/cm. What will be
the interfacial tension between the two liquids? For this system, the
interfacial tension is 1.8 dynes/cm, a fairly low number considering the
surface tensions of water and butanol. The low interfacial tension indicates
that the molecules of butanol must concentrate at the interface, decreasing
the contractile tendency of the interface. Interfacial orientation of the butanol
is favored because the hydrocarbon chain in the butanol is hydrophobic. At
20 ºC, the surface tension of ethanol is 22.39 dynes/cm and that of water is
72.80 dynes/cm. What is the interfacial tension between ethanol and water?
The answer is zero because ethanol and water are miscible. These examples
show that there is no simple relationship between the surface tensions of
liquids and their interfacial tensions. Table 6.3 lists accurately known
interfacial tensions for various organic liquids against water.
The presence of a third component of the right kind can significantly
reduce the interfacial tension between two liquids. For example, the
interfacial tension of water and iso-pentanol is 4.4 dynes/cm. If ethanol is
added to the system, the ethanol molecules will adsorb at the interface,
6-11
thereby reducing the interfacial tension between the water and the iso-
pentanol. When 25% by weight of ethanol is added, the interfacial tension is
reduced to zero. The system then becomes miscible and forms a single phase.
Figure 6.6. Apparent surface film caused by unequal attraction of molecules at the interface of two liquids.
Table 6.3. Interfacial Tension between Water and Pure Liquids
Liquid T (°C) σ (dynes/cm) n-Hexane 20 51.0 n-Octane 20 50.8 Carbon disulphide 20 48.0 Carbon tetrachloride 20 45.1 Carbon tetrachloride 25 43.7 Bromobenzene 25 38.1 Benzene 20 35.0 Benzene 25 34.71 Nitrobenzene 20 26.0 Ethyl ether 20 10.7
6-12
n-Octanol 20 8.5 n-Hexanol 25 6.8 Aniline 20 5.85 n-Pentanol 25 4.4 Ethyl acetate 30 2.9 Isobutanol 20 2.1 n-Butanol 20 1.8 n-Butanol 25 1.6
Chemicals that adsorb at interfaces are usually referred to as surface active
agents or surfactants. Such chemicals form a monolayer at the interface.
The monolayer is in a state of compression which reduces the contractile
tendency of the interface, thereby reducing the interfacial tension. The
presence of the adsorbed molecules creates a surface or spreading pressure
which reduces the interfacial tension. In fact,
oσ σ= − Π (6.5)
where σ is the reduced interfacial intension, σo is the original interfacial
tension and Π is the spreading pressure. Surfactants are often used to reduce
the interfacial tension between oil and water in order to improve oil recovery.
The interfacial tensions between reservoir water and crude oils have
been measured for a number of reservoirs and found to range from 15 to 35
dynes/cm at 70 °F, 8 to 25 dynes/cm at 100 °F, and 8 to 19 dynes/cm at 130
°F. Table 6.4 presents the results of interfacial tension measurements for
some fluid pairs.
The interfacial tension between reservoir oil and gas can be estimated
using parachors as
6-13
4
1
i NgL
i i ii L g
x yM M
ρρσ=
=
⎡ ⎤⎛ ⎞= Λ −⎢ ⎜⎜⎢ ⎥⎝ ⎠⎣ ⎦
∑ ⎥⎟⎟ (6.6)
where σ is the interfacial tension between oil and gas in dynes/cm, Λi is the
parachor of component i, xi is the mole fraction of component i in the liquid,
ML is the apparent molecular weight of the liquid, ρL is the saturated liquid
density in g/cm3, yi is the mole fraction of component i in the gas, Mg is the
apparent molecular weight of the gas, N is the total number of components in
the mixture and ρg is the saturated gas density in g/cm3.
Table 6.4. Typical Interfacial Tensions and Contact Angles for Fluid Pairs
(Archer and Wall, 1986)
Wetting Phase
Non-Wetting Phase
Conditions Contact Angle (º)
Interfacial Tension
(dynes/cm) Brine Oil Reservoir 30 30 Brine Oil Laboratory 30 48 Brine Gas Laboratory 0 72 Brine Gas Reservoir 0 50 Oil Gas Reservoir 0 4 Gas Mercury Laboratory 140 480
Many reservoir phenomena depend on the interfacial tensions between
the reservoir fluids and between the reservoir fluids and the reservoir rock.
Interfacial tensions of reservoir water and oil can be reduced significantly by
the addition of surface active agents to either the oil or to the water. Some of
these surface active agents occur naturally in crude oils.
6-14
The residual oil saturation for an immiscible displacement in a porous
medium is a function of the interfacial tension between the fluids, the
wettability, the fluid viscosities and the displacement rate. Therefore, we can
write
( )1 cos , , ,or nw wS f vσ θ μ μ= (6.7)
Using the technique of Appendix A for dimensional analysis, it can be shown
that the rank of the dimensional matrix obtained from the four variables
cosσ θ , nwμ , wμ and is two. Therefore, two independent dimensionless groups
can be derived from the four variables. It can be shown that
v
1
23
3
4
cos 0 11 1
1 00 1
nw
w
xx
4x xxxv
σ θμμ
−⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= +⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦
(6.8)
Let us choose x3 = -1, x4 = 0. The corresponding dimensionless group is given
by
1nw
w
μπμ
= (6.9)
Next, let us choose x3 = x4 = 1. The corresponding dimensionless group is
2 coswvμπ
σ θ= (6.10)
The dimensionless group in Eq.(6.10), cos
wvμσ θ
or wvμσ
in the case of perfect
wetting, is known as the capillary number. It is the ratio of the viscous to the
capillary force. Thus,
6-15
cos
wc
vN or wvμ μσ θ σ
= (6.11)
The functional relationship between the residual oil saturation and the two
dimensionless groups can be written as
2 ,cos
nw wor
w
vS f μ μμ σ θ
⎛ ⎞= ⎜
⎝ ⎠⎟ (6.12)
For a fixed viscosity ratio,
3 cosw
orvS f μ
σ θ⎛= ⎜⎝ ⎠
⎞⎟ (6.13)
The oil recovery is given by
41
1 cor wi w
wi
S S vR fS
μosσ θ
− − ⎛= = ⎜− ⎝ ⎠⎞⎟ (6.14)
Figure 6.7 shows typical correlations for residual saturations versus
capillary number for a wetting fluid displacing a nonwetting fluid and for a
nonwetting fluid displacing a wetting fluid. Such correlations are usually
referred to as capillary desaturation curves (CDC). Consider a wetting fluid
displacing a nonwetting fluid such as a waterflood in a water wet reservoir. It
can be seen that there is a critical capillary number below which the residual
nonwetting fluid saturation is constant and independent of the capillary
number. Normal waterfloods usually fall in this range of capillary number.
Above the critical capillary number, the residual oil saturation decreases as
the capillary number increases. Thus, the residual nonwetting phase can be
mobilized and displaced by increasing the capillary number of the
displacement. Capillary number can be increased by increasing wμ and v .
However, the most effective way to increase the capillary number is by
6-16
lowering the interfacial tension between the wetting and nonwetting phases
with a surfactant. Interfacial tensions less that 0.1 dyne/cm can be achieved.
In the limit, if the interfacial tension could be lowered to zero, the fluids would
become miscible and no residual saturation would be left behind. Of course,
in a practical process, the interfacial tension cannot be reduced to zero except
in a miscible process.
Consider the case of a nonwetting fluid displacing a wetting fluid, such
as a waterflood of an oil wet reservoir. The capillary desaturation curve is
similar to that for the residual nonwetting fluid except that the critical value
of the capillary number is higher. It should be noted that the numerical
values of the residual saturations given in Figure 6.7 are for illustrative
purposes. It should not be assumed that waterfloods always have a residual
oil saturation of 30%, nor should one infer from the figure that it is more
efficient to displace a wetting fluid with a nonwetting fluid. The residual oil
saturation can be greater or less than 30% depending on such factors as the
mobility ratio of the displacement and the properties of the porous medium
such as pore structure, pore size distribution, permeability and wettability to
name a few. Also, in general, it is more difficult for a nonwetting fluid to
displace a wetting fluid than for a wetting fluid to displace a nonwetting fluid.
6-17
Figure 6.7. Typical correlations of residual nonwetting and wetting phase saturations with capillary number (Lake, 1989).
Figure 6.8 shows capillary desaturation experimental data from Abrams
(1975) obtained on the same core sample but at different viscosity ratios.
Clearly, the decrease in residual oil saturation with increasing capillary
number is evident. Figure 6.9 shows the same data plotted against a modified
capillary number that includes the viscosity ratio, 0.4
w w
o
vμ μσ μ
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠. The inclusion
of the viscosity ratio appears to improve the correlation.
6-18
Figure 6.8. Capillary desaturation data (Abrams, 1975).
Figure 6.9. Capillary desaturation data with the effect of viscosity ratio included (Abrams, 1975).
6-19
6.2.3 Measurements of Surface and Interfacial Tensions
Several techniques are used to measure surface and interfacial
tensions. These include (1) capillary rise method, (2) sessile drop method, (3)
pendant drop method, (4) ring method and (5) spinning drop method.
Capillary Rise Experiment
When a capillary tube is dipped into a wetting liquid, the liquid will be
spontaneously imbibed (sucked) into the capillary tube as shown Figure 6.10.
Figure 6.10. Capillary rise experiment.
6-20
The equilibrium height is determined by the balance between the capillary
suction force and the pull of gravity. The capillary force acts upward and for a
circular capillary tube is given by
2 cosCapillary Force Force Up rπ σ θ= = (6.15)
The gravitational force acts downward and is given by
( )2 w nwGravitational Force Force Down r h gπ ρ ρ= = − (6.16)
The downward force also can be expressed in terms of the pressures on the
opposite sides of the meniscus as
( )2Pr nw wessure Force Force Down r P Pπ= = − (6.17)
At equilibrium, the force up is equal to the force down. Equating Eqs.(6.15)
and (6.16) gives
(2 cosw nwh
r) gσ θ ρ ρ= − (6.18)
Eq.(6.18) can be rearranged as
( )2cos
w nwrh gρ ρσ
θ−
= (6.19)
The surface tension, σ, can be estimated by measuring the variables on the
right side of Eq.(6.19) in a capillary rise experiment. The experiment can be
simplified by using air as the nonwetting phase and treating the capillary tube
such that it is perfectly wetted by the wetting fluid. In this case, nw wρ ρ<< , θ =
0 and cosθ = 1. Eq.(6.19) becomes
6-21
2
wrh gρσ = (6.20)
By measuring r, h and wρ , an estimate of the surface tension can easily be
obtained from Eq.(6.20).
Eq.(6.20) can be used to define a characteristic capillary length scale as
1
2w
rhg
σκρ
− = = (6.21)
For water at 25 ºC, σ = 72 dynes/cm, 1wρ = g/cm3 and of course, g = 981
cm/s2. Thus, for water,
1 72 0.27091 981x
κ − = = cm or 2.71 mm.
The significance of the capillary length is that when dealing with a system
with the characteristic length scale 1δ κ −< , the effect of gravity is negligible
and capillary effect dominates the process. For example, at the pore scale
where the characteristic length is of the order of microns, capillary effect
dominates gravity effect.
Let us define another constant for the capillary rise experiment as
2 2
w
ag
σρ
= (6.22)
The constant a2 is a property of the wetting fluid only. For water at 25 ºC, a2 =
0.1468 cm2. Eq.(6.20) can be written as
2a rh= (6.23)
6-22
Some famous scientists have studied the capillary rise experiment and have
proposed more elaborate equations to describe the capillary rise than
Eq.(6.23). Jurin (1718) gives the capillary rise equation as
2
3ra r h⎛ ⎞= +⎜
⎝ ⎠⎟ (6.24)
where the 3r corrects for the volume of the liquid in the spherical meniscus.
Hagen and Desains (19xx) proposed
2 3
22 3
0.1111 0.074113r ra rhh h h
⎛ ⎞= + − +⎜
⎝ ⎠
r⎟ (6.25)
where the last two terms on the right hand side correct for deviations of the
meniscus from sphericity. Rayleigh (1915) further refined Eq.(6.25) as
2 3
22 3
0.1288 0.131213r ra rhh h h
⎛ ⎞= + − +⎜
⎝ ⎠
r⎟ (6.26)
For our purpose, we will use the simple version of the capillary rise equation,
Eq.(6.23).
Equating Eqs.(6.15), (6.16) and (6.17) gives
(2 cosnw w w nwP P h
r) gσ θ ρ ρ− = = − (6.27)
Eq.(6.27) gives the excess pressure, (Pnw - Pw), across the curved interface
between the wetting and nonwetting phases in terms of the pertinent
variables of the capillary rise experiment. This excess pressure is known as
the capillary pressure (Pc) and will be the subject of Chapter 7. As shown in
6-23
Eq.(6.27), the excess pressure for the capillary rise experiment is positive,
which means that the pressure in the nonwetting phase is higher than that in
the wetting phase. Eq.(6.27) can now be written as
(2 cosc nw w w nwP P P h g
r)σ θ ρ ρ= − = = − (6.28)
Sessile Drop Method
The sessile drop method of determining the surface tension of a liquid
consists of measuring the number of liquid drops that fall from the capillary
end of the instrument while the surface of the liquid within the bulb is
lowered from the upper to the lower mark as shown in Figure 6.11. The
principle of the method is based on the fact that the size of the liquid drop is
proportional to surface tension of the liquid. The size of the drop is reached
when the surface tension can no longer support its weight. To a first
approximation,
2idealW rπ σ= (6.29)
where Wideal is the weight of the drop that should fall, r is the external
radius of the tube and σ is the surface tension. Figure 6.12 shows the
sequence of shapes for a drop that detaches from a tip. The detached drop
leaves behind some liquid residue. Thus, the actual weight of the drop which
is what is measured is less than the ideal weight. To account for this,
Eq.(6.29) is modified as
2actual idealW W f r fπ σ= = (6.30)
6-24
where f is a correction factor which can be expressed as a function of 13
rV
,
where V is the volume of the drop. Table 6.5 shows the correction factors for
various 13
rV
.
Figure 6.11. Sessile drop method of measuring surface tension.
Figure 6.12. Sequence of shapes for a drop.
6-25
Table 6.5. Correction Factors for Sessile Drop Method (Adamson, 1982).
Pendant Drop Method
The pendant-drop method of measuring surface or interfacial tension
depends only on the density of the fluids and the dimensions of the drop.
Figure 6.13 shows a pendant drop and the relevant dimensions. The surface
or interfacial tension is given by
( )2
e L ggdHρ ρ
σ−
= (6.31)
where σ is the surface tension, de is the maximum diameter of the drop, ρL is
the density of the liquid, ρg is the density of the vapor, H is a constant that is
a function of e
s
dd
and g is the gravitational acceleration. The constant H is
tabulated as a function of de/ds. The pendant drop method can be used to
measure surface tension or interfacial tension. It can also be adapted for
measurements at elevated temperature and pressure.
6-26
Figure 6.13. Surface tension by pendant drop method.
Ring Method
The ring method of determining surface or interfacial tension depends
on measuring the force required to pull the ring free of the interface as shown
in Figure 6.14. Theoretically, the surface or interfacial tension is given by
2FL
σ = (6.32)
where σ is the surface or interfacial tension, F is the force required to pull the
ring free of the interface and L is the circumference of the ring. The factor of 2
accounts for the fact that there are two surfaces around the ring. In practice,
corrections are needed to account for the mass of liquid lifted by the ring in
breaking through the interface as shown in Figure 6.15. Such corrections are
made available with the instrument. Figure 6.16 shows a typical instrument,
known as the du Nouy tensiometer, that employs the ring method for surface
or interfacial tension determination.
6-27
Figure 6.14. Surface tension by ring method.
6-28
Figure 6.15. Condition of liquid surface film at breaking point.
6-29
Figure 6.16. du Nouy tensiometer.
Spinning Drop Method
The spinning drop method of determining surface or interfacial tension
is based on measuring the shape of a drop of liquid or gas bubble in a more
dense liquid contained in a rotating horizontal tube. The drop or bubble is
typically rotated at speeds of 1,200 to 24,000 revolutions per minute (RPM).
Under rotation, the original spherical drop or bubble becomes elongated into a
6-30
cylindrical shape as shown in Figure 6.17. A strobe light is used to visualize
the deformed drop. A microscope is used to measure the diameter of the
drop. The interfacial tension is given by
2 314
rσ ρω= Δ (6.33)
where σ is the surface or interfacial tension, dynes/cm Δρ is the density
difference between the two fluids in g/cm3, ω is the angular velocity in
radians/s and r is the cylindrical radius of the drop in cm.
An instrument based on the spinning drop method has been designed
and patented at the University of Texas at Austin by Schechter and Wade
(Figure 6.18). The instrument is manufactured and sold by the Chemistry
Department at the University of Texas. The instrument is particularly
suitable for measuring low interfacial tensions and is therefore used
extensively in surfactant research. Interfacial tensions as low as 10-6
dyne/cm have been successfully measured with the instrument.
6.3 WETTABILITY
6.3.1 Definition
Wettability is a tendency for one fluid to spread on or adhere to a solid
surface in the presence of other immiscible fluids. The fluid that spreads or
adheres to the surface is known as the wetting fluid. In a petroleum reservoir,
the solid surface is the reservoir rock which may be sandstone, limestone, or
dolomite, together with cementing material. The fluids are water, oil and gas.
Normally, either water or oil is the wetting phase. Gas is always a nonwetting
phase.
6-31
Figure 3.17. Cylindrical liquid drops in a spinning drop apparatus. (A)
benzene-water system at 20,000 RPM, (B) octane-surfactant system at 6,000 RPM (Cayias et al., 1975).
6-32
Figure 6.18. Schematic of spinning drop tensiometer (Cayias et al., 1975).
Consider the water-oil-solid system shown in Figure 6.19. Three
interfacial tensions (specific free surface energies) arise: σos is the solid-oil
interfacial tension, σow is the oil-water interfacial tension and σws is the
water-solid interfacial tension. The angle θ is known as the contact angle and
is measured through the water (the more dense fluid). The contact angle is a
measure of the wettability of the solid.
At equilibrium, the interfacial tensions are related by the Young - Dupre
equation obtained by considering horizontal equilibrium of the point of
contact of the interfacial tensions. For Figure 6.19, this equation is given by
cosos ws owσ σ σ θ− = (6.34)
6-33
The free surface energies for the oil-solid and water-solid interfaces
cannot be measured readily. However, their difference can be determined by
measuring the oil-water interfacial tension and the contact angle. This
difference controls the movement of the interface before equilibrium is
achieved. The following may be deduced from Eq. (6.32):
Figure 6.19. Interfacial tensions in a water-oil-solid system.
1. If the free surface energies for the oil-solid (σos) and the water-solid
(σws) interfaces are equal, the left side of Eq.(6.34) is zero. Since the
oil-water interfacial (σow) tension is nonzero, cosθ on the right side of
Eq.(6.34) must be zero, giving a contact angle of 90°. A contact angle of
90° means that the solid has no preferential wettability for the oil or the
water. This is a situation of neutral or intermediate wettability.
2. If σws < σos , then θ < 90°. The solid is said to be preferentially water
wet. When the oil, water and solid are first brought in contact, water
will advance and spread on the solid surface, displacing the oil until an
equilibrium contact angle is attained according Eq.(6.34). During
6-34
spreading of the water, the free energy of the system is reduced since
σws < σos. The difference (σos − σws) is known as the adhesion tension.
3. If σws > σos , then θ > 90°. The solid is said to be preferentially oil wet.
4. Complete spreading of the oil on the surface takes place if θ = 180° and
complete spreading of water on the surface takes place if θ = 0°.
Complete spreading of crude oil or water on a surface has never been
observed with reservoir fluids.
Figure 6.20 shows the equilibrium contact angles for four wettability states.
In Figure 6.20a, the surface is preferentially oil wet; in Figure 6.20b, the
surface is of neutral wettability; in Figure 6.20c, the surface is water wet; and
in Figure 6.20d, the surface is totally water wet.
6-35
Figure 6.20. Equilibrium contact angles showing four wettability states.
6.3.2 Determination of Wettability
Reservoir wettability is usually determined either by contact angle
measurement using reservoir fluids and a pure mineral surface or by an
imbibition test on a reservoir core sample using refined oil and a synthetic
brine. No wettability determination method involves the simultaneous use of
reservoir fluids and reservoir rock.
6-36
Contact Angle Method
Contact angle is one of the earliest and still most widely used
measurement to evaluate reservoir wettability. The contact angle
measurement essentially seeks to establish whether or not the reservoir oil
contains surface active agents that could make an originally preferentially
water wet mineral surface become preferentially oil wet over time.
Accordingly, the contact angle test uses reservoir oil and brine and a pure,
clean mineral surface which is known to be preferentially water wet at the
outset. In the absence of reservoir brine, synthetic brine is used in the test
since the surface active fluid is the oil and not in the brine. The solids
normally used in the test to represent reservoir rock are pure quartz (silica)
for a sandstone reservoir, pure calcite for a limestone reservoir and pure
dolomite crystal for a dolomite reservoir. These pure minerals are known to
be preferentially water wet initially. If the oil contains surface active agents,
then these will adsorb on the mineral surface over time and increase the
degree of oil wetness. This change in the wettability of the surface can be
observed and quantified by measuring the contact angle over time until an
equilibrium contact angle is obtained. It is reasonable to assume that a
similar wetting equilibrium will be approached in the reservoir.
The contact angle measurement is performed with a contact angle cell
using an instrument known as a goniometer. The mineral surface is
immersed in the brine (or oil) and allowed to equilibrate. A drop of the oil (or
brine) is then introduced on to the surface with a hypodermic syringe as
shown in Figure 6.21. The contact angle is then measured over time. The
test can last several weeks depending on the time required to achieve
adsorption equilibrium. The equipment can be adapted for contact angle
measurements at elevated pressures and temperatures.
6-37
Figure 6.21. Contact angle cell.
Two contact angles are normally measured: the advancing and receding
contact angles. The advancing contact angle (θA) is the contact angle
obtained when water comes into equilibrium with a surface previously in
contact with oil as shown Figure 6.22. The receding contact angle (θR) is the
contact angle obtained when oil comes into equilibrium with a surface
previously in contact with water. The advancing contact angle is always
greater than the receding contact angle. Normally, it is the advancing contact
angle that is reported as the contact angle in a wettability test.
Figure 6.23 shows the results of a contact angle test. Several
interesting observations can be made. The early time contact angle
measurements showed the solid to be preferentially water wet. However, as
6-38
time passed, the degree of water wetness diminished. Eventually, after
adsorption equilibrium was achieved the solid was found to be preferentially
oil wet. Had the contact angle test been terminated prematurely, the
wettability assessment would have been wrong. Note that for this test, over
30 days of aging were needed to establish adsorption equilibrium.
Figure6.22. Advancing and receding contact angles.
Figure 6.23. Approach to equilibrium contact angle (Craig, 1971).
6-39
The major advantages of contact angle measurements are the reliability
of the results and the relative ease of obtaining uncontaminated reservoir
fluid samples compared to uncontaminated reservoir rock samples. The
following disadvantages should be noted. (1) Contact angle is measured on a
flat, clean, homogeneous mineral surface. Such a surface does not exist in the
reservoir. (2) Pure minerals are used in the test to simulate sandstone,
limestone and dolomite reservoir rocks. Pure minerals may not be
representative of actual reservoir mineralogy. (3) The test can be very long and
requires extreme cleanliness and inertness of the test system. (4) There is
evidence that the contact angle is affected by which fluid was first in contact
with the solid.
Amott Wettability Test
The Amott wettability index is obtained by a combined imbibition-
displacement test on a reservoir core sample using refined oil and synthetic
brine. After the reservoir core sample has been flushed with brine to residual
oil saturation and evacuated to remove gas, it is then subjected to the
following tests:
1. The core is immersed in oil (e.g., kerosene) and the volume of brine
displaced by the imbibition of oil is measured after 20 hours in an
imbibition cell as shown in Figure 6.24.
2. The core is centrifuged under kerosene and the additional brine
displaced by centrifuging is measured.
3. The core is immersed in brine and the volume of oil displaced by the
imbibition of brine is measured after 20 hours.
4. The core is centrifuged under brine and the additional oil displaced by
centrifuging is measured.
6-40
Figure 6.24. Imbibition cell.
The wettability indices of water (WIw) and oil (WIo) are calculated as follows:
w
Volume of oil displaced by brine imbibitionWIVolume of oil displaced by brine imbibition forced displacement
=+
(6.35)
o
Volume of brine displaced by oil imbibitionWIVolume of brine displaced by oil imbibition forced displacement
=+
(6.36)
The Amott wettability indices and dimensionless numbers that range from 0
to 1. If the rock is preferentially water-wet, WIo will be 0 and WIw > 0. The
greater the degree of water wetness, the closer will WIw be to 1. Similarly, if
6-41
the rock is preferentially oil wet, WIw will be 0 and WIo > 0. The greater the
degree of oil wetness, the closer will WIo be to 1. For a rock of intermediate or
neutral wettability, WIw and WIo will be 0 or close to 0. Sometimes, the
difference is used to as the wettability measure. In this case, the
wettability index will range from -1 to +1. An index of -1 indicates a strongly
oil wet rock whereas an index of +1 indicates a strongly water wet rock.
( wWI WI− )o
The Amott wettability index is a reliable measure of the wettability of
the core sample. However, the wettability of the core sample may not be
representative of the wettability of the reservoir rock because of the difficulty
of obtaining an unaltered core sample. The wettability of the core sample may
easily be altered by the coring operation.
United States Bureau of Mines (USBM) Wettability Index
The USBM wettability index is obtained by carry out a number of forced
water and oil displacement experiments using a centrifuge. The results of
such experiments are shown in Figure 6.25. The sample is saturated initially
with water. The water is then displaced with oil to irreducible water saturation
using the centrifuge. This is the process labeled I in each figure. Next, the
sample, which contains initial oil saturation and irreducible water saturation
is then centrifuged in water to residual oil saturation. This is the process
labeled II in each figure. The sample, which now contains water and residual
oil saturation is then centrifuged in oil to irreducible water saturation. This is
the process labeled III in each figure. The USBM wettability index is calculated
as
110
2
logwAUSBM Wettability Index IA
⎛ ⎞= = ⎜
⎝ ⎠⎟ (6.37)
6-42
where A1 and A2 are the areas under the capillary pressure curves shown in
each figure.
The area under a capillary pressure curve represents the
thermodynamic work required for the displacement. The displacement of a
nonwetting phase by a wetting phase requires less work than the
displacement of a wetting phase by a nonwetting phase. Therefore, the ratio
of the areas under the capillary pressure curves, 1
2
AA
⎛ ⎞⎜⎝ ⎠
⎟
)o
, is a measure of the
degree of wettability of the porous medium. Therefore, the USBM wettability
index for a water wet medium will be positive as shown in Figure 6.25A, that
of an oil wet medium will be negative as shown in Figure 6.25B and that of a
medium of neutral wettability will be 0 as shown in Figure 6.25C. The USBM
wettability index ranges from -1 for a strongly oil wet rock to +1 for a strongly
water wet rock. The absolute value of the index is a measure of the degree of
wettability preference. A wettability index of zero indicates no preferential
wetting by either fluid.
Figure 6.26 compares the USBM wettability index and the Amott
wettability index, , of forty three outcrop rock samples and three
reservoir rock samples. There is a strong correlation between the two
measures of wettability. In particular, both methods show the three reservoir
rock samples to be oil wet as indicated by the negative values for both
wettability indices.
( wWI WI−
6-43
Figure 6.25. Determination of USBM wettability index (Donaldson et al., 1969).
6-44
Figure 6.26. A comparison of the USBM wettability index with Amott wettability index for several core samples (Donaldson et al., 1969).
6.3.3 Wettability of Petroleum Reservoirs
The wettabilities of petroleum reservoirs span the entire spectrum from
preferentially water wet to preferentially oil wet reservoirs. Treibel et al. (1972)
measured the wettabilities of 30 sandstone and 25 carbonate reservoirs by
measuring contact angles at the reservoir temperatures using the reservoir
oils and synthetic brine. Quartz crystal was used to represent the sandstone
reservoirs whereas calcite crystal was used to represent the limestone and
dolomite reservoirs in the contact angle measurements. The wettabilities of
the reservoirs were evaluated using an arbitrary contact angle scale.
Reservoirs with contact angles from 0 to 75° were classified as water wet;
those with contact angles from 75 to 105° were classified as having
6-45
intermediate wettability and those with contact angles from 105 to 180° were
classified as preferentially oil wet.
The results showed that 27% of the reservoirs tested were preferentially
water wet, 66% were preferentially oil wet and the remaining 7% were of
intermediate wettability. It was found that 43% of the sandstones were
preferentially water wet, 50% were preferentially oil wet and 7% were of
intermediate wettability. On the other hand, 84% of the carbonate reservoirs
were preferentially oil wet, 8% were preferentially water wet and 8% were of
intermediate wettability. It would appear from the results of this study that
carbonates are more likely to be preferentially oil wet than preferentially water
wet. However, this assertion cannot be generalized because the 55 reservoirs
used in this study were not obtained by random sampling. A random sample
of reservoirs would be needed if the results of the wettability tests are to be
given statistical significance.
A similar contact angle study by Chiligarian and Chen (1983) on 161
carbonate reservoirs showed 80% of the reservoirs to be preferentially oil wet,
8% to be preferentially water wet and 12% to be of intermediate wettability.
These results are consistent with those of Treibel et al.(1972).
6.3.4 Effect of Wettability on Rock -Fluid Interactions
Wettability has a profound effect on multiphase rock-fluid interactions.
Wettability affects (a) the microscopic fluid distribution at the pore scale in the
porous medium, (b) the magnitude of the irreducible water saturation, (c) the
efficiency of an immiscible displacement in the porous medium, (d) the
residual oil saturation, (e) the capillary pressure curve of the porous medium,
(f) the relative permeability curves of the porous medium and (g) the electrical
properties of the porous medium.
Microscopic Fluid Distribution at the Pore Scale
6-46
Wettability determines the microscopic fluid distribution in a porous
medium at the pore scale. The wetting fluid occupies the small pores, coats
the surface of the solid grains and occupies the corners of the grain contacts.
The wetting phase occupies the small pores, which have high specific surface
areas (S=3(1-φ)/r) in order to minimize the specific surface free energy of the
system. The nonwetting phase occupies the large pores and are located at the
center of the pores. These pore scale fluid distributions are shown
schematically in Figure 6.27 for water wet and oil wet porous media. For the
water wet medium, at the initial state, water being the wetting phase coats the
grain surface and occupy the nooks and crannies of the medium. Oil, being
the nonwetting phase occupies the center of the pores and is surrounded by
water. After waterfooding, the residual oil globules occupy the center of the
pores. For the oil wet medium, oil being the wetting phase coats the grain
surface and occupy the nooks and crannies of the medium. The water, being
the nonwetting phase occupies the center of the pores and is surrounded by
oil. After waterflooding, the water occupies the center of the pores and the
residual oil wets the grain surface and occupies the nooks and crannies of the
medium. These microscopic fluid arrangements have implications for the
nature of the end point relative permeabilities of a water wet rock and an oil
wet rock.
Effect of Wettability on Irreducible Water Saturation
It has been observed that the irreducible water saturation in an oil wet
reservoir rock tends to be less than in a water wet reservoir rock. Craig (1971)
gives the following rule-of-thumb for irreducible water saturation for water
wet and oil wet reservoirs. For water wet reservoirs, the irreducible water
saturation is usually greater than 20 to 25% whereas for oil wet reservoirs it
is generally less than 15% and frequently less than 10% of the pore volume.
6-47
Figure 6.27. Fluid distributions as a function of wettability (adapted from Pirson, 1958).
Effect of Wettability on Electrical Properties of Rocks
Wettability affects the saturation exponent, n, in Archie's resistivity
index equation. For water wet rocks, the exponent is typically around 2.
However, for oil wet rocks, the exponent can increase to rather high values as
the water saturation decreases. Table 6.6 shows the result of laboratory
measurements of n as a function of water saturation in an oil wet sandstone.
It can be observed that below a certain water saturation, the saturation
exponent increases above the usual value of 2. An exponent as high as 9 was
measured in the experiments.
6-48
Table 6.6. Archie Saturation Exponent in Oil Wet Rocks (Mungan and Moore,
1968).
Figure 5.28 shows the effect of wettability on the resistivity index of
carbonate cores. In this study, the cores were rendered preferentially water
wet by heating up to 500 ºF and preferentially oil wet by washing with an
organic acid. Measurements were also made on cores that were of neutral
wettability. The wettability classification was based on imbibition tests. The
equations relating resistivity index to water saturation for neutral and
preferentially water wet cores are and , giving saturation
exponents of 1.92 and 1.61, respectively. The data for the preferentially oil wet
cores separated into two distinct trends described by the equations
for the first trend and for the second trend, giving
saturation exponents of 12.27 and 8.09. The separation of the oil wet data
was attributed to differences in the pore size distributions of the cores.
1.92wI S −=
0.37I =
1.61wI S −=
12.270.000027 wI −= S 8.09wS −
6-49
Figure 6.28. Effect of wettability on the resistivity index of carbonate cores (Sweeney and Jennings, 1960).
It is reasonable to expect that wettability will affect the resistivity and
hence the saturation exponent of a partially saturated porous medium. In an
oil wet medium, water being the nonwetting phase occupies the center of the
large pores. At high water saturations, the water is continuous and therefore
conducts electrical current. As the water saturation is decreased, below a
certain water saturation, the water will breakup into disconnected globules
and can no longer conduct electrical current. As a result, the resistivity of the
system will increase and this increase in resistivity is reflected in the increase
in the water saturation exponent as observed in the experiments.
6-50
Effect of Wettability on the Efficiency of an Immiscible Displacement
Wettability has a significant effect on the efficiency of an immiscible
displacement in a porous medium. Figure 6.29 shows schematically the
microscopic displacement of oil from a water wet medium and an oil wet
medium at the pore scale. In the water wet medium, the injected water is
imbibed into the medium along the pore walls in a manner that enhances the
oil displacement efficiency. The residual oil is trapped at the center of the
large pores. In the oil wet medium, the injected water channels through the
large pores leaving behind considerable residual oil in the small pores, at the
solid contacts and as coatings on the solid grains. From this pore level
picture, it is easy to see that the waterflood efficiency will be higher in the
water wet medium than in the oil wet medium everything else being equal.
The higher waterflood efficiency of the water wet rock compared to the
oil wet rock seen at the pore scale manifests itself at the macroscopic scale
(core scale) as well. Owens and Archer (1971) performed waterflood
experiments in core plugs (1.9 cm diameter and 4.4 cm length) at various
wettability conditions. The core plugs were rendered progressively oil wet by
dissolving a sulfonate in the oil phase. Figure 6.30 shows the oil recovery
curves for the waterfloods as a function of the wettability of the core. The
decline in the oil recovery efficiency with increasing oil wetness is obvious.
6-51
Figure 6.29. Microscopic displacement of oil from a pore during a waterflood: (a) strongly water wet medium, (b) strongly oil wet medium (Raza et al., 1968).
6-52
Figure 6.30. Effect of wettability on waterflood performance at an oil-water
viscosity ratio of 5 (Archer and Owens, 1971).
Peters and Hardham (1989) have conducted similar waterflood
experiments in unconsolidated sandpacks at a larger scale (4.8 cm diameter
and 54 cm length) than core plugs. The results of two such experiments are
presented here. In Experiment 1, the sandpack was first saturated with brine
and the brine was displaced by a viscous silicon-based test oil (103.4 cp) to
establish irreducible water saturation in contact with the sand grains. The
viscous oil was then displaced by brine to simulate a waterflood at an
unfavorable viscosity ratio of 85. In Experiment 2, a second sandpack was
6-53
first saturated with the same viscous silicon-based oil and the oil was then
displaced by the same brine to simulate a waterflood at the same unfavorable
mobility ratio as in Experiment 1. Although the wettabilities of the sandpacks
were not measured directly, it is believed that the first sandpack would
behave as a water wet system whereas the second sandpack would behave as
an oil wet system over the short time scale of the experiments. Both
waterfloods were imaged by X-ray CT to visualize the insitu fluid saturations
in time and space.
The oil recovery curves for the two waterflood experiments are shown in
Figure 6.31. They show the displacement in the water wet sandpack to be
more efficient than in the oil wet sandpack . These results are in agreement
with those of Archer and Owens. The low water breakthrough recoveries in
this study are due to the high oil-water viscosity ratio of 85.
Figures 6.32 and 6.33 show the water saturation images for the two
waterfloods at several pore volumes injected. The images for the water wet
sandpack show a relatively uniform and efficient displacement of the oil by
the water, with relatively high water saturations. In contrast, the images for
the oil wet sandpack show a chaotic, fragmented and inefficient displacement,
with relatively low water saturations. These images clearly show the important
role of wettability in determining the efficiency of waterfloods at the
macroscopic scale. Figures 6.34 and 6.35 show the water saturation profiles
for the experiments. They confirm the higher displacement efficiency of
Experiment 1 compared to Experiment 2.
6-54
Figure 6.31. Effect of wettability on waterflood performance at an oil-water
viscosity ratio of 91 (Peters and Hardham, 1989).
6-55
6-56
6-57
Figure 6.32. Water saturation images for a waterflood in a water wet sandpack at a viscosity ratio of 91. (A) tD = 0.05, (B) tD = 0.10, (C ) tD = 0.25, (D) tD = 0.50, (E) tD = 1.0, (F) tD = 2.0, (G) tD = 3.0 (Peters and Hardham, 1989).
6-58
6-59
6-60
6-61
Figure 6.33. Water saturation images for a waterflood in an oil wet sandpack at a viscosity ratio of 91: (A) tD = 0.05, (B) tD = 0.10, (C ) tD = 0.25, (D) tD = 0.50, (E) tD = 1.0, (F) tD = 2.0, (G) tD = 3.0 (Peters and Hardham, 1989).
6-62
Figure 6.34. Water saturation profiles for a waterflood in a water wet
sandpack.
Figure 6.35. Water saturation profiles for a waterflood in an oil wet sandpack.
6-63
6.4 THERMODYNAMICS OF INTERFACES
6.4.1 Characterization of Interfacial Tension as Specific Surface Free Energy
Consider a closed system consisting of the interface between two
immiscible fluids or between an immiscible fluid and a solid surface. The first
law of thermodynamics applied to this system gives
dU dQ dW= − (6.38)
where dU is the change in total internal energy of the system, dQ is the heat
input into the system and dW is the work done by the system. For an
equilibrium system, infinitesimal changes are reversible so that the reversible
work is given by
dW PdV daσ= − (6.39)
where da is the interfacial area. For a closed system,
dQ TdS= (6.40)
Substitution of Eqs.(6.39) and (6.40) into Eq.(6.38) gives for a closed system,
dU TdS PdV daσ= − + (6.41)
For an open system, Eq.(6.41) becomes
(6.42) 1
i N
i ii
dU TdS PdV da dnσ μ=
=
= − + + ∑
where μi is the chemical potential or molal free energy of component i in the
system, ni is the moles of component i in the system and N is the total
6-64
number of components in the system. At equilibrium, μi is the same in the
bulk fluid and at the interface.
We may obtain the interfacial tension from Eq.(6.42) in terms of the
internal energy of the system as
, , iS V n
Ua
σ ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠ (6.43)
The interfacial tension also can be expressed in terms of Helmholtz Free
Energy (A) and Gibbs Free Energy (G). The Helmholtz free energy is defined by
A U TS= − (6.44)
Differentiating Eq.(6.44) gives
dA dU TdS SdT= − − (6.45)
Substituting Eq.(6.42) into Eq.(6.45) gives
(6.46) 1
i N
i ii
dA SdT PdV da dnσ μ=
=
= − − + + ∑
We may obtain the interfacial tension from Eq.(6.46) in terms of the Helmholtz
free energy of the system as
, , iT V n
Aa
σ ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠ (6.47)
Gibbs free energy is defined by
(6.48) G A PV U TS PV H TS= + = − + = −
6-65
where H is the enthalpy of the system. Differentiating Eq.(6.48) gives
(6.49) dG dU VdP PdV TdS SdT= + + − −
Substituting Eq.(6.42) into Eq.(6.49) gives
(6.50) 1
i N
i ii
dG VdP SdT da dnσ μ=
=
= − + + ∑
We may obtain the interfacial tension from Eq.(6.50) in terms of Gibbs free
energy of the system as
, , iT P n
Ga
σ ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠ (6.51)
In general, changes in P and V accompanying surface changes are small.
6.4.2 Characterization of Microscopic Pore Level Fluid Displacements
We wish to examine the direction of energy change during an
immiscible displacement at the pore scale. The Helmholtz free energy of the
system will change as the displacement progresses. Our objective is to
determine whether the displacement will lead to a decrease or an increase in
the Helmholtz free energy of the system. A system in equilibrium always seeks
to minimize its free energy. Therefore, a displacement that leads to a decrease
in the free energy of the system is favored. Such a displacement will occur
spontaneously (without pumping) given the chance. On the other hand, a
displacement that leads to an increase in the free energy of the system is not
favored and will not occur spontaneously. Such a displacement will have to be
forced by pumping the displacing fluid.
Case 1. Displacement of a Nonwetting Phase by a Wetting Phase
6-66
Consider the displacement of a nonwetting phase by a wetting phase,
such as water displacing oil in a water wet medium, at the pore level as
shown in Figure 6.36. The interfacial forces and interfacial areas at an instant
are also shown in Figure 6.36. The Helmholtz free energy of the system is a
function of the interfacial areas. Thus,
( ), ,sw so woA f a a a= (6.52)
Let the interface move to the right by a small distance dx. The change in the
Helmholtz free energy of the system as the interfacial areas change is given by
sw sosw so wo
A A AdA da da daa a a∂ ∂ ∂
= + +∂ ∂ ∂ wo (6.53)
Figure 6.36. Displacement of a nonwetting phase by a wetting phase at the pore scale.
Suppose the wetting–nonwetting phase interface maintains the same shape
during the displacement. Then
6-67
0woda = (6.54)
Also, from the geometry of Figure 6.36,
2 swda rdx increase in the areaπ= + = (6.55)
2 soda rdx decrease in the areaπ= − = (6.56)
where r is the radius of the pore. The interfacial forces are given in terms of
the Helmholtz free energy by Eq.(6.47). Substituting Eqs.(6.47), (6.54), (6.55)
and (6.56) into Eq.(6.53) gives the change in Helmholtz free energy of the
system as
(2 )so swdA rdx
π σ σ= − − (6.57)
For the wetting–nonwetting interface to move to the right requires an
imbalance in the interfacial forces given by
cosso sw woσ σ σ> + θ (6.58)
Eq.(6.58) can be rewritten as
( ) cosso sw woσ σ σ− > θ (6.59)
The right side of Eq.(5.59) is a positive number. The left side of Eq.(6.59) is
larger than this positive number. When these facts are applied to Eq.(6.57),
we see that the change in the Helmholtz free energy during the displacement
is negative. This means that the Helmholtz free energy will decrease as the
nonwetting phase is displaced by the wetting phase. This is a favored
displacement. In fact, the displacement will occur spontaneously if given the
opportunity to do so. This is the origin of spontaneous imbibition, which
6-68
explains the spontaneous imbibition of a wetting fluid in the capillary tube
experiment. We can calculate an effective displacing force as
0effectivedAFdx
= − > (6.60)
Thus, an effective displacing force develops spontaneously to enable the
wetting phase to displace the nonwetting phase. Such a force will develop
spontaneously to enable water to displace oil in a water wet porous medium.
Case 2. Displacement of a Wetting Phase by a Nonwetting Phase
Consider the displacement of a wetting phase by a nonwetting phase,
such as water displacing oil in an oil wet medium, at the pore level as shown
in Figure 6.37. Let the interface move to the right by a small distance dx.
Because the nonwetting phase never contacts the solid,
0swda = (6.61)
Because the wetting phase is always in contact with the solid,
0soda = (6.62)
The change in the wetting–nonwetting phase interfacial area is given by
(6.63) *2woda r dxπ= +
6-69
Figure 6.37. Displacement of a wetting phase by a nonwetting phase at the pore scale.
where r* is the radius from the center of the pore to the thin wetting phase
film on the surface of the pore. Substituting Eqs.(6.47), (6.61), (6.62) and
(6.63) into Eq.(6.53) gives the change in the Helmholtz free energy of the
system as
2 wodA rdx
π σ= + (6.64)
Eq.(6.64) shows that the Helmholtz free energy of the system increases during
the displacement. This is not a favored displacement. Thus, the displacement
will not occur spontaneously. It must be forced by pumping. The effective
displacement force in this case is given by
0effectivedAFdx
= − < (6.65)
Thus, a negative displacement force arises to oppose the displacement of the
wetting phase by the nonwetting phase. This is why a positive displacement
6-70
pressure is required in order to initiate a drainage capillary pressure
measurement.
To minimize the increase in free energy of the system during the
displacement, r* will be as small as possible. Thus, the injected nonwetting
phase will channel or finger through the wetting phase leaving behind a
significant wetting phase film and residual wetting phase saturation. This
means that the waterflood efficiency of an oil wet medium will be less than the
waterflood efficiency of a water wet medium. This observation is in agreement
with the microscopic picture of the displacements shown in Figure 6.29.
NOMENCLATURE
a = interfacial area
A = Helmholtz free energy
F = force
g = gravitational acceleration
G = Gibbs free energy
h = equilibrium height in a capillary rise experiment
H = enthalpy
M = molecular weight
Mg = apparent molecular weight of gas
ML = apparent molecular weight of liquid
ni = moles of component i
N = total number of components in the mixture
Nc = capillary number
P = pressure
Pc = capillary pressure
Pnw = pressure in the nonwetting phase
Pw = pressure in the wetting phase
r = radius of capillary tube, pore radius, radius of spinning drop
Q = heat
6-71
R = oil recovery
S = entropy
Ss = surface entropy
Sor = residual oil saturation
T = temperature
U = internal energy
v = Darcy velocity
V = volume
W = work
xi = mole fraction of component i in the liquid
yi = mole fraction of component i in the gas
σ = surface or interfacial tension
osσ = oil solid interfacial tension (oil solid specific surface energy)
owσ = oil water interfacial tension (oil water specific surface energy)
wsσ = water solid interfacial tension (water solid specific surface energy)
θ = contact angle
θA = advancing contact angle
θR = receding contact angle
Lρ = saturated liquid density
gρ = saturated gas density
wρ = wetting phase density
nwρ = nonwetting phase density
iμ = chemical potential of component i
wμ = viscosity of wetting phase
nwμ = viscosity of nonwetting phase
Λ = parachor
Π = spreading pressure
6-72
REFERENCES AND SUGGESTED READINGS
Abrams, A. : "The Influence of Fluid Viscosity, Interfacial Tension, and Flow Velocity on Residual Oil Saturation Left by Waterflood," Soc. Pet. Eng. Jour. (Oct., 1975) 437-447.
Adamson, A.W. and Gast, A.P.: Physical Chemistry of Surfaces, Sixth Edition, John Wiley and Sons, Inc., New York, 1997.
Amott, E. : “Observations Relating to Wettability of Porous Rock,” Trans., AIME (1959) 216, 156-162.
Anderson, W.G. : “Wettability Literature Survey - Part 1: Rock/Oil/Brine Interactions and the Effects of Core Handling on Wettability,” J. Pet. Tech. (October 1986) 1125-1144.
Anderson, W.G. : “Wettability Literature Survey - Part 2: Wettability Measurement,” J. Pet. Tech. (November 1986) 1246-1262.
Anderson, W.G. : “Wettability Literature Survey - Part 3: The Effects of Wettability on the Electrical Properties of Porous Media,” J. Pet. Tech. (December 1986) 1371-1378.
Anderson, W.G. : “Wettability Literature Survey - Part 4: Effects of Wettability on Capillary Pressure,” J. Pet. Tech. (October 1987) 1283-1300.
Anderson, W.G. : “Wettability Literature Survey - Part 5: The Effects of Wettability on Relative Permeability,” J. Pet. Tech. (November 1987) 1453-1468.
Anderson, W.G. : “Wettability Literature Survey - Part 6: The Effects of Wettability on Waterflooding,” J. Pet. Tech. (December 1987) 1605-1622.
Archer, J.S. and Wall, C.G. : Petroleum Engineering, Graham & Trotman, London, England, 1986.
Bear, J. : Dynamics of Fluids in Porous Media, Elsevier, New York, 1972.
Benner, F.C. and Bartell, F.E. : “The Effect of Polar Impurities Upon Capillary and Surface Phenomena in Petroleum Production,” API Drilling and Production Practice (1941) 341-348.
Bobek, J.E., Mattax, C.C. and Denekas, M.O. : “Reservoir Rock Wettability - Its Significance and Evaluation,” Trans., AIME (1958) 213, 155-160.
Chatzis, I. and Morrow, N.R. : “Correlation of Capillary Number Relationships for Sandstones,” SPE 10114, Presented at the 56th Annual Fall Technical Conference and Exhibition of the Society of Petroleum Engineers, San Antonio, October 5-7, 1981.
Chilingar, G.V. and Yen, T.F. : “Some Notes on Wettability and Relative Permeabilities of Carbonate Rocks,” Energy Sources , vol. 7, No. 1 (1983) 67-75.
6-73
Choquette, P.W. and Pray, L.C. : “Geologic Nomenclature and Classification of Porosity in Sedimentary Carbonates,” AAPG Bull., Vol. 54, No. 2 (1970) 207-250.
Collins, R.E. : Flow of Fluids Through Porous Materials, Van Nostrand Reinhold Company, 1961. Reprinted by the Petroleum Publishing Company, 1976. Reprinted by Research & Engineering Consultants Inc., 1990.
Craig, F.F., Jr. : The Reservoir Engineering Aspects of Waterflooding, SPE Monograph Vol. 3, Society of Petroleum Engineers, Richardson, Texas, 1971.
Cuiec, L.E. : “Study of Problems Related to the Restoration of Natural State of Core Samples,” J. Canadian Pet. Tech (Oct. - Dec. 1977) 68-80.
de Gennes, P.G. and Quere, D. : Capillarity and Wetting Phenomena, Springer Science and Business Media, Inc., 2004.
Denekas, M.O., Mattax, C.C. and Davis, G.T. : “Effects of Crude Oil Components on Rock Wettability,” Trans., AIME (1959) 216, 330-333.
Donaldson, E.C., Thomas, R.D. and Lorenz, P.B. : “Wettability Determination and Its Effect on Recovery Efficiency,” Soc. Pet. Eng. J. (March 1969) 13-20.
Donaldson, E.C., Kendall, R.F., Pavelka, E.A. and Crocker, M.E. : “Equipment and Procedures for Fluid Flow and Wettability Tests of Geological Materials,” DOE/BETC/IC-79/5, Nat. Tech. Info. Sv, Springfield, VA 2216, 1980.
Garnes, J.M., Mathisen, A.M., Scheie, A. and Skauge, A. : "Capillary Number Relations for Some North Sea Reservoir Sandstones," SPE/DOE 20264, Presented at the SPE/DOE Seventh Symposium on Enhanced Oil Recovery, Tulsa, April 22-25, 1990.
Howard, J.J. : "Wettability and Fluid Saturations Determined From NMR T1 Distribution," Magnetic Resonance Imaging, " Vol. 12, No. 2 (1994) 197-200.
Jennings, H.Y. : “Surface Properties of Natural and Synthetic Porous Media,” Producers Monthly (March 1957) 20-24.
Lake, L.W. : Enhanced Oil Recovery, Prentice Hall, Englewood Cliffs, New Jersey, 1989.
Marle, C.M. : Multiphase Flow in Porous Media, Gulf Publishing Company, Houston, Texas, 1981.
Melrose, J.C. : "Interfacial Phenomena as Related to Oil Recovery Mechanisms," Cnd J. Chem. Eng., Vol. 48 (Dec. 1970) 638-644.
Morrow, N.R. : “Wettability and Its Effect on Oil Recovery,” J. Pet. Tech. (December 1990) 1476-1484.
6-74
Morrow, N.R., Cram, P.J. and McCaffery, F.G. : "Displacement Studies in Dolomite With Wettability Control by Octanoic Acid," SPEJ (August 1973) 221-232.
Mungan, N. : “Enhanced Oil Recovery Using Water as a Driving Fluid; Part 2 - Interfacial Phenomena and Oil Recovery: Wettability,” World Oil (March 1981) 77-83.
Mungan, N. : “Enhanced Oil Recovery Using Water as a Driving Fluid; Part 3 - Interfacial Phenomena and Oil Recovery: Capillarity,” World Oil (May 1981) 149-158.
Mungan, N. and Moore, E.J. : "Certain Wettability Effects on Electrical Resisitivity in Porous Media," J. Cdn. Pet. Tech. (Jan.-March 1968) 7, No.1, 20-25.
Owens, W.W. and Archer, D.L. : “The Effect of Rock Wettability on Oil-Water Relative Permeability Relationships,” J. Pet. Tech. (July 1971) 873-878.
Peters, E. J. and W. D. Hardham: “A Comparison of Unstable Miscible and Immiscible Displacements,” SPE 19640, Proceedings of the 64th Annual Technical Conference of the Society of Petroleum Engineers (October 1989) San Antonio.
Peters, E.J. and Hardham, W.D. : “Visualization of Fluid Displacements in Porous Media Using Computed Tomography Imaging,” Journal of Petroleum Science and Engineering, 4, No. 2, (May 1990) 155-168.
Pirson, S.J. : Oil Reservoir Engineering, McGraw-Hill Book Company, Inc., New York, 1958.
Raza, S.H., Treiber, L.E. and Archer, D.L. : “Wettability of Reservoir Rocks and Its Evaluation,” Producers Monthly (April 1968) 2-7.
Rowlinson, J.S. and Widom, B. : Molecular Theory of Capillarity, Dover Publications, Inc., Mineola, New York, 1982.
Sweeny, S.A. and Jennings, H.Y. : "Effect of Wettability on the Electrical Resistivity of Carbonate Rock from a Petroleum Reservoir," J. Phy. Chem. (May 1960) 64, 551-553.
Tiab, D. and Donaldson, E.C. : Petrophysics, Second Edition, Elsevier, New York, 2004.
Treiber, L.E., Archer, D.L. and Owens, W.W. : “A Laboratory Evaluation of the Wettability of Fifty Oil-Producing Reservoirs,” Soc. Pet. Eng. J. (December 1972) 531-540.
Wagner, O.R. and Leach, R.O. : “Improving Oil Displacement Efficiency by Wettability Adjustment,” Trans., AIME (1959) 216, 65-72.
6-75
6-76
Welge, H.J. and Bruce, W.A. : “A Restored-State Method for Determination of Oil in Place and Connate Water,” API Drilling and Production Practice (1947) 161-165.
Willhite, G. P. : Waterflooding, SPE Textbook Series Vol. 3, Society of Petroleum Engineers, Richardson, Texas, 1986.
7-1
CHAPTER 7
CAPILLARY PRESSURE
7.1 DEFINITION OF CAPILLARY PRESSURE
When two immiscible fluids are in contact, there is a pressure
discontinuity between the two fluids which depends upon the curvature of the
interface separating the two fluids. This pressure difference or excess
pressure is known as the capillary pressure. The pressure on the concave side
of the interface is higher than that on the convex side of the interface. Figure
7.1 shows a curve interface between to immisicible fluids labeled 1 and 2. The
pressure P2 is greater than P1. The capillary pressure is given by Laplace
equation (sometimes referred to as Young-Laplace equation) as
2 11 2
1 1cP P P
r rσ
⎛ ⎞= − = +⎜ ⎟
⎝ ⎠ (7.1)
In Eq.(7.1), r1 and r2 are referred to as the principal radii of curvature of the
interface. They are mutually perpendicular. The curvature of the interface is
given by
7-2
1 2
1 1Curvaturer r
⎛ ⎞= +⎜ ⎟
⎝ ⎠ (7.2)
Figure 7.1. Equilibrium at a curved interface between two immiscible fluids.
7-3
Laplace equation can be derived by considering the mechanical
equilibrium of the interface or by energy considerations. We derive it here by
energy considerations. Let the interface be expanded by a small amount in the
xyz directions, where z is vertically upward. The increase is surface area of the
interface is given by
( )( )da x dx y dy xy xdy ydx dxdy xdy ydx= + + − = + + = + (7.3)
where dxdy is considered negligibly small. The increase in surface energy is
given by
( )dE xdy ydxσ= + (7.4)
The work done in increasing the interfacial area is given by
( )2 1W P P xydzδ = − (7.5)
At equilibrium, the work done is equal to the increase in surface energy.
Thus,
( ) ( )2 1P P xydz xdy ydxσ− = + (7.6)
From similar triangles (Figure 7.2),
1 1
x dx xr dz r
+=
+ (7.7)
2 2
y dy yr dz r
+=
+ (7.8)
Substituting Eqs.(7.7) and (7.8) into Eq.(7.6) and rearranging gives
7-4
2 11 2
1 1cP P P
r rσ
⎛ ⎞= − = +⎜ ⎟
⎝ ⎠ (7.1)
which is Laplace equation. A mean radius of curvature, rm, may be defined as
1 2
1 1 1 12mr r r
⎛ ⎞= +⎜ ⎟
⎝ ⎠ (7.9)
In terms of the mean radius of curvature, Laplace equation becomes
2c
m
Prσ
= (7.10)
Figure 7.2. Fluid interface in the planes of the principal radii of curvature.
7-5
Laplace equation is the fundamental equation of capillarity. Several
special cases of Laplace equation are of interest. For a spherical liquid drop,
r1 = r2 = r, the radius of the drop. The capillary pressure or the excess
pressure of the drop is given by
2cP
rσ
= (7.11)
For a soap bubble in air, r1 = r2 = r, the radius of the bubble. The capillary
pressure or excess pressure is given by
4cP
rσ
= (7.12)
where a factor of 2 has been incorporated to account for the two gas-liquid
interfaces of a soap bubble. For a flat interface, r1 = r2 = ∞. In this case, the
capillary pressure is zero. If the two immiscible fluids are in contact with a
solid surface, the interface will intersect the solid at an equilibrium contact
angle θ given by the Young-Dupre equation. Such an interface is shown in
Figure 7.3 for the capillary rise experiment. Laplace equation holds at the
interface. Assuming the interface lies on a sphere as shown in the figure, then
r1 = r2 = r/cosθ, where r is the radius of the capillary tube. The capillary
pressure is given by
1 2
1 1 cos cos 2 coscP
r r r r rθ θ σ θσ σ
⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
(7.13)
7-6
Figure 7.3. Interface for capillary rise experiment.
For a pendular ring of wetting fluid at the contacts of two spherical
sand grains in an idealized porous medium consisting of a cubic pack of
uniform spheres as shown in Figure 7.4, the capillary pressure is given by
Laplace equation as
1 2
1 1coscPr r
σ θ⎛ ⎞
= −⎜ ⎟⎝ ⎠
(7.14)
7-7
In this case, the principal radii of curvature are on opposite sides of the
interface. By sign convention, one radius will be positive and the other will be
negative.
Figure 7.4. Immiscible fluid interface in an idealized porous medium.
If the wetting fluid saturation in the pendular ring is reduced, r1 and r2 will be
reduced. However, r1 will be reduced more than r2 as the wetting phase
recedes into the corners of the contact of the grains. As a result, the capillary
pressure will increase. If the wetting fluid saturation is increased, r1 and r2
will increase and the capillary pressure will decrease. Therefore, an inverse
relationship exists between the capillary pressure and the wetting phase
saturation for a porous medium. Low wetting phase saturation corresponds
7-8
to high capillary pressure and high wetting phase saturation corresponds to
low capillary pressure.
7.2 CAPILLARY PRESSURE - SATURATION RELATIONSHIP FOR A POROUS MEDIUM
Before considering the capillary pressure versus saturation relationship
for a porous medium, it is instructive to consider the relationship for an
idealized medium consisting of a bundle of capillary tubes of varied radii. In
this case, the capillary pressure versus wetting phase saturation relationship
can be calculated. Let the bundle of capillary tubes medium be dipped into
the wetting phase and allowed to attain capillary equilibrium as shown in
Figure 7.5. The wetting fluid will rise to a different elevation (z) above the free
wetting fluid level in each tube depending on its radius as shown in the figure.
Let the model consist of ten capillary tubes with their radii as shown in Table
7.1. Let the wetting fluid be water with a surface tension of 72 dynes/cm and
a contact angle of 0 with the solid. The non-wetting phase is air. The
equilibrium height of water in each capillary tube can be calculated with
Eq.(6.18) as
2 cos 0.1468
w
z cmr g rσ θ
ρ= = (7.15)
The capillary pressure in each tube is given by Laplace equation as
22 cos 144 /cP dynes cmr r
σ θ= = (7.16)
The wetting phase saturation as a function of elevation z is calculated from
the dimensions of the capillary tubes and is presented in Table 7.1.
7-9
Figure 7.5. Capillary rise experiment for a bundle of capillary tubes medium.
Table 7.1. Capillary Pressure versus Wetting Phase Saturation for Bundle of
Capillary Tubes Model.
Radius Pc Pc Pc z Volume Volume Water
(μm) (dynes/cm2
)
(atm) (psi) cm (cm3) Fraction Saturation
0.00 1.000
100 14400.0 0.014 0.209 14.68 0.0046 0.182 0.818
90 16000.0 0.016 0.232 16.31 0.0042 0.164 0.655
80 18000.0 0.018 0.261 18.35 0.0037 0.145 0.509
70 20571.4 0.020 0.298 20.97 0.0032 0.127 0.382
60 24000.0 0.024 0.348 24.46 0.0028 0.109 0.273
50 28800.0 0.028 0.418 29.36 0.0023 0.091 0.182
40 36000.0 0.036 0.522 36.70 0.0018 0.073 0.109
30 48000.0 0.047 0.696 48.93 0.0014 0.055 0.055
20 72000.0 0.071 1.044 73.39 0.0009 0.036 0.018
10 144000.0 0.142 2.088 146.79 0.0005 0.018 0.000
7-10
0.0254
Figures 7.6 and 7.7 show the capillary pressure versus wetting phase
saturation for this idealized medium. In Figure 7.6, the capillary pressure is
presented as height in cm above the free water level whereas in Figure 7.7,
the capillary pressure is given in psi. Both presentations are valid and can be
used for different purposes. The presentation in Figure 7.6 gives the water
saturation distribution as a function of the height above the free water level.
This type of presentation can be used to determine the water saturation
distribution in a petroleum reservoir starting from the free water level at or
below the oil water contact to the top of the reservoir. The presentation in
Figure 7.7 is useful for calculating pore size distribution.
The capillary pressure versus saturation relationship for the idealized
porous medium shown in Figures 7.6 and 7.7 captures the general trend of
capillary pressure curves for porous media. The features include a non-zero
displacement pressure at a wetting phase saturation of 1 and the inverse
relationship between capillary pressure and wetting phase saturation
mentioned previously. The capillary pressure curves have a stair-case shape
in this case because of the limited number and size of the capillary tubes
used in the experiment. The curve will approach a smooth curve if more tubes
are used and the differences in the tube diameters are made small. The only
limitation of the capillary pressure curves for the bundle of capillary tube
model is the absence of an irreducible wetting phase saturation. There is no
possibility of trapping an irreducible saturation for a model consisting of
straight and isolated capillary tubes.
7-11
Figure 7.6. Capillary pressure expressed as height of water above the free water level versus wetting phase saturation for a bundle of capillary tubes
medium.
Figure 7.7. Capillary pressure in psi versus wetting phase saturation for a bundle of capillary tubes medium.
7-12
Figure 7.8 shows capillary rise experiments for two porous media
having different grain sizes (pore sizes). The wetting phase will rise higher in
the finer grain porous medium than in the courser grain medium. Such
experiments are widely used in soil science to determine the capillary
pressure curve (referred to as water retention curve or matric suction head in
soil science) for unconsolidated soils. The soil, which is packed in a tube is
dipped into water and allowed to sit for several days or weeks to achieve
capillary equilibrium. The tube often is instrumented to measure the
resistivity of the medium in order to calculate the water saturation along the
column by Archie's equation. The capillary pressure is calculated as height of
water above the free water level as was done in the bundle of capillary tube
model.
Figure 7.8. Capillary rise experiments for two porous media of different grain sizes.
7-13
In an actual porous medium, such as reservoir rock, the complexity of
the pore structure and the fluid interface arrangements preclude the use of
Laplace equation to calculate the capillary pressure. Further, this complexity
also precludes the calculation of the wetting phase saturation from the fluid
interface arrangements as was done for the bundle of capillary tubes
experiment. Instead, the capillary pressure versus wetting phase saturation
relationship is measured experimentally. We demonstrate one possible way of
making this measurement for a porous medium using the idealized porous
medium and the experimental set up shown in Figure 7.9. Let water be the
wetting phase and air the non-wetting phase in the experiment. The porous
medium consists of one pore, which is a capillary tube with three radii as
shown in the figure. The radius and length of each segment of the pore are
shown in Table 7.2. The medium, which is strongly water wet (θ = 0º), is
initially saturated with water. The medium rests on a semi-permeable plate at
the bottom of the apparatus. This semi-permeable plate is manufactured such
that it has very fine and uniform pores. It is strongly water wet and is fully
saturated with water. It will permit water to flow through it but because of its
fine pores will prevent the air from flowing through it. Initially, the apparatus
is open to the atmosphere so that the water in the core, the semi-permeable
plate and the connecting vessel below the semi-permeable plate is at
atmospheric pressure. Gas is admitted into the apparatus at a low pressure of
Pg. If Pg is less than 2σcosθ/r1, nothing will happen. The gas pressure is not
high enough for the gas to displace the water from the largest pore. The gas
pressure is then increased to Pg1 until it is equal to 2σcosθ/r1 and the segment
of the pore with radius r1 will be drained of water. The water drainage will
stop after draining the largest pore because Pg1 is not high enough to drain
the pore with radius r2. Next, the gas pressure is increased to Pg2 equal to
2σcosθ/r2 and the segment of the pore with radius r2 will be drained.
Eventually, the pressure of the gas is increased to Pg3 equal to 2σcosθ/r3 and
the segment of the pore with radius r3 will be drained. The volume of water
7-14
drained at each capillary pressure is measured and is used to calculate the
water saturation in the medium corresponding to that capillary pressure. The
graph of Pgi versus water saturation gives the capillary pressure curve of the
porous medium. For this simple porous medium, the capillary pressure curve
can be calculated using Laplace equation and the dimensions of the pore and
is presented in Figure 7.10. Note how the shape of the capillary pressure
curve reflects the pore size distribution of the porous medium. This is the
basis for estimating the pore size distribution of a porous medium from its
drainage capillary pressure curve.
Figure 7.11 shows the effect of wettability on the capillary pressure
curve for the idealized porous medium. It compares the capillary pressure
curve for a contact angle of 0º and a contact angle of 75º. Thus, if the medium
is less water wet, the magnitude of the capillary pressure will be less at each
wetting phase saturation than when it was more water wet. Since the
wettability preference of the rock for the water is less at a contact angle of 75º
than at 0º, less work is required to desaturate the rock. Therefore, the
capillary pressure that needs to be applied to desaturate the rock will be less
at a contact angle of 75º than at 0º at any saturation level.
7-15
Figure 7.9. Capillary pressure measurement for an idealized porous medium.
Table 7.2. Capillary Pressure Curve for Idealized Porous Meium.
Capillary Radius Length Volume Fractiona
l
Pc Pc Pc Water
Tube # (μm) (μm) (cm3) Volume (dynes/cm2
)
(atm) (psi) Saturation
1 10 2 6.283E-
10
0.5747 144000 0.142 2.088 0.4253
2 4 8 4.021E-
10
0.3678 360000 0.355 5.221 0.0575
3 1 20 6.283E-
11
0.0575 1440000 1.421 20.884 0.0000
1.093E-
09
7-16
Figure 7.10. Capillary pressure curve for an idealized porous medium.
7-17
Figure 7.11. Effect of wettability on the capillary pressure curve for an idealized porous medium.
7.3 DRAINAGE CAPILLARY PRESSURE CURVE
If the idealized porous medium of Figure 7.9 were replaced by an actual
porous medium and the experiment repeated, the capillary pressure curve
would look like the one shown in Figure 7.12. This figure shows a typical
drainage capillary pressure curve obtained by displacing the wetting phase
from a porous medium with a non-wetting phase. A process in which the
wetting phase saturation decreases is known as drainage whereas the
converse process in which the wetting phase saturation increases is known as
imbibition. The drainage capillary pressure curve has several characteristic
features. The curve shows that a minimum positive pressure (Pd) must be
applied to the non-wetting phase in order to initiate the drainage. This
minimum pressure, which is known as the displacement pressure, the
threshold pressure or the entry pressure, is determined by the size of the
largest pores connected to the surface of the medium. It can be estimated
with Laplace equation where r is the largest pore radius connected to the
surface. If the rock does not have a strong wettability preference for the
initially saturating fluid, then the displacement pressure will be zero. If the
rock has a strong preference for the displacing fluid, then no pressure is
required to initiate the displacement because it will occur spontaneously. In
this case, the capillary pressure will start at the initial fluid saturation of less
than 1. As the pressure of the non-wetting phase is increased, smaller and
smaller pores are invaded by the non-wetting fluid. Eventually, the wetting
phase becomes discontinuous and can no longer be displaced from the
medium by increasing the capillary pressure. Therefore, an irreducible
wetting phase saturation is achieved for the porous medium at a high
capillary pressure. At the irreducible wetting phase saturation, the capillary
pressure curve becomes nearly vertical. The irreducible wetting phase
7-18
saturation is a function of the grain size (pore size), the wettability of the
medium and the interfacial tension between the wetting and non-wetting
fluids.
Figure 7.12. A typical drainage capillary pressure curve.
What information does the capillary pressure curve for a reservoir rock
provide about the rock? If one reservoir rock has a higher permeability than
another, we know that the higher permeability rock will permit faster fluid
flow through it, everything else being equal, than the lower permeability rock.
Thus, the higher permeability rock is more desirable than the lower
permeability rock as a petroleum reservoir rock. If one reservoir rock has a
higher porosity than another, we know that the higher porosity rock will store
more reserves than the lower porosity rock. Therefore, the higher porosity
7-19
rock is a more desirable reservoir rock than the lower porosity rock. If one
rock has a higher capillary pressure at the same wetting phase saturation
than another, what can we say about the rocks? Is the rock with the higher
capillary pressure curve more desirable or less desirable as a petroleum
reservoir rock than the rock with the lower capillary pressure curve?
Figure 7.13 shows the drainage capillary pressure curves for four rocks:
A, B, C and D. Rock A has the least displacement pressure. Therefore, it has
the largest pores connected to the surface. Its capillary pressure curve
remains essentially flat as the wetting phase saturation is decreased from
100% to 60%. This means that many of the pores are invaded by the non-
wetting fluid at essentially the same capillary pressure. This indicates that A
has uniform pores or is well sorted. Rock A also has the least irreducible
wetting phase saturation, indicating that it has relatively larger grains and
pores than the other rocks. Rock B has a higher displacement pressure than
A. Therefore, it has smaller pores than A. The capillary pressure curve at the
high wetting phase saturations is relatively flat, indicating good sorting. Rock
B has a higher irreducible wetting phase saturation than A, which is
consistent with its finer grains and pores. Rock C is even more fine grained
than B because of its higher displacement pressure. The shape of its capillary
pressure curve shows that a higher capillary pressure is required at each
wetting phase saturation to desaturate the rock. This means that C has a
wider pore size distribution than A and B. Therefore, C is poorly sorted. It
has a higher irreducible water saturation than B, which is consistent with its
finer grains and pores. Rock D is extremely fine grained, extremely poorly
sorted and would be a very poor reservoir rock. This observation is based on
its very high displacement pressure, very steep capillary pressure curve and
very high irreducible wetting phase saturation. Without being told, one can
easily infer that this rock is essentially made of clay.
7-20
Since permeability is proportional to the square of the mean grain size
(pore size), it is easy to see that Rock A has the highest permeability, followed
by B, C and D in that order. From this discussion, we conclude that the rock
with the higher capillary pressure curve is a less desirable reservoir rock than
the one with the lower capillary pressure curve.
In general, the capillary pressure curve for a porous medium is a
function of (1) pore size, (2) pore size distribution, (3) pore structure, (4) fluid
saturation, (5) fluid saturation history, (6) wettability of the rock and (7)
interfacial tension of the fluids involved.
Figure 7.13. Capillary pressure curves for four different rocks.
7-21
7.4 CONVERSION OF LABORATORY CAPILLARY PRESSURE DATA TO RESERVOIR CONDITIONS
Typically, capillary pressure curves are measured in the laboratory
using fluids other than reservoir fluids. It is not uncommon to measure the
capillary pressure curves to be used for analyzing an oil-water reservoir using
air and water or mercury and air in the laboratory. When this is done, it
becomes necessary to convert the laboratory data to reservoir conditions.
This conversion is done using Laplace equation as follows. From Laplace
equation,
( ) ( )2 coslab
c labm
Pr
σ θ= (7.17)
( ) ( )2 cosreservoir
c reservoirm
Pr
σ θ= (7.18)
where rm is the mean radius of curvature of the interface in the rock at a
particular fluid saturation. Eliminating rm from Eqs.(7.17) and (7.18) gives
( ) ( ) ( )( )
coscos
reservoirc creservoir lab
lab
P Pσ θ
σ θ= (7.19)
This ability to scale the laboratory capillary pressure data to reservoir
conditions provides the flexibility for making laboratory capillary pressure
measurements with more convenient fluids than reservoir fluids.
7.5 AVERAGING CAPILLARY PRESSURE DATA
The capillary pressure curves for rock samples from the same reservoir
having different permeabilities will be different. It is often necessary to
average the capillary pressure data for cores from the same reservoir believed
7-22
to have the same pore structure in order to obtain one capillary pressure
curve that can be used for reservoir performance analysis. This averaging can
be done using the Leverett J-function, which is a dimensionless capillary
pressure function (Leverett, 1941).
The Leverett J-function can be derived by dimensional analysis as
follows. The capillary pressure curve of a porous medium is a function of
several variables as shown in Eq.(7.20).
( ), , cos , ,c w w nwkP f S gσ θ ρ ρφ
⎛ ⎞= Γ −⎜ ⎟
⎝ ⎠ (7.20)
where Pc is the capillary pressure, Sw is the wetting phase saturation, Γ is a
dimensionless pore structure function that accounts for such things as pore
size distribution, tortuosity, cementation and dead end pores, φ is the
porosity, σ is the interfacial tension, θ is the contact angle, k is the absolute
permeability of the porous medium, ρw is the wetting phase density, ρnw is the
non-wetting phase density and g is the gravitational acceleration. The wetting
phase saturation (Sw) and the pore structure function (Γ) are dimensionless
and should be set aside from the dimensional analysis until the end. We form
the dimensionless product with the remaining variables as
( ) ( )2
31 4cos dimensionless constantx
xx xw nw c
k g Pσ θ ρ ρφ
⎛ ⎞− =⎡ ⎤⎜ ⎟ ⎣ ⎦
⎝ ⎠ (7.21)
Carrying out the dimensional analysis yields the following solution to the
dimensional analysis problem:
7-23
( )
11
2 23 4
3
4
cos1 1
11 00 1w nw
c
xk
xx x
xg
xP
σ θ
φρ ρ
⎡ ⎤− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= +⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦
(7.22)
Let us choose x3 = 1 and x4 = 0. The corresponding dimensionless group is
given by
( ) ( )1
/cos
w nw g kρ ρ φπ
σ θ−
= (7.23)
Next, let us choose x3 = 0 and x4 = 1. The corresponding dimensionless group
is given by
2/
coscP k φ
πσ θ
= (7.24)
From the dimensional analysis, we can write
( ) ( )1
// , ,cos cos
w nwcw
g kP k f Sρ ρ φφ
σ θ σ θ−⎛ ⎞
= Γ⎜ ⎟⎝ ⎠
(7.25)
The dimensionless group given as π2 in Eq.(7.23) is the ratio of gravity force to
the capillary forces at the pore scale. At the pore scale, capillary forces
dominate the gravity force. Therefore, π2 will be small and can be neglected.
For example, for an air-water capillary pressure curve, σ = 72 dynes/cm, ρw =
1 g/cm3, θ = 0º, g = 981 cm/s2. For a porous medium of 1 darcy permeability
and a porosity of 25%, π2 is of the order of 710− . For a mercury-air capillary
pressure curve, it is of the order of 610− . Eq.(7.25) then becomes
7-24
( ) ( )2/ , ,
cosc
w wP k f S J Sφσ θ
= Γ = Γ (7.26)
where ( ),wJ S Γ is a dimensionless capillary pressure function known as the
Leverett J-function. Eq.(7.26) suggests that porous media that have the same
pore structure but different permeability and porosity will have the same
Leverett J-function. Therefore, if the different capillary pressure curves of the
porous media are rescaled as a Leverett J-function, they should plot as one
curve. This curve provides the means to average capillary pressure data.
Figure 7.14 shows the Leverett J-function for nine unconsolidated
sands with widely different permeabilities ranging from 0.057 to 2160 darcies.
It is remarkable that the data plot as one curve. Figure 7.15 shows the J-
function for a carbonate reservoir. If the porous media have different pore
structures, then the Leverett J-functions for the different rocks will be
different and will not plot as one curve as may be seen in Figure 7.16.
7-25
Figure 7.14. Leverett J function for unconsolidated sands (Leverett, 1941).
7-26
Figure 7.15. Leverett J-functions for a carbonate reservoir; (a) all cores; (b) limestone cores; (c) dolomite cores; (d) microgranular limestone cores; (e)
coarse-grained limestone cores (Brown, 1951).
7-27
Figure 7.16. Leverett J-functions for different rock types (Rose and Bruce,
1949).
7-28
7.6 DETERMINATION OF THE INITIAL STATIC RESERVOIR FLUID SATURATIONS BY USE OF DRAINAGE CAPILLARY PRESSURE CURVE
Initially, the petroleum reservoir was saturated with water before oil
migrated into the reservoir and displaced the water. This displacement of a
wetting phase by a non-wetting phase is simulated in the laboratory
measurement of the drainage capillary pressure curve. The final fluid
distribution in the reservoir is determined by the equilibrium between
capillary and gravitational forces.
Consider the static equilibrium for the water and oil. From
hydrostatics, for the water,
ww
dP gdz
ρ= − (7.27)
where is z is pointed vertically upwards. Similarly, for the oil,
oo
dP gdz
ρ= − (7.28)
Assuming the fluids are incompressible, Eqs.(7.27) and (7.28) can be
integrated to obtain
( ) ( )0w w wP z P gzρ= − (7.29)
and
( ) ( )0o o oP z P gzρ= − (7.30)
Subtracting Eq.(7.29) from (7.30) gives
7-29
( ) ( ) ( )0c c w oP z P gzρ ρ= + − (7.31)
We select as datum the free water level at which the capillary pressure is zero.
With this choice of datum, Eq.(7.31) becomes
( ) ( )c w oP z gz gzρ ρ ρ= − = Δ (7.32)
where Δρ is the density of water minus the density of oil. It is remarkable that
Eq.(7.32) is the same as Eq.(6.26) for the capillary rise experiment. The free
water level occurs at a depth do below the oil-water contact given by
do
Pdgρ
=Δ
(7.33)
where Pd is the displacement pressure of the capillary pressure curve. Thus,
the elevation above the oil-water contact of any particular saturation is given
by
c dP Phgρ
−=
Δ (7.34)
If the displacement pressure of the capillary pressure curve is 0, then the free
water level and the oil-water contact will be the same. However, this is a
special case. In general, the free water level and the oil-water contact are not
the same.
Figure 7.17 shows (1) a typical static fluid distribution in a
homogeneous reservoir, (2) the oil-water contact, (3) the free water level and
(4) the oil and water pressure profiles. Note the transition zone above the oil-
water contact in which the water saturation decreases from 100% to the
irreducible water saturation. The height of this transition zone is a function
7-30
of the wettability of the rock, the oil-water density contrast, the oil-water
interfacial tension, the grain size (pore size) and sorting, which determine the
permeability of the rock.
Figure 7.17. Initial static fluid distribution in a homogeneous reservoir.
7-31
It should be noted that the capillary pressure in the reservoir is highest
at the top of the reservoir. In order to prevent escape of the hydrocarbon from
the reservoir, the cap rock must have a displacement pressure that is higher
than the maximum capillary pressure labeled Pcap in the figure. Shales
typically form the cap rock in many reservoirs. Shales are fine grained and
have very high displacement pressures. The shale, which is saturated with
water, will prevent the oil from penetrating it because its displacement
pressure is higher than the maximum capillary pressure in the reservoir. Of
course, some may think that the oil does not penetrate the shale because it
has a low permeability. However, the correct analysis is that the shale
prevents the oil from penetrating it because its displacement pressure is
much higher than the pressure in the oil phase. This is a capillary
phenomenon not a Darcy law phenomenon.
Figure 7.18 shows the initial fluid distribution in an actual petroleum
reservoir based on log analysis. The track labeled "Bulk Volume Analysis"
shows the water and oil distributions in the pay zone as a percent of the bulk
volume of reservoir rock. The dark area gives the oil content and the light area
to its left gives the water content. A careful examination of this section of the
plot shows a water saturation versus depth graph that is similar to the water
saturation versus depth graph sketched in Figure 7.17.
7-32
Figure 7.18. Initial static fluid distribution in an actual petroleum reservoir.
7-33
In a layered reservoir in which the layers have different capillary
pressure curves, the layers are in capillary equilibrium. As a result,
saturation discontinuities will occur. However, there will be only one free
water level. Figure 7.19 shows the water saturation distribution for a well
that has penetrated a layered reservoir. Given the capillary pressure curve for
each layer, it is a simple matter to apply Eqs.(7.32) and (7.33) to calculate the
water saturation distribution from the free water level to the top of the
reservoir. The steps for calculating the water saturation distribution in such a
heterogeneous reservoir is as follows.
1. Using the displacement pressure of the bottom layer, calculate the free
water level using Eq.(7.33).
2. Take a small value of z measured from the free water level.
3. Calculate the capillary pressure at that level using Eq.(7.32).
4. Determine the layer in which z occurs.
5. Using the capillary pressure curve for the layer in which z occurs, read
or calculate the water saturation for the value of capillary pressure from
step 3.
6. If z is at the boundary of two layers, there will be a saturation
discontinuity at that value of z. Two saturation values should be
calculated one from each of the capillary pressure curves of the two
layers involved.
7 Increase the value of z and repeat steps 3 through 6 until z reaches the
top of the reservoir.
7-34
This is how the saturation distribution in Figure 7.18c was calculated. If you
look closely at Figure 7.18b in which the four layers have been identified and
their capillary pressure curves have been plotted as height above the free
water level, you can mentally sketch the water saturation distribution over the
entire column of the well.
Figure 7.18. Fluid distribution for a layered reservoir; (a) well penetrating a
layered reservoir; (b) capillary pressure curves for the layers; (c) water staturation profile observed at the well (Archer and Wall, 1986).
7-35
Example 7.1
Table 7.3 gives the properties of an idealized oil reservoir consisting of four
layers with distinct petrophysical properties. The top of the reservoir is at
8000 ft below the surface and the oil water contact is at 8185 ft. Table 7.4
gives the drainage oil-water capillary pressure curve for Layer 1. All the layers
have the same pore structure but different permeabilities and porosities.
Table 7.3. Petrophysical Properties of Idealized Layered Reservoir.
Layer 1 Layer 2 Layer 3 Layer 4 Depth
(ft) 8000-8050
8050-8070
8070-8125
8125-8185
h (ft) 50 20 55 60 k (md) 144 50 10 200
φ %) 23.5 20 18 24
Table 7.4. Drainage Capillary Pressure Curve for Layer 1
Pc1
Sw (psi)
1.000 1.973
0.950 2.377
0.900 2.840
0.850 3.377
0.800 4.008
0.750 4.757
0.700 5.663
0.650 6.781
0.600 8.195
0.550 10.039
0.500 12.547
0.450 16.154
0.400 21.787
0.350 31.817
0.300 54.691
0.278 78.408
7-36
Other properties for the reservoir are as follows:
1.036wρ = g/cm3
0.822oρ = g/cm3
35σ = dynes/cm
0θ =
1. Calculate and plot the graph of the Leverett J-function for the reservoir.
2. Calculate and plot the capillary pressure curves for Layers 2, 3 and 4,
together with that of Layer 1.
3. Calculate the depth of the free water level for the reservoir.
4. Calculate and plot graphs of the initial water and oil saturations in the
reservoir from 8000 ft to the free water level assuming the reservoir is in
capillary equilibrium.
5. Calculate and plot graphs of the water and oil pressures at the initial
reservoir conditions.
6. A well drilled into the reservoir has been perforated from 8090 to 8110
ft. Determine the type of reservoir fluid that will be produced initially.
Solution to Example 7.1
1. The Leverett J-function is calculated with Eq.(7.26) using the capillary
pressure curve for Layer 1. Consistent units are required to make the
function dimensionless. For example, a set of consistent units is cP in
7-37
dynes/cm2, k in cm2 and σ in dynes/cm. For example, at 0.278wS = ,
78.408cP = psi, the J-function is calculated as
( )( )( ) ( )( ) ( )6 978.408 /14.696 1.0133 10 144 /1000 9.689 10 / 0.235
0.278, 11.90235cos0
x xJ
−
Γ = =
The calculated Leverett J-function is presented in Table 7.5 and Figure
7.19. It is the characteristic underlying dimensionless capillary
pressure curve for the reservoir.
Table 7.5. Summary of Calculated Leverett J-function and the Capillary Pressure Curves for Layers 2, 3 and 4 for Example 7.1.
Layer 1 Layer 2 Layer 3 Layer 4
Pc1 Pc2 Pc3 Pc4
Sw (psi) J(Sw) (psi) (psi) (psi)
1.000 1.973 0.299 3.089 6.553 1.692
0.950 2.377 0.361 3.721 7.894 2.038
0.900 2.840 0.431 4.446 9.432 2.435
0.850 3.377 0.513 5.287 11.215 2.896
0.800 4.008 0.608 6.275 13.311 3.437
0.750 4.757 0.722 7.447 15.799 4.079
0.700 5.663 0.860 8.866 18.807 4.856
0.650 6.781 1.029 10.616 22.520 5.815
0.600 8.195 1.244 12.830 27.217 7.027
0.550 10.039 1.524 15.717 33.341 8.609
0.500 12.547 1.905 19.643 41.670 10.759
0.450 16.154 2.452 25.290 53.649 13.852
0.400 21.787 3.307 34.109 72.357 18.683
0.350 31.817 4.830 49.812 105.668 27.283
0.300 54.691 8.302 85.624 181.635 46.898
0.278 78.408 11.902 122.755 260.402 67.235
7-38
Figure 7.19. Leverett J-function for the reservoir of Example 7.1.
2. Since all the layers have the same pore structure, they share the same
Leverett J-function. Thus, Eq.(7.26) can be solved for Pc using the
known J-function from Layer 1. However, for this example, the capillary
pressure curve for Layer j can be calculated from the data for Layer 1 as
11
1
jcj c
j
kP Pkφ
φ⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
For example, at Sw = 0.278,
2144 2078.408 122.75523.5 50cP ⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ psi
7-39
The calculated capillary pressure curves are presented in Table 7.5 and
Figure 7.20.
Figure 7.20. Capillary pressure curves for all the layers for the reservoir of Example 7.1.
3. The free water level occurs in Layer 4. Therefore, the displacement
pressure for Layer 4 is used to calculate the depth of the free water level
below the oil water contact from Eq.(7.33) as
( )( )
( )( )
61.6919 /14.696 1.0133 10555.69
1.036 0.822 981o
xd = =
− cm
555.69 / 30.48 18.23od = = ft
7-40
Eq.(7.33) can also be written in oilfield units as
[ ] [ ] [ ]3 3
144/ /
144
c co
P psi P psid ft
lb ft lb ftρ ρ= =
⎡ ⎤ ⎡ ⎤Δ Δ⎣ ⎦ ⎣ ⎦
( )( )( )( ) ( )( )
144 1.691918.24
1.036 62.4 0.822 62.4od = =−
ft
The free water level is at 8185+18.24 = 8203.24 ft.
4. The capillary pressure is zero at the free water level and increases as a
linear function of height above the free water level for incompressible
liquids. The height of any point in the reservoir above the free water
level is designated as z in Eq.(7.32) and can be used to calculate the
capillary pressure at that point in the reservoir. Clearly, at any depth D,
z is given by
z FWL D= −
For example at D = 8100 ft, which is in Layer 3,
8203.24 8100 103.24z = − = ft
The capillary pressure is calculated with Eq.(7.32) in oilfield units as
( )( ) ( )( ) ( )1.036 62.4 0.822 62.4
103.24 9.574144cP−⎡ ⎤⎣ ⎦= = psi
The water saturation at 8100 ft is calculated from the capillary pressure
curve for Layer 3 by linear interpolation as
7-41
( )9.574 9.4320.900 0.850 0.900 0.89611.215 9.432wS −⎛ ⎞= + − =⎜ ⎟−⎝ ⎠
The oil saturation is 1 1 0.896 0.104o wS S= − = − = . Table 7.6 shows the
calculated water and oil saturations from 8000 ft to 8203.24 ft (FWL).
Figure 7.21 shows the calculated saturation distributions. It can be
observed that none of the layers is at irreducible water saturation.
Therefore, each layer will produce some water if perforated.
Table 7.6. Calculated Saturations and Pressures for Example 7.1.
Depth z Pc Sw So Pw Po
ft ft psi
8000 203.24 18.847 0.426 0.574 3688.95 3670.10
8005 198.24 18.383 0.430 0.570 3689.15 3670.77
8010 193.24 17.920 0.434 0.566 3689.36 3671.44
8015 188.24 17.456 0.438 0.562 3689.57 3672.12
8020 183.24 16.992 0.443 0.557 3689.78 3672.79
Layer 1 8025 178.24 16.529 0.447 0.553 3689.99 3673.46
8030 173.24 16.065 0.451 0.549 3690.20 3674.13
8035 168.24 15.601 0.458 0.542 3690.40 3674.80
8040 163.24 15.138 0.464 0.536 3690.61 3675.47
8045 158.24 14.674 0.471 0.529 3690.82 3676.15
8050 153.24 14.210 0.477 0.523 3691.03 3676.82
8050 153.24 14.210 0.576 0.424 3691.03 3676.82
8055 148.24 13.747 0.584 0.416 3691.24 3677.49
Layer 2 8060 143.24 13.283 0.592 0.408 3691.44 3678.16
8065 138.24 12.819 0.600 0.400 3691.65 3678.83
8070 133.24 12.356 0.611 0.389 3691.86 3679.51
8070 133.24 12.356 0.823 0.177 3691.86 3679.51
8075 128.24 11.892 0.834 0.166 3692.07 3680.18
8080 123.24 11.428 0.845 0.155 3692.28 3680.85
8085 118.24 10.965 0.857 0.143 3692.49 3681.52
8090 113.24 10.501 0.870 0.130 3692.69 3682.19
7-42
Layer 3 8095 108.24 10.037 0.883 0.117 3692.90 3682.86
8100 103.24 9.574 0.896 0.104 3693.11 3683.54
8105 98.24 9.110 0.910 0.090 3693.32 3684.21
8110 93.24 8.646 0.926 0.074 3693.53 3684.88
8115 88.24 8.183 0.941 0.059 3693.73 3685.55
8120 83.24 7.719 0.957 0.043 3693.94 3686.22
8125 78.24 7.255 0.974 0.026 3694.15 3686.90
8125 78.24 7.255 0.593 0.407 3694.15 3686.90
8130 73.24 6.792 0.610 0.390 3694.36 3687.57
8135 68.24 6.328 0.629 0.371 3694.57 3688.24
8140 63.24 5.864 0.648 0.352 3694.78 3688.91
8145 58.24 5.401 0.672 0.328 3694.98 3689.58
8150 53.24 4.937 0.696 0.304 3695.19 3690.25
Layer 4 8155 48.24 4.473 0.725 0.275 3695.40 3690.93
8160 43.24 4.010 0.755 0.245 3695.61 3691.60
8165 38.24 3.546 0.791 0.209 3695.82 3692.27
8170 33.24 3.082 0.833 0.167 3696.02 3692.94
8175 28.24 2.619 0.880 0.120 3696.23 3693.61
8180 23.24 2.155 0.935 0.065 3696.44 3694.29
WOC 8185 18.24 1.691 1.000 0.000 3696.65 3694.96
Aquifer 8185 18.24 1.691 1.000 0.000 3696.65 3694.96
8190 13.24 1.228 1.000 0.000 3696.86 3695.63
8195 8.24 0.764 1.000 0.000 3697.06 3696.30
FWL 8203.24 0.00 0.000 1.000 0.000 3697.41 3697.41
7-43
Figure 7.21. Fluid saturation distribution for the reservoir of Example 7.1.
5. The water pressure is given by Eq.(7.29), which can be written in oilfield
units as
( ) ( )0144
ww w
zP z P ρ= −
7-44
where ( )0wP is the water pressure at the free water level. The water
pressure at the free water level is given by
( ) ( )( ) ( )( )( )1.036 62.4 8203.240 14.7 3697.41
144 144w
w atm
FWLP P
ρ= + = + = psia
At D = 8100 ft,
( ) ( )( ) ( )1.036 62.4103.24 3697.40 103.24 3651.06
144wP = − = psia
At the free water level, the pressure in the oil phase is equal to the
pressure in the water phase because the free water level is the reference
depth at which the capillary is zero. Thus,
( ) ( )0 0 3697.41o wP P= = psia
Of course, there is no oil in the reservoir below the water oil contact.
Therefore, there can be no oil pressure below the water oil contact. The
oil pressure starts at the water contact. However, if the oil pressure is
extrapolated to the free water level, its value will be equal to the water
pressure of 3697.41 psia. It should be noted that the difference between
the water oil pressure and the water pressure at the water oil contact is
equal to the displacement pressure of Layer 4 of 1.692 psi.
At D = 8100 ft, the oil pressure can be calculated as
( ) ( ) ( )103.24 103.24 103.24 3651.06 9.574 3641.87o w cP P P= − = − = psia
The calculated phase pressures are presented in Table 7.6 and Figure
7.22.
7-45
Figure 7.22. Water and oil phase pressures for Example 7.1.
6. The well is perforated in Layer 3, the lowest permeability layer, where
the oil saturation ranges from only 2.6% to 17.7%. This is in the
saturation range of residual oil for most reservoirs. Therefore, only
water will be produced from the well. The well should have been
perforated in Layer 1, where it would have produced a mixture of oil
and water.
7.7 CAPILLARY PRESSURE HYSTERESIS
Capillary pressure curves show a marked hysteresis depending on
whether the curve is determined under a drainage process or an imbibition
process. Figure 7.23 shows typical drainage and spontaneous imbibition
7-46
capillary pressure curves for the same porous medium. At any wetting phase
saturation, the drainage capillary pressure is higher than the imbibition
capillary pressure. At a capillary pressure of zero, the spontaneous imbibition
curve terminates at a wetting phase saturation that may or may not
correspond to the true residual non-wetting phase saturation depending on
the wettability of the rock. If the rock has a strong preference for the wetting
phase, then the wetting phase saturation at which the imbibition curve
terminates will be close to the true residual non-wetting phase saturation, Sor,
which is equal to (1-Swro). This is the case shown in Figure 7.23. If the rock
does not have a strong preference for the wetting phase, then the wetting
phase saturation at zero capillary pressure on the imbibition curve will not
correspond to the true residual non-wetting phase saturation. This means
that (1-Swro) will be larger than Sor. Additional oil can be displaced from the
rock, say be centrifuging the sample in water. This is the case shown in
Figure 7.24. The branch of the imbibition curve labeled 3 on the figure is the
forced imbition capillary pressure curve of the rock. Note that this branch
constitutes a negative capillary pressure. Note also that the true residual non-
wetting saturation (Sor) in this case can only by determined by forced
displacement not by spontaneous imbibition.
Figure 7.25 shows several cycles of capillary pressure
measurements on the same rock. The primary drainage curve labeled 1 was
performed first, followed by the spontaneous imbibition curve labeled 2. The
secondary drainage curve labeled 3 was performed after the spontaneous
imbibition measurement. It should be noted that the secondary drainage
curve will be less than the primary drainage curve at any given wetting phase
saturation. This is another aspect of capillary pressure hysteresis. If the
spontaneous imbibitition experiment is interupted and the measurement
reversed, then a different drainage curve will be followed as shown in curve 4.
If the drainage experiment is interupted and reversed, then a different
imbibition curve will be followed as shown in curve 5. Curves 4 and 5 form a
7-47
loop known as a scanning curve. Note that the area under the secondary
drainage curve was one of the areas used to define the USBM wettability
index.
Figure 7.23. Drainage and imbibition capillary pressure curves. (1) drainage
curve, (2) spontaneous imbibition curve (Killins et al., 1953).
Capillary pressure hysteresis can be explained in a variety of ways. In
Section 6.4.2, it was shown from energy considerations that more work is
required for a non-wetting phase to displace a wetting phase than for a
wetting phase to displace a non-wetting phase. This means that at any level of
saturation, more work is required during the drainage capillary pressure
measurement than during the imbibition measurement. Since work during
the capillary pressure measurement is PcΔV, where ΔV is the volume of fluid
7-48
displaced at that capillary pressure, the capillary pressure on the drainage
cycle will be greater than on the imbibition cycle to displace the same volume
of fluid.
Figure 7.24. Drainage and imbibition capillary pressure curves. (1) drainage curve, (2) spontaneous imbibition curve, (3) forced imbibition curve (Killins et
al., 1953).
7-49
Contact angle hysteresis plays a part in the capillary pressure
hysteresis. During drainage, the wetting phase recedes from the porous
medium and the contact angle is the receding contact angle, θR. During
imbibition, the wetting phase advances into the porous medium and the
contact angle is the advancing contact angle, θA. Since θR is less than θA,
2σcosθR /rm, the drainage capillary pressure, is larger than 2σcosθA /rm, the
imbibition capillary pressure at the same saturation state.
Figure 7.25. Cycles of capillary pressure measurements. (1) primary drainage, (2) spontaneous imbibition, (3) secondary drainage, (4-5) scanning curve.
7-50
The very nature of immiscible displacement plays a role in the capillary
pressure hysteresis. When the capillary pressure experiment is reversed to
measure the spontaneous imbibition curve, the pressure in the non-wetting
phase is reduced to allow the wetting phase to be imbibed. As the wetting
phase is imbibed into the rock, some non-wetting phase will be trapped in
certain pores. This trapping causes the wetting phase saturation on the
imbibition curve to be less than on the drainage curve at the same capillary
pressure.
The pore structure also plays a role in the capillary pressure hysteresis.
Consider the capillary pressure versus wetting phase saturation relationship
for the idealized pore shown in Figure 7.26 during drainage and imbibition.
During drainage, the pore is initially full of the wetting fluid at a capillary
pressure given by Laplace equation as shown in Figure 7.26a. Next, the
capillary pressure is increased to a higher value to drain some of the wetting
fluid as shown in Figure 7.26b. The higher capillary pressure versus the
wetting phase saturation is a point on the drainage capillary pressure curve.
Next, we consider the imbibition process as shown in Figures 7.26c and d. At
c, the capillary pressure is high at a wetting phase saturation of nearly zero.
After the wetting fluid has been imbibed to the equilibrium level shown in
Figure 7.26c, the imbibition capillary pressure will be approximately the same
as the drainage capillary pressure of Figure 7.26b because the mean
curvature of the interfaces at c and d are about the same. However, the
wetting phase saturation at d is considerably lower than at b. Thus, at the
same capillary pressure, the wetting phase saturation for imbibition is less
than for drainage. This is hysteresis.
7-51
Figure 7.26. Drainage and imbibition capillary pressures versus saturation for an idealized pore.
Another effect of pore structure is shown in Figure 7.27. The pore
(Figure 7.27a) is initially saturated with the wetting phase. The drainage
capillary pressure that must be applied to force the non-wetting phase into
the pore is given by
11
4 coscdrP
Deσ θ
= (7.35)
However, this capillary pressure is not sufficient to drain the entire pore
because neck De2 is smaller than De1. The interface will stop at De2. To
continue with the drainage, the applied capillary pressure must be increased
to Pcdr2 given by
7-52
22
4 coscdrP
Deσ θ
= (7.36)
Looking at the remaining necks of the pore, we see that Pcdr2 is large enough
to drain the remaining portion of the pore. Now, let us reduce the capillary
pressure to start the imbibition process. When the capillary pressure is
reduced to Pcibm1 given by
11
4 coscimbP
Dσ θ
= (7.37)
the first pore will be emptied of the non-wetting phase and the interface will
come to equilibrium at the location marked Imb1. To drain the non-wetting
phase further, the capillary pressure must be further reduced to Pcimb2 given
by
22
4 coscimbP
Dσ θ
= (7.38)
Looking at the remaining necks of the pore, we see that Pcimb2 is low enough to
empty the remaining portion of the pore of the non-wetting phase. Figure
7.27b shows the drainage and imbibition capillary pressure curves for the
experiment just described. We see capillary pressure hysteresis.
Figure 7.28 shows how the imbibition capillary pressure curve can be
used along with the drainage curve to determine the type of fluid that will be
produced at various depths in a reservoir. If the well is perforated above the
transition zone, only clean oil (water free oil) will be produced initially. If the
well is perforated in the upper part of the transition zone, both oil and water
will be produced from day one. If the well is perforated in the bottom part of
the transition zone, only water will be produced even though the zone has oil
saturation. The oil saturation in this zone is residual oil saturation.
7-53
Figure 7.27. Capillary pressure hysteresis in for an idealized pore (Dullien, 1992).
Capillary pressure hysteresis presents no problem in reservoir
engineering analysis as it is usually clear which curve should be used for a
particular analysis. The drainage curve should be used for estimating the
initial fluid saturation distribution in the reservoir whereas the imbibition
curve should be used for analyzing a waterflood performance in a water-wet
reservoir.
7-54
Figure 7.28. Drainage and imbibition capillary pressure curves showing the
depth of water free oil production (Archer and Wall, 1986).
7.8 CAPILLARY IMBIBITION
Consider a reservoir consisting of two layers with different
permeabilities and capillary pressure curves as shown in Figure 7.29 (a) and
(b). Initially, both layers are in capillary equilibrium at their respective
irreducible water saturations. Let this equilibrium be disturbed by
7-55
waterflooding the two layers. The injected water will advance further into the
more permeable layer (Figure 7.29 (c)). The oil and water pressures are
continuous across the boundary between the two layers. Thus, at the
boundary,
1 2o oP P= (7.39)
and
1 2w wP P= (7.40)
Subtracting Eq.(7.40) from (7.39) gives the condition for equilibrium as
1 2c cP P= (7.41)
Thus, at equilibrium, the capillary pressures in the two porous media will be
equal at their boundary.
In Figure 7.29c, sections A and D and C and F are in capillary
equilibrium, so no fluid exchanges will occur between these sections.
Sections B and E are not in capillary equilibrium, so fluid exchanges will
occur in an effort to achieve capillary equilibrium. Section E will loose water
to section B and gain oil from B while section B will gain water from E and
loose oil to E until a new capillary equilibrium is achieved. Thus, water will
be imbibed into the less permeable layer from the more permeable layer and
oil will be expelled from the less permeable layer into the more permeable
layer for subsequent displacement. This fluid exchange is beneficial to the oil
recovery process. However, the imbibition process is very slow. Therefore, the
water injection rate must be sufficiently slow for imbibition to assist in
waterflooding the low permeability layer.
7-56
Figure 7.29. Capillary imbibition; (a) reservoir before waterflooding; (b)
capillary pressure curves for the layers; (c) reservoir after waterflooding.
Naturally fractured reservoirs (fissured reservoirs) present another
example of capillary imbibition. The fractures have zero capillary pressure
whereas the matrix blocks have normal capillary pressure curves. When the
fractures become 100% saturated with water which comes in contact with the
oil saturated matrix blocks, the capillary equilibrium will be disturbed. Water
will be imbibed into the matrix blocks, expelling oil from the matrix blocks
into the fractures. Ultimately, the oil saturation in the matrix will be reduced
to the residual oil saturation over time.
7-57
7.9 CAPILLARY END EFFECT IN A LABORATORY CORE
7.9.1 Capillary End Effect
Another capillary phenomenon of interest is the capillary end effect
often experienced in laboratory coreflooding experiments. The end of the core
is in contact with the outside which could be viewed as a second medium with
zero capillary pressure. The condition for capillary equilibrium (Eq. 7.41)
requires that the capillary pressure inside the core at the outlet end be equal
to zero.
Consider a porous medium initially saturated with a non-wetting phase
(say oil) and irreducible wetting phase saturation (say water). The outlet end
of the core is at a higher capillary pressure than the outside. If the medium is
flooded with the wetting phase (waterflooded), initially, only the non-wetting
phase (oil) will be expelled from the outlet end at a higher capillary pressure
than the outside (Figure 7.30a). When the wetting phase arrives at the outlet
end, however, the system now has a chance to seek capillary equilibrium.
This equilibrium will be achieved by the accumulation of the wetting phase at
the outlet end of the core until the wetting phase saturation equals the
wetting phase saturation at zero capillary pressure on the imbibition capillary
pressure curve (Figure 7.30b). This saturation is marked Swro in Figure 7.30b.
Thus, the production of the wetting phase is delayed until well after the
arrival of the wetting phase at the outlet end of the core. This phenomenon is
known as capillary end effect.
This phenomenon has several undesirable consequences. The observed
breakthrough recovery of the non-wetting phase will be falsely high and the
wetting phase saturation distribution in the core will be opposite what would
normally be expected, with the wetting phase saturation being higher towards
the core outlet than in the rest of the core (Figure 7.30c). Most corefloods are
blind tests because one cannot see the fluid distribution inside the core.
7-58
Therefore, the breakthrough recovery is usually taken to be a good measure of
the displacement efficiency. In the presence of capillary end effect, the
breakthrough recovery will be too large and will give a false sense of the
displacement efficiency. Also, in the unsteady state method for relative
permeability measurement described in Chapter 8, the fractional flow of the
wetting phase versus the wetting phase saturation at the outlet end of the
core is used to calculate relative permeabilities on the assumption that there
is no capillary end effect. Therefore, if there is capillary end effect in the
experiment, the calculated relative permeabilities will be wrong.
Figure 7.30. Capillary end effect; (a) coreflood; (b) spontaneous imbibition capillary pressure curve; (c) wetting phase saturation profiles; (d) relative
permeabililty curves.
7-59
7.9.2 Mathematical Analysis of Capillary End Effect
We can derive the mathematical model for the immiscible displacement
shown in Figure 7.30a and use it to explain the capillary end effect
phenomenon. Darcy's law for the wetting and non-wetting phases is given by
w ww
w
k A Pqxμ
∂= −
∂ (7.42)
and
nw nwnw
nw
k A Pqxμ
∂= −
∂ (7.43)
where wk and nwk are the effective permeabilities to the wetting and non-
wetting phases. Let us define the relative permeabilities of the wetting and
non-wetting phases as
wrw
kkk
= (7.44)
and
nwrnw
kkk
= (7.45)
Eqs.(7.42) and (7.43) can be written in terms of the relative permeabilities as
rw ww
w
kk A Pqxμ
∂= −
∂ (7.46)
and
7-60
rnw nwnw
nw
kk A Pqxμ
∂= −
∂ (7.47)
Capillary equilibrium gives
( )nw w c wP P P S− = (7.48)
Assuming incompressible fluids, then
w nwq q q= + (7.49)
The continuity equation for the wetting phase is
0w wS qAt x
φ ∂ ∂+ =
∂ ∂ (7.50)
Finally, the saturation constraint gives
1w nwS S+ = (7.51)
Eqs.(7.46) through (7.51) constitute the complete mathematical description of
two-phase immiscible displacement in the absence of the effect of gravity.
Subtracting Eq.(7.46) from (7.47) and rearranging gives
w w nw nw nw w
rw rnw
q q P Pkk A kk A x x
μ μ ∂ ∂− = −
∂ ∂ (7.52)
Substituting Eqs.(7.48) and (7.49) into (7.52) gives upon rearrangement
7-61
1
1
rnw c
w nw
rnw w
rw nw
kk A Pq q x
kqk
μμ
μ
∂+
∂=
+ (7.53)
Let the true fractional flow of the wetting phase be defined as
ww
qfq
= (7.54)
Let an approximate fractional flow of the wetting phase be defined as
1
1w
rnw w
rw nw
F kk
μμ
=+
(7.55)
The approximate fractional flow of the wetting phase also can be defined as a
function of the mobility ratio as
111
wF
M
=+
(7.56)
Both wf and wF are functions of saturation. Substituting Eqs.(7.54) and (7.55)
into (5.53) gives the true fractional flow of the wetting phase as
1 rnw cw w
nw
kk A Pf Fq xμ
⎛ ⎞∂= +⎜ ⎟∂⎝ ⎠
(7.57)
Let the dimensionless distance from the inlet end be defined as
DxxL
= (7.58)
7-62
Let the spontaneous imbibition capillary pressure curve be given in terms of
its Leverett J-function as
( ) ( )cos ,/c w wP S J S
kσ θ
φ= Γ (7.59)
Substituting Eqs.(7.58) and (7.59) into (7.57) gives the true fractional flow of
the wetting phase as
cos1w w rnwnw D
A k Jf F kq L x
σ θ φμ
⎡ ⎤⎛ ⎞ ∂= +⎢ ⎥⎜ ⎟⎜ ⎟ ∂⎢ ⎥⎝ ⎠⎣ ⎦
(7.60)
The term in the inner bracket on the right side of Eq.(7.60) is a dimensionless
number that gives the ratio of the capillary to viscous forces in the
displacement. Let this dimensionless number be defined as
cos
capnw
A kNq L
σ θ φμ
= (7.61)
Substituting Eq.(7.61) into (7.60) gives
1w w cap rnwD
Jf F N kx
⎡ ⎤∂= +⎢ ⎥∂⎣ ⎦
(7.62)
Let the dimensionless time be defined as
Dqtt
A Lφ= (7.63)
Substituting Eq.(7.63) into (7.50) gives the continuity equation for the wetting
phase as
7-63
0w w
D D
S ft x
∂ ∂+ =
∂ ∂ (7.64)
Substituting Eq.(7.62) into (7.64) gives
0w w wcap w rnw
D w D D D
S dF S JN F kt dS x x x
⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂+ + =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠
(7.65)
Because J is a function of wS , Eq.(6.65) can be written as
0w w w wcap w rnw
D w D D w D
S dF S SdJN F kt dS x x dS x
⎛ ⎞ ⎛ ⎞∂ ∂ ∂∂+ + =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
(7.66)
Eq.(7.66) is the final form of the partial differential equation for the
wetting phase saturation for two phase immiscible displacement in a linear
porous medium. When supplemented with appropriate initial and boundary
conditions together with the rock and fluid properties, its solution can be
used to calculate the performance of the immiscible displacement such as a
waterflood or a gas flood. The solution of this equation is deferred to Chapter
8. Our concern here is the capillary end effect, not the prediction of the overall
displacement performance.
Let us examine in detail the fractional flow of the wetting phase at the
outlet end of the core. Applied to the outlet end of the core, Eq.(7.62) can be
written as
1w w cap rnwD
J Jf F N kxδ
− +⎡ ⎤⎛ ⎞−= +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦ (7.67)
where the derivative of the J-function with respect to Dx is given by the
following finite difference approximation:
7-64
D D
J J Jx xδ
− +⎛ ⎞∂ −⎜ ⎟∂ ⎝ ⎠
(7.68)
In Eq.(7.68), J + is the J-function inside the porous medium, J − is the J-
function outside the porous medium and Dxδ is a small distance in the
neighborhood of the outlet end of the porous medium as shown in Figure
7.31. Of course, J − is equal to zero. Therefore, the fractional flow of the
wetting phase at the outlet end of the porous medium becomes
1w w cap rnwD
Jf F N kxδ
+⎡ ⎤= −⎢ ⎥
⎣ ⎦ (7.69)
Depending on the values of capN , rnwk , and J + , it is possible for the following
inequality to prevail during the displacement:
1 0cap rnwD
JN kxδ
+
− ≤ (7.70)
Figure 7.31. Initial capillary barrier at the outlet end of the core.
7-65
If
1 0cap rnwD
JN kxδ
+
− = (7.71)
then
0wf = (7.72)
at the outlet end of the core. Because the fractional flow of the wetting phase
is zero at the outlet end of the core, the wetting phase cannot flow out of the
core but instead will accumulate there raising the wetting phase saturation to
an abnormal level. This is the capillary end effect phenomenon at work. If
1 0cap rnwD
JN kxδ
+
− < (7.73)
the wetting phase cannot flow out of the core either but will accumulate at the
end as before, raising the wetting saturation there to an abnormal level.
Although the inequality given by Eq.(7.73) appears to indicate that the wetting
phase will flow backwards into the core, this will not happen because there is
no supply of the wetting phase outside the outlet end of the core for it to be
imbibed into the core. Therefore, Eq.(7.70) gives the condition for capillary end
effect to occur in an immiscible displacement.
Let us now examine the physics of the displacement before and after
wetting phase breakthrough. Before the arrival of the wetting phase at the
outlet end of the core, only the non-wetting phase will be produced. The
wetting phase saturation profiles will be as shown in Figure 7.30a at t1 and t2.
At time at , the wetting phase arrives at the outlet end of the core. At this time,
J + has its maximum possible value at the irreducible wetting phase
7-66
saturation. If the values of capN , rnwk and J + are such that Eq.(7.70) holds,
then capillary end effect will occur and the wetting phase begins to
accumulate at the outlet end of the core resulting in the increase of the
wetting saturation, ( ),wS L t . As the wetting phase saturation increases, J +
decreases dramatically as shown in Figure 7.30b. When the wetting phase
saturation has increased to wroS , J + becomes zero and capillary equilibrium is
achieved between the porous medium and the outside. At this time, Eq.(7.69)
gives
0w wf F= > (7.74)
and the wetting phase flows out of the core and is produced along with the
non-wetting phase. The wetting phase saturation profile at breakthrough is
shown in Figure 7.30c at btt . After breakthrough, both phases will be
produced until residual non-wetting phase is achieved in the core after many
pore volumes of wetting phase injection. The wetting phase saturation profile
will be as shown in Figure 7.30c at t∞ . Beyond this time, only the wetting
phase will be produced.
How can capillary end effect be eliminated from the experiment? The
condition for eliminating the capillary end effect is obtained from Eq.(7.69) as
1 0cap rnwD
JN kxδ
+
− > (7.75)
or
Dcap
rnw
xNk Jδ
+< (7.76)
7-67
Thus, capN should be as small as possible in the experiment to eliminate
capillary end effect. A critical value of capN can be defined as
Dcapcritical
rnw
xNk Jδ
+= (7.77)
For capN below the critical value, capillary end effect will be eliminated. Above
the critical value, capillary end effect will occur in the displacement.
How can the value of capN be controlled in the experiment? The only
means to control capN in the experiment is through the injection rate, q.
Examination of Eq.(7.61) shows that capN can be made small by the use of a
high injection rate in the experiment. Substituting Eq.(7.61) into (7.76) gives
the condition for the injection rate to eliminate capillary end effect as
cosrnw
nw D
Ak J kqL x
σ θ φμ δ
+
> (7.78)
A critical injection rate can be defined as
cosrnw
criticalnw D
Ak J kqL x
σ θ φμ δ
+
= (7.79)
For injection rates below the critical value, capillary end effect will occur. For
rates above the critical, capillary end effect will be eliminated. In terms of
Darcy velocity, Eqs.(7.78) and (7.79) become
cosrnw
nw D
k J kqvA L x
σ θ φμ δ
+
= > (7.80)
7-68
cosrnwcritical
nw D
k J kvL x
σ θ φμ δ
+
= (7.81)
Is capillary end effect a problem in field displacements? The answer is no
because the large value of L at field scale ensures that capN is always smaller
than is required to eliminate capillary end effect at field rates.
7.9.3 Mathematical Model of Capillary End Effect During Steady State Relative Permeability Measurement.
The steady state method for relative permeability measurements
involves the simultaneous injection into the core of the wetting and non-
wetting fluids as shown in Figure 7.32. The fluids are injected at various
ratios of qw/qnw until steady state is achieved. If the total injection rate, q,
satisfies the condition of Eq.(7.76), then the injection rate is high enough to
eliminate capillary end effect and the wetting phase saturation will be uniform
throughout the core. However, if the total rate is not high enough to eliminate
capillary end effect, the wetting phase saturation will be non uniform in the
core. The wetting phase saturation will be higher at the outlet end of the core
than at the inlet end. The steady state saturation profile can be derived from
Eq.(7.62) as follows. At steady state conditions, Eq.(7.62) can be written as
1 ww w cap rnw
w D
dSdJf F N kdS dx
⎡ ⎤= +⎢ ⎥
⎣ ⎦ (7.82)
Rearranging Eq.(7.82) gives
1w
ww
Dcap rnw
w
fFdS
dJdx N kdS
⎛ ⎞−⎜ ⎟
⎝ ⎠= (7.83)
7-69
Figure 7.32. Steady state experiment.
Eq.(7.83) is the partial differential equation for the steady state wetting phase
saturation. It is a first order nonlinear equation that can easily be integrated
to obtain the steady state saturation profile. The appropriate boundary
condition for the equation is
( )1w wroS S= (7.84)
Figure 7.33 shows the steady state saturation profile obtained by solving
Eqs.(7.83) with the boundary condition given by Eq.(7.84).
7-70
Figure 7.33. Steady state wetting phase saturation profile in the presence of capillary end effect.
7.9.4 Experimental Evidence of Capillary End Effect
Perkins (1957) has presented experimental data that show capillary end
effect at work. He conducted waterfloods in laboratory cores at two rates, one
below the critical rate for capillary end effect and one above the critical rate.
The core was 12 inches in length and 1.25 inches in diameter. The oil and
water viscosities were 1.8 and 0.9 cp. The low injection rate was 2.4 ft/day
whereas the high injection rate was 36 ft/day. The injected water was 0.1
normal sodium chloride solution. The core was instrumented with two current
electrodes and nineteen potential electrodes distributed along its length.
These electrodes enabled the resistivity of the core to be measured along the
length of the core, from which the water saturation profiles were calculated
using Archie's equation.
7-71
Figure 7.34 shows the water saturation profiles for the low rate flood.
The water arrived at the outlet end of the core at tD = 0.41 pore volume
injected. However, no water was produced from the core until tD = 0.60 pore
volume injected. Thus, the flood was affected by capillary end effect. The water
saturation profile at tD = 0.60 clearly shows the capillary end effect.
Figure 7.35 shows the water saturation profiles for the high rate flood.
The water arrived at the outlet end of the core at tD = 0.60 pore volume
injected and got produced shortly thereafter at tD = 0.65 pore volume injected.
Thus, the capillary end effect was significantly reduced in the high rate
displacement compared to the low rate displacement.
Figure 7.34. Wetting phase saturation profiles at low injection rate (Perkins,
1957).
7-72
Figure 7.35. Wetting phase saturation profiles at high injection rate (Perkins,
1957).
The second example from Richardson et al. (1952) investigated capillary
end effect in steady state experiments. The core was a Berea sandstone of
length 30 cm and diameter 6.85 cm. The core was cut into 8 segments as
shown in Figure 7.36. Steps were taken to ensure capillary continuity
between the segments. The core was initially saturated with oil, which was the
wetting phase. Then helium, the non-wetting phase, and oil were injected
simultaneously at various ratios. After steady state was achieved, the core
segments were weighed to determine the wetting phase saturation in each
segment.
7-73
Figure 7.36. Segmented core used in steady state experiments (Richardson et
al., 1953).
Figures 7.37 to 7.39 show the measured and calculated wetting phase
saturation profiles at steady state for successively higher total injection rates.
The theoretical saturation profiles were calculated with Eqs.(7.83) and (7.84)
using the drainage relative permeability curves and drainage capillary
pressure curves shown in Figure 7.40. Capillary end effect occurred in the
three experiments shown in Figures 7.37 to 7.39. However, it can be seen that
the capillary end effect was reduced as the total injection rate was increased.
The lesson from these examples is that capillary end effect could
dominate laboratory scale immiscible displacements. Since such
displacements are normally used to determine relative permeability curves in
7-74
the laboratory, it is imperative that steps be taken to eliminate or at least
minimize capillary end effect in these experiments. Failure to do so will result
in the relative permeability curves so derived being wrong.
Figure 7.36. Capillary end effect in a steady state experiment. qw = 0.15
cm3/s, qnw = 0.000336 cm3/s (Richardson et al., 1953).
7-75
Figure 7.37. Capillary end effect in a steady state experiment. qw = 0.264
cm3/s, qnw = 0.0022 cm3/s (Richardson et al., 1953).
Figure 7.38. Capillary end effect in a steady state experiment. qw = 0.80
cm3/s, qnw = 0.00288 cm3/s (Richardson et al., 1953).
7-76
Figure 7.39. Drainage relative permeability curves and capillary pressure
curve for Berea sandstone (Richardson et al., 1953).
7.10 CAPILLARY PRESSURE MEASUREMENTS
Three methods are commonly used to measure capillary pressure
curves in the petroleum industry: the restored state method (porous plate
method), mercury injection method and centrifuge method.
7.10.1 Restored State Method (Porous Plate Method).
In this method, capillary pressure is measured by placing the sample,
initially saturated with a wetting fluid, in a vessel filled with the non-wetting
fluid. The bottom of the vessel consists of a semi-permeable plate, which
allows the wetting phase displaced from the sample to pass through while
blocking the passage of the non-wetting phase. Extending from the porous
7-77
plate is a graduated tube which allows the volume of the wetting phase
displaced to be measured as shown in Figure 7.41. With the sample in place,
the pressure of the non-wetting fluid is increased in steps and the system is
allowed to achieve equilibrium after each pressure change. The volume of
wetting phase displaced at each pressure is measured. The capillary pressure
is the non-wetting phase pressure minus the wetting phase pressure at each
step. The wetting phase saturation of the sample is determined from the
volume of wetting phase displaced at each pressure to obtain the capillary
pressure versus saturation relationship.
The porous plate is typically made of porcelain or fritted glass. It must
have a displacement pressure that is higher than the largest capillary
pressure to be measured. This limits the maximum capillary pressure that
can be measured with the method to about 200 psi.
The porous plate apparatus can be used to measure the imbibition
capillary pressure curve as well as the drainage curve. The method gives a
reliable estimate of the irreducible wetting phase saturation. The major
disadvantage of the porous plate method is that it takes too long to obtain the
entire capillary pressure curve. It is not unusual for the capillary pressure
experiment to take several weeks to complete.
7.10.2 Mercury Injection Method
In this method, capillary pressure is measured by injecting mercury,
which is a non-wetting phase, into the sample. The apparatus used in the
measurement is shown in Figure 7.42. It consists of a sample cell and a
mercury injection pump. A dry sample is placed in the cell and the cell is
evacuated. Mercury is injected into the cell until the mercury is level with a
graduation on the high-pressure glass capillary above the sample chamber.
Nitrogen pressure is then applied in successive increments and at each step,
7-78
Figure 7.41. Porous plate capillary pressure apparatus (Welge and Bruce,
1947).
mercury is injected to maintain the mercury level with the graduation on the
capillary. From the volume of the cell and the volume of mercury required to
fill the cell with the sample before mercury injection into the sample, the bulk
volume of the sample can be determined. The mercury-air capillary pressure
versus saturation relationship is calculated from the volume of mercury
forced into the sample pore space as a function of the applied nitrogen
pressure.
7-79
Figure 7.38. Capillary pressure cell for mercury injection (Purcell, 1949).
The mercury injection method is very fast. The capillary pressure curve
can be obtained in a matter of hours. The imbibition curve can be obtained
very easily by decreasing the nitrogen pressure and withdrawing mercury
from the system. Figure 7.43 shows typical capillary pressure curves
obtained by mercury injection, mercury withdrawal and mercury re-injection.
7-80
Figure 7.43. Mercury-air capillary pressure curves.
Brown (1951) has shown that the mercury injection method can give
essentially the same capillary pressure curve as the restored state method
except for a scaling factor. Figures 7.44 and 7.45 compare capillary pressure
curves obtained by mercury injection and the restored state method for a
sandstone and a limestone core, respectively. The results show good
agreement between the two methods. The scaling factors for the sandstone
and limestone were 7.5 and 5.5, respectively. These are different from the
7-81
scaling factor of 5.2 suggested by Purcell (1949) based on the ratio of σcosθ of
mercury-air and water-air systems.
The major disadvantage of the mercury injection method is that the core
can no longer be used for other tests after mercury injection. The method
also cannot be used to determine the irreducible wetting phase saturation.
7.10.3 Centrifuge Method.
In this method, the sample saturated with a wetting fluid is placed in a
centrifuge cup containing the non-wetting fluid as shown in Figures 7.46 and
7.47. The sample is rotated at a series of constant angular velocities and the
amount of wetting fluid displaced at equilibrium at each velocity is measured
with the aid of a stroboscopic light. The only data measured directly in this
method are the volume of wetting fluid displaced and the corresponding
rotational speed of the centrifuge. These data can be used to derive the
capillary pressure versus saturation relationship of the porous medium.
The theory underlying the method is that the centrifuge imposes a
centrifugal force (typically, over 1000 times the force of gravity) on the sample.
This causes the denser wetting fluid to be displaced outward away from the
center of rotation and the non-wetting fluid to flow into the sample through
the inlet face of the sample. Consider an oil-water system in which water is
the wetting phase and oil is the non-wetting phase. At equilibrium, the
pressure gradients in the water and oil are given by
2ww
dP rdr
ρ ω= − (7.85)
2oo
dP rdr
ρ ω= − (7.86)
7-82
Figure 7.44. A comparison of water-nitrogen and mercury-air capillary pressure curves for a sandstone core (Brown, 1951). Note the different
capillary pressure scales for the two sets of data.
7-83
Figure 7.45. A comparison of water-nitrogen and mercury-air capillary pressure curves for a limestone core (Brown, 1951). Note the different
capillary pressure scales for the two sets of data.
7-84
Figure 7.46. Positions of core and graduated tube in a centrifuge for
measurement of oil-displacing-water capillary pressure curve (Donaldson et al., 1980).
7-85
Figure 7.47. Positions of core and graduated tube in a centrifuge for
measurement of water-displacing-oil capillary pressure curve (Donaldson et al., 1980).
where ω is the angular velocity of the centrifuge in radians/s. Subtracting
Eqs. 7.85 from 7.86 gives
( ) 2 2cw o
dP r rdr
ρ ρ ω ρω= − − = −Δ (7.87)
7-86
Integration of Eq. 7.87 gives the capillary pressure at radius r as
2
2crP CρωΔ
= − + (7.88)
where C is an integration constant. Application of the Hassler-Brunner
boundary condition that Pc = 0 at r = r2 gives the capillary pressure at any r
as
( )2
2 222cP r rρωΔ
= − (7.89)
where r2 is the outlet face of the core measured from the center of rotation. At
any stage of the centrifuge experiment, the highest capillary pressure occurs
at r1, the inlet face of the core measured from the center of rotation. At the
inlet face, the capillary pressure is given by
( )2
2 21 2 12cP r rρωΔ
= − (7.90)
At any stage of the centrifuge experiment, the water saturation in the
core varies from a minimum at the core inlet, r1, to a value of 1.0 at the core
outlet, r2. Since the capillary pressure at the core inlet can be calculated
from Eq.(7.90), it is only necessary to estimate the water saturation at the
core inlet in order to obtain the required capillary pressure versus saturation
relationship. The water saturation at the core inlet can be estimated from the
average water saturation determined from the amount of water displaced at
each speed. The average water saturation in the core is given by
2
1
2 1
r
wrwav
S drS
r r=
−∫
(7.91)
7-87
Multiplying Eq.(7.91) by Δρω2r1 and rearranging gives
2
1
2 21 1
r
wav wrLr S S rdrρω ρωΔ = Δ∫ (7.92)
where L is the length of the core. Noting that Δρω2Lr1 is approximately equal
to Pc1 since L is small relative to r1 and r2 and Δρω2r1dr is equal to 1cdP− ,
Eq.(7.92) can be written as
( )1
1 0
cP
c wav w c cP S S P dP= ∫ (7.93)
Differentiating Eq.(7.93) gives the water saturation at the core inlet as
( )11
1
c wavw
c
d P SS
dP= (7.94)
or
1 11
wavw wav c
c
dSS S PdP
⎛ ⎞= + ⎜ ⎟
⎝ ⎠ (7.95)
The approximate water saturation equation, Eq.(7.94) or (7.95), is applicable
for r1/r2 equal to or greater than 0.8. When applied to experimental data,
experience shows that Eq.(7.94) gives a smoother capillary pressure curve
than Eq.(7.95).
The centrifuge method is fast and allows the capillary pressure
measurement to be completed in a day or less. The method is good for
determining the irreducible water saturation. Figure 7.48 shows a
comparison of the capillary pressure curves of the same sample from
centrifuge measurement and by the restored state method obtained by
7-88
Hassler and Brunner (1945). Although the capillary pressure curve from
centrifuge measurement is somewhat higher than from the restored state
method, the agreement between the two sets of data is reasonable. In this
figure, the open circles are data obtained by the use of Eq.(7.94) and the open
squares are data obtained by the use of Eq.(7.95) to compute the inlet water
saturation. It should be observed that both equations give somewhat different
results for the water saturations. Figure 7.49 shows comparisons of the
capillary pressure curves for the same cores measured by the restored state
method (diaphram), mercury injection and centrifuge experiments. The
agreement between the three methods is excellent.
The disadvantages of the method include (1) inability to measure the
displacement pressure since the water saturation at the core inlet is always
less than the average water saturation of the core, (2) the Hassler-Brunner
boundary condition at the core outlet may be violated at high centrifuge
speeds, (3) the calculated water saturation at the core inlet is an
approximation, and (4) inability to obtain spontaneous imbibition capillary
pressure curve. Melrose (1988) has investigated the Hassler-Brunner
boundary condition and concluded that it was unlikely to be violated under
normal core analysis conditions. Rajan (1986) has presented a more accurate
method for calculating the water saturation at the core inlet.
7-89
Figure 7.48. A comparison of capillary pressure obtained by centrifuge and by
the restored state method (Hassler and Brunner, 1945).
7-90
Figure 7.49. A comparison of capillary pressure curves obtained by centrifuge, mercury injection and by the restored state method (Hermansen et al., 1991).
Example 7.2
Table 7.7 shows the data obtained in a centrifuge experiment for determining
the air-water capillary pressure curve of a core sample. Other data from the
experiment are as follows:
Core length = 2.0 cm
Core diameter = 2.53 cm
Core pore volume = 1.73 cm3
Core permeability = 144 md
7-91
Centrifuge arm (r2) = 8.6 cm
Water-air density difference = 0.9988 gm/cm3
2 8.6r = cm
Table 7.7. Centrifuge Experimental Data Centrifuge Speed (RPM)
Volume of Water Displaced
(cc) 1300 0.30 1410 0.40 1550 0.50 1700 0.60 1840 0.70 2010 0.75 2200 0.80 2500 0.90 2740 1.00 3120 1.05 3810 1.10 4510 1.20 5690 1.25
1. Calculate the capillary pressure curves for the sample using Eq.(7.94)
(method 1) and Eq. (7.95) (method 2) for the inlet water saturation.
2. Compare the capillary pressure curves obtained from the two methods.
3. Calculate the acceleration imposed on the inlet of the core at the
centrifuge speed of 5690 RPM and compare to the acceleration due to
gravity.
7-92
Solution to Example 7.2
The results of the calculations are summarized in Table 7.8.
1. The entries in the table were computed as follows at the centrifuge
speed of 1300 revolutions per minute (RPM).
( ) ( )/ 1.73 0.30 /1.73 0.827wav p w pS V V V= − = − =
( )2 / 60 2 1300 / 60 136.14Nω π π= = = radians/s
( ) ( )( )( )2 2 2 2 2 22 1
1
0.9988 136.14 8.6 6.6281362.40
2 2c
r rP
ρωΔ − −= = = dynes/cm2
61 281362.40 /1.0133 10 0.28cP x= = atm
( )( )1 6
281362.40 14.6964.08
1.0133 10cPx
= = psi
Figure 7.50 shows the graph of 1wav cS P versus 1cP . The regression
equation is
0.62171 11.3563wav c cS P P=
Application of Eq.(7.94) gives the inlet water saturation as
( ) ( )( ) ( )( ) 0.37831
1 61
281362.40 14.6960.6217 1.3563 0.495
1.0133 10wav c
wc
d S PS
dP x
−⎡ ⎤
= = =⎢ ⎥⎣ ⎦
7-93
Table 7.8. Calculated Results for Example 7.2.
1 2 3 4 5 6 7 8 9 10 11
Centrifug
e
Method
1(Eq.7.94)
Method 2
(Eq.7.95)
Speed Vw Swav ω Pc1 Pc1 Pc1 SwavPc1 Sw1 1
wav
c
dSdP
Sw1
RPM cc radians/
s
dynes/cm2 atm psi psi psi-1 psi
1300 0.30 0.827 136.14 281362.40 0.28 4.08 3.37 0.495 -0.0739 0.525
1410 0.40 0.769 147.65 330992.07 0.33 4.80 3.69 0.466 -0.0590 0.485
1550 0.50 0.711 162.32 399984.12 0.39 5.80 4.12 0.434 -0.0455 0.447
1700 0.60 0.653 178.02 481146.36 0.47 6.98 4.56 0.404 -0.0353 0.407
1840 0.70 0.595 192.68 563657.13 0.56 8.17 4.87 0.381 -0.0283 0.364
2010 0.75 0.566 210.49 672622.63 0.66 9.76 5.53 0.356 -0.0222 0.350
2200 0.80 0.538 230.38 805795.28 0.80 11.69 6.28 0.333 -0.0173 0.335
2500 0.90 0.480 261.80 1040541.4
3
1.03 15.09 7.24 0.302 -0.0122 0.296
2740 1.00 0.422 286.93 1249915.0
1
1.23 18.13 7.65 0.282 -0.0095 0.251
3120 1.05 0.393 326.73 1620647.4
3
1.60 23.50 9.24 0.255 -0.0066 0.238
3810 1.10 0.364 398.98 2416736.5
4
2.39 35.05 12.76 0.220 -0.0038 0.231
4510 1.20 0.306 472.29 3386354.6
6
3.34 49.11 15.05 0.193 -0.0024 0.189
5690 1.25 0.277 595.86 5390187.7
2
5.32 78.17 21.69 0.162 -0.0013 0.179
Figure 7.51 shows the graph of wavS versus 1cP . The regression equation
is
0.378311.3563wav cS P−=
7-94
( )( ) ( )( ) 1.3783
61
281362.40 14.6960.3783 1.3563 0.0739
1.0133 10wav
c
dSdP x
−⎡ ⎤
= − = −⎢ ⎥⎣ ⎦
Application of Eq.(7.95) gives the inlet water saturation as
( ) ( )( ) ( )1 1 61
1.73 0.30 281362.40 14.6960.0739 0.525
1.73 1.0133 10wav
w wav cc
dSS S PdP x
− ⎡ ⎤⎛ ⎞= + = − =⎜ ⎟ ⎢ ⎥
⎝ ⎠ ⎣ ⎦
Figure 7.50. Graph of 1wav cS P versus 1cP for Example 7.2.
7-95
Figure 7.51. Graph of wavS versus 1cP for Example 7.2.
2. Figure 7.52 shows a comparison of the capillary pressure curves from
Eqs.(7.94) and (7.95). Both equations give essentially the same capillary
pressure curves. However, Eq.(7.95) shows more scatter in the
computed water saturations than Eq.(7.94).
3. The acceleration imposed at the inlet end of the core at the centrifuge
speed of 5690 RPM is given by
( ) ( )22
21 1
2 56902 6.6 2343288.1960 60
Nacceleration r rππω
⎡ ⎤⎛ ⎞= = = =⎢ ⎥⎜ ⎟⎝ ⎠ ⎣ ⎦
cm/s2
This acceleration is 2388 times the acceleration due to gravity.
7-96
Figure 7.52. Comparison of the capillary pressure curves derived from Eqs.7.94 and 7.95 for Example 7.1.
7.11 PORE SIZE DISTRIBUTION
7.11.1 Introduction
We saw in Section 7.2 that the capillary pressure versus saturation
relationship of an idealized porous medium captured the pore size
distribution of the medium. It should come as no surprise that one of the
principal applications of capillary pressure curve is for estimating the pore
size distribution of porous media. Because mercury does not wet most solids,
the capillary pressure curve derived from mercury injection is particularly well
suited for probing the pore structure of porous media. Thus, mercury
7-97
injection porosimetry is widely used in the petroleum and material science
industries to determine the pore size distribution of porous materials.
When mercury is injected into the porous medium at a low pressure,
mercury will invade those pores having pore throat radii equal to or greater
than the radius given by Laplace equation:
2 cos
cPR
σ θ= (7.96)
where R is the pore throat radius. As the capillary pressure is increased,
pores with smaller pore throat sizes are invaded by mercury. If the mercury
pressure is high enough, all the pores in the porous medium will be invaded
by mercury. Therefore, the cumulative volume of mercury injected versus the
capillary pressure can be used to determine the pore size distribution of the
medium.
Two different pore size distribution functions can be derived from
mercury porosimetry. The first is the pore volume distribution, which does not
assume a model for the porous medium such as the bundle of capillary tubes
model. The only assumption is that the pore throat is circular. The second
pore size distribution function is based on the assumption of a bundle of
capillary tubes model of the porous medium and represents the distribution of
the pore throat size assuming the pores are capillary tubes. Because they
represent different things, the two pore size distribution functions are not the
same.
7.11.2 Pore Volume Distribution
Consider the result of the mercury injection experiment as shown in
Figure 7.53. The figure shows the cumulative volume of mercury injected
expressed as a fraction of the total pore volume plotted against the capillary
7-98
pressure. This is the raw data obtained from the experiment. Note that the
cumulative volume of mercury injected expressed as a fraction of the pore
volume is the non-wetting phase saturation, Snw. Shown on the capillary
pressure axis is the corresponding pore throat size obtained from Eq.(7.96) as
2 cos
c
RP
σ θ= (7.97)
Figure 7.53. Cumulative pore volume of mercury injected versus capillary pressure.
Also shown on the figure is the wetting phase saturation versus capillary
pressure obtained from
1w nwS S= − (7.98)
7-99
It should be noted that the pore throat radius, R, decreases from left to right
in the figure. Let us replot the saturations versus pore throat size such that
the pore throat size increases from left to right as shown in Figure 7.54. It
should be observed that the Sw versus R curve now represents the cumulative
probability distribution function for the pore volume distribution whereas the
Snw versus R curve represents the expectation curve for the pore volume
distribution. At any value of R, Sw is the fraction of the pore volume occupied
by pores having pore throat size equal to R or less. If the probability density
function for the pore volume distribution is f (R), then
( )final
R
w RS f R dR= ∫ (7.99)
where
( )0 1final
R
Rf R dR =∫ (7.100)
Differentiating Eq.(7.99) using Leibnitz's rule for differentiating an integral
gives the probability density function as
( ) wdSf RdR
= (7.101)
Because of the relationship between Sw and Snw, the probability density
function for the pore volume distribution also is given by
( ) nwdSf RdR
= − (7.102)
Various alternative expressions can be derived for the probability density
function by use of Eq.(7.96). From Eq.(7.96),
7-100
2 cos a constantcP R σ θ= = (7.103)
Differentiating Eq.(7.103) gives
0c cP dR RdP+ = (7.104)
Figure 7.54. Saturations versus pore throat size from mercury injection.
Substituting Eqs.(7.96) and (7.104) into Eqs.(7.101) and (7.102) gives the
following alternative expressions for the probability density function for the
pore volume distribution:
( ) 2
2 cosw c w w
c c
dS P dS dSf RdR R dP R dP
σ θ= = − = − (7.105)
7-101
( ) 2
2 cosnw c w nw
c c
dS P dS dSf RdR R dP R dP
σ θ= − = = (7.106)
Thus, differentiating the saturation versus the pore throat radius curve from
the experiment leads directly to the probability density function for the pore
volume distribution of the rock. Further, a plot of the incremental pore
volume of mercury injected versus pore throat size also is a measure of the
pore volume distribution.
Differentiation of experimental data can be a noisy affair. The three-
point central difference approximation given previously for the calculation of
the welltest derivative function (Eq.(3.65)) can be used to differentiate the
experimental data. It should be noted that the welltest derivative function
should be divided by ti to obtain the first derivative of the function. There are
other differentiation schemes that may be less noisy than the three-point
central difference formula. The five-point central difference formula proposed
by Akima (1970) may be less noisy than the three-point central difference
formula. The computational template for the five-point central difference
formula is shown in Figure 7.55. The first derivative at x3 is given by
7-102
Figure 7.55. Computational template for calculating the first derivative using Akima's method (Akima, 1970).
3
4 3 2 2 1 3
4 3 2 1x
m m m m m mdydx m m m m
− + −⎛ ⎞ =⎜ ⎟ − + −⎝ ⎠ (7.107)
where
5 44
5 4
y ymx x
−=
− (7.108)
4 33
4 3
y ymx x
−=
− (7.109)
3 22
3 2
y ymx x
−=
− (7.110)
7-103
2 11
2 1
y ymx x
−=
− (7.111)
7.11.3 Pore Size Distribution Based on Bundle of Capillary Tubes Model
The objective is to estimate the pore size distribution of a porous
medium from the drainage capillary pressure curve based on a bundle of
capillary tubes model of the porous medium. Further, the pore size
distribution should be scaled into a pore size probability density function.
Let the probability density function for the pore size distribution be
( )Rδ , where R is the pore throat radius. Then, for the porous medium,
0
( ) 1R dRδ∞
=∫ (7.112)
The fractional number of pores with radii between R and R+dR is ( )R dRδ . The
number of pores with radii between R and R+dR is ( )N R dRδ , where N is the
total number of pores (capillary tubes) making up the porous medium. The
cross-sectional area of the pores with radii between R and R+dR is given by
2 ( )cdA R N R dRπ δ= (7.113)
The cross-sectional area occupied by all the pores can be obtained by
integrating Eq.(7.113) to obtain
2
0
( )c TA A N R R dRφ π δ∞
= = ∫ (7.114)
where AT is the total cross-sectional area of the porous medium and φ is the
porosity of the porous medium. Let
7-104
2 2
0
( ) a constant R R R dRδ∞
= =∫ (7.115)
Eq.(7.114) can be rewritten as
2c TA A NRφ π= = (7.116)
The pore volume of the porous medium is given by
2 2
0
( )p TV A L NL R R dR NLRφ π δ π∞
= = =∫ (7.117)
At capillary equilibrium, the wetting phase occupies all the pores with radii
less than R corresponding to the capillary pressure given by
2 cos
( )c wP SR
σ θ= (7.118)
The wetting phase volume is given by
2
0
( )R
wV NL R R dRπ δ= ∫ (7.119)
The wetting phase saturation is given by
2 2
0 02
2
0
( ) ( )
( )
R R
ww
p
R R dR R R dRVSV R
R R dR
δ δ
δ∞= = =∫ ∫
∫ (7.120)
Eq.(7.120) can be rearranged as
7-105
2 2 2
0 0
( ) ( )R R
wR S R R dR r r drδ δ= =∫ ∫ (7.121)
where r is a dummy integration variable, which is not to be confused with the
radius R, the upper limit of the integral. Differentiating Eq.(7.121) with
respect to R gives
2 2
0
( )R
wdS dR r r drdR dR
δ= ∫ (7.122)
The right side of Eq.(7.122) can be evaluated using Leibnitz’s rule for
differentiating a definite integral as follows:
2
2 2
0 0
( )0( ) ( ) 0 (0)R R r rd dR dr r dr R R dr
dR dR dR Rδ
δ δ δ⎡ ⎤∂ ⎣ ⎦= − +
∂∫ ∫ (7.123)
The integrand of the integral on the right side of Eq.(7.123) is zero. Therefore,
Eq.(7.123) simplifies to
2 2
0
( ) ( )Rd r r dr R R
dRδ δ=∫ (7.124)
Substituting Eq.(7.124) into Eq.(7.122) gives
2 2 ( )wdSR R RdR
δ= (7.125)
Differentiating Eq.(7.118) with respect to R gives
2
2 cos( )c wdP SdR R
σ θ= − (7.126)
7-106
Dividing Eq.(7.126) by Eq.(7.125) gives
2
4
2 cos( )( )
c w
w
RdP SdS R R
σ θδ
= − (7.127)
The probability density function for the pore size distribution can be obtained
from Eq.(7.127) as
24
2 cos( )( )c w
w
RR dP SR
dS
σ θδ= −
⎡ ⎤⎢ ⎥⎣ ⎦
(7.128)
Using Eq.(7.96), Eq.(7.128) can be rewritten in the following alternative form:
23
( )( )( )
c w
c w
w
P SRR dP SR
dS
δ= −
⎡ ⎤⎢ ⎥⎣ ⎦
(7.129)
Eqs.(7.118) and (7.129) can be used to calculate the pore size distribution
from a drainage capillary pressure curve as follows:
1. Pick a high Pc(Sw) value corresponding to a low wetting phase
saturation, Sw, and a small pore size, R.
2. Calculate the pore radius, R, using Eq.(7.118).
3. Calculate the derivative of the capillary pressure curve with respect to
the wetting phase saturation at the value of the Pc(Sw) in step 1. Note
that this is a negative quantity.
4. Calculate 2( ) /R Rδ using Eq.(7.129).
7-107
5. Pick lower values of Pc(Sw) and repeat steps 2 through 4 until the entire
capillary pressure curve has been used in the pore size distribution
calculation.
6. Plot the graph of 2( ) /R Rδ versus R. Calculate the area under the graph,
Ag. Using Ag, calculate the constant 2R so as to satisfy Eq.(7.112),
which is the requirement for a probability density function.
7. Using the value of 2R , calculate and plot the graph of ( )Rδ versus R,
which is the required probability density function for the pore size
distribution.
Eq.(7.129) was derived for a general capillary pressure curve obtained
by any method. In the case of the capillary pressure curve obtained by
mercury porosimetry, Eq.(7.129) can be transformed by substitution of
Eq.(7.104) to obtain the probability density function as
( )2
2wdSRR
R dRδ = (7.130)
or
( )2
2nwdSRR
R dRδ = − (7.131)
Using Eq.(7.130) or (7.131), the probability density function can easily be
calculated from the mercury injection data.
Figure 7.56 shows the pore size distributions of various sedimentary
rocks determined from drainage capillary pressure curves.
7-108
Figure 7.56. Pore size distributions of sedimentary rocks based on the bundle of capillary tubes model of the rock (Crocker, 1983).
7-109
Example 7.3
The first three columns of Table 7.9 give the mercury injection data for a low
permeability sandstone sample with k = 0.048 md and φ = 5.6%. Calculate
and plot the following graphs:
1. Snw and Sw versus pore throat radius, R.
2. ΔSnw versus pore throat radius, R.
3. Probability density function for pore volume distribution.
4. Probability density function for pore radius distribution assuming a
bundle of capillary tubes model of the porous medium.
Table 7.9. Mercury Injection Data and Calculated Pore Size Distributions for
Example 7.3.
1 2 3 4 5 6 7 8 9
Pore Volume Pore Radius
Distribution Distribution
Pc Pc ΔSnw R R ( ) wdSf RdR
= ( )2
2wdSRR
R dRδ =
Snw Sw (psi) (dynes/cm2
)
(cm) (μm) (1/μm) (1/μm)
0.000 1.000 124.92 8.613E+06 0.0000
0
8.538E-
05
0.853
8
0.000 0.000
0.031 0.969 150.75 1.039E+07 0.0312
1
7.075E-
05
0.707
5
0.323 0.002
0.068 0.932 175.43 1.210E+07 0.0370
6
6.080E-
05
0.608
0
0.424 0.004
0.104 0.896 200.40 1.382E+07 0.0355
3
5.322E-
05
0.532
2
0.508 0.006
0.168 0.832 249.77 1.722E+07 0.0640
4
4.270E-
05
0.427
0
0.691 0.012
7-110
0.229 0.771 300.12 2.069E+07 0.0613
1
3.554E-
05
0.355
4
0.928 0.024
0.330 0.670 400.20 2.759E+07 0.1011
0
2.665E-
05
0.266
5
1.385 0.064
0.428 0.572 499.51 3.444E+07 0.0981
5
2.135E-
05
0.213
5
2.310 0.166
0.513 0.487 599.28 4.132E+07 0.0847
8
1.780E-
05
0.178
0
2.499 0.259
0.577 0.423 699.36 4.822E+07 0.0637
2
1.525E-
05
0.152
5
2.507 0.354
0.625 0.375 799.13 5.510E+07 0.0478
7
1.335E-
05
0.133
5
2.501 0.461
0.660 0.340 899.29 6.201E+07 0.0351
3
1.186E-
05
0.118
6
2.289 0.534
0.686 0.314 999.18 6.889E+07 0.0264
7
1.067E-
05
0.106
7
2.268 0.654
0.709 0.291 1098.89 7.577E+07 0.0227
2
9.706E-
06
0.097
1
2.636 0.919
0.730 0.270 1198.44 8.263E+07 0.0212
7
8.900E-
06
0.089
0
2.637 1.094
0.748 0.252 1298.50 8.953E+07 0.0180
9
8.214E-
06
0.082
1
2.638 1.284
0.765 0.235 1398.24 9.641E+07 0.0165
3
7.628E-
06
0.076
3
2.875 1.623
0.780 0.220 1498.24 1.033E+08 0.0149
6
7.119E-
06
0.071
2
3.042 1.971
0.794 0.206 1598.66 1.102E+08 0.0141
0
6.672E-
06
0.066
7
3.245 2.394
0.806 0.194 1695.09 1.169E+08 0.0124
5
6.292E-
06
0.062
9
3.255 2.700
0.818 0.182 1797.32 1.239E+08 0.0114
5
5.934E-
06
0.059
3
3.268 3.048
0.829 0.171 1895.87 1.307E+08 0.0107
6
5.626E-
06
0.056
3
3.390 3.518
7-111
0.838 0.162 2000.98 1.380E+08 0.0095
6
5.330E-
06
0.053
3
3.539 4.091
0.856 0.144 2196.94 1.515E+08 0.0175
7
4.855E-
06
0.048
5
3.800 5.295
0.871 0.129 2396.98 1.653E+08 0.0154
9
4.450E-
06
0.044
5
3.866 6.413
0.885 0.115 2597.89 1.791E+08 0.0135
1
4.105E-
06
0.041
1
4.010 7.812
0.897 0.103 2799.03 1.930E+08 0.0121
2
3.810E-
06
0.038
1
4.038 9.133
0.907 0.093 2997.38 2.067E+08 0.0100
5
3.558E-
06
0.035
6
4.054 10.514
0.918 0.082 3248.15 2.240E+08 0.0113
2
3.284E-
06
0.032
8
4.196 12.778
0.928 0.072 3495.86 2.410E+08 0.0098
7
3.051E-
06
0.030
5
4.253 15.004
0.937 0.063 3744.60 2.582E+08 0.0087
5
2.848E-
06
0.028
5
4.213 17.054
0.944 0.056 3996.64 2.756E+08 0.0065
9
2.669E-
06
0.026
7
4.068 18.757
0.950 0.050 4246.84 2.928E+08 0.0064
0
2.511E-
06
0.025
1
4.068 21.180
0.956 0.044 4494.10 3.099E+08 0.0056
2
2.373E-
06
0.023
7
4.041 23.561
0.960 0.040 4745.57 3.272E+08 0.0047
7
2.247E-
06
0.022
5
3.725 24.214
0.965 0.035 4997.24 3.446E+08 0.0041
9
2.134E-
06
0.021
3
3.675 26.495
0.968 0.032 5245.84 3.617E+08 0.0036
7
2.033E-
06
0.020
3
3.645 28.956
0.972 0.028 5496.45 3.790E+08 0.0035
1
1.940E-
06
0.019
4
3.958 34.517
0.975 0.025 5746.20 3.962E+08 0.0036
5
1.856E-
06
0.018
6
4.128 39.346
7-112
0.978 0.022 5994.05 4.133E+08 0.0030
6
1.779E-
06
0.017
8
3.692 38.296
0.981 0.019 6246.10 4.307E+08 0.0025
8
1.708E-
06
0.017
1
3.509 39.516
0.983 0.017 6497.47 4.480E+08 0.0022
9
1.642E-
06
0.016
4
3.491 42.548
0.985 0.015 6744.53 4.650E+08 0.0022
2
1.581E-
06
0.015
8
3.308 43.446
0.987 0.013 6996.48 4.824E+08 0.0016
0
1.524E-
06
0.015
2
2.964 41.879
0.990 0.010 7497.19 5.169E+08 0.0031
6
1.423E-
06
0.014
2
2.913 47.272
0.992 0.008 7997.18 5.514E+08 0.0020
4
1.334E-
06
0.013
3
2.862 52.840
0.995 0.005 8494.95 5.857E+08 0.0025
4
1.256E-
06
0.012
6
2.734 56.951
0.997 0.003 8995.38 6.202E+08 0.0018
8
1.186E-
06
0.011
9
2.661 62.153
0.998 0.002 9495.55 6.547E+08 0.0016
3
1.123E-
06
0.011
2
2.452 63.825
0.999 0.001 9996.48 6.893E+08 0.0009
4
1.067E-
06
0.010
7
1.491 43.010
1.000 0.000 10496.0
0
7.237E+08 0.0006
5
1.016E-
06
0.010
2
0.524 16.671
1.000 0.000 10997.0
3
7.583E+08 0.0000
8
9.699E-
07
0.009
7
0.000 0.000
1.000 0.000 11495.5
6
7.926E+08 0.0000
0
9.278E-
07
0.009
3
0.000 0.000
1.000 0.000 11996.4
7
8.272E+08 0.0000
0
8.891E-
07
0.008
9
0.000
1.000 0.000 12495.5
4
8.616E+08 0.0000
0
8.536E-
07
0.008
5
0.000
1.000 0.000 12996.0
7
8.961E+08 0.0000
0
8.207E-
07
0.008
2
0.000
7-113
1.000 0.000 13495.1
1
9.305E+08 0.0000
0
7.903E-
07
0.007
9
0.000
1.000 0.000 13995.9
0
9.650E+08 0.0000
0
7.621E-
07
0.007
6
0.000
1.000 0.000 14495.6
8
9.995E+08 0.0000
0
7.358E-
07
0.007
4
0.000
1.000 0.000 14996.2
0
1.034E+09 0.0000
0
7.112E-
07
0.007
1
0.000
1.000 0.000 15495.9
8
1.068E+09 0.0000
0
6.883E-
07
0.006
9
0.000
1.000 0.000 15995.1
3
1.103E+09 0.0000
0
6.668E-
07
0.006
7
0.000
1.000 0.000 16495.5
2
1.137E+09 0.0000
0
6.466E-
07
0.006
5
0.000
1.000 0.000 16995.2
8
1.172E+09 0.0000
0
6.276E-
07
0.006
3
0.000
1.000 0.000 17495.3
3
1.206E+09 0.0000
0
6.096E-
07
0.006
1
0.000
1.000 0.000 17995.5
0
1.241E+09 0.0000
0
5.927E-
07
0.005
9
0.000
1.000 0.000 18495.9
5
1.275E+09 0.0000
0
5.766E-
07
0.005
8
0.000
1.000 0.000 18996.3
7
1.310E+09 0.0000
0
5.615E-
07
0.005
6
0.000
1.000 0.000 19495.3
3
1.344E+09 0.0000
0
5.471E-
07
0.005
5
0.000
1.000 0.000 19995.7
3
1.379E+09 0.0000
0
5.334E-
07
0.005
3
0.000
1.000 0.000 20995.3
1
1.448E+09 0.0000
0
5.080E-
07
0.005
1
0.000
1.000 0.000 21995.7
9
1.517E+09 0.0000
0
4.849E-
07
0.004
8
0.000
1.000 0.000 22995.6
4
1.586E+09 0.0000
0
4.638E-
07
0.004
6
0.000
7-114
1.000 0.000 23995.7
2
1.655E+09 0.0000
0
4.445E-
07
0.004
4
0.000
1.000 0.000 24996.3
4
1.724E+09 0.0000
0
4.267E-
07
0.004
3
0.000
1.000 0.000 25994.9
6
1.792E+09 0.0000
0
4.103E-
07
0.004
1
0.000
1.000 0.000 26995.6
5
1.861E+09 0.0000
0
3.951E-
07
0.004
0
0.000
1.000 0.000 27995.6
6
1.930E+09 0.0000
0
3.810E-
07
0.003
8
0.000
1.000 0.000 28996.0
8
1.999E+09 0.0000
0
3.678E-
07
0.003
7
0.000
1.000 0.000 29995.4
9
2.068E+09 0.0000
0
3.556E-
07
0.003
6
0.000
1.000 0.000 30996.2
1
2.137E+09 0.0000
0
3.441E-
07
0.003
4
0.000
1.000 0.000 31995.5
3
2.206E+09 0.0000
0
3.333E-
07
0.003
3
0.000
1.000 0.000 32996.2
0
2.275E+09 0.0000
0
3.232E-
07
0.003
2
0.000
1.000 0.000 33996.5
7
2.344E+09 0.0000
0
3.137E-
07
0.003
1
0.000
1.000 0.000 34996.1
7
2.413E+09 0.0000
0
3.048E-
07
0.003
0
0.000
1.000 0.000 35996.3
2
2.482E+09 0.0000
0
2.963E-
07
0.003
0
0.000
1.000 0.000 36995.5
3
2.551E+09 0.0000
0
2.883E-
07
0.002
9
0.000
1.000 0.000 37996.3
9
2.620E+09 0.0000
0
2.807E-
07
0.002
8
0.000
1.000 0.000 38996.3
8
2.689E+09 0.0000
0
2.735E-
07
0.002
7
0.000
1.000 0.000 39995.7
5
2.758E+09 0.0000
0
2.667E-
07
0.002
7
0.000
7-115
1.000 0.000 41995.3
9
2.896E+09 0.0000
0
2.540E-
07
0.002
5
0.000
1.000 0.000 43995.2
2
3.034E+09 0.0000
0
2.424E-
07
0.002
4
0.000
1.000 0.000 45993.7
9
3.171E+09 0.0000
0
2.319E-
07
0.002
3
0.000
1.000 0.000 47991.1
8
3.309E+09 0.0000
0
2.222E-
07
0.002
2
0.000
1.000 0.000 49990.5
5
3.447E+09 0.0000
0
2.134E-
07
0.002
1
0.000
1.000 0.000 51989.8
4
3.585E+09 0.0000
0
2.051E-
07
0.002
1
0.000
1.000 0.000 53989.7
5
3.723E+09 0.0000
0
1.975E-
07
0.002
0
0.000
1.000 0.000 55989.4
0
3.861E+09 0.0000
0
1.905E-
07
0.001
9
0.000
1.000 0.000 57988.7
6
3.998E+09 0.0000
0
1.839E-
07
0.001
8
0.000
Solution to Example 7.3
The calculated results are shown in columns 4 to 9 of Table 7.9. For
mercury, σ = 480 dynes/cm and θ = 140º. Figure 7.57 shows the graphs of
non-wetting phase saturation, wetting phase saturation and incremental pore
volume of mercury injected versus pore throat size. It should be observed that
the incremental pore volume of mercury injected provides a rough estimate of
the pore volume distribution for the porous medium.
7-116
Figure 7.57. Graphs of Snw, Sw and ΔSnw versus pore throat radius for Example 7.3.
Columns 8 and 9 of Table 7.9 show the probability density functions for the
pore volume and pore throat radius distributions calculated with the 5-point
central difference formula of Akima. For the pore radius distribution, 2R was
found to be 0.003284. Figures 7.58 and 7.59 show the graphs of the
probability density functions for pore volume and pore radius distributions.
7-117
Figure 7.58. Probability density function for pore volume distribution for Example 7.3.
Figure 7.59. Probability density function for pore throat radius distribution assuming a bundle of capillary tubes model for the sample of Example 7.3.
7-118
7.11.4 Mercury Injection Porosimeter
Several different types of equipment are used for mercury injection
porosimetry. The mercury injection equipment used by Purcell (1949) was
described in Section 7.9.2. However, more modern mercury porosimeters are
now available. One such porosimeter is the Autopore IV 9520, manufactured
by Micromeritics. Figure 7.60 shows this equipment. It is a compact, high
precision, computer-controlled instrument capable of injecting mercury into a
sample at capillary pressures up to 60,000 psi. Pressure is measured with
transducers that produce an electrical signal that is proportional to the
applied pressure. This analog signal is converted into digital code for
processing by the monitoring computer.
The transducer that detects the volume of mercury injected is
integrated into the sample holder assembly as shown in Figure 7.61. The
sample cup has a capillary stem, which serves both as the mercury reservoir
and as an element of the mercury volume transducer. The capillary stem is
made of glass, which is an insulator but the outer face of the capillary is
plated with metal, which is a conductor. The sample is placed in the sample
cup and the cup along with the capillary stem is filled with mercury, which is
a conductor. The combination of two concentric electrical conductors
separated by an insulator produces a co-axial capacitor, whose capacitance is
a function of the length of mercury in the capillary. As mercury is injected
into the sample, the length of mercury in the capillary decreases and this in
turn changes the capacitance of the penetrometer assembly. The capacitance
of the assembly is measured at each step of mercury injection and used to
calculate the volume of mercury injected into the sample.
7-119
Figure 7.60. Autopore IV 9520 mercury injection porosimeter (from Micromeritics).
7-120
Figure 7.61. Penetrometer for the Autopore IV 9520 mercury injection
porosimeter (from Micromeritics).
7-121
7.12 CALCULATION OF PERMEABILITY FROM DRAINAGE CAPILLARY PRESSURE CURVE
7.12.1 Calculation of Absolute Permeability from Drainage Capillary Pressure Curve
The objective is to estimate the permeability of a porous medium from
the drainage capillary pressure curve based on the bundle of capillary tubes
model of the porous medium. This problem was addressed in Section 3.7.3
using the probability density function for the pore throat diameter. Here, we
present the estimate of the absolute permeability of the porous medium based
on the probability density function for the pore throat radius obtained from
mercury injection porosimetry.
Let the probability density function for the pore throat radius be ( )Rδ ,
where R is the pore throat radius. Then, for the porous medium,
0
( ) 1R dRδ∞
=∫ (7.132)
The fractional number of pores with radii between R and R+dR is ( )R dRδ . The
number of pores with radii between R and R+dR is ( )N R dRδ , where N is the
total number of pores making up the porous medium. The cross-sectional
area of the pores with radii between R and R+dR is given by
2 ( )cdA R N R dRπ δ= (7.133)
The cross-sectional area occupied by all the pores can be obtained by
integrating Eq.(7.133) to obtain
2
0
( )c TA A N R R dRφ π δ∞
= = ∫ (7.134)
7-122
where AT is the total cross-sectional area of the porous medium and φ is the
porosity of the porous medium. Let
2 2
0
( ) a constant R R R dRδ∞
= =∫ (7.135)
Eq.(7.135) can be written as
2c TA A NRφ π= = (7.136)
The volumetric flow rate for pores with radii between R and R+dR is
given by Hagen-Poiseuille’s law as
( )2
21 ( )8 ( )
R Pdq R N R dRR L
π δμ τ
⎡ ⎤ Δ= ⎢ ⎥
⎣ ⎦ (7.137)
where ( )Rτ is the tortuosity of the porous medium and L is the length of the
medium. Note that the tortuosity is a function of the pore throat size, R. The
wetting phase occupies the pores with radii less than R. The volumetric flow
rate of the wetting phase is obtained by integrating Eq.(7.137) as
4
0
( )8 ( )
R
ww
N P R Rq dRL R
π δμ τ
Δ= ∫ (7.138)
The effective permeability to the wetting phase is given by Darcy’s law as
4
0
( )8 ( )
Rw w
wT T
q L N R Rk dRA P A R
μ π δτ
= =Δ ∫ (7.139)
The pore volume of the porous medium is given by
7-123
2 2
0
( )p TV A L NL R R dR NLRφ π δ π∞
= = =∫ (7.140)
At capillary equilibrium, the wetting phase occupies all the pores with radii
less than R corresponding to the capillary pressure given by
2 cos
( )c wP SR
σ θ= (7.141)
The wetting phase volume is given by
2
0
( )R
wV NL R R dRπ δ= ∫ (7.142)
The wetting phase saturation is given by
2 2
0 02
2
0
( ) ( )
( )
R R
ww
p
R R dR R R dRVSV R
R R dR
δ δ
δ∞= = =∫ ∫
∫ (7.143)
Eq.(7.143) can be rearranged as
2 2 2
0 0
( ) ( )R R
wR S R R dR r r drδ δ= =∫ ∫ (7.144)
where r is a dummy integration variable, which is not to be confused with the
pore throat radius R, the upper limit of the integral. Differentiating Eq.(7.144)
with respect to R gives
2 2
0
( )R
wdS dR r r drdR dR
δ= ∫ (7.145)
7-124
The right side of Eq.(7.145) can be evaluated using Leibnitz’s rule for
differentiating a definite integral as follows:
2
2 2
0 0
( )0( ) ( ) 0 (0)R R r rd dR dr r dr R R dr
dR dR dR Rδ
δ δ δ⎡ ⎤∂ ⎣ ⎦= − +
∂∫ ∫ (7.146)
The integrand of the integral on the right side of Eq.(7.146) is zero. Therefore,
Eq.(7.146) simplifies to
2 2
0
( ) ( )Rd r r dr R R
dRδ δ=∫ (7.147)
Substituting Eq.(7.147) into Eq.(7.145) gives
2
2
( )w
R R dRdSR
δ= (7.148)
Substituting Eqs.(7.136) and (7.148) into Eq.(7.139) gives the effective
permeability to the wetting phase as
2
08 ( )
wS
w wRk dSR
φτ
= ∫ (7.149)
The R in Eq.(7.149) is the largest pore size filled with the wetting phase. This
can be obtained from the capillary pressure curve as
2 cos
( )c w
RP Sσ θ
= (7.150)
Let us approximate the tortuosity by a function of the form
7-125
( ) aRRατ = (7.151)
where a and α are numerical constants. This function satisfies the
requirement that the tortuosity increases as the pore size decreases.
Substituting Eqs.(7.150) and (7.151) into Eq.(7.149) gives the effective
permeability to the wetting phase as
( )2
20
2 cos8
wSw
wc
dSka P
α
α
σ θφ
+
+= ∫ (7.152)
The absolute permeability of the porous medium can be obtained from
Eq.(7.152) for the medium fully saturated by the wetting phase as
( )2 1
20
2 cos8
w
c
dSka P
α
α
σ θφ
+
+= ∫ (7.153)
Let α = 0 and 1/a = F1. Substituting these values into Eq.(7.153) gives the
absolute permeability of the porous medium as
( )2 1
1 20
2 cos8
w
c
dSk FP
σ θφ= ∫ (7.154)
If the capillary pressure is in psi, the permeability is in millidarcy and the
surface tension is in dynes/cm, Eq.(7.154) becomes, after the units
conversion,
( )1
2
1 20
10.6566 cos w
c
dSk FP
σ θ φ= ∫ (7.155)
7-126
In the case of mercury injection, σ = 480 dynes/cm and θ = 140º. Substituting
these values into Eq.(7.155) gives the permeability as
1
61 2
0
1.441 10 w
c
dSk x FP
φ= ∫ (7.156)
Eq.(7.156) is Purcell's equation in which F1 is a lithology factor to
account for tortuosity of the porous medium. Purcell (1949) performed
measurements on core samples and computed the lithology factors that made
the measured permeabilities to be equal to the calculated permeabilities from
the drainage capillary pressure curves of the samples. His results are shown
in Table 7.10. The permeabilities of samples 17 and 18 were too low to
measure but were estimated to be less than 0.1 md. The calculated lithology
factors ranged from 0.085 to 0.363, with an average value of 0.216. He
suggested the use of the average value of 0.216 for estimating the
permeability of a rock from its drainage capillary pressure curve. However, as
shown in Figure 7.62, there is a positive correlation between the lithology
factor and permeability. Clearly, the higher the permeability, the higher the
lithology factor. Similarly, the lower the permeability, the lower the lithology
factor. The lithology factor is related to the reciprocal of the tortuosity of the
porous medium. Therefore, the positive correlation between the lithology
factor and permeability is to be expected. Because of this correlation, the use
of the average lithology factor of 0.216 to calculate the permeability of all
porous media can lead to poor estimates of permeability. Figure 7.63
compares the measured permeability with the calculated permeability using
the average lithology factor of 0.216 plotted on log scales. The correlation
between the two sets of data appears very strong. However, when the same
data are plotted on linear scales as shown in Figure 7.64, the correlation is
weaker than in Figure 7.63.
Table 7.10. Lithology factors for various core samples (Purcell, 1949).
7-127
Lithology Calculated
Permeability
Sample Permeability Factor Using F1 = 0.216
No (md) (md)
1 1.2 0.085 3.04
2 12 0.122 21.2
3 13.4 0.168 17.3
4 36.9 0.149 53.5
5 57.4 0.200 61.9
6 70.3 0.165 91.6
7 110 0.257 92.3
8 116 0.256 97.5
9 144 0.191 163
10 336 0.107 680
11 430 0.216 430
12 439 0.273 348
13 496 0.276 388
14 772 0.185 902
15 1070 0.282 816
16 1459 0.363 865
17 0.003
18 0.1
19 35.7 0.182 42.2
20 40.2 0.158 54.9
21 184 0.231 172
22 235 0.276 183
23 307 0.215 308
24 320 0.163 422
25 506 0.284 383
26 634 0.272 502
27 1150 0.338 734
7-128
Figure 7.62. Correlation between lithology factor and permeability (Purcell, 1949).
Figure 7.63. A comparison of measured permeability and calculated permeability using the average lithology factor of 0.216 plotted on log scales
(Purcell, 1949).
7-129
Figure 7.64. A comparison of measured permeability and calculated permeability using the average lithology factor of 0.216 plotted on linear
scales (Purcell, 1949).
Example 7.4
Estimate the permeability of the sample of Example 7.3 using the mercury
injection capillary pressure data of Table 7.9.
Solution to Example 7.4
Table 7.11 presents the calculated results. Figure 7.65 shows the graphs of cP
and 2
1
cP versus wS . The integration called for in Eq.(7.156) was performed with
the trapezoidal rule to obtain
1 6 220
8.355 10 1/w
c
dS x psiP
−=∫
7-130
Using the average lithology factor of 0.216 in Eq.(7.156) gives
( )( )( )6 61.441 10 0.216 0.056 8.355 10 0.146k x x −= = md, compared to the measured
permeability of 0.048 md. However, if the correlation of the lithology factor
with permeability is recognized and taken into account, the lithology factor for
the very low permeability sample should be much less than 0.216. Using the
lowest lithology factor of 0.085 in Table 7.10 gives the permeability as
( )( )( )6 61.441 10 0.085 0.056 8.355 10 0.057k x x −= = md, compared to the measured
value of 0.048 md. Thus, the low lithology factor of 0.085 gives a better
estimate of the permeability than the average value of 0.216 for this tight
sample.
Figure 3.30 gives a correlation between tortuosity and porosity based on
the experimental data of Winsauer et al. (1952). The equation of the
regression line is
27.35 10.987τ φ= − +
For the sample with a porosity of 0.056, the tortuosity predicted by the
regression line is 9.4554. Therefore, the lithology factor (1/τ) predicted by the
correlation is 0.1058. Using the this value of the lithology factor in Eq.(7.156)
gives the estimated permeability as
( )( )( )6 61.441 10 0.1058 0.056 8.355 10 0.071k x x −= = md, compared to the measured
permeability of 0.048 md. This estimate also is better than that based on the
average lithology factor of 0.216.
Table 7.11. Calculated Results for Example 7.4.
7-131
cP 2
1
cP 21
wS w
c
dSP∫
wS psi ( )21/ psi ( )21/ psi
1.000 124.919 6.408E-
05
0.000E+0
0
0.969 150.747 4.401E-
05
1.687E-06
0.932 175.428 3.249E-
05
3.104E-06
0.896 200.401 2.490E-
05
4.124E-06
0.832 249.768 1.603E-
05
5.434E-06
0.771 300.121 1.110E-
05
6.266E-06
0.670 400.203 6.244E-
06
7.143E-06
0.572 499.515 4.008E-
06
7.646E-06
0.487 599.278 2.784E-
06
7.934E-06
0.423 699.359 2.045E-
06
8.088E-06
0.375 799.133 1.566E-
06
8.174E-06
0.340 899.290 1.237E-
06
8.223E-06
0.314 999.184 1.002E-
06
8.253E-06
0.291 1098.894 8.281E-
07
8.274E-06
0.270 1198.443 6.962E-
07
8.290E-06
0.252 1298.503 5.931E- 8.302E-06
7-132
07
0.235 1398.241 5.115E-
07
8.311E-06
0.220 1498.239 4.455E-
07
8.318E-06
0.206 1598.665 3.913E-
07
8.324E-06
0.194 1695.092 3.480E-
07
8.328E-06
0.182 1797.317 3.096E-
07
8.332E-06
0.171 1895.866 2.782E-
07
8.335E-06
0.162 2000.983 2.498E-
07
8.338E-06
0.144 2196.945 2.072E-
07
8.342E-06
0.129 2396.980 1.740E-
07
8.345E-06
0.115 2597.892 1.482E-
07
8.347E-06
0.103 2799.027 1.276E-
07
8.349E-06
0.093 2997.379 1.113E-
07
8.350E-06
0.082 3248.154 9.478E-
08
8.351E-06
0.072 3495.865 8.183E-
08
8.352E-06
0.063 3744.603 7.132E-
08
8.353E-06
0.056 3996.642 6.261E-
08
8.353E-06
0.050 4246.843 5.545E- 8.353E-06
7-133
08
0.044 4494.100 4.951E-
08
8.354E-06
0.040 4745.567 4.440E-
08
8.354E-06
0.035 4997.241 4.004E-
08
8.354E-06
0.032 5245.841 3.634E-
08
8.354E-06
0.028 5496.453 3.310E-
08
8.354E-06
0.025 5746.203 3.029E-
08
8.354E-06
0.022 5994.055 2.783E-
08
8.355E-06
0.019 6246.104 2.563E-
08
8.355E-06
0.017 6497.474 2.369E-
08
8.355E-06
0.015 6744.532 2.198E-
08
8.355E-06
0.013 6996.476 2.043E-
08
8.355E-06
0.010 7497.188 1.779E-
08
8.355E-06
0.008 7997.178 1.564E-
08
8.355E-06
0.005 8494.954 1.386E-
08
8.355E-06
0.003 8995.378 1.236E-
08
8.355E-06
0.002 9495.550 1.109E-
08
8.355E-06
0.001 9996.479 1.001E- 8.355E-06
7-134
08
0.000 10496.00
0
9.077E-
09
8.355E-06
0.000 10997.02
9
8.269E-
09
8.355E-06
0.000 11495.55
9
7.567E-
09
8.355E-06
0.000 11996.47
0
6.949E-
09
8.355E-06
0.000 12495.53
8
6.405E-
09
8.355E-06
0.000 12996.07
4
5.921E-
09
8.355E-06
0.000 13495.10
8
5.491E-
09
8.355E-06
0.000 13995.90
2
5.105E-
09
8.355E-06
0.000 14495.68
0
4.759E-
09
8.355E-06
0.000 14996.19
5
4.447E-
09
8.355E-06
0.000 15495.97
6
4.164E-
09
8.355E-06
0.000 15995.13
0
3.909E-
09
8.355E-06
0.000 16495.52
0
3.675E-
09
8.355E-06
0.000 16995.27
7
3.462E-
09
8.355E-06
0.000 17495.33
2
3.267E-
09
8.355E-06
0.000 17995.50
4
3.088E-
09
8.355E-06
0.000 18495.94 2.923E- 8.355E-06
7-135
7 09
0.000 18996.36
5
2.771E-
09
8.355E-06
0.000 19495.32
6
2.631E-
09
8.355E-06
0.000 19995.73
2
2.501E-
09
8.355E-06
0.000 20995.31
3
2.269E-
09
8.355E-06
0.000 21995.79
1
2.067E-
09
8.355E-06
0.000 22995.64
3
1.891E-
09
8.355E-06
0.000 23995.72
1
1.737E-
09
8.355E-06
0.000 24996.34
4
1.600E-
09
8.355E-06
0.000 25994.96
3
1.480E-
09
8.355E-06
0.000 26995.65
4
1.372E-
09
8.355E-06
0.000 27995.66
0
1.276E-
09
8.355E-06
0.000 28996.07
8
1.189E-
09
8.355E-06
0.000 29995.49
4
1.111E-
09
8.355E-06
0.000 30996.21
1
1.041E-
09
8.355E-06
0.000 31995.52
7
9.768E-
10
8.355E-06
0.000 32996.19
9
9.185E-
10
8.355E-06
0.000 33996.57 8.652E- 8.355E-06
7-136
4 10
0.000 34996.16
8
8.165E-
10
8.355E-06
0.000 35996.32
0
7.718E-
10
8.355E-06
0.000 36995.53
1
7.306E-
10
8.355E-06
0.000 37996.39
1
6.927E-
10
8.355E-06
0.000 38996.37
9
6.576E-
10
8.355E-06
0.000 39995.75
4
6.251E-
10
8.355E-06
0.000 41995.39
1
5.670E-
10
8.355E-06
0.000 43995.21
9
5.166E-
10
8.355E-06
0.000 45993.78
5
4.727E-
10
8.355E-06
0.000 47991.17
6
4.342E-
10
8.355E-06
0.000 49990.55
5
4.002E-
10
8.355E-06
0.000 51989.83
6
3.700E-
10
8.355E-06
0.000 53989.75
4
3.431E-
10
8.355E-06
0.000 55989.39
8
3.190E-
10
8.355E-06
0.000 57988.76
2
2.974E-
10
8.355E-06
0.000 59988.26
6
2.779E-
10
8.355E-06
7-137
Figure 7.65. cP and 2
1
cP versus wS for Example 7.4.
7.12.2 Calculation of Relative Permeabilities from Drainage Capillary Pressure Curve
The relative permeability to the wetting phase is given by
7-138
2
01
20
( )
wSw
cwrw w
w
c
dSPkk S
k dSP
α
α
+
+
= =∫
∫ (7.157)
The relative permeability to the non-wetting phase is given by
1
2
1
20
( ) w
w
cSnwrnw w
w
c
dSPkk S
k dSP
α
α
+
+
= =∫
∫ (7.158)
If α = 0, Eqs.(7.157) and (7.158) become
2
01
20
( )
wSw
cwrw w
w
c
dSPkk S
k dSP
= =∫
∫ (7.159)
and
1
2
1
20
( ) w
w
cSnwrnw w
w
c
dSPkk S
k dSP
= =∫
∫ (7.160)
The weakness of the above relative permeability models is that (krw+krnw) = 1,
which does not agree with experimental observations of relative permeability
functions. Experiments show that, in general, (krw+krnw) < 1. This lack of
agreement of the models with experiments is because the tortuosity of the
7-139
porous medium has been neglected. In fact, the tortuosity of the medium in
the presence of multiphase fluids is a function of saturation. Furthermore, the
models do not allow for residual saturations of the wetting and non-wetting
phases. These defects of the models will be corrected in Chapter 8 to derive
more realistic relative permeability curves from drainage capillary pressure
curves.
7.13 EMPIRICAL CAPILLARY PRESSURE MODELS
Often, it is desirable to fit analytical models to capillary pressure curves
to simply reservoir performance calculations, especially in numerical
simulations. Capillary pressure appears in the immiscible displacement model
as a derivative, either as c
D
Px
∂∂
or c
w
dPdS
. Therefore, an analytical model will allow
the derivatives to be calculated without numerical noise. Several empirical
analytical models are available for this purpose. We present two such models
from Brooks and Corey (1966) and van Genuchten (1980).
7.13.1 Brooks-Corey Capillary Pressure Models
A popular capillary pressure model in the petroleum industry and soil
physics is the Brooks-Corey model (Brooks and Corey, 1966). Based on
evaluations of several drainage capillary pressure curves for consolidated
porous media, Brooks and Corey observed that all the drainage capillary
pressure curves they examined could be represented by linear functions of the
form
*ln ln lnw c eS P Pλ λ= − + (7.161)
or
7-140
*1ln ln lnc w eP S Pλ
= − + (7.162)
by an appropriate choice of irreducible wetting phase saturation, where *wS is
the reduced wetting phase saturation defined by
*
1w wirr
wwirr
S SSS
−=
− (7.163)
In Eqs.(7.161) and (7.162), Pe is a constant given by the value of Pc on the
straight lines at *wS = 1, and λ is the pore size distribution index obtained from
the slopes of the straight lines. It should be observed from Eq.(7.162) that λ
controls the slope of the linear capillary pressure plot. A large value of λ gives
a small slope, which corresponds to the capillary pressure curve with a
narrow pore size distribution whereas a small value of λ gives a large slope,
which corresponds to the capillary pressure curve for a wide pore size
distribution. Thus, small values of λ indicate a wide pore size distribution
whereas large values of λ indicate a narrow pore size distribution. A porous
medium with a uniform pore size corresponds to λ = ∞ . In view of these
observations, Brooks and Corey called λ the pore size distribution index.
Eqs.(7.161) and (7.162) give a drainage capillary pressure model of the form
( )1
*c e wP P S λ
−= (7.164)
Brooks and Corey also proposed an imbibition capillary pressure model of the
form
( )1
1c e eP P S λ−⎡ ⎤
= −⎢ ⎥⎣ ⎦
(7.165)
7-141
where Se is the effective wetting phase saturation defined by
1
w wirre
wirr nwr
S SSS S
−=
− − (7.166)
where Snwr is the residual non-wetting phase saturation. To fit this model to
measured drainage capillary pressure data, a log-log plot of the drainage
capillary pressure data is made either as *ln wS versus ln cP or as ln cP versus
*ln wS . If the plot is nonlinear, then wirrS is adjusted until the plot is linear. Pe is
determined from the linear log-log plot at * 1wS = and λ is determined from the
slope of the straight line. The two parameters are then substituted into
Eqs.(7.164) and (7.165) to calculate the drainage and imbibition capillary
pressure curves.
Example 7.5
1. Fit the Brooks-Corey model to the air-water drainage capillary pressure
data given in Table 7.12 and compare the result of the model to the
original data.
2. Derive the spontaneous imbibition capillary pressure curve for the
sample using the Brooks-Corey spontaneous imbibition capillary
pressure model.
Table 7.12. Air-Water Drainage Capillary Pressure Data for Example 7.5.
Drainag
e
7-142
Pc
Sw psi
1.000 1.973
0.950 2.377
0.900 2.840
0.850 3.377
0.800 4.008
0.750 4.757
0.700 5.663
0.650 6.781
0.600 8.195
0.550 10.039
0.500 12.547
0.450 16.154
0.400 21.787
0.350 31.817
0.300 54.691
0.278 78.408
Solution to Example 7.4
1. Using Swirr = 0.278 as indicated by the drainage capillary pressure data
of Table 7.12, a log-log graph of *wS versus cP was plotted as shown in
Figure 7.66. Clearly, the graph is nonlinear. The curve fitting procedure
calls for the adjustment of Swirr until the graph is linear. Figure 7.67
shows the log-log plot with Swirr = 0.100. It is linear. Shown in Figure
7.68 is the corresponding log-log plot of cP versus *wS for Swirr = 0.100.
As expected, it too is linear.
7-143
Figure 7.66.Log-log plot of *wS versus cP for 0.278wirrS = for Example 7.5.
Figure 7.67. Log-log plot of *wS versus cP for 0.100wirrS = for Example 7.5.
7-144
Figure 7.68. Log-log plot of cP versus *wS for 0.100wirrS = for Example 7.5.
From Figure 7.68, the resulting Brooks-Corey drainage capillary
pressure straight line is
*ln 2.1443ln ln 2.2238c wP S= − +
Therefore,
1 2.1443λ
− = −
0.4664λ =
2.2238eP =
7-145
The Brooks-Corey drainage capillary pressure equation for this example
is
( ) ( )1 2.1443* *2.2238c e w wP P S Sλ
− −= =
Figure 7.69 shows a comparison of the Brooks-Corey capillary pressure
model with the measured capillary pressure curve. The agreement is
good. The Brooks-Corey log-log plot extended the capillary pressure
data to an irreducible wetting phase saturation of 0.10. It also changed
the displacement pressure from 1.973 to 2.224 psi. Figure 7.70 shows
the same comparison as in Figure 7.68 but on an enlarged capillary
pressure scale. The overall agreement between the model fit and the
original data is good.
Figure 7.69. A comparison of the Brooks-Corey drainage capillary pressure curve with the measured data for Example 7.5.
7-146
Figure 7.70. A comparison of the Brooks-Corey drainage capillary pressure curve with the measured data on an enlarged capillary pressure scale for
Example 7.5.
2. Figure 7.71 shows the predicted Brooks-Corey drainage and imbibition
capillary pressure curves for Snwr = 0.25 for the sample of Example 7.5.
Table 7.13 shows the original capillary pressure data along with the
calculated results with the Brooks-Corey model.
7-147
Figure 7.71. A comparison of the Brooks-Corey drainage and imbibition capillary pressure curves for the sample of Example 7.5.
Table 7.13. Calculated Results with the Brooks-Corey Model for Example 7.5.
Original
Data Brooks-Corey Model
Drainage Drainage Imbibition
Pc Pc Pc
Sw psi *wS psi Se psi
1.000 1.973 1.000 2.224
0.950 2.377 0.944 2.514
0.900 2.840 0.889 2.863
0.850 3.377 0.833 3.288
0.800 4.008 0.778 3.812
0.750 4.757 0.722 4.468 1.000 0.000
0.700 5.663 0.667 5.305 0.923 0.416
7-148
0.650 6.781 0.611 6.393 0.846 0.958
0.600 8.195 0.556 7.843 0.769 1.679
0.550 10.039 0.500 9.831 0.692 2.669
0.500 12.547 0.444 12.656 0.615 4.075
0.450 16.154 0.389 16.851 0.538 6.163
0.400 21.787 0.333 23.452 0.462 9.448
0.350 31.817 0.278 34.672 0.385 15.032
0.300 54.691 0.222 55.947 0.308 25.620
0.278 78.408 0.198 71.829 0.274 33.524
0.250 0.167 103.678 0.231 49.374
0.200 0.111 247.331 0.154 120.867
0.150 0.056 1093.394 0.077 541.933
0.100 0.000 0.000
The Brooks-Corey capillary pressure model cannot adequately fit a
capillary pressure curve with an inflection (S-shaped curved) such as shown
in Figure 7.13 for sample C. In fact, fitting the Brooks-Corey model to such a
capillary pressure curve will eliminate the inflection and turn the capillary
curve into a hyperbola. The shape of the drainage capillary pressure curve of
Example 7.5 was that of a hyperbola. As a result, the Brooks-Corey model
gave a good fit of the data. If the original data had an inflection as typically
observed in poorly sorted porous media, the Brooks-Corey model would have
given a poor fit, especially at high wetting phase saturations. Since the shape
of the drainage capillary pressure curve is a reflection of the pore throat size
distribution, other capillary pressure models have been devised that preserve
the shape of the drainage capillary pressure curve, especially at high wetting
phase saturation.
7-149
7.13.2 van Genuchten Capillary Pressure Model
An empirical capillary pressure model that preserves the shape of the
capillary pressure curve at high wetting phase saturations was proposed by
van Genuchten (1980). This model is widely used in soil physics and in
hydrology. The model is given by
( )
* 11
m
w nc
SPα
⎡ ⎤= ⎢ ⎥
+⎢ ⎥⎣ ⎦ (7.167)
where α, n and m are the fitting parameters.
Example 7.6
Fit the van Genuchten model to the air-water drainage capillary pressure data
given in Table 7.14 and compare the result of the model to the original data.
Solution to Example 7.6
Table 7.14 shows the calculated results for the van Genuchten model using
the following parameters:
α = 0.018
m = 45
n = 6
Figure 7.72 compares the van Genuchten model with the measured data. The
agreement is good especially at high values of the wetting phase saturation. It
should be observed that this model can match capillary pressure curves with
inflections, unlike the Brooks-Corey model. It can also match a capillary
pressure curve with a hyperbolic shape.
7-150
Table 7.14. Calculated Results for van Genuchten model.
Measured Data
van Genuchten
Model
Pc
Sw (psi) *wS *
wS Sw
0.20 50.0 0.000 0.000 0.200
0.24 39.0 0.050 0.006 0.205
0.25 38.0 0.063 0.012 0.210
0.28 35.0 0.100 0.065 0.252
0.30 34.0 0.125 0.100 0.280
0.35 32.0 0.188 0.199 0.359
0.40 30.0 0.250 0.332 0.466
0.50 28.5 0.375 0.444 0.555
0.56 28.0 0.450 0.481 0.585
0.60 27.8 0.500 0.496 0.597
0.68 27.0 0.600 0.555 0.644
0.70 26.8 0.625 0.569 0.655
0.70 26.8 0.625 0.569 0.655
0.75 26.0 0.688 0.625 0.700
0.80 24.8 0.750 0.701 0.761
0.85 23.0 0.813 0.798 0.838
0.90 21.0 0.875 0.877 0.902
0.95 18.0 0.938 0.949 0.959
1.00 13.5 1.000 0.991 0.993
7-151
Figure 7.72. A comparison of van Genuchten model with measured capillary pressure data for Example 7.6.
7.14 CAPILLARY TRAPPING IN POROUS MEDIA
Capillary trapping ensures that an immiscible displacement at normal
interfacial tensions and rates is never complete. There is always a residual
phase that is trapped. Several models have been proposed to explain capillary
trapping. We examine two such models here: the pore doublet model and the
snap-off model.
7.14.1 Pore Doublet Model of Capillary Trapping
Figure 7.73 shows the pore doublet model, which consists of two pores
that are joined at the inlet and outlet ends, with one pore larger than the
other. The pore doublet is initially filled with a non-wetting phase. A wetting
phase is then injected at a rate q to displace the non-wetting phase from both
7-152
branches of the pore doublet. The problem is to determine which of the two
interfaces in the capillary tubes will arrive at the outlet first (Point B). We will
assume that once the interface in one of the capillary tubes has arrived at B,
the non-wetting phase in the other tube will be trapped. To determine which
interface will arrive at B first, we need to derive the expressions for the
velocities of the interfaces as a function of the relevant parameters of the
model. Although it is not necessary to do so, let us assume that the wetting
and non-wetting phases have the same viscosity to simplify the analysis.
Figure 7.73. Pore doublet model. (a) in a porous medium; (b) capillary tubes
approximation.
The pressure drop across each capillary tube is given by
7-153
A B A w w nw nw BP P P P P P P P− = − + − + − (7.168)
where Pw and Pnw are the pressures on either side of the interface. From
Hagen-Poisseuille's law,
( )41
118
A wP PrqL
πμ
−= (7.169)
( )( )
41
118
nw BP PrqL L
πμ
−=
− (7.170)
where L1 is the distance of the interface from the inlet end and L is the total
length of the pore doublet from A to B. Substituting Eqs.(7.169) and (7.170)
into Eq.(7.168) and noting that (Pnw - Pw) is the capillary pressure gives
114
1
8A B c
q LP P Prμ
π− = − (7.171)
Similarly, for the second capillary tube,
224
2
8A B c
q LP P Prμ
π− = − (7.172)
Equating Eqs.(7.171) and (7.172) and rearranging gives
( )1 2 2 14 41 2 2 1
8 8 1 12 cosc cL Lq q P P
r r r rμ μ σ θ
π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞
− + = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(7.173)
where the Laplace equation has been used to replace the capillary pressures.
Assuming incompressible fluids,
1 2q q q+ = (7.174)
7-154
Eqs.(7.173) and (7.174) are two linear simultaneous equations in q1 in q2,
which can easily solved to obtain
4
2 2 11
4 41 2
8 1 12 cos
8 8
L qr r r
qL L
r r
μ σ θπ
μ μπ π
⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠=⎛ ⎞ ⎛ ⎞
+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(7.175)
4
1 2 12
4 41 2
8 1 12 cos
8 8
L qr r r
qL L
r r
μ σ θπ
μ μπ π
⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠=⎛ ⎞ ⎛ ⎞
+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(7.176)
Diving Eq.(7.176) by (7.175) and simplifying gives
4 42 2
1 2 124
1 2
2 1
cos 1 14cos 1 1
4
r rqr L r rq
q rqL r r
π σ θμ
π σ θμ
⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠=⎛ ⎞
− −⎜ ⎟⎝ ⎠
(7.177)
The interface velocities are given by
11 2
1
qvrπ
= (7.178)
22 2
2
qvrπ
= (7.179)
Let
2
1
rr
β⎛ ⎞
= ⎜ ⎟⎝ ⎠
(7.180)
7-155
31 cosvcapq LN
rμ
π σ θ= (7.181)
Substituting Eqs.(7.178) through (7.181) into (7.177) gives the ratio of the
interface velocities as
2
212
14 1
4 1 1
vcap
vcap
Nv
Nvβ
ββ β
⎛ ⎞+ −⎜ ⎟
⎝ ⎠=⎛ ⎞
− −⎜ ⎟⎝ ⎠
(7.182)
Eq.(7.182) can be used to determine the conditions under which the
non-wetting phase will be trapped in the smaller pore or in the larger pore. If
2
1
1vv
> (7.183)
the non-wetting phase will be trapped in the smaller pore. Substituting
Eq.(7.182) into (7.183) gives the condition for trapping in the smaller pore as
( )( )
2 14 1vcapN
β β
β
+>
+ (7.184)
If
2
1
1vv
< (7.185)
the non-wetting phase will be trapped in the larger pore. Substituting
Eq.(7.182) into (7.185) gives the condition for the non-wetting phase to be
trapped in the larger pore as
7-156
( )( )
2 14 1vcapN
β β
β
+<
+ (7.186)
The critical value of vcapN for trapping in either pore is given by
( )( )
2 14 1vcapcriticalN
β β
β
+=
+ (7.187)
For fixed values of r1, r2, σcosθ, μ and L, vcapN will depend on the rate q. If q is
low, the displacement will be dominated by capillary forces and the non-
wetting phase will be trapped in the larger pore resulting in a low
displacement efficiency. If q is high, the displacement will be dominated by
the viscous forces and the non-wetting phase will be trapped in the smaller
pore resulting in a high displacement efficiency. These observations are in
qualitative agreement with macroscopic observations in corefloods. Figure
7.74 shows the breakthrough oil recovery as a function of vμL obtained by
Rapoport and Leas (1953). The breakthrough oil recovery here is the oil
recovery at the time of water arrival at the outlet end of the core. As such, it is
a measure of the displacement efficiency. It should be observed that vμL is
directly proportional to the macroscopic version of vcapN . Clearly, the
displacement efficiency increases with an increase in vμL or vcapN in agreement
with the prediction of the pore doublet model.
7-157
Figure 7.74. Breakthrough oil recovery versus Rapaport and Leas scaling coefficient, vμL (Rapoport and Leas, 1953).
If
2
1
0vv
< (7.188)
the domination of the displacement by the capillary forces will be strong that
the interface in the larger pore will retract resulting in the trapping of the
non-wetting phase in the larger pore and low displacement efficiency.
Substituting Eq.(7.182) into (7.188) gives the condition for this to happen as
7-158
1 114vcapN
β⎛ ⎞
< −⎜ ⎟⎝ ⎠
(7.189)
It is interesting to compare vcapN with capN , which is the relevant
dimensionless number for diagnosing the problem of capillary end effect.
From Eq.(7.61),
1coscap
q LN A k
μσ θ φ
= (7.190)
A comparison of Eqs.(7.181) and (7.190) shows that both dimensionless
numbers give the ratio of viscous to capillary forces or vice versa. capN gives the
ratio of capillary to viscous forces at the macroscopic scale whereas vcapN gives
the ratio of the viscous to capillary forces at the pore scale. Once again, we
see the competition between viscous and capillary forces that pervades
immiscible displacements in porous media.
Chatzis and Dullien (1983) and Laidlaw and Wardlaw (1983) provide
more detailed theoretical analyses of the pore doublet model along with
experimental verifications.
7.14.2 Snap-Off Model of Capillary Trapping
When a non-wetting phase is forced through a pore constriction, it can
suffer from capillary instability and snap off (breakup) after exiting the
constriction. This phenomenon is controlled by the ratio of the pore body to
the pore throat size. Figure 7.75 shows oil being displaced in two pores, one
with a low aspect ratio and the other with a high aspect ratio. The aspect ratio
is define as
7-159
1
2
DAspect RatioD
= (7.191)
where 1D and 2D are the pore body diameter and the pore throat diameter,
respectively. In Figure 7.75a, the aspect ratio is low and the oil is displaced
through the pore without trapping. In Figure 7.75b, the aspect ratio is high
and the oil suffers capillary instability and snaps off at the pore throat and be
trapped.
Figure 7.75. Capillary trapping by snap-off mechanism in a single pore.(a) low
aspect ratio; (b) high aspect ratio (Chatzis et al., 1983).
7-160
Figure 7.76 shows a drop of non-wetting fluid passing through a pore
constriction is a uniform pack of spheres. The drop will become unstable and
breakup when the capillary pressure at the pore neck exceeds the capillary
pressure at the leading edge of the drop. The condition for snap-off is given by
1 1 2cn
n t f
Pr r r
σσ⎛ ⎞
= − >⎜ ⎟⎝ ⎠
(7.192)
Figure 7.76. Snap-off in a porous medium (Stegemeier, 1976).
7-161
Figure 7.77 shows a sequence of mercury injection and withdrawal into
a pore system. Figure 7.77A shows the capillary pressure hysteresis loops
whereas Figure 7.77B shows mercury trapping by snap-off corresponding to
the hysteresis loops.
7.14.3 Mobilization of Residual Non-Wetting Phase
Once a phase has been trapped, the pressure gradient required to
mobilize it can be significantly higher than can be generated under the
flooding conditions. Let us calculate the pressure gradient required to
mobilize a trapped oil blob in a waterflood as shown in Figure 7.78. For the
blob to pass through the pore throat, the pressure drop across the leading
edge must exceed the entry pressure or the displacement pressure of the pore
throat. Thus, the condition for the blob to pass through the pore throat is
given by
'
1
2 cosB BP P
rσ θ
− ≥ (7.193)
From Laplace equation,
'
2
2 cosA AP P
rσ θ
− = (7.194)
Subtracting Eq.(7.194) from (7.193) gives
( ) ( )' '
1 2
1 12 cosB B A AP P P Pr r
σ θ⎛ ⎞
− − − ≥ −⎜ ⎟⎝ ⎠
(7.195)
7-162
Figure 7.77. Snap-off in mercury injection-withdrawal experiment. A: capillary pressure scanning curves; B: corresponding mercury trapping by snap-off
(Stegemeier, 1976).
7-163
or
( ) ( )' '
1 2
1 12 cosA B B AP P P Pr r
σ θ⎛ ⎞
− + − ≥ −⎜ ⎟⎝ ⎠
(7.196)
Because ' 'B AP P= , Eq.(7.196) becomes
( )1 2
1 12 cosA BP Pr r
σ θ⎛ ⎞
− ≥ −⎜ ⎟⎝ ⎠
(7.197)
The pressure gradient required to mobilize the blob is given by
( )1 2
2 cos 1 1A BP PL L r r
σ θ− ⎛ ⎞≥ −⎜ ⎟
⎝ ⎠ (7.198)
Figure 7.78. Trapped oil blob.
Let us estimate the pressure gradient required to mobilize a trapped oil
droplet in a normal waterflood in a reservoir rock using typical values of the
relevant parameters. Let
1 10r mμ=
7-164
2 50r mμ=
50L mμ=
30 /dynes cmσ =
0θ =
500 wk md=
1 w cpμ =
1 /wv ft day=
The pressure gradient required to mobilize the droplet is given by
( )( )( )( ) ( )( ) ( )( )
( )( )66 6 6
2 30 1 1 1 14.696 30.48 /1.0133 1050 10 100 10 10 100 50 10 100
A BP P psi ftL xx x x− − −
⎡ ⎤− ⎛ ⎞⎢ ⎥≥ − ⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
or
4243.73 /A BP P psi ftL−
≥
This is an enormous pressure gradient requirement. The pressure gradient
generated by the waterflood is given by Darcy's law as
( )( )( )( )( )
1 10.32 /
0.001127 5.615 500P psi ftL
Δ= =
We see that the pressure gradient generated by the waterflood is not sufficient
to mobilize the oil droplet. Therefore, it will remain trapped.
7-165
7.14.4 Oil Migration
Figure 7.79 shows an upward migrating oil bubble from a source rock
into a reservoir initially fully saturated with water. The migrating bubble has
encountered a restriction at a pore throat of radius rH. Obviously, in order for
migration to continue, the leading end of the bubble (A) must squeeze through
the pore throat. Assuming that ends A and B of the bubble are hemispherical
with B having a radius rB, one can calculate the length of the oil blob required
for the blob to pass through the restriction and continue its upward
migration. To pass through the restriction, the capillary pressure at the
leading edge of the blob must exceed the displacement pressure of the
restriction. Therefore, the condition for upward migration through the
restriction is
( ) 2oA wA
H
P Prσ
− > (7.199)
Laplace equation gives
( ) 2oB wB
B
P Prσ
− = (7.200)
From hydrostatics through the water,
wB wA wP P ghρ= + (7.201)
7-166
Figure 7.79. Migrating oil filament.
7-167
From hydrostatics through the oil,
oB oA oP P ghρ= + (7.202)
Subtracting Eq.(7.201) from (7.202) gives
( ) ( ) ( )oB wB oA wA w oP P P P ghρ ρ− = − − − (7.203)
Substituting Eq.(7.200) into (7.203) and rearranging gives
( ) ( ) 2oA wA w o
B
P P ghrσρ ρ− = − + (7.204)
Substituting Eq.(7.204) into (7.199) gives the condition for upward migration
as
( ) 2 2w o
B H
ghr rσ σρ ρ− + > (7.205)
or
( )
2 1 1
w o H B
hg r r
σρ ρ
⎛ ⎞> −⎜ ⎟− ⎝ ⎠
(7.206)
Let
10Hr mμ=
50Br mμ=
30 /dynes cmσ =
ρw = 1.00 g/cm3
7-168
ρo = 0.70 g/cm3
The minimum length of the blob to continue its upward migration is given by
( )( )( )( ) ( )( ) ( )( )6 6
2 30 1 1 163.1 1 0.70 981 10 10 100 50 10 100
h cmx x− −
⎡ ⎤⎢ ⎥> − =
− ⎢ ⎥⎣ ⎦
7.15 EFFECTS OF WETTABILITY AND INTERFACIAL TENSION ON CAPILLARY PRESSURE CURVES
The effects of wettability and interfacial tension on capillary pressure
curves can easily be deduced from Laplace equation. The effect of wettability
on capillary pressure curve is shown qualitatively in Figure 7.80. As the
degree of preferential wettability is reduced, the capillary pressure curve will
decrease. The effect of interfacial tension on capillary pressure is shown
qualitatively in Figure 7.81. Capillary pressure decreases as the interfacial
tension of the fluids decrease.
7-169
Figure 7.80. Effect of wettability on capillary pressure curve.
7-170
Figure 7.81. Effect of interfacial tension on capillary pressure curve.
NOMENCLATURE
A = cross sectional area in the flow direction
Bo = oil formation volume factor
Bw = water formation volume factor
do = depth of free water level below water oil contact
fw = fractional flow of wetting phase
fw = fractional flow of water
fnw = fractional flow of non-wetting phase
fnw2 = fractional flow of non-wetting phase at the outlet end of porous medium
7-171
fo = fractional flow of oil
f(R) = probability density function for pore volume distribution
Fw = approximate fractional flow of wetting phase
1F = lithology factor
g = gravitational acceleration
h = height above water oil contact
J = Leverett J-function
k = absolute permeability of the medium
ko = effective permeability to oil
kw = effective permeability to wetting phase
knw = effective permeability to non-wetting phase
kwr = end-point relative permeability to wetting phase
kg = effective permeability to gas
kro = relative permeability to oil
krw = relative permeability to water
krg = relative permeability to gas
krw = relative permeability to wetting phase
krnw= relative permeability to non-wetting phase
knwr= end-point relative permeability to non-wetting phase
L = length
M = mobility ratio
ME = end-point mobility ratio
N = centrifuge speed in revolutions per minute
NpD = dimensionless cumulative production
Ncap = dimensionless capillary to viscous force ratio
P = pressure
Pc = capillary pressure
Pc1 = capillary pressure at the inlet end of core in a centrifuge
Pd = displacement pressure
Pe = displacement pressure for Brooks-Corey model
7-172
Pg = pressure in the gas phase
Pnw= pressure in the non-wetting phase
Po = pressure in the oil phase
Pw = pressure in the water phase
Pw = pressure in the wetting phase
q = total volumetric injection rate
qo = volumetric flow rate of oil
qg = volumetric flow rate of gas
qnw= volumetric flow rate of non-wetting phase
qw = volumetric flow rate of water
qw = volumetric flow rate of wetting phase
r = radius of capillary tube
mr = mean radius of curvature of an interface
1 2,r r = radii of curvature of an interface
1 2,r r = distance of inlet end and outlet end of core from the center of rotation in a centrifuge
R = pore throat radius
Se = effective wetting phase saturation
Sg = gas saturation
So = oil saturation
Sor = residual oil saturation
Sw = water saturation
Sw = wetting phase saturation
Swirr = irreducible wetting phase saturation
Swro = wetting phase saturation at which the imbibition capillary pressure is zero
Sw1 = wetting phase saturation at the inlet end of porous medium in a centrifuge
Snw= non-wetting phase saturation
Snwr = residual non-wetting phase saturation
7-173
Swav = average wetting phase saturation
Swav = average water saturation
Swf = frontal saturation
*wS = normalized wetting phase saturation
t = time
Dt = dimensionless time
v = flux vector, Darcy velocity vector
vw = Darcy velocity for the wetting phase
vnw = Darcy velocity for the non-wetting phase
x = distance in the direction of flow
Dx = dimensionless distance
z = height above free water level
wρ = density of water
wρ = density of wetting phase
nwρ = density of non-wetting phase
σ = interfacial tension
θ = contact angle
λ = pore size distribution index
μ = viscosity
μg = gas viscosity
μο = oil viscosity
μw = water viscosity
μw = wetting phase viscosity
μnw= non-wetting phase viscosity
δ(R) = probability density function for pore throat size distribution based on bundle of capillary tubes model of porous medium
φ = porosity, fraction
τ = tortuosity
γ = liquid specific gravity
7-174
ω = angular velocity of centrifuge
ΔP = pressure drop
ΔPw = pressure drop in the wetting phase
ΔPnw = pressure drop in the non-wetting phase
Γ = pore structure
REFERENCES AND SUGGESTED READINGS
Abrams, A. : "The Influence of Fluid Viscosity, Interfacial Tension, and Flow Velocity on Residual Oil Saturation Left by Waterflood," Soc. Pet. Eng. Jour. (Oct., 1975) 437-447.
Adamson, A.W. and Gast, A.P.: Physical Chemistry of Surfaces, Sixth Edition, John Wiley and Sons, Inc., New York, 1997.
Afzal, N.: Investigation of Heterogeneities and Fluid Displacements in Porous Media Using CT, MS Thesis, The University of Texas at Austin, May 1992.
Akima, H.: "A New Method of Interpolation and Smooth Curve Fitting Based on Local Procedures," J. ACM, Vol.17, No. 4 (Oct. 1970) 589-602.
Anderson, W.G. : “Wettability Literature Survey - Part 4: Effects of Wettability on Capillary Pressure,” J. Pet. Tech. (October 1987) 1283-1300.
Anderson, W.G. : “Wettability Literature Survey - Part 6: The Effects of Wettability on Waterflooding,” J. Pet. Tech. (December 1987) 1605-1622.
Archer, J.S. and Wall, C.G. : Petroleum Engineering, Graham & Trotman, London, England, 1986.
Arriola, A., Willhite, G.P. and Green, D.W. : "Trapping of Oil Drops in a Noncircular Pore Throat and Mobilization Upon Contact With a Surfactant," SPEJ (Feb. 1983) 99-114.
Bear, J. : Dynamics of Fluids in Porous Media, Elsevier, New York, 1972.
Benner, F.C. and Bartell, F.E. : “The Effect of Polar Impurities Upon Capillary and Surface Phenomena in Petroleum Production,” API Drilling and Production Practice (1941) 341-348.
Bethel, F.T. and Calhoun, J.C. : "Capillary Desaturation in Unconsolidated Beads," Trans. AIME (1953) Vol. 198, 197-202.
Brooks, R.H. and Corey, A.T. : “Properties of Porous Media Affecting Fluid Flow,” Jour. Irrigation and Drainage Div., Proc. Amer. Soc. Of Civil Engr. (June, 1966) 61-88.
7-175
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8-1
CHAPTER 8
RELATIVE PERMEABILITY
8.1 DEFINITION OF RELATIVE PERMEABILITY
In a petroleum reservoir, it is possible for two or three fluids to flow
simultaneously. Examples are (a) the flow of gas and water in a gas reservoir,
(b) the flow of oil and water in an oil reservoir, (c) the flow of oil and gas in an
oil reservoir and (d) the flow of oil, water and gas in an oil reservoir. In
multiphase flow situations, the absolute permeability of the porous medium is
no longer sufficient to calculate the flow rate of each fluid type or to calculate
the total flow rate of all the fluids.
In order to make quantitative predictions for multiphase flow, we need
to know the permeability to each fluid in the presence of the other fluids in
the rock. The permeability of one fluid in the presence of the other immiscible
fluids is known as the effective permeability to that fluid. To calculate the
flow rate of each fluid in multiphase flow, we extend Darcy’s Law to
multiphase flow. For example, for simultaneous flow of oil, water and gas in
an inclined linear system, Darcy’s Law is applied to each phase as follows:
sino oo o
o
k A Pq gx
ρ αμ
∂⎛ ⎞= − +⎜ ⎟∂⎝ ⎠ (8.1)
sinw ww w
w
k A Pq gx
ρ αμ
∂⎛ ⎞= − +⎜ ⎟∂⎝ ⎠ (8.2)
8-2
sing gg g
g
k A Pq g
xρ α
μ∂⎛ ⎞
= − +⎜ ⎟∂⎝ ⎠ (8.3)
where α is the angle of inclination with the horizontal. Eqs.(8.1) to (8.3) show
that using the concept of an effective permeability, Darcy’s Law is applied to
each phase as if the other phases did not exist. Capillary equilibrium between
the phases gives
( )/o w c ow wP P P S− = (8.4)
( )/g o c go oP P P S− = (8.5)
/ / /c ow c go c gwP P P+ = (8.6)
It is often more convenient to work with a dimensionless effective
permeability known as the relative permeability obtained by dividing the
effective permeability by a base permeability such as the absolute
permeability of the porous medium. Thus, for the three phase example, using
the absolute permeability of the porous medium as the base permeability, the
relative permeabilities to oil, water and gas are given by
oro
kkk
= (8.7)
wrw
kkk
= (8.8)
grg
kk
k= (8.9)
In terms of relative permeabilities, Eqs.(8.1) through (8.3) become
sinro oo o
o
kk A Pq gx
ρ αμ
∂⎛ ⎞= − +⎜ ⎟∂⎝ ⎠ (8.10)
8-3
sinrw ww w
w
kk A Pq gx
ρ αμ
∂⎛ ⎞= − +⎜ ⎟∂⎝ ⎠ (8.11)
sinrg gg g
g
kk A Pq g
xρ α
μ∂⎛ ⎞
= − +⎜ ⎟∂⎝ ⎠ (8.12)
Sometimes, the effective permeability to the non-wetting phase at the
irreducible wetting phase saturation is used as the base permeability for
defining the relative permeability. In this case, the end point relative
permeability to the non-wetting phase will be 1.0. As the base pressure
appears in Darcy's law as shown in Eqs.(8.10) to (8.12), it is necessary to
ascertain the base permeability used to define a relative permeability curve
before such a curve is used in performance calculations. Failure to do so will
lead to wrong results.
Figure 8.1 shows typical imbibition relative permeability curves for a
two-phase system. The following observations can be made about the key
features of the relative permeability curves.
1. The relative permeability curves are nonlinear functions of fluid
saturation.
2. The sum of the relative permeabilities at each saturation is always less
than 1.0.
3. There is an irreducible wetting phase saturation (Swirr) at which the
relative permeability to the wetting phase is zero and the relative
permeability to the non-wetting phase attains a maximum end point
value (knwr)
4. There is a residual non-wetting phase saturation (Snwr) at which the
relative permeability to the non-wetting phase is zero and the relative
8-4
permeability to the wetting phase attains a maximum end point value
(kwr).
Figure 8.1. Typical imbibition relative permeability curves.
5. The relative permeability curves are not defined in the saturation ranges
given by ( )1 1nwr wS S− < < and 0 w wirrS S< < .
6. Two phase flow occurs over the saturation range ( )1wirr w nwrS S S< < − .
Imbibition relative permeability curves typically are used to perform the
following reservoir performance calculations:
• Waterflood calculations in a water wet reservoir in which water
displaces oil and/or gas.
8-5
• Natural water influx calculations in a water wet reservoir in which water
displaces oil and/or gas.
• Oil displaces gas, which occurs when oil is forced into a gas cap.
Figure 8.2 shows typical drainage relative permeability curves with
features that are very similar to the imbibition curves of Figure 8.1. The
obvious differences between the drainage and imbibition curves are that the
drainage curve for the wetting phase starts at a wetting phase saturation of
1.0 and that of the non-wetting phase is zero at the wetting phase saturation
of 1.0. This is because at the start of the drainage relative permeability
measurements, the porous medium was fully saturated with the wetting
phase. The permeability of the wetting phase is then equal to the absolute
permeability of the porous medium. Of course, at the start of the experiment,
there was no non-wetting phase in the medium. Therefore, the relative
permeability to the non-wetting phase must be zero. This is the only occasion
in which the sum of relative permeabilities is equal to 1.0 because there was
only one phase present. As will be shown later, because of capillary pressure
hysteresis, the drainage and imbibition relative permeability curves will be
different.
Drainage relative permeability curves typically are used to perform the
following reservoir performance calculations:
• Solution gas drive calculations in which gas displaces oil.
• Gravity drainage calculations in which gas replaces drained oil.
• Gas drive calculations in which gas displaces oil and/or water.
• Oil or gas displacing water in tertiary recovery processes.
8-6
Figure 8.2. Typical drainage relative permeability curves.
8.2 LABORATORY MEASUREMENT OF TWO-PHASE RELATIVE PERMEABILITIES BY THE STEADY STATE METHOD
The most straight-forward laboratory measurement technique for
relative permeabilities is the steady state method. For imbibition relative
permeability measurement, the test starts with the core initially saturated
with an irreducible wetting phase saturation (Swirr) and a non-wetting phase
saturation of (1-Swirr). Then a mixture of the two phases is injected into the
inlet face of the core at a fixed ratio of nw
w
⎛ ⎞⎜ ⎟⎝ ⎠
until steady state is achieved.
Steady state is achieved when the pressure drop across the core no longer
changes with time and the ratio of the produced fluids is the same and the
ratio of the injected fluids. The steady state pressure drop across the core and
8-7
the injection rate of each phase are measured. The relative permeabilities are
calculated with the integrated forms of Darcy's law for two phase flow as
shown later. The saturations are usually calculated by material balance.
At steady state, the continuity equations for the wetting and non-
wetting phases for horizontal flow are
0wvx
∂=
∂ (8.13)
0nwvx
∂=
∂ (8.14)
Darcy's law applied to each phase gives
a constantw ww
w
k Pvxμ
∂= − =
∂ (8.15)
a constantnw nwnw
nw
k Pvxμ
∂= − =
∂ (8.16)
From capillary equilibrium,
( )nw w c wP P P S− = (8.17)
where Pc(Sw) in this case is the imbibition capillary pressure. If the
saturations, Sw and Snw, are uniform throughout the porous medium, then
kw, knw and Pc are independent of x. Then, Eqs.(8.15) and (8.16) can be
integrated to give
w w w ww
w w
v L q LkP A P
μ μ= =
Δ Δ (8.18)
nw nw nw nwnw
nw nw
v L q LkP A P
μ μ= =
Δ Δ (8.19)
Since ( )c wP S is uniform, wPΔ and nwPΔ are equal and the pressure drop across
the core can be measured in either phase and used to calculate the effective
8-8
permeabilities with Eqs.(8.18) and (8.19). The steady state saturation
distribution in the core can be calculated with Eq.(7.83), which is reproduced
here for convenience:
1w
ww
Dcap rnw
w
fFdS
dJdx N kdS
⎛ ⎞−⎜ ⎟
⎝ ⎠= (7.83)
with a specified inlet boundary condition. Of course, to do so, rwk , rnwk and
( )c wP S must be known.
If the saturation distribution in the core is not uniform because of
capillary end effect, then Eqs.(8.18) and (8.19) are not valid and cannot be
used to calculate the effective permeabilities. As there are no alternative
equations to use, the relative permeability experiment will be a failure.
Therefore, in the steady state experiment, the total injection rate ( )w nwq q q= +
should be sufficiently high to minimize capillary end effect as outlined in
Chapter 7.
Figure 8.3 shows an apparatus that can be used for the steady state
experiment. A typical sequence of steps for obtaining the imbibition relative
permeability curves might be as follows:
1. Install the clean, dry core sample in the Hassler apparatus as shown in
Figure 8.3. Evacuate the core and saturate with the wetting phase.
Determine the absolute permeability of the core by wetting phase flow.
2. Displace the wetting phase with the non-wetting phase until no more
wetting phase flows from the core. Calculate the irreducible wetting
phase saturation and the initial non-wetting phase saturation.
Measure the steady state pressure drop and the non-wetting phase
8-9
injection rate and calculate the relative permeability to the non-wetting
phase at the irreducible wetting phase saturation by use of Eq.(8.19) as
nw nwrnw
nw
q LkkA Pμ
=Δ
(8.20)
3. Inject a mixture of the wetting and non-wetting phases at rates qnw and
qw such that the ratio, qw/qnw, is very much less than 1 until steady
state is achieved. Steady state is achieved when the injected and
produced qw/qnw ratios are equal and the pressure drop no longer
changes with time.
4. Measure the pressure drop and calculate the wetting phase saturation
by material balance. Calculate the relative permeabilities to the non-
wetting and wetting phases at the latest wetting phase saturation using
Eq.(8.20) and (8.18) as
w wrw
w
q LkkA Pμ
=Δ
(8.21)
5. Increase the ratio qw/qnw and repeat steps 3 and 4 to calculate the
relative permeabilities at higher and higher wetting phase saturations.
6. Finally, inject only the wetting phase until no more non-wetting phase
flows from the core. Calculate the residual non-wetting phase
saturation. Measure the steady state pressure drop and the wetting
phase injection rate and calculate the relative permeability to the
wetting phase at residual non-wetting phase saturation. This completes
the relative permeability measurements.
In the steady state relative permeability experiment, it is necessary to
minimize capillary end effect. This can be accomplished by injecting at a
sufficiently high total rate or by other means as discussed by Richardson et
al. (1952). Figures 8.4 and 8.5 show the pressure profiles in the gas and oil
8-10
Figure 8.3. Hassler’s apparatus for relative permeability measurement (Osoba
et al., 1951).
phases for a gas-oil steady state relative permeability experiment conducted at
two rates by Richardson et al. (1952). At the lower injection rate (Figure 8.4),
capillary end effect is apparent whereas at the higher rate (Figure 8.5), there
is little or no capillary end effect. Note also, that when the capillary end effect
has been eliminated, the pressure drop in each phase is the same. Therefore,
the pressure drop measured in either phase is sufficient for calculating both
relative permeabilities.
8-11
The various steady state methods such as the Penn State method,
single core dynamic method, dispersed feed method, Hafford method and
Hassler method differ primarily in the techniques used to minimize or
eliminate capillary end effect (Richardson et al., 1952). When capillary end
effect has been eliminated, all the steady state methods give the same results
as shown in Figures 8.6 and 8.7.
Figure 8.4. Steady state oil and gas pressure profiles at a relatively low injection rate (Richardson et al., 1952).
8-12
Figure 8.5. Steady state oil and gas pressure profiles at a relatively high injection rate (Richardson et al., 1952).
8-13
Figure 8.6. Relative permeability curves from six steady state methods, short core section (Richardson et al., 1952).
8-14
Figure 8.7. Relative permeability curves from six steady state methods, long core section (Richardson et al., 1952).
The major problem with the steady state method for relative
permeability measurements is that it takes too long to complete. It is not
unusual for a steady state experiment to take several weeks to complete. An
alternative and much faster technique is the unsteady state method or the
dynamic displacement method based on immiscible displacement theory.
Because the calculation of relative permeabilities from unsteady state
experiment is based on the solution of two-phase immiscible displacement
equation, we must first solve the two-phase immiscible displacement problem
before we can discuss the unsteady state relative permeability measurements.
8-15
8.3 THEORY OF ONE DIMENSIONAL IMMISCIBLE DISPLACEMENT IN A POROUS MEDIUM
8.3.1 Mathematical Model of Two-Phase Immiscible Displacement
Consider the displacement of a non-wetting phase by a wetting phase in
a linear inclined core as shown in Figure 8.8. Darcy’s Law applied to each
phase gives
sinrnw nwnw nw
nw
kk A Pq gx
ρ αμ
∂⎛ ⎞= − +⎜ ⎟∂⎝ ⎠ (8.22)
sinrw ww w
w
kk A Pq gx
ρ αμ
∂⎛ ⎞= − +⎜ ⎟∂⎝ ⎠ (8.23)
Capillary equilibrium gives
( )nw w c wP P P S− = (8.17)
Assuming incompressible fluids, mass conservation requires that
w nwq q q= + (8.24)
The true fractional flows of the wetting and non-wetting phases are defined as
follows:
w ww
w nw
q qfq q q
= =+
(8.25)
1nw nwnw w
w nw
q qf fq q q
= = = −+
(8.26)
The continuity equation for the wetting phase is
0w wS qAt x
φ ∂ ∂+ =
∂ ∂ (8.27)
8-16
Figure 8.8. Displacement of a non-wetting phase by a wetting phase in an inclined core.
The saturation constraint gives
1w nwS S+ = (8.28)
Subtracting Eq.(8.22) from (8.23) and rearranging gives
( ) sinw w nw nw nw ww nw
rw rnw
q q P P gkk A kk A x x
μ μ ρ ρ α∂ ∂− = − − −
∂ ∂ (8.29)
8-17
Substituting Eqs.(8.17) and (8.24) into (8.29) gives upon rearrangement
( )1 sin
1
rnw cw nw
w nw
rnw w
rw nw
kk A P gq q x
kqk
ρ ρ αμ
μμ
∂⎡ ⎤+ − −⎢ ⎥∂⎣ ⎦=+
(8.30)
Let an approximate fractional flow of the wetting phase be defined as
1
1w
rnw w
rw nw
F kk
μμ
=+
(8.31)
Substituting Eqs.(8.30) and (8.31) into (8.25) gives the true fractional flow of
the wetting phase as
( )1 sinrnw cw w w nw
nw
kk A Pf F gq x
ρ ρ αμ
⎧ ⎫∂⎡ ⎤= + − −⎨ ⎬⎢ ⎥∂⎣ ⎦⎩ ⎭ (8.32)
Let the dimensionless distance from the inlet end be defined as
DxxL
= (8.33)
Let the spontaneous imbibition capillary pressure curve be given in terms of
its Leverett J-function as
( ) ( )cos ,/c w wP S J S
kσ θ
φ= Γ (8.34)
Substituting Eqs.(8.83) and (8.34) into (8.32) gives the true fractional flow of
the wetting phase as
( ) sincos1 w nww w rnw rnw
nw D nw
kA gA k Jf F k kq L x q
ρ ρ ασ θ φμ μ
⎡ ⎤⎛ ⎞ −⎛ ⎞∂= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ∂⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦
(8.35)
Eq.(8.35) can be written as
8-18
1w w cap rnw rnw gD
Jf F N k k Nx
⎡ ⎤∂= + −⎢ ⎥∂⎣ ⎦
(8.36)
where capN is given by
coscap
nw
A kNq L
σ θ φμ
= (7.61)
and gN is given by
( ) sinw nwg
nw
kA gN
qρ ρ α
μ−
= (8.37)
capN is the same dimensionless number we encountered in the analysis of
capillary end effect. It represents the ratio of capillary to viscous forces in the
displacement. gN is a new dimensionless number, which represents the ratio
of gravity to viscous forces in the displacement. The mobility ratio of the
displacement is given by
( ) rw nww
rnw w
kM Sk
μμ
= (8.38)
The mobility ratio as defined in Eq(8.38) is a function of saturation and will be
different at each point in the porous medium depending on the saturation. A
characteristic mobility ratio for the displacement can be defined in terms of
the end-point relative permeabilities as
wr nwE
nwr w
kMk
μμ
= (8.39)
where wrk and nwrk are the end-point relative permeabilities for the wetting and
non-wetting phases. The mobility ratio given in Eq.(8.39) is a characteristic
dimensionless number for the displacement that is independent of saturation.
Eq.(8.36) can also be written as
8-19
1
11
cap rnw rnw gD
w
JN k k Nxf
M
∂+ −
∂=
⎛ ⎞+⎜ ⎟⎝ ⎠
(8.40)
where the approximate fractional flow of the wetting phase is given by
111
wF
M
=⎛ ⎞+⎜ ⎟⎝ ⎠
(8.41)
In order to maximize the displacement efficiency of the non-wetting
phase, we need to minimize the fractional flow of the wetting phase at each
point in the porous medium. Much can be deduced about the immiscible
displacement from the fractional flow equation, Eq.(8.40). Examination of this
equation leads to the following qualitative deductions about immiscible
displacements in porous media:
1. The fractional flow of the wetting phase is a strong function of
saturation.
2. The displacement behavior can be rate-sensitive if the effect of
capillarity or gravity is significant.
3. Capillarity is detrimental to the displacement efficiency as it increases
the fractional flow of the wetting phase at a given saturation.
4. Gravity is beneficial to the displacement efficiency for up-dip
displacement of the lighter non-wetting phase by the heavier wetting
phase as it reduces the fractional flow of the wetting phase at a given
saturation. Conversely, gravity will be detrimental to the displacement
efficiency for down-dip displacement of the lighter non-wetting phase by
the heavier wetting phase.
5. The displacement efficiency can be increased by reducing the mobility
ratio. This can be accomplished in practice by increasing the viscosity
8-20
of the wetting phase (the injected fluid) by use of a polymer. This is the
basis for polymer flooding as an improved oil recovery technique.
6. The effects of gravity and capillarity on the displacement can be reduced
by increasing the injection rate.
There are additional facts about the immiscible displacement that are
not apparent from the fractional flow equation. The fractional flow equation
indicates that the displacement efficiency can be improved by injecting the
wetting phase at a high enough rate to minimize capillary smearing of the
displacement front. This is generally true for a favorable mobility ratio
displacement. If the mobility ratio is unfavorable, an increase in rate can
result in viscous instability which reduces the displacement efficiency. The
fractional flow equation suggests that the effect of gravity will be eliminated if
the porous medium is horizontal. This is misleading because, in practice, if
there is a density contrast between the fluids, the injection rate is sufficiently
low and the core has a vertical dimension (which it does), gravity segregation
will occur even in a horizontal medium. In this case, the one dimensional
displacement model is inadequate to describe the displacement. A
multidimensional model is needed to correctly describe the gravity-dominated
displacement. The only fail proof way to eliminate the effect of gravity is to
eliminate the density contrast between the fluids or perform the displacement
in outer space. One can create a gravity number for displacement in a
horizontal core by replacing sinα in Eq.(8.37) by the aspect ratio (d/L).
The partial differential equation for the wetting phase saturation can be
derived as follows. Let
( )w w rnw gS F k NΨ = − (8.42)
( )w rnww
dJS kdS
Ω = (8.43)
Eq.(8.36) then becomes
8-21
( ) ( ) ( ) ww w w cap w
D
Sf S S N Sx
∂= Ψ + Ω
∂ (8.44)
Eq.(8.27) can be written in dimensionless form as
0w w
D D
S ft x
∂ ∂+ =
∂ ∂ (8.45)
where Dt is given by
Dqtt
A Lφ= (7.63)
Substituting Eq.(8.44) into (8.45) gives the partial differential equation for the
wetting phase saturation as
( ) 0w w wcap w
D w D D D
S S Sd N St dS x x x
⎡ ⎤∂ ∂ ∂Ψ ∂+ + Ω =⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦
(8.46)
We have reduced the immiscible displacement problem to the solution of a
second order, nonlinear, parabolic partial differential equation for the wetting
phase saturation. When supplemented with appropriate initial and boundary
conditions, Eq.(8.46) can be solved, usually numerically, to obtain the wetting
phase saturation in time and space.
8.3.2 Buckley-Leverett Approximate Solution of the Immiscible Displacement Equation
Eq.(8.46) cannot be solved analytically for the saturation profiles. Here,
we examine the approximate solution obtained by Buckley and Leverett
(1941). The continuity equation, Eq.(8.45), can be written as
0w w w
D w D
S df St dS x
∂ ∂+ =
∂ ∂ (8.47)
where the true fractional flow of the wetting phase for horizontal
displacement is given by
8-22
1
11
wcap rnw
w Dw
SdJN kdS xf
M
∂+
∂=
⎛ ⎞+⎜ ⎟⎝ ⎠
(8.48)
It should be observed that the true fractional flow function contains w
D
Sx
∂∂
,
which is unknown. Buckley and Leverett (1941) obtained an approximate
solution to Eq.(8.47) by making a key simplifying assumption. They dropped
the capillary pressure term from Eq.(8.48) and as a result, they approximated
the fractional flow of the wetting phase as
w wf F (8.49)
Substituting Eq.(8.49) into (8.47) gives the partial differential equation for the
wetting phase saturation as
0w w w
D w D
S dF St dS x
∂ ∂+ =
∂ ∂ (8.50)
Eq.(8.50) is known as the Buckley-Leverett equation in the petroleum
industry. The Buckley-Leverett approximation changes the original second
order parabolic partial differential equation for the wetting phase saturation to
a first order, hyperbolic partial differential equation. This is a radical change
in the structure of the mathematical problem. However, the change allows an
approximate analytical solution to be obtained for the wetting phase
saturation profiles that is adequate for making gross performance calculations
for the immiscible displacement.
Eq.(8.50) is a nonlinear, first order, hyperbolic partial differential
equation that can be solved by the method of characteristics. From calculus,
the total time derivative of ( ),D DS x t is given by
w w wD
D D D D
S S dSdxt dt x dt
⎛ ⎞∂ ∂+ =⎜ ⎟∂ ∂⎝ ⎠
(8.51)
8-23
Subtracting Eq.(8.50) from (8.51) gives
w w wD
D w D D
dF S dSdxdt dS x dt
⎛ ⎞ ∂− =⎜ ⎟ ∂⎝ ⎠
(8.52)
Eq.(8.52) can be decomposed into the following two simultaneous equations:
0wD
D w
dFdxdt dS
− = (8.53)
0w
D
dSdt
= (8.54)
Eq.(8.53) gives the characteristic path for the hyperbolic partial differential
equation given by Eq(8.54). Eq.(8.54) shows that along the characteristic
path given by Eq.(8.53), the saturation is a constant.
Eq.(8.53) can be integrated to determine the distance traveled by a
constant saturation at a given time as
( ) ( ) ( )00w D w
wD DDS t DS
w
dFx x t tdS
− = − (8.55)
If there was no prior injection, 0Dt will be zero and all the saturations from Swi
to (1 - Snwr) will be located at the inlet end of the system, making ( )0wDSx equal
to zero. In this case, Eq.(8.55) becomes
( )w D
wDDS t
w
dFx tdS
= (8.56)
Eq.(8.56) can be written as
wD D
w
dFx tdS
= (8.57)
8-24
where Dx is the dimensionless distance traveled by a given saturation at time
Dt . Eq.(8.57) can be written in dimensional form as
( )i w
w
Q t dFxA dSφ
= (8.58)
Eq.(8.58) is usually referred to in the petroleum industry as the Buckley-
Leverett frontal advance equation. It should be emphasized that Eq.(8.57) or
(8.58) applies to a particular wetting phase saturation. To determine the
dimensionless distance traveled by a particular saturation 1wS at time Dt , we
use Eq.(8.57) to compute the distance as
1w
wD D
w S
dFx tdS
⎛ ⎞= ⎜ ⎟
⎝ ⎠ (8.59)
where the derivative of the approximate fractional flow curve is evaluated at
1wS . Eq.(8.57) can be used to derive a similarity transformation for an
immiscible displacement. The similarity transformation is given by
w wD
D w w
dF dfxzt dS dS
= = = (8.60)
If the saturation profiles for an immiscible displacement are plotted as wS
versus D
D
xt
, all the saturation profiles will collapse into one curve. If the
saturation profiles in an immiscible displacement are imaged say by CT or by
NMR, then Eq.(8.60) can be used to calculate the true fractional flow curve,
including the effect of capillarity, as
( ) w
wirr
SD
w w wSD
xf S dSt
= ∫ (8.61)
Given the relative permeability curves and the viscosity ratio, the
approximate fractional flow function and its derivative can be computed and
8-25
plotted as shown in Figure 8.9. In this figure, the S-shaped curve ADBC is
the approximate fractional flow curve (Fw) obtained from the relative
permeability curves and the viscosity ratio. The curve AFE is the derivative of
this function w
w
dFdS
⎛ ⎞⎜ ⎟⎝ ⎠
. Using Eq.(8.59) and this derivative function, the distance
traveled by each wetting phase saturation between Swirr and (1-Snwr) at a
given time Dt can be computed. Figure 8.10 shows the saturation profile that
will be obtained before wetting phase breakthrough by use of Eq.(8.59) and
the approximate derivative function. We see that the Buckley-Leverett
approximation gives rise to multiple-valued saturations at various Dx which is
physically impossible. This multiple-valued solution is caused by neglecting
the capillary term in the fractional flow equation. It is no accident that the
multi-valued solution occurs in the saturation range wirr w wfS S S< < where the
capillary pressure gradient is high and should not have been neglected.
To eliminate the multiple-valued solution, we appeal to physical reality
as follows. At time t, Qi(t) of wetting phase has been injected and the flood
front has traveled a distance fx into the medium. A volumetric balance of the
injected wetting phase can be used to calculate fx as follows:
( ) ( )0
fx
i w wirrQ t A S S dxφ= −∫ (8.62)
Integration of Eq.(8.62) by parts and substitution of Eq.(8.58) gives
( ) ( )1
wf
nwr
S wi f wf wirr i wS
w
dFQ t Ax S S Q dSdS
φ−
= − − ∫ (8.63)
Upon performing the integration in Eq.(8.63) and rearranging, one obtains
8-26
Figure 8.9. Approximate fractional flow function and its first derivative. Note the tangent construction.
Figure 8.10. Calculated water saturation distribution based on the Buckley-Leverett approximation showing the discontinuity in saturation as required by
a material balance.
8-27
( ) ( )w wff i
wf wirr
F SAx Q t
S Sφ =
− (8.64)
From the Buckley-Leverett frontal advance equation, Eq.(8.58), one can also
obtain
( )wf
wf i
w S
dFAx Q tdS
φ⎛ ⎞
= ⎜ ⎟⎝ ⎠
(8.65)
A comparison of Eqs.(8.64) and (8.65) gives
( )
wf
w wfw
w wf wirrS
F SdFdS S S
⎛ ⎞=⎜ ⎟ −⎝ ⎠
(8.66)
The saturation distribution in Figure 8.10 will be single valued if all the
saturations between Swirr and the frontal saturation, Swf, are eliminated.
Eq.(8.66) shows that the frontal saturation (Swf) is the saturation at which the
straight line passing through the point Sw = Swirr and Fw = 0 is tangent to the
approximate fractional flow curve, Fw. This line is shown in Figure 8.9 as AB.
This tangent construction was first suggested by Welge (1952). The effect of
the tangent construction is to correct the approximate fractional flow curve Fw
for the capillary term that was neglected to obtain the true fractional flow
curve fw. Such a correction is needed at the front (low wetting phase
saturation) where the capillary pressure gradient is high and should not have
been neglected. With the tangent construction correction in place, the true
fractional flow curve, fw, is now given by the curve ABC (Fig. 8.9) thereby
eliminating the S-shaped lower portion of Fw, which led to the tripple-valued
saturation solution of Figure 8.10. With this correction, the derivative of the
true fractional flow curve w
w
dfdS
⎛ ⎞⎜ ⎟⎝ ⎠
used in the solution is given by the curve
EFG (Fig. 8.9). After the tangent construction, the true fractional flow curve
and its derivative are given by
8-28
Figure 8.11. Similarity transformation for an immiscible displacement.
Figure 8.12. Integration of the transformed saturation data to calculate the true fractional flow curve including capillarity.
8-29
( )( ) ( )
( )
for
for 1.0 wf
w wirr ww wf w wf wirr w wf
wf wirr w Sw w
w w wf w
S S dFF S S S S S SS S dSf S
F S S S
⎧⎛ ⎞ ⎛ ⎞−= − ≤ ≤⎪⎜ ⎟ ⎜ ⎟⎪⎜ ⎟−= ⎝ ⎠⎨⎝ ⎠
⎪≤ ≤⎪⎩
(8.67)
and
= a constant for
for 1.0
wf
w
wwirr w wf
w Sw
w wwf w
w S
dF S S SdSdf
dS dF S SdS
⎧⎛ ⎞≤ ≤⎪⎜ ⎟
⎝ ⎠⎪= ⎨
⎛ ⎞⎪ ≤ ≤⎜ ⎟⎪⎝ ⎠⎩
(8.68)
The similarity transformation for the immiscible displacement is given
by the curve EFGH (Fig. 8.9) and is shown in Figure 8.11. Figure 8.12 shows
how the transformed saturation data can be integrated to obtain the true
fractional flow curve that includes the effect of capillarity.
We now show that the intersection of the tangent line with the line Fw =
1 (point J in Fig. 8.9) gives the constant average wetting phase saturation
behind the front before and at wetting phase breakthrough. The slope of the
tangent line can be written as
( )1
wf
w wfw
w wav wfS
F SdFdS S S
−⎛ ⎞=⎜ ⎟ −⎝ ⎠
(8.69)
which can be rearranged as
( )1
wf
w wfwav wf
w
w S
F SS S
dFdS
−= +
⎛ ⎞⎜ ⎟⎝ ⎠
(8.70)
Before breakthrough, the average wetting phase saturation behind the front is
given by
8-30
1
0
nwrS
wwav
f
AxdSS
Ax
φ
φ
−
= ∫ (8.71)
Substituting Eq.(8.58) into (8.71) and integrating gives the average wetting
phase saturation behind the front as
( )1
wf
w wfwav wf
w
w S
F SS S
dFdS
−= +
⎛ ⎞⎜ ⎟⎝ ⎠
(8.72)
Eq.(8.72) is identical to Eq.(8.70), which confirms that the intersection of the
tangent line and the line Fw = 1 gives the wetting phase saturation (Swav)
corresponding to point J in Figure 8.9. Thus, the average wetting phase
saturation at breakthrough can be obtained graphically from the tangent
construction.
The average wetting phase saturation after breakthrough can be
obtained by a tangent construction at the outlet wetting phase saturation
between Swf and 1 - Snwr. The intersection of the tangent line and the line
Fw = 1 gives the average wetting phase saturation in the porous medium
corresponding to the outlet wetting phase saturation. By this tangent
construction, the Buckley-Leverett approximation can be used to predict the
performance of the one-dimensional immiscible displacement after wetting
phase breakthrough. Before breakthrough, the amount of non-wetting phase
recovered is equal to the amount of fluid injected. Thus, the entire
displacement performance can be predicted for a given set of wetting and non-
wetting relative permeability curves and wetting and non-wetting viscosity
ratio.
8-31
8.3.3 Waterflood Performance Calculations from Buckley–Leverett Theory
We now apply Buckley-Leverett theory to calculate a waterflood
performance from beginning to end. It is assumed that the true fractional flow
curve and its derivative have been computed using the relative permeability
curves, the viscosity ratio and the Welge tangent construction. Therefore, the
equations in this section are written in terms of the true fractional flow curve.
The methodology presented also applies to the calculation of the performance
of a gas flood using gas-oil drainage relative permeability curves.
Oil Recovery at any Time
The oil recovery at any time after the initiation of water injection is
given by
1wav wirr
wirr
S SRS−
=−
(8.73)
where R is the oil recovery as a fraction of the initial oil in place, Swav is the
average water saturation in the porous medium at the time of interest and
Swirr is the initial water saturation in the porous medium before water
injection which is assumed to be the irreducible water saturation. Thus, in
principle, the oil recovery can be calculated at any time by first calculating the
average water saturation in the porous medium at that time and applying
Eq.(8.73). However, depending on the stage of water injection, Eq.(8.73) may
not offer the most direct way to calculate the oil recovery. Let us examine the
waterflood performance at various stages of the flood.
Oil Recovery Before Water Breakthrough
Consider a constant rate water injection project. Assuming
incompressible fluids, the amount of oil recovered before water breakthrough
must equal the amount of water injected. Thus, at reservoir conditions,
8-32
( ) ( )w i oqB t Q t Q t= = (8.74)
where q is the constant water injection rate, in surface units, Bw is the water
formation volume factor, t is the time of interest before water breakthrough,
Qi is the cumulative water injected at time t in reservoir units and Qo is
the cumulative oil produced at time t in reservoir units. The cumulative oil
produced at surface conditions is
( ) ( )Cumulative Oil Produced i ow
o o o
Q t Q tqB tB B B
= = = (8.75)
where Bo is the current oil formation volume factor. The oil recovery as a
fraction of the initial oil in place is given by
( )
( )( ) ( )1 1 1
iw i
wirr wirr wirr
Q tqB t WRAL S AL S Sφ φ
= = =− − −
(8.76)
where Wi is the pore volume of water injected.
Oil Recovery at Water Breakthrough
From the Buckley–Leverett frontal advance equation, Eq.(8.58), the
distance traveled by a given saturation is given by
( )w w
iw w w
w wS S
Q tqB t df dfxA dS A dSφ φ
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ (8.77)
Let us apply Eq.(8.77) to the frontal water saturation Swf to get
( )wf wf
iw w wf
w wS S
Q tqB t df dfxA dS A dSφ φ
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ (8.78)
At the moment of water breakthrough, the frontal saturation arrives at the
outlet end of the porous medium and fx equals L. At the moment of water
breakthrough, Eq.(8.78) then becomes
8-33
( )wf wf
iw w w
w wS S
Q tqB t df dfLA dS A dSφ φ
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ (8.79)
Eq.(8.75) can be rearranged as
( ) 1
wf
iwi
w
w S
Q tqB tWAL AL df
dSφ φ
= = =⎛ ⎞⎜ ⎟⎝ ⎠
(8.80)
where Wi is the pore volume of water injected. The cumulative oil recovery at
water breakthrough is equal to the cumulative water injected in reservoir
volumes. The fractional oil recovery at water breakthrough is obtained from
Eq.(8.80) as
( )
( )( ) ( )
11 1 1
1wf
ii w
wi wi wi wwi
w S
Q tW qB tRS AL S AL S dfS
dSφ φ
= = = =− − − ⎛ ⎞
− ⎜ ⎟⎝ ⎠
(8.81)
The breakthrough time, tbt, can be obtained from Eq.(8.80) as
wf
btw
ww S
ALtdfqBdS
φ=
⎛ ⎞⎜ ⎟⎝ ⎠
(8.82)
or in dimensionless form as
1
wf
Dbtw
w S
tdfdS
=⎛ ⎞⎜ ⎟⎝ ⎠
(8.83)
The average water saturation in the porous medium behind the displacement
front before and at water breakthrough is given by
1
0
orS
wwav
f
AxdSS
Ax
φ
φ
−
= ∫ (8.71)
8-34
Figure 8.13 shows a typical water saturation distribution at time t before
breakthrough. From Figure 8.13, we see that the integral (area under the
curve) in Eq.(8.71) can be split into two parts as follows:
1 or
wf
S
wf f wSwav
f
AS x AxdSS
Ax
φ φ
φ
−+
=∫
(8.84)
Substituting Eq.(8.58) into (8.84) gives the average water saturation as
( )
1 or
wf
S
i wSwav wf
f
Q t dFS S
Axφ
−
= +∫
(8.85)
Performing the integration in Eq.(8.85) gives
( ) ( ) ( )1i w or w wf
wav wff
Q t f S f SS S
Axφ
⎡ ⎤− −⎣ ⎦= + (8.86)
But Fw at Sw = (1 – Sor) is equal to 1.0. Thus, Eq.(8.81) can rewritten as
( ) ( )1i w wf
wav wff
Q t f SS S
Axφ
⎡ ⎤−⎣ ⎦= + (8.87)
Substituting Eq.(8.78) into (8.87) gives the average water saturation behind
the front as
( )1
wf
w wfwav wf
w
w S
f SS S
dfdS
⎡ ⎤−⎣ ⎦= +⎛ ⎞⎜ ⎟⎝ ⎠
(8.72)
It should be observed in Figure 8.9 that the average water saturation behind
the front up until water breakthrough as given in Eq.(8.72) is the same as the
water saturation at which the tangent to the fractional flow curve intersects
the Fw = 1 axis. Thus, the average water saturation in the porous medium at
water breakthrough can easily be determined graphically. The average water
8-35
saturation can then be substituted into Eq.(8.73) to calculate the oil recovery
at water breakthrough. We can easily show that the result obtained by this
approach will be the same as that obtained by Eq.(8.81). Substituting
Eq.(8.72) into Eq.(8.73) gives
Figure 8.13. Typical water saturation profile at time t before water breakthrough.
8-36
( )1
1wf
w wfwf wirr
w
w S
wirr
f SS S
dfdS
RS
⎡ ⎤−⎣ ⎦+ −⎛ ⎞⎜ ⎟⎝ ⎠
=−
(8.88)
From the equation of the tangent line in Figure 8.9, we find that
( )
wf
w wfw
w wf wirrS
f SdfdS S S
⎛ ⎞=⎜ ⎟ −⎝ ⎠
(8.66)
Substituting Eq.(8.66) into (8.88) gives the oil recovery at water breakthrough
as
( )
1
1wf
wwi
w S
RdfSdS
=⎛ ⎞
− ⎜ ⎟⎝ ⎠
(8.89)
which is identical to Eq.(8.81).
Oil Recovery After Water Breakthrough
After water breakthrough, Eq.(8.77) applied to the outlet end of the
porous medium gives
( )2 2w w
iw w w
w wS S
Q tqB t df dfLA dS A dSφ φ
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ (8.90)
where Sw2 is the water saturation at the outlet end of the porous medium
which now lies between Swf and (1 – Sor). Rearrangement of Eq.(8.90) gives
the pore volumes of water injected as
( )
2
1
w
iwi
w
w S
Q tqB tWAL AL df
dSφ φ
= = =⎛ ⎞⎜ ⎟⎝ ⎠
(8.91)
8-37
where Wi is the pore volumes of water injected since the initiation of water
injection. Eq.(8.91) is analogous to Eq.(8.80) before breakthrough.
A material balance for the water after water breakthrough gives
( ) ( ) ( )0
L
i w wirr wQ t A S S dx Q tφ− − =∫ (8.92)
The integral in Eq.(8.92) can be performed using integration by parts.
Performing integration by parts, Eq.(8.92) can be written as
( ) ( ){ } ( )2
1
0
or
w
SLi w wirr w wS
Q t A S S x AxdS Q tφ φ−
− − − =⎡ ⎤⎣ ⎦ ∫ (8.93)
Substituting the limits for the first integral gives
( ) ( ){ } ( )2
1
2or
w
S
i w wirr w wSQ t AL S S AxdS Q tφ φ
−− − − =∫ (8.94)
Substituting Eq.(8.77) into (8.94) gives
( ) ( ) ( ){ } ( )2
1
2or
w
S
i w wirr i w wSQ t AL S S Q t df Q tφ
−− − − =∫ (8.95)
Performing the integration in Eq.(8.95) gives
( ) ( ) ( ) ( ) ( ){ } ( )2 21i w wirr i w or w w wQ t AL S S Q t f S f S Q tφ− − − − − =⎡ ⎤⎣ ⎦ (8.96)
or
( ) ( ) ( ) ( ){ } ( )2 21i w wirr i w w wQ t AL S S Q t f S Q tφ− − − − =⎡ ⎤⎣ ⎦ (8.97)
since fw at Sw = 1–Sor is equal to 1. Eq.(8.97) can be rearranged as
( ) ( ) ( ) ( )2 21i w iw wirr w w
Q t Q t Q tS S f S
AL ALφ φ−
= + − −⎡ ⎤⎣ ⎦ (8.98)
8-38
( )2 21w wirr pD i w wS S N W f S= + − −⎡ ⎤⎣ ⎦ (8.99)
( )2 21pD w wi i w wN S S W f S= − + −⎡ ⎤⎣ ⎦ (8.100)
where NpD is the oil recovery as a fraction of the total pore volume. We observe
that the sum of the first two terms on the right hand side of Eq.(8.99) is the
average water saturation in the porous medium after water breakthrough.
Thus, Eq.(8.99) can be rewritten as
( )2 21w wav i w wS S W f S= − −⎡ ⎤⎣ ⎦ (8.101)
Substituting Eq.(8.91) into (8.101) and rearranging gives the average water
saturation in the porous medium after water breakthrough as
( )
2
22
1
w
w wwav w
w
w S
f SS S
dfdS
−⎡ ⎤⎣ ⎦= +⎛ ⎞⎜ ⎟⎝ ⎠
(8.102)
which is analogous to Eq.(8.72) at water breakthrough. Figure 8.14 shows
that the average water saturation after water breakthrough as given by
Eq.(8.102) is equal to the water saturation where the tangent line to the
fractional flow curve at the outlet water saturation intersects the Fw = 1 axis.
The average water saturation in the porous medium after water
breakthrough could also have been derived using Eq.(8.71) and the water
saturation profile shown in Figure 8.15. The average water saturation is then
given by
2
1
2or
w
S
w wSwav
AS L AxdSS
AL
φ φ
φ
−+
=∫
(8.103)
Substituting Eq.(8.77) into (8.103) gives the average water saturation as
8-39
( )
2
1
2
or
w
S
i wSwav w
Q t dfS S
ALφ
−
= +∫
(8.104)
Figure 8.14. Average water saturation after water breakthrough.
Performing the integration in Eq.(8.104) gives
( ) ( ) ( )2
2
1i w or w wwav w
Q t f S f SS S
ALφ− −⎡ ⎤⎣ ⎦= + (8.105)
But fw at Sw = 1 – Sor is equal to 1.0. Thus, Eq.(8.105) can be rewritten as
8-40
Figure 8.15. Typical water saturation profile at time t after water breakthrough.
( ) ( )2
2
1i w wwav w
Q t f SS S
ALφ−⎡ ⎤⎣ ⎦= + (8.106)
Substituting Eq.(8.90) into (8.106) gives the average water saturation after
water breakthrough as
( )
2
22
1
w
w wwav w
w
w S
F SS S
dFdS
−⎡ ⎤⎣ ⎦= +⎛ ⎞⎜ ⎟⎝ ⎠
(8.107)
8-41
which is identical to Eq.(8.102).
Water Production
There is no water production before water breakthrough. After water
breakthrough, the water oil ratio is given by
1
w
ww o w
o w wo
o
fBq B FWOR
q B FfB
⎛ ⎞⎜ ⎟ ⎛ ⎞⎝ ⎠= = = ⎜ ⎟−⎛ ⎞ ⎝ ⎠⎜ ⎟⎝ ⎠
(8.108)
The pore volumes of water produced is given by material balance on the water
as
Water produced Cumulative water injected Water stored= − (8.109)
Substituting appropriate symbols into Eq.(8.109) gives
( )p i wav wirrW W S S= − − (8.110)
Substituting Eq.(8.91) into (8.110) gives the pore volumes of water produced
as
( )
2
1
w
p wav wiw
w S
W S SdfdS
= − −⎛ ⎞⎜ ⎟⎝ ⎠
(8.111)
Example 8.1
A waterflood is to be performed in a linear reservoir. The relative permeability
curves for the reservoir are adequately described by the following analytical
models:
3rw wr ek k S= (8.112)
8-42
( )21rnw nwr ek k S= − (8.113)
where Se is defined as
1
w wirre
wirr nwr
S SSS S
−=
− − (8.114)
The other pertinent data are as follows:
0.20wirrS =
0.30nwrS =
0.95nwrk =
0.35wrk =
10nw oμ μ= = cp
1wμ = cp
1.20oB = RB/STB
1.0wB = RB/STB
1. Calculate and plot graphs of the relative permeability curves.
2. Calculate and plot graphs of the approximate fractional flow curve
( )wF and its derivative w
w
dFdS
⎛ ⎞⎜ ⎟⎝ ⎠
.
3. Perform the Welge tangent construction and from it determine the
frontal water saturation ( )wfS , the average water saturation at water
breakthrough ( )wavS and the true fractional flow curve ( )wf and its
derivative w
w
dfdS
⎛ ⎞⎜ ⎟⎝ ⎠
.
8-43
4. Plot the graphs of the true fractional flow curve and its derivative.
5. Calculate the end point mobility ratio for the waterflood.
6. Calculate and plot graphs of the water saturation profiles at tD = 0.20,
0.30 and 1.0.
7. Calculate the dimensionless breakthrough time.
8. Calculate the breakthrough oil recovery as a fraction of the initial oil in
place.
9. Calculate and plot the graph of oil recovery versus pore volume of water
injected before and after water breakthrough.
10. Calculate and plot the graph of water oil ratio versus oil recovery.
Solution to Example 8.1
The results of the calculations are summarized in Table 8.1.
1. The relative permeability curves calculated with Eqs.(8.112) and (8.113)
are shown in Figure 8.16.
2. Figure 8.17 shows the approximate fractional flow curve calculated with
Eq.(8.41) and its derivative calculated by differentiating Fw with respect
to Sw analytically.
Table 8.1. Calculated Results for Example 8.1.
tD tD tD
0.20 0.30 1.00
Sw krw krnw Fw w
w
dFdS
fw w
w
dfdS
xD xD xD Wi R WOR
0.200 0.00000 0.950 0.00000 0.000 0.000 2.775 0.555 0.833 2.775 0.000 0.000 0.000
0.210 0.00000 0.912 0.00003 0.009 0.028 2.775 0.555 0.833 2.775 0.008 0.023 0.000
0.220 0.00002 0.876 0.00026 0.039 0.056 2.775 0.555 0.833 2.775 0.016 0.045 0.000
0.230 0.00008 0.839 0.00090 0.094 0.083 2.775 0.555 0.833 2.775 0.025 0.068 0.000
0.240 0.00018 0.804 0.00222 0.176 0.111 2.775 0.555 0.833 2.775 0.033 0.091 0.000
8-44
0.250 0.00035 0.770 0.00453 0.290 0.139 2.775 0.555 0.833 2.775 0.041 0.113 0.000
0.260 0.00060 0.736 0.00815 0.441 0.167 2.775 0.555 0.833 2.775 0.049 0.136 0.000
0.270 0.00096 0.703 0.01348 0.632 0.194 2.775 0.555 0.833 2.775 0.057 0.158 0.000
0.280 0.00143 0.670 0.02094 0.866 0.222 2.775 0.555 0.833 2.775 0.066 0.180 0.000
0.290 0.00204 0.639 0.03097 1.147 0.250 2.775 0.555 0.833 2.775 0.074 0.202 0.000
0.300 0.00280 0.608 0.04403 1.473 0.278 2.775 0.555 0.833 2.775 0.082 0.223 0.000
0.310 0.00373 0.578 0.06057 1.844 0.305 2.775 0.555 0.833 2.775 0.090 0.243 0.000
0.320 0.00484 0.549 0.08103 2.254 0.333 2.775 0.555 0.833 2.775 0.098 0.263 0.000
0.330 0.00615 0.520 0.10575 2.693 0.361 2.775 0.555 0.833 2.775 0.106 0.281 0.000
0.340 0.00768 0.492 0.13496 3.150 0.389 2.775 0.555 0.833 2.775 0.115 0.299 0.000
0.350 0.00945 0.466 0.16875 3.607 0.416 2.775 0.555 0.833 2.775 0.123 0.315 0.000
0.360 0.01147 0.439 0.20703 4.044 0.444 2.775 0.555 0.833 2.775 0.131 0.330 0.000
0.370 0.01376 0.414 0.24949 4.439 0.472 2.775 0.555 0.833 2.775 0.139 0.343 0.000
0.380 0.01633 0.389 0.29560 4.772 0.500 2.775 0.555 0.833 2.775 0.147 0.355 0.000
0.390 0.01921 0.365 0.34465 5.024 0.527 2.775 0.555 0.833 2.775 0.156 0.365 0.000
0.400 0.02240 0.342 0.39576 5.181 0.555 2.775 0.555 0.833 2.775 0.164 0.374 0.000
0.410 0.02593 0.320 0.44794 5.238 0.583 2.775 0.555 0.833 2.775 0.172 0.381 0.000
0.420 0.02981 0.298 0.50019 5.195 0.611 2.775 0.555 0.833 2.775 0.180 0.388 0.000
0.430 0.03407 0.277 0.55153 5.058 0.638 2.775 0.555 0.833 2.775 0.188 0.393 0.000
0.440 0.03871 0.257 0.60109 4.842 0.666 2.775 0.555 0.833 2.775 0.197 0.398 0.000
0.450 0.04375 0.238 0.64815 4.561 0.694 2.775 0.555 0.833 2.775 0.205 0.403 0.000
0.460 0.04921 0.219 0.69216 4.234 0.722 2.775 0.555 0.833 2.775 0.213 0.407 0.000
0.470 0.05511 0.201 0.73274 3.879 0.749 2.775 0.555 0.833 2.775 0.221 0.411 0.000
0.480 0.06147 0.184 0.76969 3.511 0.777 2.775 0.555 0.833 2.775 0.229 0.416 0.000
0.490 0.06829 0.168 0.80296 3.144 0.805 2.775 0.555 0.833 2.775 0.237 0.421 0.000
0.491 0.06900 0.166 0.80608 3.107 0.808 2.775 0.555 0.833 2.775 0.246 0.423 0.000
0.492 0.06971 0.164 0.80917 3.071 0.810 2.775 0.555 0.833 2.775 0.254 0.426 0.000
0.493 0.07043 0.163 0.81222 3.035 0.813 2.775 0.555 0.833 2.775 0.262 0.428 0.000
0.494 0.07115 0.161 0.81524 2.999 0.816 2.775 0.555 0.833 2.775 0.270 0.430 0.000
0.495 0.07188 0.160 0.81822 2.964 0.819 2.775 0.555 0.833 2.775 0.278 0.432 0.000
0.496 0.07262 0.158 0.82117 2.928 0.822 2.775 0.555 0.833 2.775 0.287 0.434 0.000
0.497 0.07335 0.157 0.82408 2.893 0.824 2.775 0.555 0.833 2.775 0.295 0.436 0.000
0.498 0.07410 0.155 0.82695 2.857 0.827 2.775 0.555 0.833 2.775 0.303 0.438 0.000
0.499 0.07485 0.154 0.82979 2.822 0.830 2.775 0.555 0.833 2.775 0.311 0.440 0.000
0.500 0.07560 0.152 0.83260 2.788 0.833 2.775 0.555 0.833 2.775 0.319 0.442 0.000
8-45
0.500 0.07568 0.152 0.83288 2.784 0.833 2.775 0.555 0.833 2.775 0.328 0.444 0.000
0.500 0.07575 0.152 0.83316 2.781 0.833 2.775 0.555 0.833 2.775 0.336 0.445 0.000
0.500 0.07583 0.152 0.83343 2.777 0.833 2.775 0.555 0.833 2.775 0.344 0.447 0.000
0.500 0.07583 0.152 0.83343 2.777 0.833 2.775 0.555 0.833 2.775 0.352 0.449 0.000
0.500 0.07586 0.151 0.83357 2.775 0.834 2.775 0.555 0.833 2.775 0.360 0.450 6.010
0.501 0.07636 0.150 0.83537 2.753 0.835 2.753 0.551 0.826 2.753 0.363 0.451 6.089
0.502 0.07712 0.149 0.83810 2.718 0.838 2.718 0.544 0.816 2.718 0.368 0.452 6.212
0.503 0.07789 0.147 0.84081 2.684 0.841 2.684 0.537 0.805 2.684 0.373 0.453 6.338
0.504 0.07866 0.146 0.84347 2.650 0.843 2.650 0.530 0.795 2.650 0.377 0.454 6.466
0.505 0.07944 0.144 0.84611 2.616 0.846 2.616 0.523 0.785 2.616 0.382 0.455 6.598
0.506 0.08023 0.143 0.84871 2.583 0.849 2.583 0.517 0.775 2.583 0.387 0.456 6.732
0.507 0.08102 0.142 0.85127 2.549 0.851 2.549 0.510 0.765 2.549 0.392 0.457 6.868
0.508 0.08181 0.140 0.85380 2.516 0.854 2.516 0.503 0.755 2.516 0.397 0.458 7.008
0.509 0.08261 0.139 0.85630 2.483 0.856 2.483 0.497 0.745 2.483 0.403 0.459 7.151
0.510 0.08341 0.137 0.85877 2.450 0.859 2.450 0.490 0.735 2.450 0.408 0.460 7.297
0.520 0.09175 0.123 0.88169 2.137 0.882 2.137 0.427 0.641 2.137 0.468 0.469 8.943
0.530 0.10062 0.110 0.90160 1.850 0.902 1.850 0.370 0.555 1.850 0.540 0.479 10.995
0.540 0.11005 0.097 0.91878 1.591 0.919 1.591 0.318 0.477 1.591 0.628 0.489 13.575
0.550 0.12005 0.086 0.93351 1.360 0.934 1.360 0.272 0.408 1.360 0.736 0.499 16.849
0.560 0.13064 0.074 0.94606 1.154 0.946 1.154 0.231 0.346 1.154 0.866 0.508 21.048
0.570 0.14183 0.064 0.95668 0.974 0.957 0.974 0.195 0.292 0.974 1.027 0.518 26.502
0.580 0.15364 0.055 0.96561 0.816 0.966 0.816 0.163 0.245 0.816 1.226 0.528 33.693
0.590 0.16609 0.046 0.97306 0.678 0.973 0.678 0.136 0.203 0.678 1.474 0.537 43.348
0.600 0.17920 0.038 0.97923 0.559 0.979 0.559 0.112 0.168 0.559 1.788 0.546 56.589
0.610 0.19298 0.031 0.98430 0.456 0.984 0.456 0.091 0.137 0.456 2.191 0.555 75.235
0.620 0.20745 0.024 0.98841 0.368 0.988 0.368 0.074 0.110 0.368 2.716 0.564 102.358
0.630 0.22262 0.019 0.99171 0.292 0.992 0.292 0.058 0.088 0.292 3.420 0.573 143.471
0.640 0.23852 0.014 0.99430 0.228 0.994 0.228 0.046 0.068 0.228 4.392 0.581 209.224
0.650 0.25515 0.009 0.99629 0.172 0.996 0.172 0.034 0.052 0.172 5.798 0.589 322.295
0.660 0.27254 0.006 0.99777 0.126 0.998 0.126 0.025 0.038 0.126 7.966 0.597 537.909
0.670 0.29070 0.003 0.99882 0.086 0.999 0.086 0.017 0.026 0.086 11.663 0.605 1020.015
0.680 0.30966 0.002 0.99951 0.052 1.000 0.052 0.010 0.016 0.052 19.193 0.612 2444.665
0.690 0.32942 0.000 0.99988 0.024 1.000 0.024 0.005 0.007 0.024 42.067 0.619 10402.648
0.700 0.35000 0.000 1.00000 0.000 1.000 0.000 0.000 0.000 0.000
8-46
Figure 8.16. Relative permeability curves for Example 8.1.
8-47
Figure 8.17. Approximate fractional flow curve and its derivative for Example
8.1.
3. The Welge tangent construction is shown in Figure 8.17. From the
tangent construction,
0.500035wfS =
2.775wf
w
w S
dfdS
⎛ ⎞=⎜ ⎟
⎝ ⎠
0.5603wavS =
4. The true fractional flow curve and its derivative obtained from the
tangent construction are shown in Figure 8.18.
8-48
Figure 8.18. True fractional flow curve and its derivative for Example 8.1.
5. The end point mobility ratio for the waterflood is given by
( ) ( ) ( ) ( )/ / / 0.35 /1 / 0.95 /10 3.68E rw w rnw nwM k kμ μ= = =
6. The water saturation profiles calculated with Eq.(8.57) are shown in
Figure 8.19.
8-49
Figure 8.19. Water saturation profiles for Example 8.1.
7. The dimensionless breakthrough time is calculated with Eq.(8.83) as
1 1 0.4602.775
wf
Dbtw
w S
tdfdS
= = =⎛ ⎞⎜ ⎟⎝ ⎠
pore volume injected.
8. The breakthrough oil recovery as a fraction of the initial oil in place is
calculated with Eq.(8.81) as
( ) ( )( )
1 1 0.4501 0.20 2.775
1wf
btw
wirrw S
RdfSdS
= = =−⎛ ⎞
− ⎜ ⎟⎝ ⎠
9. Before water breakthrough, the oil recovery is a linear function of the
pore volume injected and can be calculated with Eq.(8.86). After water
8-50
breakthrough, the oil recovery is calculated with Eq.(8.100) as
1pD
wirr
NR
S=
−.
Figure 8.20 shows the calculated oil recovery curve.
Figure 8.20. Oil recovery curve for Example 8.1.
10. The producing water oil ratio is zero before water breakthrough. After
water breakthrough, the producing water oil water ratio is calculated
with Eq.(8.108). After breakthrough, the producing water oil ratio
increases rapidly as shown in Figure 8.21.
8-51
Figure 8.21. Producing water oil ratio for Example 8.1.
8.4 LABORATORY MEASUREMENT OF TWO-PHASE RELATIVE PERMEABILITIES BY THE UNSTEADY STATE METHOD
The major problem with the steady state method for relative
permeability measurements is that it is too slow. An alternative and much
faster technique is the unsteady state method or the dynamic displacement
method (Welge, 1952; Johnson et al., 1959; Jones and Roszelle, 1978). In
this method, for an imbibition test, the core is first saturated with the non-
wetting phase at irreducible wetting phase saturation as in the steady state
method. However, only the wetting phase is injected into the core to displace
the non-wetting phase. As the experiment progresses, the wetting phase
breaks through at the outlet end of the core and over time a higher and higher
fraction of the total produced fluid is the wetting phase.
8-52
By measuring the produced fractions of the wetting and non-wetting
phases at the outlet end of the core and the pressure drop across the core
versus time, the relative permeability curves can be calculated from the
production and pressure data using the theory of immiscible displacement in
porous media. This method is much faster than the steady state method,
usually requiring a few hours to complete compared to several weeks for the
steady state method. If adequate precautions are taken, the dynamic
displacement method will give relative permeability curves that are
comparable to those obtained by the steady state method.
Figure 8.22 shows the experimental setup and the measured data.
Because the point of observation is the outlet end of the core, it is necessary
that capillary end effect be minimized otherwise the calculated relative
permeability-saturation relationship will be wrong. It should be noted that
relative permeability curves can only be obtained over the saturation range Swf
to 1-Snwr. Therefore, it is necessary to choose the fluid viscosities that will give
the widest possible saturation window. This is obtained by using performing
and adverse mobility ratio displacement. A favorable mobility ratio
displacement will be unsuitable because for such a displacement, Swf is equal
to (1-Snwr) and there is no saturation window for calculating the relative
permeability curves. The relative permeability to the wetting phase below Swf
can only be obtained by extrapolating the data above Swf.
The technique for calculating relative permeability curves from
unsteady state measurements was developed by Welge (1952) and Johnson,
Bossler and Neumann (JBN, 1959). The fractional flow of the non-wetting
phase at the outlet end of the core is given by
21
1nw
rw nw
rnw w
f kk
μμ
=+
(8.115)
8-53
Figure 8.22. Unsteady state method for determining two-phase relative permeability curves; (a) coreflood; (b) measured data.
It should be noted that for saturations above Swf, Eq.(8.115) gives the true
fractional flow of the non-wetting phase because above Swf, the true fractional
flow and the approximate fractional curves are equal. Eq.(8.115) can be
rearranged to calculate the wetting-non-wetting phase relative permeability
ratio as
2
1 1rw w
rnw nw nw
kk f
μμ
⎛ ⎞= −⎜ ⎟
⎝ ⎠ (8.116)
The fractional flow of the non-wetting phase at the outlet end of the core is
also given by
8-54
( )( )2
pDnwnwnw
i i
dNdQ tqfq dQ t dW
= = = (8.117)
where Qnw(t) and Qi(t) are the cumulative non-wetting phase produced and
the cumulative wetting phase injected and pDN and iW are their dimensionless
counterparts as fractions of the total pore volume. Eqs.(8.116) and (8.117)
were first presented by Welge (1952). It should be noted that these equations
give no useful information before breakthrough because the fractional flow of
the non-wetting phase at the outlet end of the core is 1 and the relative
permeability to wetting phase is zero. This is why the unsteady state relative
permeability method is limited to only post breakthrough wetting phase
saturations between Swf and 1-Snwr.
After wetting phase breakthrough, we need to associate the computed
relative permeability ratio with the wetting phase saturation at the outlet end
of the core, the point of observation. To determine the wetting phase
saturation at the outlet end of the core, we perform a material balance for the
wetting phase after breakthrough to obtain
( )2 21w wirr pD i w wS S N W F S= + − −⎡ ⎤⎣ ⎦ (8.99)
Eq.(8.99) can be written in terms of the fractional flow of the non-wetting
phase as
2 2w wirr pD i nwS S N W f= + − (8.118)
Using Eqs.(8.116) and (8.118), rw
rnw
kk
versus 2wS can be computed.
Johnson, Bossler and Neumann (JBN, 1959) presented equations for
calculating the individual relative permeability curves by the unsteady state
method by incorporating the pressure drop into the computations. The
pressure drop across the porous medium at time t is given by
8-55
0
L PP dxx
∂Δ = −
∂∫ (8.119)
Darcy’s Law for the non-wetting phase gives
rnwnw
nw
kk A Pqxμ
∂= −
∂ (8.120)
Dividing Eq.(8.120) by q and rearranging gives the pressure gradient as
nwnw
rnw
qP fx kk A
μ⎛ ⎞∂= −⎜ ⎟∂ ⎝ ⎠
(8.121)
Substituting Eq.(8.121) into (8.119) gives
0
Lnw nw
rnw
q fP dxkA kμ⎛ ⎞Δ = ⎜ ⎟
⎝ ⎠ ∫ (8.122)
Applying the Buckley-Leverett frontal advance equation, Eq.(8.77), at the
outlet end of the core after breakthrough gives
( )2w
i w
w S
Q t dfLA dSφ
⎛ ⎞= ⎜ ⎟
⎝ ⎠ (8.90)
Dividing Eq.(8.77) by (8.90) gives
'
'2
w
w
fxL f
= (8.123)
where 'wf and '
2wf are the derivatives of the fractional flow functions at any
distance and at the core outlet, respectively. Differentiating Eq.(8.123) with
respect to 'wf gives
''2
ww
Ldx dff
= (8.124)
8-56
Substituting Eq.(8.124) into (8.122) and rearranging gives
'
2'
' 20
wf nw ww
rnw nw
f PkAfdfk q Lμ
Δ=∫ (8.125)
Let
a constants nw
q kAP Lμ
⎛ ⎞ = =⎜ ⎟Δ⎝ ⎠ (8.126)
Substituting Eq.(8.126) into (8.125) gives
'
2
'2
'
0
wwf nw s
wrnw
q ff Pdf
qkP
⎛ ⎞⎜ ⎟Δ⎝ ⎠=
⎛ ⎞⎜ ⎟Δ⎝ ⎠
∫ (8.127)
Let a relative injectivity ratio be defined as
r
s
qPI
qP
⎛ ⎞⎜ ⎟Δ⎝ ⎠=⎛ ⎞⎜ ⎟Δ⎝ ⎠
(8.128)
Substituting Eq.(8.128) into (8.127) gives
'
2'
' 20
wf nw ww
rnw r
f fdfk I
=∫ (8.129)
Differentiating Eq.(8.129) with respect to '2wf gives
'
2 2'2
nw w
rnw w r
f fdk df I
⎛ ⎞= ⎜ ⎟
⎝ ⎠ (8.130)
Substituting Eq.(8.91) into (8.130) gives
2 11
nw
rnw i r
i
f dk W I
dW
⎛ ⎞= ⎜ ⎟⎛ ⎞ ⎝ ⎠
⎜ ⎟⎝ ⎠
(8.131)
8-57
Eq.(8.131) can be used to calculate the relative permeability of the non-
wetting phase as
2
11
nwrnw
i r
i
fkd
W Id
W
=⎛ ⎞⎜ ⎟⎛ ⎞ ⎝ ⎠
⎜ ⎟⎝ ⎠
(8.132)
Knowing the relative permeability of non-wetting phase, the relative
permeability of the wetting phase can be calculated from Eq.(8.116) as
2
1 1wrw rnw
nw nw
k kf
μμ
⎛ ⎞= −⎜ ⎟
⎝ ⎠ (8.133)
The advantage of the unsteady method over the steady state method of
relative permeability measurement is that it is considerably faster. Because
the method is based on the Buckley-Leverett displacement model, the
unsteady state method can only be used to calculate relative permeability
curves between Swf and the wetting phase saturation at the residual
nonwetting phase saturation (1-Snwr) as previously noted. If Swf is high as in
the case of a favorable mobility ratio displacement, then much of the relative
permeability curves cannot be obtained because one is limited to a very small
saturation observation window. To solve this problem, unfavorable mobility
ratio displacements are typically used to determine relative permeability
curves by the unsteady state method. Further, in order to minimize capillary
end effect, high displacement rates are also typically used. The combination
of high rate and adverse mobility ratio can lead to viscous instability that will
make the displacement performance to be rate sensitive. If this happens, the
relative permeability curves obtained by the unsteady state method will be
rate sensitive and can be quite different from the relative permeability curves
of the same porous medium obtained by the steady state method (Peters and
Khataniar, 1987).
8-58
Eqs.(8.117) and (8.132) call for differentiating the measured
experimental data. The challenge in calculating the relative permeability
curves from these equations is to ensure that the curves are smooth. Any type
of finite difference approximation of the derivatives will result in numerical
noise leading to noisy relative permeability curves. The best way to process
the experimental data is by fitting well behaved functions to the experimental
data and then differentiating the functions. Peters and Khataniar (1987) have
suggested the following curve fits, which they have shown to work well.
( ) ( )21 2 3ln lnpD i iN A A W A W= + + (8.134)
2
1 2 31 1 1ln ln lni r i i
B B BW I W W
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ (8.135)
Example 8.2
Table 8.2 gives the experimental data for an unsteady state relative
permeability measurement for a sandpack. In the experiment, water was used
to displace a viscous oil at a constant injection rate. The pore volume of water
injected (Wi), the cumulative oil produced (Qo) and the pressure drop across
the sandpack (ΔP) were measured as functions of time.
Table 8.2. Experimental Data for Unsteady State Relative Permeability
Measurements.
Wi Qo ΔP
PV %IOIP psi
0.339 38.28 9.02
0.351 38.95 8.30
0.395 40.10 6.91
0.439 40.91 6.07
0.502 41.92 5.42
0.587 42.95 4.87
8-59
0.670 43.77 4.55
0.840 45.11 4.00
1.137 46.55 3.32
1.604 47.96 2.78
2.029 48.96 2.52
2.624 50.08 2.42
3.225 50.78 2.30
4.346 51.78 2.13
5.719 52.67 1.99
7.092 53.23 1.90
8.464 53.67 1.83
10.516 54.16 1.79
11.203 54.34 1.75
12.578 54.60 1.74
13.271 54.71 1.70
14.644 54.82 1.70
16.016 54.90 1.70
Other data for the experiment are as follows:
Injection rate = 100 cc/hr
Irreducible water saturation = 11.90%
Length of porous medium = 54.7 cm
Diameter of porous medium = 4.8 cm
Average porosity of porous medium = 30.58%
Absolute permeability of porous medium = 3.42 Darcies
Oil viscosity = 108.37 cp
Oil density = 0.959 gm/cm3
Water viscosity = 1.01 cp
Water density = 0.996 gm/cm3
Oil-water interfacial tension = 26.7 dynes/cm
8-60
Effective permeability to oil at irreducible water saturation = 3.16 Darcies
Oil recovery at water breakthrough = 38.28 % IOIP
Final oil recovery at termination of experiment = 54.9 % IOIP
1. Plot graphs of the raw experimental data.
2. Perform the curve fits suggested in Eqs.(8.134) and (8.135) and display
the results graphically.
3. Calculate the oil-water relative permeability curves for the porous
medium using the Johnson-Bossler-Neumann (JBN) method.
4. Plot graphs of the relative permeability curves.
5. Plot the graph of the true fractional flow curve measured in the
experiment.
6. How long was this test?
Solution to Example 8.2
The results of the calculations are summarized in Table 8.3.
1. Figure 8.23 shows the graphs of the raw experimental data.
2. Figures 8.24 and 8.25 show the curve fits of pDN versus ln iW and
1lni rW I
⎛ ⎞⎜ ⎟⎝ ⎠
versus 1lniW
⎛ ⎞⎜ ⎟⎝ ⎠
. The curve fit equations are
( )20.4026 0.0474ln 0.0066 lnpD i iN W W= + +
2
1 1 1ln 2.3600 1.5798ln 0.1130 lni r i iW I W W
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
These equations can be differentiated analytically to obtain
8-61
( )( )2
0.0474 2 0.0066 lnpD inw
i i
dN Wf
dW W−
= =
( )( ) 2
1 12.3600 1.5798ln 0.1130 ln2
1 12 0.1130 ln1.5798
1 1 1i iW Wi r inw
rnw
i i i
dW I Wf e
kd
W W W
⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞⎜ ⎟− + + ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎝ ⎠
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥= = +⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
Table 8.3. Calculated Results for Example 8.2.
Wi NpD ΔP
PV PV psi ln(Wi) 1ln
iW
⎛ ⎞⎜ ⎟⎝ ⎠
fnw2 Sw2 Ir 1
i rW I 1
lni rW I
⎛ ⎞⎜ ⎟⎝ ⎠
2nw
rnw
f
k krnw krw
0.119 0.924 0.000
0.339 0.337 9.02 -1.082 1.082 0.182 0.395 4.335 0.680 -0.385 0.368 0.494 0.021
0.351 0.343 8.30 -1.047 1.047 0.174 0.401 4.711 0.605 -0.503 0.356 0.490 0.022
0.395 0.353 6.91 -0.929 0.929 0.151 0.413 5.659 0.447 -0.804 0.319 0.473 0.025
0.439 0.360 6.07 -0.823 0.823 0.133 0.421 6.442 0.354 -1.040 0.290 0.457 0.028
0.502 0.369 5.42 -0.689 0.689 0.113 0.432 7.214 0.276 -1.287 0.258 0.436 0.032
0.587 0.378 4.87 -0.533 0.533 0.093 0.443 8.029 0.212 -1.550 0.226 0.411 0.037
0.67 0.386 4.55 -0.400 0.400 0.079 0.452 8.594 0.174 -1.751 0.203 0.388 0.042
0.84 0.397 4.00 -0.174 0.174 0.059 0.467 9.775 0.122 -2.106 0.170 0.349 0.052
1.137 0.410 3.32 0.128 -0.128 0.040 0.483 11.778 0.075 -2.595 0.136 0.295 0.066
1.604 0.423 2.78 0.473 -0.473 0.026 0.500 14.065 0.044 -3.116 0.108 0.237 0.084
2.029 0.431 2.52 0.708 -0.708 0.019 0.512 15.517 0.032 -3.449 0.094 0.199 0.097
2.624 0.441 2.42 0.965 -0.965 0.013 0.526 16.158 0.024 -3.747 0.082 0.162 0.113
3.225 0.447 2.30 1.171 -1.171 0.010 0.534 17.001 0.018 -4.004 0.074 0.135 0.125
4.346 0.456 2.13 1.469 -1.469 0.006 0.547 18.358 0.013 -4.379 0.064 0.100 0.144
5.719 0.464 1.99 1.744 -1.744 0.004 0.559 19.649 0.009 -4.722 0.057 0.074 0.162
7.092 0.469 1.90 1.959 -1.959 0.003 0.566 20.580 0.007 -4.983 0.053 0.057 0.175
8.464 0.473 1.83 2.136 -2.136 0.002 0.573 21.367 0.006 -5.198 0.050 0.045 0.185
10.516 0.477 1.79 2.353 -2.353 0.002 0.580 21.844 0.004 -5.437 0.047 0.033 0.197
11.203 0.479 1.75 2.416 -2.416 0.001 0.582 22.344 0.004 -5.523 0.047 0.030 0.200
12.578 0.481 1.74 2.532 -2.532 0.001 0.586 22.472 0.004 -5.644 0.045 0.025 0.206
8-62
13.271 0.482 1.70 2.586 -2.586 0.001 0.588 23.001 0.003 -5.721 0.045 0.022 0.208
14.644 0.483 1.70 2.684 -2.684 0.001 0.590 23.001 0.003 -5.820 0.044 0.019 0.213
16.016 0.484 1.70 2.774 -2.774 0.001 0.592 23.001 0.003 -5.909 0.043 0.016 0.217
Figure 8.23. Raw experimental data for the unsteady state relative permeability measurements of Example 8.2.
8-63
Figure 8.24. Curve fit of pDN versus ln iW for Example 8.2.
Figure 8.25. Curve fit of 1lni rW I
⎛ ⎞⎜ ⎟⎝ ⎠
versus 1lniW
⎛ ⎞⎜ ⎟⎝ ⎠
for Example 8.2.
3. The oil-water relative permeability data calculated with Eqs.(8.132) and
(8.133) are presented in Table 8.2.
4. Figure 8.26 shows the oil-water relative permeability curves from the
unsteady state experiment. It should be noted that the relative
permeability curves are obtained over the limited saturation range of
0.395 0.592wS≤ ≤ . The relative permeability curves between 0.119wirrS =
and 0.395wfS = cannot be obtained from the experiment. They can only
be obtained by extrapolation of the computed data to the conditions at
the irreducible water saturation where 0.000rwk = and 0.924nwrk = . The
8-64
experiment predicts a residual oil saturation of 40% in this
homogeneous high permeability sand.
Figure 8.26. Computed relative permeability curves for Example 8.2.
5. The true fractional flow curve measured in the experiment is shown in
Figure 8.27. It is interesting to note that it is only that portion of the
true fractional flow curve that is equal to the approximate fractional
flow curve that can be measured in the experiment. If the saturation
profiles in the experiment could be imaged, then it would possible to
calculate the true fractional flow curve between wirrS and wS by the
similarity transformation and the integration outlined in Figures 8.11
and 8.12.
6. The unsteady state experiment lasted 48.48 hours compared to several
weeks for the steady state experiment.
8-65
Figure 8.27. True fractional flow curve measured in the unsteady state experiment of Example 8.2.
8.5 FACTORS AFFECTING RELATIVE PERMEABILITIES
The factors that affect or could affect relative permeability curves
include (1) fluid saturation, (2) fluid saturation history, (3) Wettability, (4)
injection rate, (5) viscosity ratio, (6) interfacial tension, (7) pore structure, (8)
temperature and (9) heterogeneity.
8.5.1 Fluid Saturation
Relative permeabilities are strongly dependent on fluid saturations. The
higher the fluid saturation, the higher the relative permeability to that fluid.
In general, relative permeabilities are nonlinear functions of fluid saturation
as shown in Figures 8.6 and 8.7.
8-66
8.5.2 Saturation History
Like capillary pressure curves, relative permeability curves show
saturation hysteresis. Figure 8.28 shows typical relative permeability curves
for drainage and imbibition. The imbibition non-wetting phase relative
permeability curve is generally lower than the drainage curve at the same
saturations. The imbibition wetting phase relative permeability curve is
slightly greater than the drainage curve. These differences can easily be
explained. During drainage, the non-wetting phase flows through the large
pores displacing the wetting phase along the way. The thin film of wetting
phase that coats the grain surface acts as a lubricant for the flow of the non-
wetting phase. Therefore, the relative permeability to the non-wetting phase
will be high during drainage. That of the wetting phase also will be high
because it starts from 1 and decreases as the non-wetting phase begins to
occupy some of the pores that were previously occupied by the wetting phase.
During imbibition, some of the non-wetting phase will be trapped in the large
pores. This capillary trapping reduces the amount of non-wetting phase
available to flow during imbibition compared to during drainage. It also
reduces the cross-sectional area of the medium occupied by the connected
non-wetting phase. As a result, the imbibition relative permeability to the
non-wetting phase is reduced compared to that during drainage. Because of
capillary trapping of the non-wetting phase during imbibition, the wetting
phase is forced to occupy and flow through pore sizes that are larger than it
would otherwise have flowed if there was no trapping of the non-wetting
phase. This forcing of the wetting phase to flow through larger pores than it
would otherwise have done in the absence of trapping enhances the relative
permeability of the wetting phase on the imbibition cycle compared to the
drainage cycle. These observations are in accord with the experimental results
shown in Figure 8.28.
8-67
Figure 8.28. Relative permeability hysteresis (Osoba et al., 1951).
8.5.3 Wettability
Relative permeability curves are markedly affected by the wettability of
the medium. Jennings (1957) measured steady state oil water relative
permeability curves on a core that was initially strongly water wet. He then
treated the core with a surface active agent (organo chlorosilane) that
rendered the core oil wet and repeated the relative permeability
measurements. The results are shown Figure 8.29. In general, the relative
permeability to oil decreases while the relative permeability to water increases
as the medium changes from a strongly water wet to a strongly oil wet
8-68
medium. It is interesting to replot the relative permeability curves of Figure
8.29 as functions of wetting phase saturation instead of water saturation. The
replotted curves are shown in Figure 8.30. We see that when plotted against
the wetting phase saturation, the relative permeability curves for the oil wet
core and the water wet core are close to each other. They are not identical
because the degree of wettability preference in the two experiments may be
different. However, the relative permeability curves for the wetting phase and
the non-wetting phase from the two experiments are essentially the same.
Based on experimental observations, Craig (1971) gives the following
rules of thumb about the relative permeabilities for water wet and oil wet
media. (1) The irreducible water saturation for a water wet medium is usually
greater than 20% to 25% whereas that of an oil wet medium it is generally
less than 15%, and frequently less than 10%. (2) The water saturation at
which the oil and water relative permeabilities are equal is greater than 50%
for a strongly water wet medium whereas it is less than 50% for a strongly oil
wet medium. (3) The end-point relative permeability to water is generally less
than 30% for a strongly water wet medium and greater than 50% and
approaches the oil end point for a strongly oil wet medium. These
observations are consistent with the effect of wettability on the fluid
distribution and displacement discussed in Section 6.3.4.
It should be emphasized that the above rules of thumb are applicable
only to systems that show a strong preferential wettability to either water or
oil. In general, one cannot infer the wettability of a porous medium based
solely on the relative permeability curves. For example, relative permeability
curves that intersect at a water saturation of 50% does not mean that the
medium is of “neutral” wettability.
8-69
Figure 8.29. Effect of strong preferential wettability on steady state relative permeability curves (Jennings, 1957).
Figure 8.30. Relative permeability curves from Figure 8.28 replotted as functions of wetting phase saturation (adapted from Jennings, 1957).
8-70
At a given saturation, the relative permeability for a phase is higher
when that phase is the non-wetting phase than when it is the wetting phase.
This is observation can be seen in Figure 8.29. At any water saturation, the
relative permeability to water is higher when the water was the non-wetting
phase than when it was the wetting phase. Similarly, at any water saturation,
the relative permeability to oil is higher when the oil was the non-wetting
phase than when it was the wetting phase.
Owens and Archer (1971) measured relative permeability curves of
sandstones that were rendered progressively oil wet with a surface active
agent. Figure 8.31 shows their results for contact angles ranging from 0 to
180°. Note the general decrease in the oil relative permeabilities and increase
in the water relative permeabilities as the system was made progressively
more oil wet. Note also, that the strongly preferentially wet systems with
contact angles of 0 and 180° generally obey Craig’s rules of thumb regarding
the end-point water relative permeability and the water saturation at which
the water and oil relative permeabilities are equal. The rules of thumb do not
strictly apply to the intervening degrees of wettability.
8.5.4 Injection Rate
Injection rate usually does not affect relative permeabilities obtained by
the steady state method provided the rate is sufficiently high to minimize
capillary end effect. However, Peters and Khataniar (1987) have shown that
relative permeabilities obtained by the unsteady state displacement method
can show rate sensitivity due to viscous instability. Figures 8.32 and 8.33
show the effects of rate and viscosity ratio on relative permeability curves for
oil wet and water wet sandpacks. Both the curves for the oil wet medium and
the water wet medium shift to lower water saturations as the injection rate
(stability number) is increased. It should be noted that for the water wet
system, the relative permeability curves obtained by the unsteady state
method deviate from the steady state curves as the degree of instability is
8-71
increased. The water curve increases and the oil curve decreases away from
the steady state curves as the degree of instability of the displacement
experiment increases.
Figure 8.31. Relative permeabilities for range of wetting conditions (Owens
and Archer, 1971).
8-72
Figure 8.32. Effect of stability number on unsteady state relative permeability curves for oil wet sandpacks (Peters and Khataniar, 1987).
8-73
Figure 8.33. Effect of stability number on unsteady state relative permeability
curves for water wet sandpacks (Peters and Khataniar, 1987).
8.5.5 Viscosity Ratio
Viscosity ratio usually does not affect relative permeabilities obtained by
the steady state method since there is no displacement involved. Figure 8.34
shows the relative permeability curves obtained with the steady state method
at various viscosity ratios. Clearly, no viscosity ratio effect is apparent.
However, Peters and Khataniar (1987) have shown that relative permeabilities
obtained by the unsteady state displacement method at adverse viscosity
ratios can show sensitivity to injection rate and viscosity ratio due to viscous
instability (Figures 8.32 and 8.33).
8-74
Figure 8.34. Effect of viscosity ratio on relative permeability curves obtained
by the steady state method (Leverett, 1939).
8.5.6 Interfacial Tension
Relative permeability curves are affected by interfacial tension only at
interfacial tensions lower than 0.1 dyne/cm. Above this value, relative
permeabilities are unaffected by interfacial tension.
Bardon and Longeron (1978) studied the effect of interfacial tension on
gas-oil relative permeabilities using methane and normal heptane
displacement experiments. In their study, interfacial tensions were calculated
using parachors and the gas-oil relative permeabilities were calculated using
relative permeability models and a numerical simulator to history match the
displacement data. Figure 8.35 shows the relative permeability curves that
gave the best fit to the displacement recovery data at the calculated interfacial
tensions. The results show that the relative permeabilities to gas and oil
8-75
increased as the interfacial tension decreased. The residual fluid saturations
decreased as the interfacial tension decreased as expected from the effect of
capillary number on residual fluid saturations. In the limit, at ultra-low
interfacial tensions, the relative permeability curves were approximately
straight lines. These general trends in the effect of interfacial tensions on
relative permeability curves have been confirmed by Amaefule and Handy
(1981).
Figure 8.35. Effect of interfacial tension on gas-oil relative permeability curves
(Bardon and Longeron, 1978).
8.5.7 Pore Structure
Morgan and Gordon (1970) have presented results that show that rocks
with large pores and correspondingly small specific surface areas have low
irreducible water saturations that leave a relatively large amount of pore
8-76
space available for multiphase flow. Therefore, for such rocks, end point
relative permeabilities are high and a large saturation change may occur
during two phase flow. By contrast, rocks with small pores have larger
specific surface areas and larger irreducible water saturations that leave less
room for multiphase flow. As a result, the end point relative permeabilities
are lower and the saturation range for two phase flow is smaller than in rocks
with large pores. Finally, rocks having some relatively large pores connected
by small pores have a large surface area, resulting in high irreducible water
saturation and relative permeability behavior that is similar to rocks with
small pores only. These observations are summarized in Figure 8.36.
8.5.8 Temperature
There are data in the literature that suggest that relative permeability
curves are affected by temperature. Poston et al., (1970) found that
temperature causes residual oil saturation to decrease and irreducible water
saturation to increase, with corresponding increases in relative permeability
curves (Figure 8.37). On the other hand, there are data in the literature that
also show that relative permeabilities are not temperature dependent (Miller
and Ramey, 1985). Apparently, the effect of temperature on relative
permeabilities is still and open question. This situation is understandable
because temperature can affect rock and fluid properties which in turn can
affect relative permeability curves. For example, high temperature can change
the wettability of the rock which affects relative permeabilities. It can also
reduce interfacial tensions, which can affect relative permeabilities and the
irreducible saturations. Because of the effect of temperature on the other
properties of the system that can affect relative permeabilities, it is difficult to
categorically determine the effect of temperature on relative permeabilities.
8-77
Figure 8.36. Effect of pore structure on relative permeability curves; (a)
sandstone with large, well-connected pores with k = 1314 md; (b) sandstone with small, well-connected pores with k = 20 md; (c) sandstone with a few
large pores connected with small pores with k = 36 md (Morgan and Gordon, 1970).
8-78
Figure 8.37. Effect of temperature on relative permeability curves (Poston et
al., 1970).
8.5.9 Heterogeneity
Relative permeabilities are typically measured on homogeneous core
samples. These curves are then used in numerical simulators to model the
performance of heterogeneous reservoirs. It is often necessary to adjust the
laboratory measured relative permeability curves in order to successfully
history match the performance of the heterogeneous reservoirs. Gharbi and
Peters (1993) simulated the waterflood performance of a heterogeneous
reservoir using a set of input relative permeability curves and then used the
simulated oil recovery versus pore volumes of water injected and the
simulated pressure drop to calculate the equivalent relative permeability
curves for the heterogeneous medium by the JBN method. Figure 8.38
8-79
compares the input relative permeabilities with the computed equivalent
relative permeabilities for the heterogeneous medium. The effect of the
heterogeneity is to shift the oil and water relative permeabilities to low water
saturations thereby increasing the water relative permeability curve and
decreasing the oil relative permeability curve. Thus, the relative
permeabilities for the heterogeneous medium are similar to the relative
permeabilities for a strongly oil wet medium.
Figure 8.38. Effect of heterogeneity on relative permeability curves (Gharbi
and Peters, 1993).
8.6 THREE-PHASE RELATIVE PERMEABILITIES
Three phase relative permeabilities are required to predict the
performance of three phase flow of oil, water and gas. There are considerably
less experimental data in the literature on three phase relative permeabilities
than two phase relative permeabilities. Figure 8.39 shows, on a ternary
8-80
saturation diagram, the approximate regions of single phase flow, two phase
flow and three phase flow in an oil, water and gas system ( Leverett and Lewis,
1941). It can be seen that the three phase flow region is small compared to
single phase and two phase flow regions. Figures 8.40, 8.41 and 8.42 show
the three phase water, oil and gas relative permeabilities measured by
Leverett and Lewis (1941). They found that the relative permeability to water
was only a function of the water saturation. However, the relative
permeabilities to oil and gas were functions of all three fluid saturations.
Figure 8.39. Approximate limits of saturations giving 5 per cent or more of all
components in flow stream for the flow of nitrogen, kerosene and brine. Arrows point to increasing fraction of respective components in stream
(Leverett and Lewis, 1941).
8-81
Figure 8.40. Three phase relative permeability to water (Leverett and Lewis,
1941).
Figure 8.41. Three phase relative permeability to oil (Leverett and Lewis,
1941).
8-82
Figure 8.42. Three phase relative permeability to gas (Leverett and Lewis,
1941).
Three phase relative permeabilities are not routinely measured in the
laboratory as two phase relative permeabilities. Instead, three phase relative
permeabilities are usually calculated from two phase relative permeability
data using various relative permeability models. Delshad and Pope (1989)
have reviewed the various three phase relative permeability models and found
that some of them do not always agree with the available experimental three
phase relative permeability data.
8.7 CALCULATION OF RELATIVE PERMEABILITIES FROM DRAINAGE CAPILLARY PRESSURE CURVE
In Section 7.12.2, we derived the following approximate drainage
relative permeability curves for wetting and non-wetting phases from the
drainage capillary pressure curve:
8-83
2
01
20
( )
wSw
cwrw w
w
c
dSPkk S
k dSP
= =∫
∫ (7.159)
and
1
2
1
20
( ) w
w
cSnwrnw w
w
c
dSPkk S
k dSP
= =∫
∫ (7.160)
We found that these models were defective in two respects: (1) they do not
include trapped residual saturations and (2) the sum of the relative
permeabilities is equal to 1, which is contrary to experimental observations.
These deficiencies result from the fact that the models neglect certain facts
about the nature of two phase flow in porous media. First, the cross-sectional
area open to the flow of the wetting phase is not a constant as assumed in the
models but is a function of the wetting phase saturation. Second, the
tortuosity for the flow of the wetting phase, which was neglected in the
models, is also a function of the wetting phase saturation. Burdine (1953)
proposed the following normalized drainage relative permeability models,
which account for these saturation dependencies in the cross-sectional area
and the tortuosity for two phase flow:
( )
*
*2* 2* * 0
1**
20
1( )( )
( 1) 1
wS
wcw w
rw w ww w
wc
dSPk Sk S S
k SdS
P
= ==
∫
∫ (8.136)
( ) *
1*
2* 2* *1*
*2
0
1
( )( ) 1( 0) 1
w
wcSnw w
rnw w wnw w
wc
dSPk Sk S S
k SdS
P
= = −=
∫
∫ (8.137)
8-84
where *wS is the normalized wetting phase saturation given by
*
1w wirr
wwirr
S SSS
−=
− (8.138)
In Eqs.(8.136) and (8.137), the ratios of the integrals on the right side
account for the cross-sectional area changes with saturations and the terms
( )2*wS and ( )2*1 wS− account for the tortuosity changes with saturations. It
should be noted that the base permeability used in Eq.(8.136) to define the
normalized relative permeability to the wetting phase is equal to the absolute
permeability of the medium, whereas the base permeability used to define the
normalized relative permeability to the non-wetting phase in Eq.(8.137) is
equal to the effective permeability to the non-wetting phase at the irreducible
wetting phase saturation. Thus, the normalized wetting phase relative
permeability given by Eq.(8.136) is also the true relative permeability to the
wetting phase. However, the normalized non-wetting phase relative
permeability given by Eq.(8.137) must be multiplied by the end point relative
permeability to the non-wetting phase (knwr) in order to obtain the true non-
wetting phase relative permeability. Furthermore, the normalized non-wetting
phase relative permeability of Eq.(8.137) starts at a wetting phase saturation
of 1.0 or a non-wetting phase saturation of zero. Normally, a critical non-
wetting phase saturation is required before the non-wetting phase can flow.
Thus, the end-point non-wetting phase relative permeability and a critical
non-wetting phase saturation must be introduced into Eq.(8.137) to obtain
the true relative permeability for the non-wetting phase. Given the drainage
capillary pressure curve, the integrals in Eqs.(8.136) and (8.137) can easily be
calculated numerically to obtain the normalized drainage relative permeability
curves.
An alternative approach to evaluating the integrals in Eqs.(8.136) and
(8.137) is to fit the Brooks-Corey (1966) model to the drainage capillary
8-85
pressure curve and then integrate the resulting linear function. As discussed
in Section 7.13.1, the Brooks-Corey drainage capillary pressure model is given
by
*ln ln lnw c eS P Pλ λ= − + (7.161)
or
*1ln ln lnc w eP S Pλ
= − + (7.162)
and
( )1
*c e wP P S λ
−= (7.164)
where λ is the pore size distribution index obtained from the straight line
given by Eq.(7.161) or (7.162). Substituting Eq.(7.164) into Eqs.(8.136) and
(8.137) and performing the integrations gives the normalized drainage relative
permeability curves as
( )2 3
*( )rw w wk S Sλ
λ+
= (8.139)
and
( ) ( )22* *( ) 1 1rnw w w wk S S S
λλ+⎡ ⎤
= − −⎢ ⎥⎣ ⎦
(8.140)
A critical saturation can be introduced into the relative permeability model for
the non-wetting phase as
( )2 2
*( ) 1 1w wirrrnw w w
m wirr
S Sk S SS S
λλ+⎛ ⎞ ⎡ ⎤−
= − −⎜ ⎟ ⎢ ⎥− ⎣ ⎦⎝ ⎠ (8.141)
8-86
where Sm is the wetting phase saturation corresponding to the critical non-
wetting phase saturation. Finally, the true relative permeability curve for the
wetting and non-wetting phases are given by
( )2 3
*( )rw w wk S Sλ
λ+
= (8.142)
( )2 2
*( ) 1 1w wirnw w nwr w
m wi
S Sk S k SS S
λλ+⎛ ⎞ ⎡ ⎤−
= − −⎜ ⎟ ⎢ ⎥− ⎣ ⎦⎝ ⎠ (8.143)
where nwrk is the non-wetting phase relative permeability at the irreducible
wetting phase saturation.
Example 8.3
Use the air-water capillary pressure data of Table 8.4 to calculate the
drainage relative permeability curves by the method of Brooks and Corey for a
core sample.
Table 8.4. Drainage Capillary Pressure Curves for Example 8.3.
Saturation Capillary Pressure
(psi)
1.000 1.973
0.950 2.377
0.900 2.840
0.850 3.377
0.800 4.008
0.750 4.757
0.700 5.663
0.650 6.781
0.600 8.195
0.550 10.039
0.500 12.547
0.450 16.154
0.400 21.787
0.350 31.817
8-87
0.300 54.691
0.278 78.408
Solution to Example 8.3
Figure 8.43 shows the graph of lnPc versus *ln wS for Swirr = 0.10. The
equation of the resulting straight line is given by
*ln 2.1443ln ln 2.2238c wP S= − +
Therefore,
1 2.1443λ
− = −
0.4664λ =
2.2238eP =
Figure 8.44 shows the graph of *ln wS versus lnPc for Swi = 0.10. It also is
linear and could have been used for the subsequent calculations. The Brooks-
Corey drainage capillary pressure equation is given by
( ) ( )1 2.1443* *2.2238c e w wP P S Sλ
− −= =
Figure 8.45 shows the normalized drainage relative permeability curves
calculated with Eqs.(8.139) and (8.140). Figure 8.46 shows the true drainage
relative permeability curves for Sm = 0.95 and knwr = 0.961. The results of the
calculations are summarized in Table 8.5.
8-88
Figure 8.43. Log-log graph of Pc versus *wS for Example 8.3.
Figure 8.44. Log-log graph of *wS versus Pc for Example 8.3.
8-89
Figure 8.45. Normalized drainage relative permeability curves for Example 8.3.
Figure 8.46. True drainage relative permeability curves for Example 8.3.
8-90
Table 8.5. Results of Drainage Relative Permeability Calculations for Example
8.3
Original Data
Brooks-Corey
Model Drainage Relative Permeability Curves
Pc Pc
Sw psi *wS psi ( )rw wk S ( )rnw wk S rwk rnwk
1.000 1.973 1.000 2.224 1.000 0.000 1.000 0.000
0.950 2.377 0.944 2.514 0.659 0.001 0.659 0.000
0.900 2.840 0.889 2.863 0.424 0.006 0.424 0.002
0.850 3.377 0.833 3.288 0.265 0.017 0.265 0.008
0.800 4.008 0.778 3.812 0.160 0.036 0.160 0.022
0.750 4.757 0.722 4.468 0.093 0.063 0.093 0.044
0.700 5.663 0.667 5.305 0.052 0.098 0.052 0.073
0.650 6.781 0.611 6.393 0.028 0.140 0.028 0.111
0.600 8.195 0.556 7.843 0.014 0.189 0.014 0.156
0.550 10.039 0.500 9.831 0.006 0.244 0.006 0.208
0.500 12.547 0.444 12.656 0.003 0.304 0.003 0.266
0.450 16.154 0.389 16.851 0.001 0.371 0.001 0.331
0.400 21.787 0.333 23.452 0.000 0.443 0.000 0.401
0.350 31.817 0.278 34.672 0.000 0.521 0.000 0.479
0.300 54.691 0.222 55.947 0.000 0.605 0.000 0.562
0.278 78.408 0.198 71.829 0.000 0.643 0.000 0.601
0.250 0.167 103.678 0.000 0.694 0.000 0.652
0.200 0.111 247.331 0.000 0.790 0.000 0.749
0.150 0.056 1093.394 0.000 0.892 0.000 0.852
0.100 0.000 0.000 1.000 0.000 0.962
8-91
NOMENCLATURE
A = cross sectional area in the flow direction
Bo = oil formation volume factor
Bw = water formation volume factor
fw = fractional flow of wetting phase
fw = fractional flow of water
fnw = fractional flow of non-wetting phase
fnw2 = fractional flow of non-wetting phase at the outlet end of porous medium
fo = fractional flow of oil
Fw = approximate fractional flow of wetting phase
g = gravitational acceleration
Ir = relative injectivity
J = Leverett J-function
k = absolute permeability of the medium
ko = effective permeability to oil
kw = effective permeability to water
kwr = end-point relative permeability to wetting phase
kg = effective permeability to gas
kro = relative permeability to oil
krw = relative permeability to water
krg = relative permeability to gas
krw = relative permeability to wetting phase
krnw= relative permeability to non-wetting phase
knwr= end-point relative permeability to non-wetting phase
L = length
M = mobility ratio
ME = end-point mobility ratio
Ncap = dimensionless capillary to viscous force ratio
Ng = gravity number
NpD = dimensionless cumulative production
8-92
Nvcap = capillary number
P = pressure
Pc = capillary pressure
Pe = displacement pressure for Brooks-Corey model
Pg = pressure in the gas phase
Pnw= pressure in the non-wetting phase
Po = pressure in the oil phase
Pc/ow = oil-water capillary pressure curve
Pc/go = gas-oil capillary pressure curve
Pc/gw = gas-water capillary pressure curve
Pw = pressure in the water phase
Pw = pressure in the wetting phase
q = total volumetric injection rate
qo = volumetric flow rate of oil
qg = volumetric flow rate of gas
qnw= volumetric flow rate of non-wetting phase
qw = volumetric flow rate of water
qw = volumetric flow rate of wetting phase
Qi = cumulative injection
Qnw = cumulative non-wetting phase produced
Qo = cumulative oil produced
R = oil recovery as a fraction of initial oil in place
Se = effective wetting phase saturation
Sg = gas saturation
So = oil saturation
Sor = residual oil saturation
Sw = water saturation
Sw = wetting phase saturation
Swirr = irreducible wetting phase saturation
8-93
Swro = wetting phase saturation at which imbibition capillary pressure is zero
Sw2 = wetting phase saturation at the outlet end of porous medium
Snw= non-wetting phase saturation
Snwr = residual non-wetting phase saturation
Swav = average wetting phase saturation
Swav = average water saturation
Swf = frontal saturation
*wS = normalized wetting phase saturation
t = time
tbt = breakthrough time
Dt = dimensionless time
v = flux vector, Darcy velocity vector
vw = Darcy velocity for the wetting phase
vnw = Darcy velocity for the non-wetting phase
x = distance in the direction of flow
fx = distance to the displacement front
Dx = dimensionless distance
Dfx = dimensionless distance to the displacement front
iW = dimensionless pore volume injected
pW = cumulative water produced
WOR = water oil ratio
δx = small length in the neighborhood of the outlet end of porous medium
gρ = density of gas
oρ = density of oil
wρ = density of water
wρ = density of wetting phase
nwρ = density of non-wetting phase
σ = interfacial tension
8-94
θ = contact angle
λ = pore size distribution index
μ = viscosity
μg = gas viscosity
μο = oil viscosity
μw = water viscosity
μw = wetting phase viscosity
μnw= non-wetting phase viscosity
φ = porosity, fraction
τ = tortuosity
γ = liquid specific gravity
ω = angular velocity of centrifuge
ΔP = pressure drop
ΔPw = pressure drop in the wetting phase
ΔPnw = pressure drop in the non-wetting phase
Γ = pore structure
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8-96
Gharbi, R.: Numerical Modeling of Fluid Displacements in Porous Media Assisted by Computed Tomography Imaging, PhD Dissertation, The University of Texas at Austin, Austin, Texas, August 1993.
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APPENDIX A
A Systematic Approach To Dimensional Analysis
Summary
This appendix presents a systematic method for obtaining the
complete set of independent dimensionless groups pertinent to a
problem. The algebraic theory underlying the method is presented. It is
shown that the dimensionless groups occupy the null space of the
dimensional matrix. The eigenvectors of this null space, which constitute
the bases vectors for the null space, contain a complete set of
independent dimensionless π groups. Other complete and independent
dimensionless π groups can be obtained systematically by a careful
navigation through this vector space. An example problem is presented
in detail to demonstrate the method.
Introduction
Dimensional analysis is a powerful tool for solving engineering
problems. It can be used to design cost effective experimental programs,
organize presentations of experimental as well as numerical simulation
results, and to scale the results of model experiments to predict the
performance of a large scale system, the prototype system.
Dimensional analysis is typically covered in a cursory manner in
fluid mechanics courses in which ad hoc techniques are presented for
A-1
deriving the dimensionless groups. Perhaps, because of this cursory
treatment, most engineers have failed to appreciate the power of the
technique and have therefore not taken advantage of it in their work.
The objective of this appendix is to present a systematic method for
obtaining the complete set of independent dimensionless groups
pertinent to a problem. The algebraic theory underlying the method is
presented. An example problem is presented in detail to demonstrate the
method.
Algebraic Theory of Dimensional Analysis
Let a physical process be described by n variables .
Suppose each of these variables can be expressed dimensionally in terms
of the primary variables of mass (M), length (L) and time (T) such that
dimensionally,
1 2, ,..., nu u u
[ ] i i ia b ciu M L T= (A.1)
The dimensional analysis problem may now be stated as follows. We wish
to determine the exponents of real numbers 1 2, ,..., nx x x such that the
power product 1 21 2 ... nxx x
nu u u is dimensionless; that is, has the dimensions of
M0L0T0.
The dimension of the power product is given by
( ) ( ) ( )1 1 2 2 1 1 2 2 1 1 2 21 2 ... ... ...1 2 ... n n n n n nn a x a x a x b x b x b x c x c x c xxx x
nu u u M L T+ + + + + + + + +⎡ ⎤ =⎣ ⎦ (A.2)
The power product will be dimensionless if and only if the following
conditions are met:
A-2
1 1 2 2 ... 0n na x a x a x+ + + = (A.3)
1 1 2 2 ... 0n nb x b x b x+ + + = (A.4)
1 1 2 2 ... 0n nc x c x c x+ + + = (A.5)
Eqs. (A.3) to (A.5) are derived from Eq.(A.2) by setting the dimensions of
mass (M), length (L) and time (T) to zero, respectively. Eqs.(A.3) to (A.5)
constitute a homogeneous system of linear algebraic equations for the n
unknowns 1 2, ,..., nx x x
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
. The system of equations is homogeneous because
the right-hand side is zero. Dimensional analysis always gives rise to a
homogeneous system of linear algebraic equations similar to Eqs.(A.3) to
(A.5).
Eqs.(A.3) to (A.5) may be written in matrix notation as
(A.6)
1
21 2 3 4
31 2 3 4
41 2 3 4
... 0
... 0
... 0...
n
n
n
n
xx
a a a a ax
b b b b bx
c c c c c
x
⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
or
0Ax = (A.7)
where the dimensional matrix A is given by
1 2 3 4
1 2 3 4
1 2 3 4
...
...
...
n
n
n
a a a a aA b b b b b
c c c c c
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
(A.8)
A-3
and x is the vector of the unknowns. In general, because there are more
unknowns than equations, the dimensional matrix is usually an
matrix in which . A homogeneous system of linear algebraic
equations with more unknowns than equations usually has an infinite
number of nontrivial solutions. It is these nontrivial solutions that give
rise to the dimensionless π groups we seek.
mxn m n<
If the rank of the dimensional matrix A is r, then we can solve for r
of the variables in terms of the remaing ( )n r− variables. Such a solution
will result in independent dimensionless groups, usually called
dimensionless π groups. Thus, to predict the number of independent
dimensionless groups in advance, we need to form the dimensional
matrix and then determine its rank. The rank r of a matrix A is the
largest submatrix of A with a nonzero determinant. Thus, for the mx
dimensional matrix A, the rank is m or less. We can easily determine the
rank of A by calculating the determinants of all possible
submatrices. If we fail to find an submatrix with a nonzero
determinant, then we calculate the determinants of all possible
submatrices until we determine the rank of A.
(n r−
)1
)
rxr
(1 m
n
mxm
mxm
( )m x− −
Equation (A.6) or (A.7) can be solved by simple row and column
operations on the dimensional matrix A. We know from linear algebra
that an system of homogeneous linear algebraic equations does not
have a unique nontrivial solution. Put another way, we know that an mx
system of homogeneous linear equations has an infinitely large number
of solutions. In the case in which the dimensional matrix has a rank of 3,
it is possible for the system of equations to be reduced to the following
row echelon form by row and column operations:
mxn
n
A-4
(A.9)
1
2
14 15 1 3
24 25 2 4
34 35 3 5
1 0 0 ... 00 1 0 ... 00 0 1 ... 0
...
n
n
n
n
xx
a a a xa a a xa a a x
x
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤⎢ ⎥⎢ =⎢ ⎥⎢⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎥ ⎢ ⎥⎥ ⎢ ⎥
⎢ ⎥⎣ ⎦
Thus, the initial objective of the solution procedure is to reduce the
dimensional matrix using row and column operations such that an rxr
unit submatrix appears at the beginning of the transformed matrix as
shown in Eq.(A.9).
We can now solve Eq.(A.9) to obtain the following result:
1 14 4 15 5 1... n nx a x a x a x= − − − − (A.10)
2 24 4 25 5 2... n nx a x a x a x= − − − − (A.11)
3 34 4 35 5 3... n nx a x a x a x= − − − − (A.12)
We supplement Eqs.(A.10) to (A.12) with the following additional solutions:
4 4x x= (A.13)
5 5x x= (A.14)
...
n nx x= (A.15)
A-5
Eqs.(A.10) to (A.15), which constitute the required solution to the
dimensional analysis problem, can be rearranged as a linear combination
of ( vectors as follows: )n r−
1 14 15 1
2 24 25 2
3 34 35 3
4 4 5
5
...1 0 00 1 0
... ... ... ...0 0 1
n
n
n
n
n
x a a ax a a ax a a ax x xx
x
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦
x
)
(A.16)
Eq.(A.16) shows that the solution to the system of linear homogeneous
algebraic equations arising in dimensional analysis consists of a linear
combination of vectors. We can identify these vectors (n r−1 2, ,..., n re e e −
00
1
aaa
⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
)
in
Eq.(A.16) as
(A.17)
14 15 1
24 25 2
34 35 3
1 2, ,...,1 00 1... ... ...0 0
n
n
n
n r
a aa aa a
e e e −
− − −⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢− − −⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢− − −⎢ ⎥ ⎢ ⎥ ⎢= = =⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣
These vectors are the eigenvectors of the null space of the
dimensional matrix. These eigenvectors play a critical role in the
dimensional analysis. They contain a complete set of
(n r−
( )n r−
dimensionless π groups (Buckingham’s π theorem). Because, these
eigenvectors are linearly independent, the dimensionless π groups
derived from them are also linearly independent. Because the
A-6
eigenvectors are the bases vectors for the null space of the dimensional
matrix, their linear combinations can be used to access every part of the
vector space thereby allowing us to derive other ( )n r− complete and
independent π groups. The π groups derived from the eigenvectors are
complete in the sense that any other π group that we derive will be
found to be a combination of this initial set of π groups. The π groups
are independent in the sense that each π group cannot be obtained by
algebraic manipulation of the other π groups.
Eq.(A.16) shows that solutions to the dimensional analysis problem
can be constructed by arbitrarily choosing the values for 4 5, ,..., nx x x . Such
choices can be used to obtain an initial set of complete and independent
dimensionless π groups. The choices can also be used to transform this
initial complete set of independent dimensionless π groups into new
complete and independent sets, which may be more convenient for
applications than the initial set.
An initial complete set of π groups can be obtained systematically
as follows. Let us choose 4 5 61, ... 0.nx x x x= = = = The solution given by
Eq.(A.16) becomes
1 14
2 24
3 34
4 1
5
10
... ...0n
x ax ax ax ex
x
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
(A.18)
The corresponding dimensionless π group is given by
A-7
3414 24 1 0 01 1 2 3 4 5 ...aa a
nu u u u u uπ −− −=
Next, we choose The solution given by Eq.(A.16)
becomes
5 4 61, ... 0.nx x x x= = = =
1 15
2 25
3 35
4 2
5
01
... ...0n
x ax ax ax ex
x
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
(A.19)
The corresponding dimensionless π group is given by
15 25 35 0 1 02 1 2 3 4 5...
a a anu u u u u uπ − − −=
We continue in this fashion until we finally choose 1nx = ,
The solution given by Eq.(A.16) becomes 4 5 1... 0.nx x x −= = = =
1 1
2 2
3 3
4
5
00
... ...1
n
n
n
n r
n
x ax ax ax ex
x
−
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
(A.20)
The last dimensionless π group is then given by
1 2 3 0 0 11 2 3 4 5 ...n n na a a
n r nu u u u u uπ − − −− =
A-8
Transformations of the Dimensionless π Groups
The π groups π1, π2, ..., πn–r are complete and independent. However,
they are not unique. By appropriate choice of the values for x4, x5, ..., xn,
other complete and independent dimensionless π groups can be derived.
However, these new complete sets will be found to be transformations of
1 2 n–r
( )
independent, we must use a different eigenvector in forming each of the
1 2 n–r 1 1
π , π , ..., π . To ensure that the new set of π groups is complete, we
must derive of them. To ensure that the new π groups are
dimensionless groups. For example, if the new set of π groups is
designated as , ..., Π , then Π must contain the eigenvector
n r−
π
Π , Π e
either alone or as a linear combination with one or more of the other
eigenvectors; must contain Π2 2e either alone or in combination with the
other eigenvectors and Πn–r must contain n re − either alone or in
e or more of the other eigenvectors.
am
coefficient is a function of the
fluid properties, the porous medium and the displacement conditions.
et us write the functional relationship as
combination with on
Ex ple Problem
We wish to investigate experimentally the factors that affect
dispersion in porous media. A preliminary analysis shows that dispersion
in porous media in the direction of flow can be characterized by a
longitudinal dispersion coefficient. The preliminary analysis further
indicates that the longitudinal dispersion
L
( )1 , , , , , , ,L s p o o s oD f D u D gμ μ ρ ρ= (A.21)
A-9
The objective is to derive a complete and independent set of
dimensionless π groups that affect the longitudinal dispersion coefficient
by dimensional analysis.
Procedure
Step 1.
The first step is to determine if the number of variables can be
reduced by combining some of them. For example, gravity segregation of
two fluids always involve the combination of variables (ρo–ρs)g or Δρg. So,
we can replace the three variables ρo , ρs, and g by the one variable Δρg.
Step 2.
Next, we form the power product of all the variables involved in the
problem to obtain
(A.22) ( ) 63 5 71 2 4 a dimensionless constantxx x xx x xs p o o LD u D g Dμ μ ρΔ =
Step 3.
Third, we express each of the variables in Eq.(A.22) in terms of the
primary dimensional variables of mass (M), length (L) and time (T). The
dimensions of each variable in Eq.(A.22) are shown in Table 1.
Step 4.
Fourth, we substitute the dimensions of each variable from Table 1
into Eq.(A.22) and insist that the power product become dimensionless
(i.e., has dimensions of M0L0T0). This step gives rise to the following
A-10
system of linear homogeneous algebraic equations for the powers x1, x2,
x3, x4, x5, x6 and x7:
1 5 6 0x x x+ + = (A.23)
(A.24) 1 2 3 4 5 6 72 2 2x x x x x x x− + + + − − + = 0
0 (A.25) 1 3 4 5 6 72x x x x x x− − − − − − =
Table 1 Dimensions of the Variables in Example Problem
Variable Symbol Dimensions
Mean grain diameter Dp M0L1T0
Interstitial velocity u M0L1T–1
Diffusion coefficient Do M0L2T–1
Solvent viscosity μs M1L–1T–1
Oil viscosity μo M1L–1T–1
Buoyancy group Δρg M1L–2 T–2
Dispersion coefficient DL M0L2T–1
Eq.(A.23) comes from insisting that the power product have a mass
dimension of M0. Similarly, Eq.(A.24) comes from L0 and Eq.(A.25) comes
from T0. Let us organize Eqs.(A.23), (A.24) and (A.25) into a matrix
equation as follows:
A-11
(A.26)
1
2
3
4
5
6
1 0 0 0 1 1 0 01 1 1 2 1 2 2 01 0 1 1 1 2 1 0
n
xxxxxxx
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤⎢ ⎥⎢− − − ⎢ ⎥⎢⎢ ⎥⎢ ⎥− − − − − −⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎥ ⎢ ⎥=⎥ ⎢ ⎥
⎢ ⎥⎣ ⎦
⎥⎥
Eq.(A.26) is a matrix equation similar to Eq.(A.7) where the dimensional
matrix is given in this case by
(A.27) 1 0 0 0 1 1 01 1 1 2 1 2 21 0 1 1 1 2 1
A⎡ ⎤⎢= − − −⎢⎢ ⎥− − − − − −⎣ ⎦
We could have derived the dimensional matrix directly from Step 2.
However, Steps 2 and 3 were presented to show the origin of the
dimensional matrix. Let us combine Steps 3 and 4 into one step and call
it Step 3(revised).
Step 3 (revised).
Next, we derive the dimensional matrix using the information in
Table 1. The result is shown in Table 2.
Table 2. Dimensional Matrix for Example Problem
x1 x2 x3 x4 x5 x6 x7
μs Dp u Do μo Δρg DL
M 1 0 0 0 1 1 0
L –1 1 1 2 –1 –2 2
T –1 0 –1 –1 –1 –2 –1
A-12
It should be noted that the first row of the dimensional matrix is the
power of M in each variable, the second row is the power of L and the
third row is the power of T. This is what the first column of Table 2 is
designed to show. It should also be noted that the variables and their
powers in Eq.(A.2) have been placed on top of the dimensional matrix to
remind us of the relationship between the variables and the entries in
the dimensional matrix.
Step 4 (revised).
Next, we solve Eq.(A.26) to obtain the powers of the variables that
will make the power product dimensionless. An initial reaction to solving
Eq.(A.26) is to propose the solution x1 = x2 = x3 = x4 = x5 = x6 = x7 = 0.
This is the trivial solution in which we are not interested. Our interest is
in the nontrivial solutions.
There is extensive theory in linear algebra about the solutions of a
system of homogeneous linear algebraic equations such as Eq.(A.26).
Although we cannot review all of this theory at this point, we will use
some of it to enable us to solve Eq.(A.26) to obtain a complete set of
independent dimensionless π groups. Eq.(A.26) has more unknowns (7
unknowns) than equations (3 equations). A system of linear
homogeneous algebraic equations with more unknowns than equations
has many (infinite number of) solutions. These are the solutions we
want.
The dimensional matrix A has a rank of r where r is the largest rxr
submatrix of A with a nonzero determinant. Buckingham’s π–theorem
states that the number of complete and independent dimensionless π
groups is equal to the number of variables minus the rank of the
dimensional matrix, (n–r). Therefore, we can predict the number of
A-13
complete and independent dimensionless π groups in advance by
determining the rank of the dimensional matrix without actually solving
the system of equations.
To determine the rank of the dimensional matrix in this example, we
compute the determinant of the first 3x3 submatrix of A. This gives a
determinant of –1 as shown in Eq.(A.27).
1 0 0
1 1determinant 1 1 1 1 determinant 1
0 11 0 1
x− =−
− −= − (A.28)
Because the determinant of a 3x3 submatrix of A is nonzero, the rank, r,
of A is 3. Therefore, there will be 7–3 or 4 complete and independent
dimensionless π groups in this example problem. In this example, the
rank of the dimensional matrix , r, is equal to the number of equations,
m. However, we should be aware that it is possible for the rank of the
dimensional matrix to be less than the number of equations if the rows of
the dimensional matrix are linearly dependent. Whatever the rank of A is,
we can find it by examining all possible 3x3 submatrices and if need be,
all possible 2x2 submatrices. If we examine all possible 3x3 submatrices
and fail to find a nonzero determinant, then the rank is obviously less
than 3. We will then examine all possible 2x2 submatrices to see if the
rank is 2.
Let us proceed to solve Eq.(A.26) by Gaussian elimination, which is
a standard method for solving a system of linear algebraic equations.
Gaussian elimination involves only simple row and column operations. In
this example, the variables in the dimensional matrix have been carefully
arranged initially to avoid column exchanges during the solution steps.
Therefore, only simple row operations will be required to solve the system
A-14
of equations. Our initial objective in the solution process is to perform
the elimination to the point that the first 3x3 submatrix of A is
transformed into a unit matrix of the form
1 0 00 1 00 0 1
(28)
as shown in Eq.(A.9). Because the right hand side of Eq.(A.26) is zero
and will always remain zero no matter how we manipulate the equations,
we can ignore it in the solution procedure until the end and concentrate
on performing the row and column operations on the dimensional matrix
(A) alone.
Let us solve Eq.(A.26) systematically as follows. Add rows 1 and 2
and place the outcome in row 2. The result is shown in Table 3.
Table 3
x1 x2 x3 x4 x5 x6 x7
μs k u Do μo Δρg DL
M 1 0 0 0 1 1 0
L 0 1 1 2 0 –1 2
T –1 0 –1 –1 –1 –2 –1
Add rows 1 and 3 of Table 3 and place the outcome in row 3. The result
is shown in Table 4.
Table 4
x1 x2 x3 x4 x5 x6 x7
μs k u Do μo Δρg DL
A-15
M 1 0 0 0 1 1 0
L 0 1 1 2 0 –1 2
T 0 0 –1 –1 0 –1 –1
Add rows 2 and 3 of Table 4 and place the outcome in row 2. The result
is shown in Table 5.
Table 5
x1 x2 x3 x4 x5 x6 x7
μs k u Do μo Δρg DL
M 1 0 0 0 1 1 0
L 0 1 0 1 0 –2 1
T 0 0 –1 –1 0 –1 –1
Multiply row 3 of Table 5 by –1 and place the outcome in row 3. The
result is shown in Table 6.
Table 6
x1 x2 x3 x4 x5 x6 x7
μs k u Do μo Δρg DL
M 1 0 0 0 1 1 0
L 0 1 0 1 0 –2 1
T 0 0 1 1 0 1 1
We have achieved our initial objective of obtaining a 3x3 unit
submatrix at the beginning of the modified dimensional matrix as shown
in Table 6. Let us now bring the right hand side of Eq.(A.26) into the
picture to complete our solution. The right hand side of Eq.(A.26) has
remained zero and has been unaffected by the row operations on the
A-16
dimensional matrix A. With the right hand side of Eq.(A.26) brought into
the picture, we can now solve for the 7 unknowns as follows:
1 5 6x x x= − − (A.29)
2 4 62 7x x x x= − + − (A.30)
3 4 6 7x x x x= − − − (A.31)
4 4x x= (A.32)
5 5x x= (A.33)
6 6x x= (A.34)
7 7x x= (A.35)
The solutions presented in Eqs.(A.29) to (A.35) appear somewhat
confusing and unclear. Let us reorganize them into a linear combination
of the eigenvectors of the null space of the dimensional matrix as follows:
1
2
3
4 4 5 6
5
6
7
0 1 1 01 0 2 11 0 1 1
1 0 0 00 1 0 00 0 1 00 0 0 1
s
p
o
o
L
xxDxux x x xDxxgxD
μ
μρ
− −⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥⎢ ⎥
−⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + +⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
7x⎥⎥⎥⎥⎥
(A.36)
The variables associated with the solution have been placed in column 1
of Eq.(A.36) for orientation purposes. We can readily identify the (n–r)
eigenvectors of the null space of A in Eq.(A.36) as
A-17
(A.37) 1 2 3 4
0 1 11 0 21 0 1
, , ,1 0 00 1 0 00 0 10 0 0 1
e e e e
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
011
0
0
−
We are now ready to obtain the predicted 4 (n–r) complete and
independent dimensionless π groups from the solutions given in
Eq.(A.36).
Step 5.
Next, we obtain the predicted (n–r) complete and independent
dimensionless π groups by constructing nontrivial solutions of Eq.(A.26)
as follows. Eq.(A.36) shows that nontrivial solutions can be constructed
by arbitrarily choosing numerical values for x4, x5, x6 and x7. Let us
choose x4 = 1, x5 = 0, x6 = 0 and x7 = 0. The solution becomes
1
2
3
4 1
5
6
7
011
1000
s
p
o
o
L
xxDxux eDxxgxD
μ
μρ
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.38)
The corresponding dimensionless π group is obtained by substituting this
solution into Eq.(A.25) to obtain
( )00 1 1 1 0 0 01 0s p o L
p
DD u D g DuD
π μ μ ρ− −= Δ = (A.39)
A-18
The dimensionless π group could have been obtained directly by
inspection of Eq.(A.38). The first column of Eq.(A.38) contains the
variables and the last column contains the powers of the variables in the
dimensionless π group. Therefore, 01
p
DuD
π = could have been written down
directly by inspection. Next, let us choose x5 = 1, x4 = 0, x6 = 0 and x7 =
0. The solution then becomes
1
2
3
4 2
5
6
7
1000100
s
p
o
o
L
xxDxux eDxxgxD
μ
μρ
−⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.40)
The corresponding dimensionless π group is
( )01 0 0 0 1 0 02 0s p o L
s
D u D g D μπ μ μ ρμ
−= Δ = (A.41)
Next, let us choose x6 = 1, x4 = 0, x5 = 0 and x7 = 0. The solution then
becomes
1
2
3
4 3
5
6
7
121
0010
s
p
o
o
L
xxDxux eDxxgxD
μ
μρ
−⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.42)
The corresponding dimensionless π group is
A-19
( )2
11 2 1 0 0 03 0
ps p o L
s
D gD u D g D
uρ
π μ μ ρμ
− − Δ= Δ = (A.43)
Finally, let us choose x7 = 1, x4 = 0, x5 = 0 and x6 = 0. The solution given
by Eq.(A.36) then becomes
1
2
3
4 4
5
6
7
011
0001
s
p
o
o
L
xxDxux eDxxgxD
μ
μρ
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.44)
The corresponding dimensionless π group is
( )00 1 1 0 0 14 0
Ls p o L
p
DD u D g DuD
π μ μ ρ− −= Δ = (A.45)
The dimensionless groups π1, π2, π3 and π4 constitute a complete set
of independent dimensionless groups. The π groups are complete
because any other dimensionless groups derived from the solutions will
be some combination of π1, π2, π3 and π4. The groups are independent
because each of the π groups cannot be formed by combining the
remaining π groups. For example, π1 cannot be formed by combining π2,
π3 and π4. Similarly, π2 cannot be formed by combining π1, π3 and π4.
Further, π3 cannot be formed by combining π1, π2 and π4. Finally, π4
cannot be formed by combining π1, π2 and π3. A physical argument to
convince oneself that π1, π2, π3 and π4 are independent is to observe that
Do appears only in π1; μo appears only in π2; Δρg appears only in π3; and
DL appears only in π4. Since Do appears only in π1, it is impossible to
A-20
manipulate π2, π3 and π4 to get π1. A similar reasoning establishes the
independence of π2, π3 and π4.
Transformations of the Dimensionless π Groups for Example
Problem
The dimensionless π groups π1, π2, π3 and π4 are complete and
independent but are not unique. Other legitimate complete sets of π
groups can be formed. However, these other groups will be found to be
transformations of the original basic group π1, π2, π3 and π4.
There are occasions in which the transformations of the original π
groups into new ones are necessary. Let us demonstrate the need for
transformations of the π groups in our example problem. The result of
the dimensional analysis obtained so far is given by
( )4 2 1 2 3, ,fπ π π π= (A.46)
or
2
2 , , po oL
p p s s
D gDD fuD uD u
ρμμ μ
⎛ ⎞Δ= ⎜⎜
⎝ ⎠⎟⎟ (A.47)
Suppose we wish to conduct an experimental study to investigate the
variation of π4 as a function of π1, π2 and π3. In other words, we wish to
establish the nature of the function f2 experimentally. A good
experimental strategy requires that we vary one π group at a time to
determine its influence. For example, we should vary π1 while keeping π2
and π3 constant, then vary π2 while keeping π1 and π3 constant and
finally, vary π3 while keeping π1 and π2 constant.
A-21
For a given fluid pair and porous medium, μo, μs, Do, Δρ, and of
course, g are fixed. The only variable available for varying π1 is u.
However, we see in Eq.(A.47) that u occurs in π4, π1 and π3. The
occurrence of u in π4 poses no problem. However, its occurrence in π1
and π3 poses a serious problem in controlling the experiments if we were
to plan the experiments based on the current π groups. A change in u
will simultaneously change π1 and π3. Therefore, it would impossible to
tell if an observed change in π4 was caused by the change in π1 or by the
change in π3. What we need is a new set of complete and independent π
groups that allows us to exert more experimental control by being able to
vary one π group while keeping the other π groups constant. Such a new
set of dimensionless π groups can be obtained by transforming the
current set of π1, π2, π3 and π4.
pD
To perform the transformation systematically, we return to the
vector space spanned by the eigenvectors 1e , 2e , 3e and 4e . We recall that
the complete set of dimensionless π groups was obtained by our
judicious choice of the values of x4, x5, x6 and x7 to obtain the solutions
to Eq.(A.26). We can choose these values in some fashion to access
different parts of the vector space spanned by 1e , 2e , and . This
vector space contains an infinite set of complete and independent (n–r)
dimensionless π groups. The challenge in the transformations is to find
these complete sets of dimensionless π groups without including the
linearly dependent ones. Careful navigation through the vector space
allows us to meet this challenge.
3e 4e
Let us form a new set of complete and independent dimensionless π
groups that we shall call Π1, Π2, Π3 and Π4 by choosing the values x4, x5,
A-22
x6 and x7 in the following manner. Let us choose x4 = –1, x5 = 0, x6 = 0
and x7 = 0. The solution given by Eq.(A.36) becomes
1
2
3
4 1
5
6
7
0111
000
s
p
o
o
L
xxDxux eDxxgxD
μ
μρ
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.48)
The corresponding dimensionless π group is
( )00 1 1 1 0 01 0
0 1
1ps p o L
uDD u D g D
Dμ μ ρ
π−Π = Δ = = (A.49)
Let us choose x5 = 1, x4 = x6 = x7 = 0. The solution then becomes
1
2
3
4 2
5
6
7
1000100
s
p
o
o
L
xxDxux eDxxgxD
μ
μρ
−⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.50)
The corresponding dimensionless π group is
( )01 0 0 0 1 0 02 0s p o L
s
D u D g D 2μμ μ ρ πμ
−Π = Δ = = (A.51)
Let us choose x6 = 1, x4 = –1, x5 = 0 and x7 = 0. The solution becomes
A-23
1
2
3
4 1 3
5
6
7
1301
010
s
p
o
o
L
xxDxux e eDxxgxD
μ
μρ
−⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= − + = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.52)
The corresponding dimensionless π group is
( )3
11 3 0 1 0 0 33 0
1
ps p o L
o s
D gD u D g D
Dρ πμ μ ρμ π
− − ΔΠ = Δ = = (A.53)
Finally, let us choose x7 = 1, x4 = –1, x5 = 0 and x6 = 0. The solution then
becomes
1
2
3
4 1 4
5
6
7
0001
001
s
p
o
o
L
xxDxux e eDxxgxD
μ
μρ
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= − + = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
(A.54)
The corresponding dimensionless π group is
( )00 0 0 1 0 1 44 0
1
Ls p o L
o
DD u D g DD
πμ μ ρπ
−Π = Δ = = (A.55)
The dimensionless groups Π1, Π2, Π3 and Π4 constitute a new set of
complete and independent dimensionless groups. However, note that
they are transformations of the original complete set of π1, π2, π3 and π4.
The new set was obtained by a careful manipulation of the vector space
A-24
as a linear combination of the eigenvectors. To ensure that the π groups
are complete, we formed the theoretically required (n–r) new groups. To
ensure that the groups are linearly independent, we used a different
eigenvector to form the different π groups. We used 1e to form Π1, we
used to form Π2, we used 2e 3e (in combination with 1e ) to form Π3, and
we used (in combination with 4e 1e ) to form Π4. These choices of
eigenvectors and eigenvalues preserve the linear independence of the
dimensionless π groups.
The transformations lead to the result
( )4 3 1 2 3, ,fΠ = Π Π Π (A.56)
or
3
3 , ,p poL
o o s o
uD D gD fD D D
ρμμ μ
⎛ ⎞Δ= ⎜⎜
⎝ ⎠s⎟⎟ (A.57)
We see that the new set of dimensionless π groups allows us more
experimental control. Note that u now appears in Π1 only. It is now
possible to vary Π1 while keeping Π2 and Π3 constant. Similarly, we can
vary Π2 while keeping Π1 and Π3 constant. We can also vary Π3 while
keeping Π1 and Π2 constant. Therefore, we are now in a position to
experimentally investigate the separate effects of Π1, Π2 and Π3 on Π4.
In the laboratory experiment for determining the dispersion
coefficient, the viscosity of the injected solvent must be matched with
that of the displaced fluid. Therefore, o
s
μμ
⎛ ⎞⎜⎝ ⎠
⎟ is a constant equal to 1. The
effect of this dimensionless group is eliminated from the experiment.
A-25
Also, the density of the injected solvent must be matched with that of the
displaced fluid. Therefore, the dimensionless group 3p
o s
D gD
ρμ
⎛ ⎞Δ⎜⎜⎝ ⎠
⎟⎟ is a
constant equal to zero. Its effect is eliminated from the experiment. With
these restrictions imposed on the experiments, Eq.(A.57) becomes
4pL
o o
uDD fD D
⎛ ⎞= ⎜
⎝ ⎠⎟ (A.58)
A similar equation can be written for the transverse dispersion coefficient
as
5pT
o o
uDD fD D
⎛ ⎞= ⎜
⎝ ⎠⎟ (A.59)
The experimental program will then consist of determining the functions
f4 and f5 by varying the Peclet Number, p
o
uDD
⎛ ⎞⎜⎝ ⎠
⎟ . The results of such
experimental programs are shown in Figures 5.27 and 5.28, which are
reproduced here for convenience.
Figure 5.27. Correlations for dimensionless longitudinal and transverse
A-26
dispersion coefficients plotted on the same scale (Perkins and Johnson, 1963).
Figure 5.28. Correlation for dimensionless longitudinal dispersion coefficient by various authors (Pfannkuch, 1963; Saffman, 1960).
Some Practical Considerations
There are several interesting observations that can be made about
solving the dimensional analysis problem.
1. In general, there are usually more unknowns than equations. In
the example problem, there are 7 unknowns and 3 equations. This
is an over-determined system with an infinite number of nontrivial
solutions.
2. Reduction of the dimensional matrix to row echelon form without
any row becoming zero shows that the rank r of the dimensional
A-27
matrix is equal to the number of equations. In the example
problem, the rank of the dimensional matrix was determined in
advance to be equal to 3. This is confirmed by the fact that upon
reduction to row echelon form (Table 6), none of the rows became
zero. Thus, the rank of the dimensional matrix in this case is equal
to the number of equations. If one of the rows had become zero
during the reduction to row echelon form, then the rank of the
dimensionless matrix would have been 3-1 or 2. A row will become
zero during the row operations if that row is a linear combination
of the other rows. In such a case, no new information is contained
in the extra row. These are important considerations because
according to Buckingham’s π-theorem, the number of independent
dimensionless groups is equal to the number of variables minus
the rank of the dimensional matrix.
3. Because there are n unknowns and r equations, r unknowns can
be solved for in terms of the remaining (n–r) unknowns. In the
example problem, we solved for x1, x2 and x3 in terms of x4, x5, x6
and x7.
4. The choice of which r variables to solve for in terms of the
remaining (n–r) variables can be used to good advantage. The
variables whose powers are given by the remaining (n–r) unknowns
will each appear in only one dimensionless group. Thus, this
choice can be used to isolate the most important variables of the
problem into one dimensionless group. In the example problem, in
forming the initial dimensionless π groups, we solved for x1, x2 and
x3 in terms of x4, x5, x6 and x7. Therefore, in the initial set of
dimensionless π groups, Do whose power is x4 appeared only in π1;
μo whose power is x5 appeared only in π2; Δρg whose power is x6
appeared only in π3; and DL whose power is x7 appears only in π4.
A-28
On the other hand, the variables whose powers are given by the r
variables solved for can appear in more than one dimensionless
group.
5. The initial arrangement of the variables is important in
constructing the dimensional matrix. Since the initial objective of
the solution of the homogeneous system of the linear algebraic
equations arising in dimensional analysis is to reduce the first
mxm or rxr submatrix of the dimensional matrix to a unit matrix,
it is advantageous to arrange the variables in such a way that the
first mxm or rxr submatrix of A is a diagonal matrix if possible.
There are two significant advantages from such an arrangement.
The first advantage is that the rank of the matrix can be
determined easily by inspection because the determinant of a
diagonal matrix is equal to the product of the diagonal terms. The
second advantage is that solution of the resulting system of
homogeneous system of algebraic equations is considerably
simplified if the first rxr submatrix of the dimensional matrix is a
diagonal matrix. Reduction to row echelon is then easily achieved
by dividing each row of the dimensional matrix by the first nonzero
entry of that row. Of course, the dimensional analysis can still be
performed with the variables arranged in any order. In such a case,
both row and column operations may be necessary to solve the
resulting homogeneous system of equations. Finally, since the
dependent variable of the problem needs to be isolated into only
one dimensionless π group, this variable should be placed in the
rightmost column of the dimensional matrix. In the example
problem, the dependent variable DL was placed in the last column
of the dimensional matrix to ensure that it appeared in one
dimensionless π group only.
A-29
Concluding Remarks
A systematic procedure has been presented to derive the complete
and independent set of dimensionless π groups pertinent to a given
problem. The theoretical basis of the method has been presented and
discussed. It is shown that a complete set of independent dimensionless
π groups is given by the eigenvectors of the null space of the dimensional
matrix. These eigenvectors contain a set of complete and independent
dimensionless π groups. By appropriate choice of eigenvalues, the initial
set of dimensionless π groups can be transformed systematically to
obtain other complete and independent dimensionless π groups that may
be more convenient to use in experiments than the initial set. It is hoped
that the systematic approach to dimensional analysis presented here will
inspire those who have hitherto neglected this powerful tool to employ it
in solving their engineering and scientific problems.
Nomenclature
DL = longitudinal dispersion coefficient
Dp = mean grain diameter of the porous medium
f1 = an unknown function we wish to determine experimentally
f2 = an unknown function we wish to determine experimentally
f3 = an unknown function we wish to determine experimentally
f4 = an unknown function we wish to determine experimentally
f5 = an unknown function we wish to determine experimentally
Do = binary diffusion coefficient between the solvent and oil
g = gravitational acceleration
u = interstitial displacement velocity
μo = viscosity of the displaced fluid
A-30
μs = solvent viscosity
ρo = density of the displaced fluid
ρs = solvent density
π = initial set of dimensionless groups (has nothing to do with 3.141..)
Π = transformed set of dimensionless groups
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A-31
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A-32
A-33
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