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Applied PetroleumReservoir EngineeringSecond Edition

B. C. CRAFTandM. F. HAWKINSLouisiana Stale University

Revised byRONALD E. TERRY

Brigham Young University

Prcnticc Hall PTREnglcwood Cliffs, NJ 07632

Library of Congress Cataloging-in-Publication Data

Craft B C (Benjamin Cole)Applied petroleum reservoir engineering I B C Craft and M F

hawkins — 2nd ed /rcviscd by Ronald F. Terryp cm

Includes bibliographical references and indexISBN1 Oil reservoir engineering I Hawkins. Murray F11 Terry. Ronald E III Title

TN871 C07 199062T 338—dc2O 90—47806

CIP

Editonal/production supervision andinterior design Fred DahI and Rose Kernan

Pre-press buyer Kelly RehrPress Manufacturing Buyer Susan Brunke

0 1991 by Prentice hall PTRPrentice-Hall, IncA Simon & Schuster CompanyEnglesiood Cliffs, New Jersey 01632

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Printed in thee United States of America10 9 8 7

ISBN

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To Rebecca

Contents

PREFACE, ixNOMENCLATURE, ix

Chapter 1INTRODUCTION TO RESERVOIR ENGINEERING, I

1. History of Reservoir Engineenng. 12. Petroleum Reservoirs and Production from Petroleum Reservoirs. 43. Reservoir Types Defined with Reference to Phase Diagrams, 64. Review of Rock Properties. 95. Review of Gas Properties, 126. Review of Crude Oil Properties, 317. Review of Reservoir Water Properties, 458. Summary, 48

Problems, 48

References, 53

Chapter 2THE GENERAL MATERIAL BALANCE EQUATION, 56

1. Introduction, 56

2. Derivation of Material Balance Equation, 56

vi Contents

3. Uses and Limitations of the Material Balance Method. 644. The Havlcna and Odeh Method of Applying the Matenal Balance Equation, 66

References, 67

Chapter 3SINGLE-PHASE GAS RESERVOIRS, 69

1. Introduction. 692. Calculating Gas in Place by the Volumetric Method. 703. Cakulation of Unit Recovery from Volunictnc Gas Reservoirs. 764. Calculation of Unit Recovery from Gas Reservoirs Under Water Dnvc, 785. Material Balance, 836. 'Ihe Gas Equivalent of Produced Condensate and Water, 917. Gas Reservoirs as Storage Reservoirs, 958. Abnormally Pressured Gas Reservoirs, 979. Limitations of Equations and Errors, 99

Problems, 100References, 105

Chapter 4GAS-CONDENSATE RESERVOIRS, 107

1. Introduction, 107

2. Calculation of Initial Gas and Oil, 1103. Performance of Volumctnc Reservoirs, 1184. Usc of Material Balance, 1275. Comparison Between thc Predicted and Actual Production Histories of

Volumetric Reservoirs, 1306. Lean Gas Cycling and Water Drive. 1347. Use of Nitrogen for l'ressurc Maintenancc, 139

Problems, 140Rcfcrcnces, 144

Chapter 5UNDERSATURATED OIL RESERVOIRS, 146

1. Introduction, 1462. Calculation of Initial Oil in Place by the Volumetric Method and Estimation of

Oil Recoveries, 1483. Material Balance in Undersaturated Reservoirs. 1534. Kelly-Snyder Held, Canyon Reef Reservoir, 158

Contents vii

5. The Gloyd-Mitchell Zone of the Rodessa Field, 1636. Calculations Including Formation and Water Comprcssibilities, 170

Problems, 177References. 182

Chapter 6SATURATED OIL RESERVOIRS 184

1. Introduction, 1842. Matenal Balance in Saturated Reservoirs, 1863. Matenal Balance as a Straight Line, 1914. Flash and Differential Gas Liberation, 194S. The Calculation of Formation Volume Factor and Solution Ratio from

Differential Vaporization and Separator Test, 2006. Volatile Oil Reservoirs, 2017. Maximum Efficient Rate (MER), 203


Chapter 7SiNGLE-PHASE FLUID FLOW IN RESERVOIRS, 210

I. Introduction, 2102. 1)arcy's Law and Permeability, 2103. The Classification of Reservoir Flow Systems. 2134. Steady-State Flow Systems, 2185. Development of' the Radial Differential Equation, 2316. iransient Flow Systems, 2337. Pseudostcady-State Flow Systems, 2428. Productivity Index (P!). 2469. Superposition, 249

10. Introduction to Pressure Transient Testing, 253Problems, 263References, 271

Chapter 8WATER INFLUX, 273

1. Introduction, 273

2. Steady-State Models, 2753. Unsteady State Models. 280

vii Contents

4. Pseudosteady-Statc Models, 325Problems, 329References, 333

Chapter 9THE DISPLACEMENT OF OIL AND GAS, 335

I. Introduction, 3352. Recovery Efficiency. 335

3. Immiscible Displacement Processes. 3474. Introduction to Waterflooding, 3755. Introduction to Enhanced Oil Recovery Processes. 380


Chapter 10HISTORY MATCHING, 391

1. Introduction. 3912. Development of the Model, 3923. An Example of a History Match. 394

Problems, 419

References. 424

INDEX, 425

Preface

Shortly after undertaking the project of revising the text Applied PetroleumReservoir Engineering by Ben Craft and Murray Hawkins, several colleaguesexpressed the wish that the revision retain the flavor and format of the originaltext. I am happy to say that I have attempted to do just that. The text containsmany of the field examples that made the original text so popular and still morehave been added. The revision includes a reorganization of the material as wcllas updated material in several chapters.

1 have attempted to reorganize the chapters following a sequence that Ihave used for several years in teaching an undergraduate class in reservoirengineering. The first chapter contains a review of fluid and rock properties.Several new correlations arc presented in this chapter that will assist thosedoing computer modeling Chapter 2 contains a development of the generalmaterial balance equation. The ncxt four chapters present information on thedifferent reservoir types which the original text treated in the first four chap-ters. New material has been added in several places throughout these fourchapters. Chapter 7 presents a discussion of one phase fluid flow. The radialdiffusivity equation is derived and pressure transient analysis is introduced.Chapter 8 contains new material on water influx. Both edgewater and bottom-water drives arc discussed. Chapter 9 is an update of the original Chapter 7 but

lx

x Preface

also contains some new material on waterflooding and cnhanced oil recoverytechniques Chapter 10 is a new chapter on history matching. This is a conceptto which each reservoir engineer should have some exposure. The approachtaken in the chapter is to provide an example of a history match by combiningthe Schilthuis material balance equation with a fluid flow equation.

There were some problems in the original text with units. 1 havetempted to eliminate these problems by using a consistent definition of terms.For example, formation volume factor is expressed in reservoir volume/surfacecondition volume. A consistent set of units is used throughout the text. Theunits used are ones standardized by the Society of Petroleum Engineers.

I would like to express my sincere appreciation to all those who have insome part contributed to the text. For their encouragement and helpful sugges-tions, I give special thanks to the following colleagues: John Lee at Texas A.M.,James Smith formerly of Thxas Tech, Don Green and Floyd Preston of theUniversity of Kansas, and David Whitman and Jack Evers of the University ofWyoming.

Ronald E. Terry

Nomenclature

Normal Definition UnitsSymbolA areal extent of reservoir or well acres or ft2Ar cross sectional area perpendicular to fluid flow ft2B' water influx constant bbVpsiaBR gas formation volume factor ft3/SCF or bbl/SCF

gas formation volume factor at initial reservoir ft3/SCF or bbl/SCFpressuregas formation volume factor at abandonment ft3/SCF or bbl/SCFprcssurc

B,, formation volume factor of injected gas ft '/SCF or bbIISCFoil formation volume factor bblISTB or ft'l

STBBorn oil formation volume factor at bubble point from bbl/STB or ft 'I

separator test STI3B,,, oil formation volume factor at initial reservoir bbl/STB or ft)/

pressure STBB,,,, oil formation volume factor at bubble point bbl/STB or ft3/

pressure STBBodb oil formation volume factor at bubble point from hbl/STB or ft3I

differential test STB

xi

xil Nomenclature

Normal Definition UnitsSymbol8, two phase oil formation volume factor bbl/STB Of ft31

STBwater formation volume factor bbl/STB or ft31

Smc isothermal compressibility psi -CA reservoir shape factor unitlessC, formation isothermal compressibility psi -

gas isothermal compressibility - psi 1

c0 oil isothermal compressibility psi -'c, reduced isothermal compressibility fraction, unitless

total compressibility psi 1

total compressibility at initial reservoir pressurewater isothermal compressibility PSi_I

E overall recovery efficiency fraction, unitlessEd microscopic displacemcnt efficiency fraction, unitless

vertical displacement efficiency traction, unitlessE0 expansion of oil (Havlcna and Odeh method) bbl/STBE1,. expansion of formation and water (I lavlena and hbIISTB

Odeh method)expansion of gas (ilaviena and Odch method) bbl/STB

E, areal displacement efficiency fraction, unitlessE, macroscopic or volumetnc displacement efficiency fraction, unitless

gas cut of reservoir fluid flow fraction. unitlcsswatercut of reservoir fluid flow fraction, unitlcss

F net production from reservoir (Haviena and Odeh bblmethod)ratio of vertical to horizontal permeability unitless

(J initial reservoir gas volume SCFGa remaining gas volume at abandonment pressure SCF

volume of free gas in reservoir SCF0, volume of injected gas SCF

gas from primary separator SCFgas from secondary separator SCFgas from stock tank SCE

GE gas equivalent of one STB of condensate liqtud SCFGE,. gas equivalent of one STB of produced waterGOR gas-oil ratio SCFISI13h formation thickness ftI injectivity index STB/day-psiI productivity index STB/day-psiI, specific productivity index STBlday-psi-ft

productivity index for a standard well STB/day-psik permeability mdk water influx constant bbllday-psiakQ,R average permeability md

permeability to gas phase mdk0 permeability to oil phase md

Nomenclature xiii

Definition

permeability to water phaserelative permeability to gas phaserelative permeability to oil phaserelative permeability to water phaselength of linear flow regionratio of initial reservoir free gas volume to initialreservoir oil volume

m(p) real gas pseudo pressurem(pi) real gas pseudo pressure at initial reservoir

pressurem(pwf) real gas pseudo pressure1 flowing wellM mobility ratioM,. molecular weightM,.,, molecular weight of oiln molesN initial volume of oil in reservoir

cumulative produced oilcapillary number

p pressurePb pressure at bubble point

pressure at critical pointcapillary pressure

PD dimensionless pressurePr prcssure at outer boundary

pressure at initial reservoir pressurepressure at one hour from transient time period onsemilog plotpseudocntical pressurereduced pressure

PR pressure at a reference pointPic pressure at standard conditionsPw pressure at welibore radius

prcssure at wellbore for flowing wellpressure of flowing well just prior to shut in duringa pressure build up testshut in pressure at wellborevolumetric average reservoir pressurechange in volumetnc average reservoir pressure

q flow rate in standard condition unitstotal flow rate in the reservoir in reservoir volumeunits

r distance from center of well (radial dimension)rD dimensionless radius

distance from center of well to outer boundarydistance from center of well to oil reservoirboundarydistance from center of well to wellbore

NormalSymbolk,,k,,k,0kr,1Lm

Units

mdfraction, unitlcssfraction, unitlessfraction, unitlessftratio, unitless

psia2/cppsia2/cp

psia2/cpratio, unitlesslb/lb-molelb/lb-molelb-moleSTBSTBratio, unitlesspsiapsiapsiapsiaratio, unitlesspsiapsiapsia

psiaratio. unitlesspsiapsiapsiapsiapsia

psiapsiapsiaSTh/day (liqwd)bbl/day

ftratio, unitlessftft

ft

xiv Nomenclature

Normal Definition UnitySymbolR instantaneous produced gas-oil ratio SCF/STBR' universal gas constantR, cumulative produced gas-oil ratio SCF/STBR,0 solution gas-oil ratio SCF/STB

solution gas-oil ratio at bubble point pressure SCF/STBsolution gas-oil ratio, from differential liberation SCF/STBtestsolution gas-oil rdtio at bubble point pressure, SCF/STBfrom differential liberation test

R,01b solution gas-oil ratio, sum of separator gas and SCF/STBstock tank gas from scparator testsolution gas-oil ratio at initial rcscrvoir pressure SCE/STBsolution gas-water ratio for brine SCFISTBsolution gas-water ratio for deionized water SCF/STBsolution gas-oil ratio for liquid stream out of SCF/STBseparator

R, solution gas-oil ratio for liquid stream out of stock SCF/STBtank

RF recovery factor fraction. unitlessR V relative volume from a flash liberation test ratio, unitlessS fluid saturation fraction, unitlessS1 gas saturation fraction, unitless5,,, residual gas saturation fraction, unitless5,. total liquid saturation fraction. unitlessS0 oil saturation fraction, unitlessS,, water saturation fraction, unitless

water saturation at initial reservoir conditions fraction, unitlesstime hour

A: time of transient test hourto dimensionless timc ratio. unitless

time of constant rate production prior to well shut hourintime to reach pseudosteady state flow region hour

T temperature °F or °Rtemperature at cntical point °F or °R

T,T,, pseudocritical temperature °F or DRT,,, reduced temperature fraction, unitless

pseudo reduced temperature fraction, unitlesstemperature at standard conditions °F or DR

V volume ft3

V6 bulk %olume of reservoir ft3 or acre-ftV, pore volume of reservoir ft3

V, relative oil volume ft'VR volume at some reference point ft'W width of fracture ft

water influx bbl

Nomenclature

Normal Definition UnitsSymbol

dimcnsionlcss water influx ratio, unitlcssencroachable water in place at initial reservoir bblconditions

W, volume of injected water STB14',, cumulative produced water STBz gas deviation factor or gas compressibility factor ratio, unitiess

gas deviation factor at initial reservoir pressure ratio, unitless

Greek Definkion UnitsSymbola 90°-dip angle degrees

4, porosity fraction, unitlessspecific gravity ratio, unitlessgas specific gravity ratio, unitlessoil specific gravity ratio, unitlesswelt fluid specific gravity ratio, uniflessfluid specific gravity (always relative to water) ratio, unitlessspecific gravity of gas coming from separator ratio, unitlessspecific gravity of gas coming from stock tank ratio, unitlessformation diffusivity ratio. unitless

A mobility (ratio of permeability to viscosity) md/cpAR mobility of gas phase md/cpA. mobility of oil phase md/cpA. mobility of water phase md/cpIL VtSCOSIty cp

PR gas viscosity cpviscosity at initial reservoir pressure cpoil viscosity cpoil viscosity at bubble point cpdead oil viscosity cpwater viscosity cpwater viscosity at 14 7 psia and reservoir cptemperatureviscosity at 14 7 psia and reservoir temperature cp

p apparent fluid velocity in reservoir bbl/day-ft2apparent gas velocity in reservoir bbl/day-ft2apparent total velocity in reservoir bhl/day-ft2

o contact angle degreesp density lb/It'

gas density lb/ft3p. reduced density ratio, unitless

oil density °APIoil-bnne interfacial tension dynes/cm

Chapter 1

Introductionto Reservoir Engineering

1. HISTORY OF RESERVOIR ENGINEERING

Crude oil, natural gas, and water are the substances that are of chief concernto petroleum engineers. Although these substances sometimes occur as solidsor scmjsolids, usually at lower temperatures and pressures, as paraffin, gas-hydrates, ices, or high pour-point crudes, in the ground and in the wells theyoccur mainly as fluids, either in the vapor (gaseous) or in the liquid phase or,quite commonly, both. Even where solid materials are used, as in drilling.cementing, and fracturing, they arc handled as fluids or slurries. The divisionof the well and reservoir fluids between the liquid and vapor phases dependsmainly on the temperature and pressure. The state or phase of a fluid in thereservoir usually changes with pressure, the temperature remaining substan-tially constant In many cases the state or phasc in the reservoir is quiteunrelated to the state of the fluid when it is produced at the surface. Theprecise knowledge of the behavior of crude oil, natural gas, and water, singlyor in combination, under static conditions or in motion in the reservoir rockand in pipes and under changing temperature and pressure. is the main con-cern of petroleum engineers.

As early as 1928 petroleum engineers were giving serious considerationto gas-energy relationships and recognized the need for more precise infor-

2 Introduction to Reservoir Engineering

mation concerning physical conditions in wells and underground reservoirs.Early progress in oil recovery methods made it obvious that computationsmade from wcllhead or surface data were generally misleading. Sciater andStephenson described the first recording bottom-hole pressure gauge and thieffor sampling fluids under pressure in wells. is interesting that this refer-ence defines bottom-hole data as to positive measurements of pres-sure, temperature, gas-oil ratios, and the physical and chemical nature of thefluids. The need for accurate bottom-hole pressures was further emphasizedwhen Millikan and Sidwell described the first precision pressure gauge andpointed out the fundamental importance of bottom-hole pressures to petro-leum engineers in dctermining the most efficient methods of recovery andlifting procedures.2 With this contribution the engineer was able to measurethe most important basic information for reservoir performance calculations:reservoir pressure.

The study of the prqpcrties of the rocks and their relationship to thefluids they contain in both the static and flowing states is called petro -physics.Porosity, permeability, fluid saturations and distributions, electrical conduc-tivity of both the rock and the fluids, pore structure, and radioactivity aresome of the more important petrophysical properties of rocks. Fancher, Lewis,and Barnes made one of the earliest pctrophysical studies of reservoir rocksin 1933, and 1934 Wycoff, Botset, Muskat, and Reed developed a method formeasuring the permeability of reservoir rock samples based on the fluid flowequation discovered by Darcy in 1856 and Botset made a significantadvance in thcir studies of the simultaneous flow of oil and water, and of gasand water in unconsolidated sands.5 This work was later extended to consoli-dated sands and other rocks, and in 1940 Leverctt and Lewis reported researchon the three-phase flow of oil, gas, and water.6

It was early recognized by the pioneers in reservoir engineering thatbefore the volumes of oil and gas in place could be calculated, the change inthe physical properties of bottom-hole samples of the reservoir fluids withpressure would be required. Accordingly in 1935 Schilthuis described abottom-hole sampler and a method of measuring the physical properties ofthe samples obtained.7 'These measurements included the pressure-volume-temperature relations, the saturation or bubble-point pressure. the total quan-tity of gas dissolved in the oil, the quantities of gas liberated under variousconditions of temperature and pressure, and the shrinkage of the oil resultingfrom the release of its dissolved gas from solution These data made thedevelopment of certain useful equations feasible, and they also provided anessential correction to the volumetric equation for calculating oil in place

The next significant development was thc recognition and measurementof connate water saturation, which was considered indigenous to the forma-tion and remained to occupy a part of the pore space after oil or gas accumula-

throughout thc text arc given at the end of chapter

1. History of Reservoir Engineering 3

tion.8 ' This development further explained the poor oil and gas recoveries inlow permeability sands with high connate water saturation, and introduced theconcept of water, oil, and gas saturations as percentages of the total porespace. The measurement of water saturation provided another important cor-rection to the volumetric equation by correcting the pore volume to hydro-carbon pore space.

Although temperature and geothermal gradients had been of interest togeologists for many years, engineers could not make USC of these importantdata until a precision subsurface recording thermometer was developed. Mliii-kan pointed out the significance of temperature data in applications to reser-voir and well studies.'° From these basic data Schilthuis was able to derive auseful equation, commonly called the Scbilthuis material balance equation "It is a modification of an eariicr equation presented by Coleman, Wilde, andMoore and is one of the most important tools of reservoir engineers 12 Basi-cally it is a statement of the conservation of matter and is a method of account-ing for the volumes and quantities of fluids initially present in, produced from,injected into, and remaining in a reservoir at any state of depletion. Odeh andHavlcna have shown how the material balance equation can be arranged intoa form of a straight line and solved.'3 In reservoirs under water drive thevolume of water which encroaches into the reservoir also enters into the ma-terial balance on the fluids. Although Schilthuis proposed a method of calcu-lating water encroachment using the material-balance equation, it remainedfor Hurst and, later, van Everdingcn and Hurst to develop methods for calcu-lating water encroachment independent of the material balance equation,which applies to aquifers of either limited or infinite extent, in either steady-state or unsteady-state fIow.1M4 IS The calculations of van Everdingen andHurst have been simplified by Fetkovich.'6 Following these developments forcalculating the quantities of oil and gas initially in place or at any stage ofdepletion Tamer and Buckley and Leverett laid the basis for calculating theoil recovery to be expected for particular rock and fluid characteristics.'7 18Tamer and, later, Muskat'9 presented methods for calculating recovery by theinternal or solution gas drive mechanism, and Buckley and Leverett presentedmethods for calculating the displacement of oil by external gas cap drive andwater drive. These methods not only provided means for estimating recoveriesfor economic studies; they also expiaincd the cause for disappointingly lowrecoveries in many fields. This discovery in turn pointed the way to improvedrecoveries by taking advantage of the natural forces and energies, and bysuppl}ing supplemental energy by gas and water injection, and by unitizingreservoirs to offset the losses that may be caused by competitive operations.

During the I960s, the terms reservoir simulation and reservoir mathe-matical modeling became popular.2° 21 fl These terms arc synonomous andrefer to the ability to usc mathematical formulas to predict the performanceof an oil or gas reservoir. Reservoir simulation was aided by the developmentof large-scale, high-speed digital computers. Sophisticated numerical methods


were also developed to allow the solution, of a large number of equations byfinite-difference or finite-element techniques.

With the development of these techniques, concepts, and equations,reservoir engineering became a powerful and well-defined branch of petro-leum engineering. Reservoir engineering may be defined as the application ofscientific principles to the drainage problems ansing during the developmentand production of oil and gas reservoirs. It has also been defined as "the artof developing and producing oil and gas fluids in such a manner as to obtaina high economic recovery." The working tools of the reservoir engineer aresubsurface geology, applied mathematics, and the basic laws of physics andchemistry governing the behavior of liquid and vapor phases of crude oil,natural gas, and water in reservoir rocks Because reservoir engineering is thescience of producing oil and gas, it includes a study of all the factors affectingtheir recovery. Clark and Wessely urge a joint application of geological andengineering data to arrive at sound field development programs!A Ultimatelyreservoir engineering concerns all petroleum engineers, from the drilling en-ginccr who is planning the mud program to the corrosion engineer who mustdesign the tubing string for the producing life of the well.

2. PETROLEUM RESERVOIRS AND PRODUCTION FROMPETROLEUM RESERVOIRS

Oil and gas accumulations occur in underground traps formed by structuraland/or stratigraphic Fortunately they usually occur in the moreporous and permeable portions of beds, which are mainly sands, sandstones,limestones, and dolomites, in the intergranular openings, or in pore spacescaused by joints, fractures, and solution activity. A reservoir is that portion ofa trap which contains oil and/or gas as a single hydraulically connected system.Many hydrocarbon reservoirs are hydraulically connected to various volumesof water-bearing rock called aquifers. Many reservoirs are located in largesedimentary basins and share a common aquifer. In this case the productionof fluid from one reservoir will cause the piessure to decline in other reservoirsby fluid communication through the aquifer. In some cases the entire trap isfilled with oil or gas, and in this case the trap and the reservoir are the same.

Under initial reservoir conditions, the hydrocarbon fluids are in either asingle-phase or a two-phase state. The single phase may be a liquid phase inwhich all the gas present is dissolved in the oil. There are therefore dissolvednatural gas reserves as well as crude oil reserves to be estimated. On the otherhand, the single phase may he a gas phase. If there are hydrocarbons vapor-ized in this gas phase that are recoverable as natural gas liquids on the surface,the reservoir is called gas-condensate, or gas-distillate (the older name). Inthis case there are associated liquid (condensate or distillate) reserves as wellas the gas reserves to be estimated. Where the accumulation is in a two-phasestate, the vapor phase is called the gas cap and the underlying liquid phase, the

• 2. Petroleum Reservoirs and Production from Petroleum Reservoirs 5

oil zone. In this case there will be four types of reserves to be estimated:the free gas or associated gas, the dissolved gas, the oil in the oil zone, and therecoverable natural gas liquid from the gas cap.

Although the hydrocarbons in place are fixed quantities, which arereferred to as the resource, the reserves depend on the method by which thereservoir is produced. In 1986 the Society of Petroleum Engineers (SPE)adopted the following definition for reserves:

Reserves are estimated volumes of crude oil, condensate, natural gas, natural gasliquids, and associated marketable substances anticipated to be commerciallyrecoverable and marketable from a given date forward, under existing economicconditions, by established operating practices, and under current government

The amount of reserves is calculated from available engineering and geologicdata. Thc estimate is updated over the producing life of the reservoir as moredata become available. The SPE definition is further broken down into provedand unproved reserves. These definitions are fairly lengthy, and we encourageyou to obtain a copy of the reference if you desire further information.

The initial production of hydrocarbons from the underground reservoiris accomplished by the use of natural reservoir energy and is referred to asprimary production. The oil and gas are displaced to production wells underprimary production by (a) fluid expansion, (b) fluid displacement, (c) gravita-tional drainage, and/or (d) capillary expulsion When there is no aquifer andno fluid is injected into the reservoir, thc hydrocarbon recovery is broughtabout mainly by fluid expansion; however, in the case of oil it may be materi-ally aided by gravitational drainage. When there is water influx from theaquifer or when, in lieu of this, water is injected into selected wells, recoveryis accomplished by the displacement mechanism, which again may be aided bygravitational drainage or capillary expulsion. Gas is injected as a displacingfluid to help in the recovery of oil and is also used in gas cycling to recovergas-condensate fluids.

The use of either a natural gas or a water injection scheme is called asecondary recovery operation. When a water injection sheme is used as asecondary recovery process, the process is referred to as waterflooding. Themain purpose of either a natural gas or a water injection process is to maintainthe reservoir pressure. Hence, the term pressure mainrenace program is alsoused to describe a secondary recovery process.

Other displacement processes called tertiary recovery processes havebeen developed for application in situations in which secondary processes havebecome ineffective However, the same processes have also been consideredfor reservoir applications when secondary recovery techniques were not usedbecause of low recovery potential. In this latter case, the word tertiary is amisnomer. For some reservoirs, it is advantageous to begin a secondary or atertiary process before primary production is completed. For these reservoirs.


the term improved recovery was introduced and has become popular in re-fernng to, in general, any recovery process that improves the recovery overwhat the natural reservoir energy would be expected to yield.

In many reservoirs several recovery mechanisms may be operating simul-taneously, but generally one or two predominate. During the producing life ofa reservoir, the predominance may shift from one mechanism to another,either naturally or because of operations planned by engineers. For example,the volumetric reservoir (no aquifer) may produce initially by fluid expansion.When its pressure is largely depleted, it may produce to the wells mainly bygravitational drainage, the fluid being lifted to the surface by pumps. Stilllater, water may be injected in some wells to drive additional oil to other wellsIn this case, the cycle of the mechanisms is expansion-gravitational drainage-displacement. There are many alternatives in these cycles, and it is the objectof reservoir engineering to plan these cycles for maximum recovery, usually inminimum time.

3. RESERVOIR TYPES DEFINED WITH REFERENCETO PHASE DIAGRAMS

From a technical point of view, the various types of reservoirs can be definedby the location of the initial reservoir temperature and pressure with respectto the two-phase (gas and liquid) region as commonly shown on pressure-temperature (PT) phase diagrams. Figure 1.1 is the PT phase diagram of aparticular reservoir fluid. 1he area enclosed by the bubble-point and dew-point lines to the lower left is the region of pressure-temperature combinationsin which both gas and liquid phases will exist. The curves within the two-phaseregion show the percentage of the total hydrocarbon volume that is liquid forany temperature and pressure. Initially each hydrocarbon accumulation willhave its own phase diagram, which depends only on the composition of theaccumulation.

Consider a reservoir containing the fluid of Fig. 1.1 initially at 300°F and3700 psia, point A. Since this point lies outside the two-phase region, it isoriginally in a one-phase state, commonly called gas as located at point A.Since the fluid remaining in the reservoir during production remains at 300°F,it is evident that it will remain in the single-phase or gaseous state as thepressure declines along path AA1. Furthermore, the composition of the pro-duced well fluid will not change as the reservoir is depleted. This is true forany accumulation of this composition whcrc the reservoir temperature ex-ceeds the cricondentherm, or maximum two-phase temperature (250°F for thepresent example). Although the fluid left in the reservoir remains in onephase, the fluid produced through the wellborc and into surface separators,although the same composition, may enter the two-phase region owing to thetemperature decline, as along line AA2. This accounts for the production ofcondensate liquid at the surface from a gas in the reservoir. Of course, if the

3 Reservoir Types Defined with Reference to Phase Diagrams 7

Fig. 1.1. P

cricondenthcrm of a fluid is below say 50°F, then only gas will exist on thesurface at usual ambient temperatures, and the production will be called drygas. Nevertheless, it may contain liquid fractions that can he rcmoved bylow-temperature separation or by natural gasoline plants.

Next, consider a reservoir containing the same fluid of Fig. 1.1 but at atemperature of 180°F and an initial pressure of 3300 psia, point B. Herc thefluid is also initially in the one-phase state, commonly called gas, where thereservoir temperature exceeds the critical temperature. As pressure declinesbecause of production, the composition of the produced fluid will bc the sameas for reservoir A and will remain constant until the dew-point pressure isreached at 2700 psia, point B1 Below this pressure a liquid condenses out ofthe reservoir fluid as a fog or dew, and this type of reservoir is commonlycalled a dew-point reservoir. This condensation leaves the gas phase with alower liquid content. Because the condensed liquid adheres to the walls of thepore spaces of the rock, it is immobile. Thus the gas produced at the surfacewill have a lower liquid content, and the producing gas-oil ratio thereforerises. This process of retrograde condensation continues until a point of maxi-mum liquid volume is reached, 10% at 2250 psia, point B2. The term reiro-

temperature phase diagram of a reservoir fluid


grade is used because generally vapon?ation, rather than condensation, occursduring isothermal expansion. Actually, after the dewpoint is reached, becausethe composition of the produced fluid changes. the composition of the remain-ing reservoir fluid also changes, and the phase envelope begins to shift. Thephase diagram of Fig 1.1 represents one and only one hydrocarbon mixture.Unfortunately for maximum liquid recovery, this shift is toward the right, andthis further aggravates the retrograde liquid loss within the pores of the reser-voir rock.

Neglecting for the moment this shift in the phase diagram, for qualitativepurposes vaporization of the retrograde liquid occurs from to the abandon-ment pressurc B3. This revaponzation aids liquid recovery and may be evi-denced by decreasing gas-oil ratios on the surface. The overall retrograde losswill evidently be greater (a) for lower reservoir temperatures, (b) for higherabandonment pressures, and (c) for greater shift of the phase diagram to theright—this latter being a property of the hydrocarbon system. The retrogradeliquid in the reservoir at any time is composed to a large extent of methane andethane by volume, and so it is much larger than the volume of stable liquid thatcould be obtained from it at atmospheric temperature and pressure. Thecomposition of this retrograde liquid is changing as pressure declines so that4% retrograde liquid volume at, say, 750 psia might contain as much stablesurface condensate as, say, 6% retrograde liquid volume at 2250 psia.

If the accumulation occurred at 2900 psia and 75°F, point C, the reser-voir would be in a one-phase state, now called liquid, because the temperatureis below the critical temperature. This type is called a reservoir;as pressure dcclines, the bubble point will be reached, in this case at 2550 psia,point C1. Below this point, bubbles, or a free-gas phase, will appear. Even-tually the free gas evolved begins to flow to the well bore, and in ever increas-ing quantities. Conversely, the oil flows in ever decreasing quantities, and atdepiction much unrecovered oil remains in the reservoir. Other names for thistype of liquid (oil) reservoir are depletion, dissolved gas, solution gas drive,expansion, and internal gas drive.

Finally, if this same hydrocarbon mixture occurred at 2000 psia and150°F, point D, it would be a two-phase reservoir, consisting of a liquid or oilzone overlain by a gas zone or cap. Because the composition of the gas andoil zones are entirely different from each other, they may be representedseparately by individual phase diagrams that bear little relation to each otheror to the composite. The liquid or oil zone will be at its bubble point and willbc produced as a bubble-point reservoir modified by the presence of the gascap. The gas cap will be at the dew point and may be either retrograde asshown in Fig. 1.2 (a) or nonrctrograde, Fig. 1 2 (b).

From this technical point of view, hydrocarbon reservoirs are initiallyeither in a single-phase state (A, B, and C) or a two-phase state (0), depend-ing on their temperatures and pressures relative to their phase envelopes. Onvolumetric depletion (no water influx), these various one-phase reservoirs

4 Review of Rock Properties 9

'a'

II'a.

'a'

'I'a.

TTEMPERATURE TEMPERATURE

io) Ibi

Fig. 1.2. Phase diagrams of a cap gas and oil zone fluid showing (a) retrograde cap gasand (b) nonretrograde cap gas

may behave as simple, single-phase gas reservoirs (A), in which reservoirtemperature exceeds thc cricondentherrn, or retrograde condensate (dew-point) reservoirs (B), in which reservoir temperature lies between the criticaltemperature and the cricondcntherm; or dissolved gas (bubble-point) reser-voirs (C), in which reservoir temperature is below the critical temperature.When the pressure and temperature lie within the two-phase region, an oilzone with an overlying gas cap exists. The oil zone produces as a bubble-pointoil reservoir and the gas cap either as a single-phase gas reservoir (A) or aretrograde gas reservoir (B)

4. REVIEW OF ROCK PROPERTIES

Properties discussed in this section include porosity, isothermal compressi-bility, and fluid saturation. Although permeability is a property of a rockmatrix, because of its importance in fluid flow calculations, a discussion ofpermeability is postponed until Chapter 7, in which single phase fluid flow isconsidered.

4.1. Porosity

The porosity of a porous medium is given the symbol of 4, and is defined asthe ratio of void space, or pore volume, to the total bulk volume of the rock.This ratio is expressed either as a traction or in pecent. When using a value ofporosity in an equation it is nearly always expressed as a fraction. The termhydrocarbon porosity refers to that part of the porosity that contains hydro-carbon. It is the total porosity multiplied by the fraction of the pore volumethat contains hydrocarbon.

'rhe value of porosity is usually reported as either a total or an effectivcporosity, depending on the type of measurement used. The total porosityrepresents the total void space of the medium. The effective porosity is the


amount of the void space that contributes to the flow of fluids. This is the typeof porosity usually measured in the laboratory and used in calculations of fluidflow.

The laboratory methods of measuring porosity include Boyle's law,water-saturation, and organic-liquid saturation methods. Dotson, Slobod,McCreery, and Spurlock have described a porosity-check program made byfive laboratories on 10 samples The average deviation of porosity from theaverage values was ±0.5 porosity %. The accuracy of the average porosity ofa reservoir as found from core analysis depcnds on the quality and quantity ofthe data available and on the uniformity of the reservoir. The average porosityis seldom known more precisely than to I porosity % e.g., to 5% accuracy at20% porosity. The porosity is also calculated from electric logs and neutronlogs, often with the assistance of some core measurements. Logging tech-niques have the advantagc of averaging larger volumes of rock than in coreanalysis. When calibrated with core data, they should provide average poros-ity figures in the same range of accuracy as core analysis. When there arevariations in porosity across the reservoir, the average porosity should befound on a weighted volume basis.

4.2. Isothermal CompressibIlIty

The isothermal compressibility for a substance iS given by the followingequationS

I dv(1.1)

where,

c isothermal compressibilityv = volume

p = pressure

The equation describes the change in volume that a substance undergoesduring a change in pressure while the temperature is held constant. The unitsare in reciprocal pressure units When the internal fluid pressure within thepore spaces of a rock, which is subjected to a constant external (rock oroverburden) pressure, is reduced, the bulk volume of the rock decreases whilethe volume of the solid rock material (e g., the sand grains of a sandstone)increases Both of these volume changes act to reduce the porosity of therock slightly, of the order of 0 5% for a 1000 psi change in the internal fluidpressure (e.g , at 20% porosity to 19.9%).

Studies by van der Knaap indicate that this change in porosity for a givenrock depends only on the difference between the internal and external pres-sures, and not on the absolute value of the pressures.28 As with the volume of

4. Review of Rock Properties 11

reservoir coils above the bubble point, however, the change in pore volume isnonlinear and the pore volume compressibility is not constant. The porevolume compressibility C, at any value of external-internal pressure differcnccmay be defined as the change in pore volume per unit of pore volume per unitchange in pressure. The values for limestone and sandstone reservoir rocks liein the range of 2 x to 25 X 10-6 psi_i If the compressibility is given interms of the change in pore volume per unit of hulk volume per unit changein pressure, dividing by the fractional porosity places it on a pore volumebasis For example, a compressibility of 1 Ox 10-6 pore volume per bulkvolume per psi for a rock of 20% porosity is 50 x 6 pore volume per porevolume per psi

Newman measured isothermal compressibility and porosity values in 79samples of consolidated sandstones under hydrostatic When he fitthe data to hyperbolic equation, he obtained the following correlation

— 97.3200(!0)_6(12)

This correlation was developed for consolidated sandstones having a range ofporosity values from 0 02<4 <0.23. The average absolute error of the cor-relation over the entire range of porosity values was found to be 2.60%.

Newman also developed a similar correlation for limestone formationsunder hydrostatic The range of porosity values included in thecorrelation was 0.02 < d <0.33, and the average absolute error was found tobe 11.8% The correlation for limestone formations is as follows:

0.853531Cj = (14 2

(1.3)

Even though the rock compressihilities are small figures. their effect maybe important in some calculations on reservoirs or aquifers that contain fluidsof compressibilites in the range of 3 to 25(10) 6 l• One application is givenin Chapter 5 involving calculations above the bubble point. Oeertsma pointsout that when the reservoir is not subjected to uniform external pressure, asare the samples in the laboratory tests of Newman, the effective value in thereservoir will be less than the measured value.30

4.3. Fluid Saturations

The ratio of the volume that a fluid occupies to the pore volume is called thesaturation of that fluid. The symbol for oil saturation is S0. where S refers tosaturation and the subscript o refers to oil Saturation is expressed as eithera fraction or a percentage, but it is used as a fraction in equations. The satu-rations of all fluids present in a porous medium add to 1.

There are, in general, two ways of measuring original fluid saturations:


the direct approach and the indirect approach. The direct approach involvcseither the extraction of the reservoir fluids or the leaching of the fluids froma sample of the reservoir rock. The indirect approach relies on a measurementof some other property, such as capillary pressure, and the derivation of amathematical relationship between the measured property and saturation.

Direct methods include retorting the fluids from the rock, distilling thefluids with a modified ASTM (American Society for Testing and Materials)procedure, and centrifuging the fluids. Each method relics on some procedureto remove the rock sample from the reservoir. Experience has found that it isdifficult to remove the sample without altering the state of the fluids and/orrock. The indirect methods use logging or capillary pressure measurements.With either method, errors are built into the measurement of saturationHowever, under favorable circumstances and with careful attention to detail,saturation values can be obtained within useful limits of accuracy.

5. REViEW OF GAS PROPERTIES

5.1 Ideal Gas Law

Relationships that descnbe the pressure-volume-temperature (PVT) behaviorof gases arc called equations of state. The simplest equation of state is calledthe ideal gas law and is given by:

pVnR'T (1.4)

where,

p is absolute pressureV is total volume

n is moles1' is absolute temperature

R' is the gas constant

When R' = 10.73, p must be in pounds per square inch absolute (psia), V incubic feet, n in pound-moles (lb-moles), and Tin degrees Rankine. The idealgas law was developed from Boyle's and Charles's laws, which were formedfrom experimental observations.

The petroleum industry works with a set of standard conditions—usually14 7 psia and 60°F. When a volume of gas is reported at these conditions, itis given the units of SCE (standard cubic feet) Sometimes the letter M willappear in the units (e.g., MCFur MSCF). This refers to 1000 standard cubicfeet. The volume that I lb-mole occupies at standard conditions is 379.4 SCF.A quantity of a pure gas can be expressed as the number of cubic feet at a

5 Review of Gas Properties 13

specified temperature and pressure, the number of moles, the number ofpounds, or the number of molecules. For practical measurement, the weighingof gases is difficult, so that gases are metered by volume at measured tem-peratures and pressures, from which the pounds or moles may be calculated.Example 1.1 illustrates the calculations of the contents of a tank of gas in eachof four units.

Example 1.1. Calculating the contents of a tank of ethane in moles,pounds, molecules, and SCF.

Given. A 500 cu ft tank of ethanc at 100 psia and

Sou.rrloN: Assuming ideal gas behavior:

100 x500Moles

= x560 = 8.32

Pounds = 8.32 X 30.07 250.2

Molecules =8 32 x 2.733 x = 22.75 x

At 14.7 psia and 60°F:

SCF=8.32 x3794=3157

Alternate solution using Eq. (1.4):

— nR ' T — 8.32 x 10.73 x 520 —SCF—

— 14.73158

5.2 SpecIfic Gravity

Because the density of a substance is defined as mass per unit volume, thedensity of gas, at a given temperature and pressure can be derived asfollows.

massdensity

volume V

V

Ri VIpM1,,

density =

where,

= molecular weight


Because it is more convenient to measure the specific gravity of gases than thegas density, specific gravfty is more commonly used. Specific gravity is definedas the ratio of the density of a gas at a gWen temperature and pressure to thedensity of air at the same temperature and pressure, usually near 60°F andatmospheric pressure. Whereas the density of gases varies with temperatureand pressure, the specific gravity is independent of temperature and pressurewhen the gas obeys the ideal gas law. By the previous equation, the densityof air is

p x28.97R'T

Then the specific gravity, y,, of as gas is

R'T(1.5)

R'T

Equation (1.5) might also have been obtained from the previous state-ment that 379 4 cu ft of any ideal gas at 14.7 psia and 60°F is one mole, andtherefore a weight equal to the molecular weight. Thus, by definition ofspecific gravity:

Weight of 379.4 cu ft of gas at 14.7 and 60°F —

— Weight of 379.4 cu ft of air at 14.7 and 60°F — 28.97

Thus if the specific gravity of a gas iS 0 75, its molecular weight is 21.7 poundsper mole.

5.3 Real Gas Law

All that has just been said applies to a perfect or ideal gas. Actually there areno perfect gases; however, many gases near atmospheric temperature andpressure approach ideal behavior. All molecules of real gases have two ten-dencies (1) to fly apart from each other because of their constant kineticmotion, and (2) to come together because of electrical attractive forces be-tween the molecules. Because the molecules are quite far apart, the attractiveforces arc , and the gas behaves close to ideal. Also at high tem-peratures the kinetic motion, being greater, makes the attractive forces com-paratively negligible, and, again, the gas approaches ideal behavior.

Since the volume of a gas will be less than what the ideal gas volumewould be, the gas is said to he supercompressibie. The number, which is amcasure of the amount the gas deviates from perfect behavior, is sometimescalled the supercompressibility factor, usually shortened to the compressibility


factor. More commonly it is called the gas deviation factor, symbol z. Thisdimensionless quantity varies usually between 0.70 and 1.20, a value of 1.00representing ideal behavior.

At very high pressures, above about 5000 psia, natural gases pass froma supercompressible condition to one in which compression is more difficultthan in the ideal gas The explanation is that, in addition to the forces men-tioned earlier, when the gas is highly compressed, the volume occupied by themolecules themselves becomes an appreciable portion of the total volume.Since it is really the space between the molecules which is compressed, andthere is less compressible space, the gas appears to be more difficult to com-press. In addition, as the molecules get closer together (i.e., at high pressure),repulsive forces begin to develop between the molecules. This is indicated bya gas deviation factor greater than unity. The gas deviation factor is by defini-tion the ratio of the volume actually occupied by a gas at a given pressure andtemperature to the voiume it would occupy if it behaved ideally, or:

_V1_Acttialvoluineofn moles of gas at Tandp16

M ldcalvolumcofn molesatsame Tandp . )

These theories qualitatively explain the behavior of nonidcal or realgases. Equation (1.6) may be substituted in the ideal gas law, Eq. (1.4)to givean equation for use with nonideal gases

(1.7)

where V1 is the actual gas volume. The gas deviation factor must be deter-mined for every gas and every combination of gases and at the desired tem-perature and pressure, for it is different for each gas or mixture of gases andfor each temperature and pressure of that gas or mixture of gases. The omis-sion of the gas deviation factor in gas reservoir calculations may introduceerrors as large as Figure 1 3 shows the gas deviation factors of twogases, one of 0 90 specific gravity and the other of 0 665 specific gravity. Thesecurves show that the gas deviation factors drop from unity at low pressures toa minimum value near 2500 psia. They rise again to unity near 5000 psia andto values greater than unity at still higher pressures. In the range of 0 to 5000psia, the deviation (actors at the same temperature will be lower for theheavier gas, and for the same gas they will be lower at the lower temperature.

The deviation factor of natural gas is commonly measured in the labora-tory on samples of surface gases. If there is condensate liquid at the point ofsampling, the sample must be taken in such a way as to represent the single-phase reservoir gas. This may be accomplished with a special sampling nozzleor by recombining samples of separator gas, stock tank gas, and stock tankliquid in the proportions in which they arc produced. The deviation factor of


NI

04I..

-JU)U)U)liia.

0U

Fig. 1.3. Effect oIpressurc, temperature, and composition on the gas deviation factor

solution gas is measured on samples evolved from solution in oil during theliberation process.

The gas deviation factor is commonly determined by measuring thevolume of a sampic at desired pressures and temperatures, and then mea-suring the volume of the same quantity of gas at atmospheric pressure and ata temperature sufficiently high so that all the material remains in the vaporphase. For example, a sample of the Bell Field gas has a measured volume of364.6 cu cm at 213°F and 3250 psia. At 14.80 psia and 82°F it has a volume of70,860 cu cm. Then by Eq. (1.7), assuming a gas deviation factor of unity atthe lower pressure, the deviation factor at 3250 psia and 213°F is:

_3250x364.6 1.OOx(460+82)_Z — 460+213

X14.80x 70,860

—0.910

If the gas deviation factor is not measured, it may be estimated from itsspecific gravity Example 1.2 shows the method for estimating the gas devi-ation factor from its specific gravity. The method uses a correlation to estimatepseudocritical tempcrature and pressure values for a gas with a given specificgravity. The correlation was developed by Sutton on the basis of 264 differentgas samples and is shown in Fig. 1 Sutton also conducted a regressionanalysis on the raw data and obtained the following equations over thc rangeof specific gas gravities with which he worked—O.57 < <1.68:

Pc756.8 (1.8)

(1.9)

Having obtained the pseudocritical values, the pseudoreduced pressure andtemperature are calculated. The gas deviation factor is then found by using thecorrelation chart of Fig. 1.5.

RESERVOIR PRESSURE iN PSI

4U,a.

Ui

Cl)U)UI

0.-J4C.,

U0a:,UICl)a.Cz4

UI

Ui0.

Lu

-J4U

U0aLu'I)a.

5. Review of Gas Properties 17

700

650

600

550

500

450

400

350

300

05 06 07 08 09 10 11 12 13 14 15 16 17 18

7g. GAS SPECIFIC GRAVITY

Fig. 1.4. Pscudocritical propertics of natural gases (After Sutton

Example 1.2. Calculating the gas deviation factor of the Bell Field gasfrom its specific gravity.

Given:

Specific gravity =0 665Reservoir temperature = 2 13°F

Reservoir pressure = 3250 psia

5. Ileview of Gas Properties 19

SourrioN: From Fig. 1 4 the pseudocritical pressure and temperatureare

668 psia and = 369°R

Using Eq. (1 8) and (1 9), the pseudocritical values are

Ppr = 756 8— 131.O(.665) — = 668 psia

1;. = 169 2 — 349.5( 665) — 74.0( 665)2 369°R

For 3250 psia and 213°F, the pseudoreduced pressure and temperature are

3250 460+ 2131;,

369-= 1.82

Enter Fig. 1.5 with these values. lind z =0 918.In many reservoir cngincering calculations, it is necessary to use the

assistance of a computer, and the chart of Standing and Katz then becomesdifficult to use. Dranchuk and fitted an equation of state to thedata of Standing and Katz in order to estimate the gas deviation factor incomputer routines.34Dranchuk and Abou-Kasscm used 1500 data points andfound an average absolute error of 0.486% over ranges of pseudoreducedpressure and temperature of:

0.2 30

and for

0

The Dranchuk and Abou-Kassem equation of state gives poor results for1,,= 0 The form of the Dranchuk and Abou-Kassem equationof state is as follows:

z = I (110)

where,

p,=O.27pp,/(Z 7,,) (1.lOa)

c1 (7;,) = A1 4- + + A4/T, + (1 .lOb)

c3(7,) (l.lOd)

cxp (l.lOc)


where the constants —

A1 =03265 A2=—1.0700 A3=—0.5339

A4=0.0l569 A5=—O.05165 A6=0.5475

117 = —0.7361 AR = 0.1844 A9 = 0 1056

A10=0.6134 A11 =0 7210

Because the z-factor is on both sides of the equation, a trial-and-errorsolution is necessary to solve the Dranchuk and Abou-Kasscrn equation ofstate Any one of the number of iteration techniques can be used to assist inthe trial-and-error procedure. One that is frequently used is the secantmethod35 which has the following iteration formula

1(LII)

tin In IJ

To apply the secant method to the foregoing procedure, Eq. (1.10) isrearranged to the form:

F(z) = z — (I + — + c4(p,,1;,)) = 0 (1.12)

The left-hand side of Eq. (1.12) becomes the function, f, and the z -factorbecomes x The iteration procedure is initiated by choosing two values of thei-factor and calculating the corresponding values of the function, f. Thesecant method provides the new guess for z, and the calculation is repeateduntil the function,f, is zero or within a specified tolerance (i.e., ± Thissolution procedure is fairly easy to program on a computcr.

Another popular iteration technique is the Newton-Raphson method,which has the following iteration formula:

14Xn

in

As can be seen in Eq. (1.13), the derivative,f'(z). is required for the Newton-Raphson technique. The derivative of Eq. (1.12) with respect to z is as follows:

= 1 + + —

(114)


A more accurate estimation of the deviation factor can be made whenthe analysis of the gas is available. This calculation assumes that each com-ponent contributes to the pseudocritical pressure and temperature in propor-tion to its volume percentage in the analysis and to the critical pressure andtemperature. respectively, of that component. Table 1.1 gives the criticalpressures and temperatures of the hydrocarbon compounds and others com-monly found in natural It also gives some additional physical proper-ties of these compounds. Example 1.3 shows the method of calculating the gasdcviation factor from the composition of the gas.

TABLE 1.1.Physical properties of the paraffin hydrocarbons and other compounds(After Ei!erts36)

Es:Part 1.s:

Volume Parta: Volume

Liquid Density OO°b. atBoiling Critical Constants 60°F, 14 14 7 60°FPoiiuat G psia, 14,4

Mokc- 14 7 (Grams) Lb Gal psia.ular psia Pressure, Temperature per per per M Gal per

Compound Weight °F p,,psia T,,°R cc Gal SCF Lb-mok

Methane 1604 —2387 6731 3432 0348 290 146 553Ethane 3007 —1275 7083 5499 '0485 404 196 744Propane 4409 —43.7 6174 6660 "05077 4233 27.46 10417Isohutane 5812 109 5291 7346 "05631 4695 3264 12380n-Butanc 5812 311 550! 7657 b05844 4872 3144 11929Isopcntane 72 15 82 1 483 5 829 6 0 6248 5 209 36 50 13 851

n-Pentanc 7215 969 4898 8462 06312 5262 3614 13710n-Hcxane 8617 1557 4401 9142 06641 5536 4103 15565n-Ilcptane 1002 2092 3959 9724 06882 5738 4603 17463n-Octane 1142 2582 3622 10249 07068 5892 5109 19385n-Nonanc 1283 3034 334 1073 07217 6017 5619 21314n-Decane 1423 3454 3J2 1115 07341 6121 6127 23245Air 2897 3177 547 239Carbon

dioxide 4401 —1093 10702 5475Helium 4.003 —452 1 33 2 9 5Hydrogen 2 106 —423 0 189 0 59 8Hydrogen

sulfide 3408 —766 13065 6724Nitrogen 2802 —3204 4922 2270Oxygen 3200 —2974 7369 2786Water 1802 2120 32095 11652 09990 8337

Basis partial volumc in solutionbAt bubble-point pressure and 60°F


Example 1.3. Calculating the gas dcviation factor of the Bell Field gasfrom its composition.

Given: The composition Cal. (2), and the physical data Cols (3) to (5)taken from Table 1.1.

(2)

(I)Component

Comp,MoleFract.

(3)Mo!Wi

('4)

p,(5)i

(6)(2) X (3)

(7)(2) X(4)

(8)(2) x

Methane 08612 1604 673 343 13.81 579.59 295.39Ethanc 0.0591 30.07 708 550 1.78 41.84 32 51Propane 00358 44.09 617 666 1.58 22.09 23.84Butane 0.0172 58 12 550 766 1 00 9.46 13 18Pentanes 0.0050 72 15 490 846 0.36 2.45 4 23CO2 0.0010 44.01 1070 548 0.04 1.07 0.55N2 0.0207

1.000028.02 492 227

19.1510 18

666.68 374.40

SOLUTION: The specific gravity may be obtained from the sum of Cal.(6), which is the average molecular weight of thc gas,

19.15

The sums of Cols (7) and (8) are the pseudocntical pressure and temperature,respectively. Then at 3250 psia and 213°F, the pseudorcduced pressure andtemperature are

3250 6737;r3744 1.80

'ihe gas deviation factor using Fig. 1.5 is z = 0.91.Wichert and Aziz have developed a correlation to account for inaccu-

racies in the Standing and Kat? chart when the gas contains significant frac-tions of carbon dioxide (C02) and hydrogen sulfide (H2S).37 The Wichert andAziz correlation modifies the values of the pseudocritical constants of thenatural gas Once the modified constants are obtained, they are used to calcu-late pseudorcduced properties as described in Ex. 1 2 and the z-factor is deter-mined from Fig. 1.5 or Eq. (1.10). The Wichert and Aziz correlation equationis as follows.

5 120(A°9—A'6)+15(B°'—B4) (1.15)where,

A sum of the mole fractions of CO2 and H2S in the gas mixtureB = mole fraction of H2S in the gas mixture

5. Review of Gas Properties

The modified pseudocritical properties are given by:

(LISa)

g—-—5b(1.1 )

Wichert and Aziz found their correlation to have an average absolute crror of0.97% over the following ranges of data: L54<p (psia) < 7026 and 40< T(°F) <300. The correlation is good for concentrations of CO2 <54.4% (mole%) and H2S<73.8% (mole %).

5.4. Formation Volume Factor and Density

Gas voiwne factors, symbol B,, relate the volume of gas in the reservoir to thevolume on the surface (I e., at standard conditions, Pw and 7w). They are gen-erally expressed in either cubic feet or barrels of reservoir volume per standardcubic foot of gas. Assuming a gas deviation factor of unity for the standardconditions, the reservoir volume of one standard cubic foot = 1.00) atreservoir pressure p and temperature T by Eq. (1.7) is

(1.16)

Where Psc is 14.7 psia and is 60°F

8, cu fIJSCF

bbl/SCF (1.17)

The constants in Eqs. (1.17) are for 14.7 psia and 60°F only, and differentconstants must be calculated for other standards. Thus for the Bell Field gasat a reservoir pressure of 3250 psia, a temperature of 213°F, and a gas devi-ation factor of 0.910, tbc gas volume factor is

0.02829 x 0.910 x 673B, =

3250 —= 0.00533 cu ft/SCF

These gas volume factors mean that I std cu ft (at 14.7 psia and 60°F) willoccupy 0.00533 cu ft of space in the reservoir at 3250 psia and 213°F. Becauseoil is usually expressed in barrels and gas in cubic feet, when calculations aremade on combination reservoirs containing both gas and oil, either the ohvolume must be expressed in cubic feet or the gas volume in barrels. Theforegoing gas volume factor expressed in barrels is 0.000949 bbl/SCF. Then


1000 cu ft of reservoir pore volume in the Bell Field gas reservoir at 3250 psiacontains

Ci = 1000 cu ft ÷ 0.00533 cu ft/SCF -= 188 M SCF

Equation (1.7) may also be used to calculate the dcnsity of a reservoirgas. The moles of gas in I cu ft of reservoir gas pore space is p/ZRT. By Eq(1.5) the molecular weight of a gas is 28.97 X lb per mole. Therefore thepounds contained in I cu ft, that is the reservoir gas density, symbol PR. is

— 28.97

zR'!' )

For example, the density of the Bell Field reservoir gas with a gas gravity of0.665 is

28.97 x 0.665 x 3250= 0.910 x 1i73 x 673

9.530 lb/cu ft

5.5 Isothermal Compressibility

The change in volume with pressure for gases under isothermal conditions,which is closely realized in reservoir gas flow, is expressed by the real gas law:

znR'T zV=—-'or V=constantx— (1.18)p p

Somctimes it is useful to introduce the concept of gas compressibility. Thismust not he confused with the gas deviation factor, which is also referred toas the gas compressibility factor 'l'he above equation may be differentiatedwith respect to pressure at constant temperature to give

dV nRTdz ztiR'Tdp p dp p2

fznR'T\J dz fznR'I\ I

I dV ldz I

Finally, because1 dV

1 ldz(1.19)

p zdp


For an ideal gas z = 1.00 and dz/dp =0 and the compressibility is simply thereciprocal of the pressure. An ideal gas at 1000 psia. then, has a com-pressibility of 111000 or 1000 x 10-6 psi• . Example 1.4 shows the calculationof the compressibility of a gas from the gas deviation factor curve of Fig. 1.6using Eq. (1.19).

Example 1.4. To find the compressibility of a gas from thc gas deviationfactor curve.

Given: The gas deviation (actor curve for a gas at 150°F, Fig. 1.6.

SoluTIoN At 1000 psia, the slope dzldp is shown graphically in Fig. 1.6as — 127 x 10_6 Note that this is a negative slope Then, because z 0.83:

I I

1(Y6)

6=llS3xlO6psi

At 2500 psiathe slope dz/dp is zero, so the compressibility is simply:

Cg = = 400 X 10

Fig. 1.6. Gas compressibility from the gas deviation factor versus pressure plot (SecExample 1 4)

26 introduction to Reservoir Engineenng

At 4500 psia the slope dz/dp is positive and as shown in Fig. 1 6 is equal to110 X 106psi '.Since z 0.90 at 4500 psia:

1 1 10-6)

lOOx lO6psi'

Trube has replaced the pressure in (1.19) by thc product of the pseu-docntical and the pscudoreduced pressures, or p = and dp =This obtains

1 1 dz(1.20)

PprPpr ZPpc

Multiplying through by the pseudocritical pressure, the product isobtained, which Trubc defined as the pseudoreduced compressibility, c,, or

1 1 dz(1.21)

Ppr Z

Mattar, Brar, and Aziz developed an analytical expression for calculating thepseudoreduced compressibility. The expression is

1 0.27 [( . )

z

Taking the derivative of Eq. (110), the equation of state developed byDranchuk and the following are obtained

Ci(1,,) + — 5c3( + (1.23)

and

+ — exp (1.24)

Using Eqs (1.22) to (1 24) and the definition of the pseudorcduced com-pressibility, the gas compressibility can be calculated for any gas as long as thegas pressure and temperature are within the ranges specified for the Dranchukand Abou-Kassem correlation Using these equations. Blasingame. Johnston,and Poe generated Figs. 1 7 and 1 8 4° in these figures. the product of is

plotted asa function 1,,. Example 1.5illustrates how to use these figures. Because they are logarithmic in nature,better accuracy can be obtained by using the equations directly

REDUCED PRESSURE, P,,,

Fig. 1.7. Variation in c,1; for natural gases for I 05 I 4 (After Blasingamc

Example 1.5. To find the compressibility of a 0 90 specific gravity gas at150°F and 4500 psia using the Mattar, Brar,and A71L method.

Souru From Fig. 1.4, = 636 psia -= 4240 R

4500 150 + 4607r424 1.44

10


C.,

01

0 01

0 30 100

28 Introduction to Reservoir Engineenng

U

01

001100

REDUCED PRESSURE,

Fig. 1.8. Variation in for natural gases for 14 3.0 (After Blasingame)

From Fig. 1.8, c,1,0.088.

Then

C 7, 1.44

CCP_.Oót_9591w_bPpc

10

01 1 10


5.6 ViscosItyThe viscosity of natural gas depends on the temperature, pressure, and com-position of the gas. It has units of centipoise (cp) It is not commonly measuredin the laboratory because it can be estimated with good precision. Carr,Kobayashi, and Burrows have developed correlation charts, Fig. 1.9 and 1.10,for estimating the viscosity of natural gas from the pseudoreduced criticaltemperature and pressure.4' The pseudoreduced temperature and pressuremay be estimated from the gas gravily or calculated from the composition ofthe gas. The viscosity at one atmosphere and reservoir temperature (Fig. 1.9)is multiplied by the viscosity ratio (Fig. 1.10) to obtain the viscosity at reser-voir temperature and pressure. The inserts of Fig. 1.9 are corrections to beadded to the atmospheric viscosity when the gas contains nitrogen. carbondioxide, and/or hydrogen sulfide. Example 1.6 illustrates the use of the esti-mation charts.

Example 1.6. Usc of correlation charts to estimate reservoir gas vis-cosity.

Given.

Reservoir pressure = 2680 psia

Reservoir temperature = 212°F

Well fluid specific gravity 0.90 (Air = 1.00)

Pseudocritical temperature = 420°R

Pseudocritical pressure = 670 psia

Carbon dioxide content =5 mole %

5*5 0001U.S

00tt

t

C'5 50 25 30 35 40 45 50 55 60 0 3 10 3

WOLCCUL*R iTEUGHT MCI. S

Fig. 1.9. The viscosity of hydrocarbon gases at one atmosphere and reservoir tem-peratures. with corrections for nitrogen, carbon dioxide, and hydrogen sulfide (AfterCarr, Kobayashi, and Burrows,4' Trans. AIME)


60

40

o'3

•

0o •

2 4 16 ie LO 22 24 26 28 5.0

TEMPERATURE.

Fig. 1.10. Viscosity ratio as a function of pseudoreduced temperature and pressure.(After Carr. Kobayashi,and Burrows,4' Trans AIME.)

Sot UTION:

= 0.0117 cp at one atmosphere (Fig. 1.9)

Correction for CO2 = 0.0003 cp (Fig. 1.9, insert)

Iii =0.0117+0.0003=0.0120 cp (corrected for C02)

672 2680 —1;,

— 4201 60

6704.00

110)

= 1.60 x 0 0120 = 0.0192 cp at 212°F and 2608 psia

Lee, Gonzalez, and Eakin developed a semiempirical method that givesan accurate estimate of gas viscosity for most natural gases if the z-factor hasbeen calculated to include the effect of contaminants.42 For the data fromwhich the correlation was developed, the standard deviation in the calculatedgas viscosity was 2.69%. The ranges of variables used in the correlation werel00<p(psia)<8000, 100< T(°F)<340, and0.90<C02 (mole %)<3.20. inaddition to the gas temperature and pressure, the method requires the z-


factor and molecular weight of the gas. The following equations are used in thecalculation for the gas viscosity:

(1O4)K exp (1 25)

where

p (l.25a)zT

1 25b( )

986X=3.5+T (1.25c)

Y=2.4—O.2X (1.25d)

where.

p = gas density, g/cc

p pressure, psia

T = temperature, °R

M,,, = gas molecular weight

6. REVIEW OF CRUDE OIL PROPERTIES

The next few sections contain information on crude oil properties, includingseveral correlations that can be used to estimate values for the properties.However, these crude oil property correlations are in general, not as reliableas the correlations that have been presented earlier for gases. There are twomain reasons for the oil correlations being less reliable. The first is that oilsusually consist of many more components than do gases. Whereas gases arcmostly made up of alkanes, oils can be made up of several different classes ofcompounds (e.g., aromatics and paraffins). The second reason is that mixturesof liquid components exhibit more nonidealities than do mixtures of gascomponents. These nonidealities can lead to errors in extrapolating correla-tions that have been developed for a certain database of samples to particularapplications outside the database. Before using any of the correlations, theengineer should make sure that the application of interest fits within the rangeof parameters for which a correlation was developed. As long as this is done,then, the correlations used for estimating liquid properties will be adequateand can be expected to yield accurate results.


6.1. Solution Gas-Oil Ratio,

The solubility of natural gas in crude oil depends on the pressure, the tem-perature, and the composition of thc gas and the crude oil. For a particular gasand crude oil at constant temperature, the quantity of solution gas increaseswith pressure; and at constant pressure the quantity decreases with incrcasingtemperature. For any temperature and pressure, the quantity of solution gasincreases as the compositions of the gas and crude oil approach each other;that is, it will be greater for higher specific gravity gases and higher APi gravitycrudes Unlike the solubility of, say, sodium chloride in water, gas is infinitelysoluble in crude oil, the quantity being limited only by the pressure or by thequantity of gas available

A crude oil is said to be saturated with gas at any pressure and tem-perature if on a slight reduction in pressure some gas is released from solutionConversely, if no gas is released from solution, the crude oil is said to beundersazurated at that pressure. The undersaturated state implies that there isa deficiency of gas present and that had there been an abundance of gaspresent, the oil would be saturated at that pressure. The undersaturatcd statefurther implies that there is no free gas in contact with the crude oil (i.e., thereis no gas cap).

Gas solubility under isothermal conditions is generally expressed interms of the increase in solution gas per unit of oil per unit increase in pressure(e.g.. SCF/STI3/psi or Although for many reservoirs this soluhilityfigure is approximately constant over a considerable range of pressures, forprecise reservoir calculations the soluhility is expressed in terms of the totalgas in solution at any pressure (e g.. SCF/STB or R30). It will be shown thatthe volume of crude oil increases appreciably because of the solution gas, andfor this reason the quantity of solution gas is usually referred to as a unit ofstock tank oil, and the solution gas-oil ratio, Rk,. is expressed in standard cubicfeet per stock tank barrel. Figure 111 shows the variation of solution gas withpressure for the Big Sandy reservoir fluid at reservoir temperature of 160°F.At the initial reservoir pressure of 3500 psia there is 567 SCF/STB of solutiongas. The graph indicates that no gas is evolved from solution when the pres-sure drops from the initial pressure at 2500 psia. Thus the oil is undersaturatedin this region, and there is no free gas phase (cap) in the reservoir. Thepressure 2500 psia is called the bubble-point pressure. for at this pressurebubbles of free gas first appear. At 1200 psia the solution gas is 337 SCF/STB,and the average soluhility betwccn 2500 and 1200 psia is.

Average soluhility = = 0 177 SCF/STB/psi

The data of Fig. 1.11 were obtained from a laboratory PVT study of a bottom-hole sample of the Big Sandy reservoir fluid using the flash liberation process.

6. Review of Crude Oil Properties 33

Fig. 1.11. Solution gas-oil ratio of the Big Sandy Field reservoir oil, by flash liberationat reservoir temperature of 160°F

In Chapter 6 it will be shown that the solution gas-oil ratio and other fluidproperties depend on the manner by which the gas is liberated from the oil.The nature of the phenomenon is discussed together with the complications itintroduces into certain reservoir calculations. For the present, however, forthe sake of simplicity this phenomenon is ignored, and a stock tank barrel ofoil is identified with a barrel of residual oil following a flash, liberation pro-cess, and the solution gas-oil ratios by flash liberation is used.

When laboratory analyses of the reservoir fluids arc not available, it isoften possible to estimate with reasonable accuracy the solution gas-oil ratio.Standing gives a correlation method from which the solution gas-oil ratio maybe estimated from the reservoir pressure, the reservoir temperature, the APIgravity of the tank oil, and the specific gravity of the produced gas.4 Bcggspresents Standing's correlation in equation form for pressures less than orequal to the bubble point That equation is:

where,

= 0.00091 T —0 Ol25Po API

I p \1104

(1.26)

T = temperature, °F

p = pressure, psia

PRESSURE, PSIA


For 105 samples tested, the absolute error found in this correlation was 4.8%.The database contained the following ranges of parameters:

130<Po (psia) <7000100< T(°F)<25820 < gas-oil ratio (SCF/STI3) < 1425

16.5< Po APi (°API)< 63.80 59<y,<O.95

1.024 <oil formation volume factor (bbl/STB) <2.05

6.2. FormatIon Volume Factor, B0

It was observed earlier that the solution gas causcs a considerable increase inthe volume of the crude oil. Figure 1.12 shows the variation in the volume ofthe reservoir liquid of the Big Sandy reservoir as a function of pressure atreservoir temperature of 160°F. Because no gas is released from solution whenthe pressure drops from thc initial pressure of 3500 psia to the bubble-pointpressure at 2500 psia, the reservoir fluid remains in a single (liquid) state;however, because liquids are slightly compressible, the volume increases from1.310 bbl/STB at 3500 psia to 1.333 bbl/STB at 2500 psia. Below 2500 psia thisliquid expansion continues but is masked by a much larger effect: the decreasein the liquid volume due to the release of gas from solution. At 1200 psia thevolume has decreased to 1.210 hbl/STB, and at atmospheric pressure and160°F to 1.040 bbl/STB. The coefficient of temperature expansion for the 30°

Fig. 1.12. Formation volumc factor of the Big Sandy Field reservoir oil, by flashliberation at reservoir temperature of 160°?

1500 2000PRESSURE, PSIA

6 Review of Crude Oil Properties 35

API stock tank oil of the Big Sandy reservoir is close to 0.00040 per °F:therefore, one stock tank barrel at 60°F will expand to about 1.04 bbl at 160°Fas calculated

= 1 04 bbl (1.27)

where 13 is the temperature coefficient of expansion of the oil.The formation volume factor, symbol B0 and the abbreviation FVF, at

any pressure may be defined as the volume in barrels that one stock tankbarrel occupies in the formation (reservoir) at reservoir temperature and withthe solution gas which can be held in the oil at that pressure. Because both thetemperature and the solution gas increase the volume of the stock tank oil, thefactor will always be greater than 1. When all the gas present is in solution inthc oil (i.e., at the bubble-point pressure), a further increase in pressuredecreases the volume at a rate that depends on the compressibility of theliquid. One obvious implication of the formation volume factor is that forevery 1.310 bbl of reservoir liquid in the Big Sandy reservoir, only 1.000 bbl,or 76.3%, can reach the stock tank. This figure, 76.3%, or 0.763, is the recipro-cal of the formation volume factor and is called the shrinkage factor. Just asthe formation volume factor is multiplied by the stock tank volume to find thereservoir volume, the shrinkage factor is multiplied by the reservoir volume tofind the stock tank volume. Although both terms are in use, petroleum engi-neers have almost universally adopted formati on volume factor. As mentionedpreviously, the formation volume factors depend on the type of gas liberationprocess, a phenomenon that we ignore until Chapter 6.

In some equations it has been found convenient to use a term called thetwo-phase formation volume factor, symbol 8,, which may be defined as thevolume in barrels one stock tank barrel and its initial complement of dissolvedgas occupies at any pressure and reservoir temperature. In other words, itincludes the liquid volume, plus the volume of the difference between theinitial solution gas-oil ratio, R,,,,, and the solution gas-oil ratio at the specifiedpressure, If Bg is the gas volume factor in barrels per standard cubic footof the solution gas, then the two-phase formation volume factor may beexpressed as

B,= (1.28)

Above the bubble-point, pressure = R,,, and the single-phase and two-phase factors are equal. Below the bubble point, however, while the single-phase factor decreases as pressure decrcases, the two-phase factor increasesowing to the release of gas from solution and the continued expansion of thegas released from solution.

The single-phase and two-phase volume factors for the Big Sandy rcser-


voir fluid may be visualiicd by referring to Fig. 1.13, which is based on datafrom Figs. 1.11 and 1.12. Fig. 1.13 (A) shows a cylinder fitted with a pistonthat initially contains 1.310 bbl of the initial reservoir fluid (liquid) at the initialpressure of 3500 psia and 160°F. As the piston is withdrawn, the volumeincreases and the pressure consequently must decrease. At 2500 psia, whichis the bubble-point pressure, the liquid volume has expanded to 1.333 bbl.Below 2500 psia, a gas phase appears and continues to grow as the pressuredeclines, owing to the release of gas from solution and the expansion of gasalready released; conversely, the liquid phase shrinks because of loss of solu-tion gas, to 1.210 bbl at 1200 psia. At 1200 psia and 160°F the liberated gas hasa deviation factor of 0.890, and therefore the gas volume factor with rcferenccto standard conditions of 14.7 psia and 60°F is

znR 'T — 0.890 x 10.73 x 620B1

— p — 379.4 x 1200

=0.01300 cu ft/SCF

-=0.002316 bbl/SCF

Figure 1.11 shows an initial solution gas of 567 SCF/STI3 and at 1200 psia 337SCF/STB, the difference 230 SCF being the gas liberated down to 1200 psia.The volume of these 230 SCF is

%ç=230xo01300=2.990cu ft

This free gas volume, 2.990 Cu ft or 0.533 bbl, plus the liquid volume, 1.210 bbl,is the total volume, or 1.743 bhl/STB—thc two-phase volume factor at 1200psia It may also be obtained by Eq. (1.28) as:

B, = 1.210 + 0.002316 (567—337)

= 1.210+0.533= 1.743 bbl/STI3

P 1L4i

P F 1...& IFREEGASI'- •

PSIA P51* P.1200 PSIA P*r14j PSIATj,1'160F T01sI60F Tcn'IGOF

A B C D

Fig. 1.13. Visual conception of the change in single-phase and in two-phase formationvolume factors for the Big Sandy reservoir fluid


Figure 113 (C) shows these separate and total volumes at 1200 psia At 14 7psia and 160°F (D), the gas volume has increased to 676 cu ft and the oilvolume has decreased to 1 040 bbl. The total liberated gas volume, 676 cu ft.at 160°F and 14 7 psia, is converted to standard cubic feet at 60°F and 14 7 psiausing the ideal gas law to give 567 SCF/STB as shown in (E). Correspondingly,1.040 bhl at 160°F is converted to stock tank conditions of 60°F as shown byEq. (1.27) to give 1 000 STL3, also shown in (13).

The single-phase formation volume factor may be estimated from thesolution gas, the gravity of the solution gas. the API gravity of the tank oil,and reservoir temperature using a correlation prepared by Standing.43 Heggs"presents Standing's correlation for the oil formation volume factor in equationform as:Forp �pb:

B0 0.972 + 0.000147F1 (1.29)

where,

141.5= oil specific gravity =

L.)L.J -'- PO.AP


The average error determined from this correlation with the same databaseused by Standing and Beggs in the solution gas-ratio correlation was 1.17%.Forp >pb;

B0=BObCXP[c0(pb—p)J (130)

where,

B0b = oil formation volume factor at the bubble-point pressurec0 = oil compressibility, psi

Col (2) of Table 1.2 shows the variation in the volume of a reservoir fluidrelative to the volume at the bubble point of 2695 psig, as measured in thelaboratory. These relative volume factors (RVF) may be converted to for-mation volume factors if the formation volume factor at the bubble point isknown. For example, if B0b = 1.391 bbl/STB, then the formation volume fac-tor at 4100 psig is

at 4100 psig = 1.391(0.9829) = 1.367 bbl/STB


TABLE 1.2.Relative volume data

(1) (2)Pressure, psig RVF°V,

5000 0 97394700 0.97684400 097994100 0.98293800 0.98623600 098863400 0.99093200 0.99343000 0.99602900 0.99722800 0.9985

- 2695 10000

V. — volume yelativc to the volume atthe bubble-point pressure Vb. laboratorydata

6.3. Isothermal Compressibility

Sometimes it is desirable to work with values of the liquid compressibilityrather than the formation or relative volume factors. The compressibility, orthe bulk modulus of elasticity of a liquid. is defined

I dV(1.1)

Because dV/dp is a negative slope, the negative sign converts the compres-sibility into a positive number. Because the values of the volume Vand theslope of dV/dp arc different at each pressure. the compressibility is differentat each pressure, being higher at the lower pressure Average compressibilitiesmay bc used by writing Eq. (1 1) in the difference form as:

(1.31)V (pt—m)

The reference volume V in bq (1 31) may be V. V2. or an average of V1 andIt is commonly reported for reference to the smaller volume—that is, the

volume at the higher pressure. Then the average compressibility of the fluidof Table 1.2 between 5000 psig and 4100 psig is

— 09829—09739 —

— 09739 (5000 —4100)— 10.27 x 10 psi

6. Aeview of Crude Oil Properties 39

Between 4100 psig and 3400 psig,

0.9909—0.982911l.63x 106ps1

And between 3400 psig and 2695 psig,

1.0000—0.99091C009909 13.03x

A compressibility of 13.03 X 10 1 means that the volume of Imillion barrels of reservoir fluid will increase by 13.03 bbls for a reduction of1 psi in pressure. The compressibility of undersaturated oils ranges from S to100 x 10-6 psi', being higher for the higher API gravities, for the greaterquantity of solution gas, and for higher temperatures.

Villcna-Lanzi developed a correlation to estimate c0 for black oils.45 Thecorrelation is good for pressures below the bubble-point pressure and is givenby:

ln(c0) = — 0.664 — 1.430 In(p) — 0.395 Ifl(pb) + 0.390 ln(7)

+0.455 lfl(Po API) (1.32)

where,

T=°F

The correlation was developed from a database containing the followingranges:

31.0(l0)6<c0 (psia) <6600(10)-b500 <p (psig) <5300763 <Ph (psig) <530078< T(°F)<3301.5< GOR, gas-oil ratio (SCF/SiB) < 1947

API (°API)<52.0

°•58<Yg< 1.20

Vasquei and Heggs presented a correlation for estimating the compressibilityfor pressures above the bubble point pressure.TM This correlation is

c0(5 17.2T Il8O*lg+ 1261PQAP1 x (1.33)

Introduction to Reservoir Engineenng

The correlation gives good results for the following ranges of data:

126 <p (psig) <95001.006< (bbl/STB) <2.226

9.3< GOR, gas-oil ratio (SCF/STB) <219915.3 < PO.API (°API) <59.5

0.511 <Yg< 1.351

6.4. ViscosIty

The viscosity of oil under reservoir conditions is commonly measured in thelaboratory. Fig. 1.14 shows the viscosities of four oils at reservoir tempera-ture, above and below bubble-point pressure. Below the bubble point, theviscosity decreases with increasing pressure owing to the thinning effect of gasentering solution, but above the bubble point, the viscosity increases withincreasing pressure.

When it is necessary to estimate the viscosity of reservoir oils, correla-tions have been developed for both above and below the bubble-point pres-sure. Egbogah presented a correlation which is accurate to an average errorof 6.6% for 394 different oils.'7 The is for what is referred to as"dead" oil, which simply means it does not contain solution gas. A secondcorrelation is used in conjuction with the Egbogah correlation to include the

I ig. 1.14. Ihc viscosity of four crude oil samples under rcscrvoir conditions.

6 Review of Crude 011 Properties 41

effect of solution gas. Egbogah's correlation for dead oil at pressures less thanor equal to the bubble-point pressure is

+ 1)] 1.8653 — O'O2SO86Po — 0.5644log(T) (1.34)

where,

Pod = dead oil viscosity, cp


The correlation was developed from a database containing the followingranges:

59<T(°F)< 176—58 < (°F) <59

5.0 < P0 API (°API) < 58.0

Beggs and Robinson"" developed the live oil viscosity correlation thatis used in conjunction with the dead oil correlation given in Eq (1.34)

(1.35)

where,

A 10.715 (R30+

B 5.44 (R30 +

The average absolute error found by Beggs and Robinson while working with2073 oil samples was 1.83%. The oil samples contained the following

O<p (psig) <525070< 7 (°F) <29520< GOR, gas-oil ratio (SCF/STB) <207016< Po API (°AP1) <58

For pressures above the bubble point, the oil viscosity can he estimatedby the following correlation developed by Vasquez and Beggs:

Po Pob (P"Pb)m (1.36)

where,

5pJ

Pob = oil viscosity at the bubble point-pressure, cp


Vasqucz and Bcggs found an average absolute error for this correlation of7.54% for 3143 oil samples that involved the following ranges:

l26<p (psig)<9500O.1l7<p0 (cp)< 148.0

9 3 <GOR, gas-oil ratio (SCF/STB) < 2199

15 3<p0 API (°API)<59.50.511 1.351

The following example problcm illustrates the use of the correlations that havebeen presentcd for the various oil properties.

Example 1.7. Use ofcorrelations to estimate values for liquid propertiesat pressures of 2000 and 4000 psia.

Given:T= 180°F

Pb = 2500 psia

11= 0.80

Po.AP1 = 40°API

0.85

SoLuTIoN: Solution gas-oil ratio,

p =4000 psia (p >pb)

For pressures greater than the bubble-point prcssure, = R,0,,, thereforefrom Eq. (1 26):

R)Ob =

Yg 0.00091 T —0 0125Po. API = 0.00091(180) —0 0125(40) — 0.336

1 2500 1204

R50b = = 772 SCF/STB

p=2000psia(p<p6)1 1 2000 1120'

590SCF/STB


Isothermal compressibility, c0:

p =4000 psia (p >Pb)

From Eq (1.33):

C0 = + 17.2T — "8OYg + 12.61 PO.AP1 — 1433)/(p(IO)')

c0 = (5(772) + 17.2(180) — 1180(0.80) + 12.61(40) — 1433

= 12.7(10)6

From Eq. (1.32):

lnc0=—0.664—1.430 ln(p)—O.39Sln(pb)+O.39O ln(T)

f 0.455 ln(R30b) + 0.262 ln(p0 API)

In c0 — 0.664 — 1 430 In (2000) — 0.395 In(2500) -f 0.390 ln(180)

+ 0.455 ln(772)+0.262 In(40)

c0=183(10)6psi I

Formation volume factor, B0:

p =4000 psia (p >p6)

From Eq. (1.30).

= exp 1c0( Pb

B0b is calculated from Eq. (1.29)

B0b = 0.972 + 0.000147F1

where,

/ \05+125T

F 772(080)05 + 1.25(180) = 985

B0b = 0.972 + 0 000147(985)' 175 = 1.456 bbl/STB

B0 = 1.456 exp [12 7(I0) 6(2500 — 4000)1 1.429 bbl/STI3

p =2000 psia (p <pb)

44 introduction to Reservoir Engineering

From Eq. (1.29):

B0=0.972 +000(J147F'

F = 1 .25T 590 806

= 0.972 + 0.000147(806)1175= J .354 bhl/STB

Viscosity, p.°.

p — 4000 psia (p >Pb)

From Eq. (1.36)

— ( JP'o

m

m = 2 6(4000)' exp[— 11.513 — 8.98(10) = 0.342

From Eq. (1 34):

+ 1)] = 1.8653— O.025086PO.API — 0.56441og(T)

+ 1)] = 1.8653 — 0.025086(40) —0 56441og(180)

P'obd= 1.444 Cp

From Eq. (1.35)

BP-ob — P.obd

A 10.715 (R,ob + 100) 10.715(772+

B =544 150) 44(772+150)

= 0 328(1 444)0 = 0.400 cp

= 0 400(4000/2500)° 342 0.470 cp

p -2000psia(p<pb)

From Eq. (1 34), EJ.0a will be the same as 11ob4.

I.todl 444cp= A

4 = 10.715 100)_hJ515= 100) O51')....0 370

B =544 (R,0 44(590+ 150)

= 0.458 cp

7. Review of Reservoir Water Properties 45

7. REVIEW OF RESERVOIR WATER PROPERTIES

The properties of formation waters, like crude oils but to a much smallerdegrec, are affected by temperaturc, pressure, and the quantity of solution gasand dissolved solids. The compressibility of the formation, or connate, watercontnhutes materially in some cases to the production of volumetric reservoirsabove the bubble point and accounts for much of the water influx in water-drive reservoirs. When the accuracy of other data warrants it, the propertiesof the connate water should he entered into the material-balance calculationson reservoirs. The following sections contain a number of correlations ade-quate for use in engineering applications

7.1. FormatIon Volume Factor

5° developed the following correlation for the water formation vol-ume factor, B1,, (hbl/STB):

(1.37)

where,

l.000lOx 10-2+133391 x T+5.50654x 10 T2

—195301 x 109pT— 1.72834x 10'3p2T—3.58922x

—2.25341x lO'°p2T = temperature, °F

p = pressure, psia

For the data used in the deveiopment of the correlation, the correlation wasfound to be accurate to within 2%. The correlation does not account for thesalinity of normal reservoir brines explicitly, but McCain observed that vari-ations in salinity caused offsetting errors in the terms and Theoffsetting errors cause the correlation to be within cngineering accuracy forthe estimation of the of reservoir brines.

7.2. Solution Gas-Water Ratio

McCain has also developed a correlation for the solution gas-water ratio,(SCF/STB).49 5° The correlation is.

S1. 38


where

S = salinity, % by weight solidsT temperature, °F

= solution gas to pure water ratio, SCF/STH

R,,, is given by another correlation developed by McCain as

+Rp 4 CpZ (1.39)

where,

A =8.15839—6.12265x102 T+I.91663x10 4T2—2.1654x107 T3

R=1.01021x102—7.44241 T+3.05553x10712

1010T3

C = — 10 T ÷ 8.53425 x T2

—2.34122x106V +237049x10 9T4)


The correlation of Eq. (1.39) was developed for the following range of data

and found to he within 5% of the published data:

I000<p (psia)< 10,000

100< T(°F)< 340

Eq. (1 38) was developed for the following range of data and found to beaccurate to within 3% of published data:

70 < T (°F) < 250

7.3. Isothermal Compressibility

Osif developed a correlation for the water isothermal compressibility, c%, forpressures greater than the bubble point pressure.51 The equation is

I 1

= [7.033p + 403,3001(1.40)

where,

CN.(I = salinity, g NaCI/liter

T — temperature, °F

7 Review of Reservoir Water Properties 47

The correlation was developed for the following range of data:

1000<p (psig)<20,000

0< CN.c, (g NaC1/liter) <200

200 < T (°F) < 270

The water isothermal compressibility has been found to be strongly affectedby the presence of free gas. Therefore, McCain proposed using the followingexpression for estimating c,, for pressures below or equal to the bubble-pointpressure:

c (141)8p Jr

The first term on the right-hand side of Eq. (1.41) is simply the expression force,, in Eq. (1.40) Tbc second term on the right-hand side is found by differ-entiating Eq. (1.39) with respect to pressure, or:

=B+2Cp\ ap

where

B and C are defined in Eq. (1.39)

In proposing Eq. (1.41). McCain suggested that BR should be estimatedusing a gas with a gas gravity of 0.63, which represents a gas composed mostlyof methane and a small amount of cthanc. McCain could not verify this ex-pression by comparing calculated values of cN with published data, so there isno guarantee of accuracy This suggests that Eq (1.41) should be used onlyfor gross estimations of

7.4. Viscosity

The viscosity of water increases with decreasing temperature and in generalwith increasing pressure and salinity. Pressure below about 70°F causes areduction in viscosity, and some salts (e.g., KC1) reduce the viscosity at someconcentrations and within some temperature ranges. The effect of dissolvedgases is believed to cause a minor reduction in viscosity McCain developedthe following correlation for water viscosity at atmospheric pressure and res-ervoir temperature:

(1.42)


where

A = 109.574—8 40564 S + 0.313314 S2 + 8.72213 x S3

B-—1.12166+2.63951X102S—6.79461x 10 4S2—5.47119x10 5S3+ l55586x106S4


S = salinity, % by weight solids

Eq. (1.42) was found to be accurate to within 5% over the following range ofdata'

100 < T (°F) <4000<S(%)<26

The water viscosity can be adjusted to reservoir pressure by the followingcorrelation, again developed by McCain:

(1.43)

This correlation was found to be accurate to within 4% for pressures below10,000 psia and within 7% for pressures between 10,000 and 15,000 psia Thetemperature range for which the correlation was developed was between 86and 167 °F.

8. SUMMARY

The correlations presented in this chapter are valid for estimating propertiesproviding the parameters fall within the specified ranges for the particularproperty in question The correlations were presented in the form of equationsto facilitate their implementation into computer programs.

PROBLEMS

1.1 Calculate the volume 1 lb-mole of ideal gas will occupy at'

(a) 14.7 psia and 60°F(b) 14 7 psia and 32°F

(c) 14 7 pIus 10 oz and 80°F

(d) 15 025 psia and 60°F

Problems 49

1.2 A 500 cu ft tank contains 10 lb of methane and 20 lb of ethanc at 90°F.

(a) How may moles arc in the tank9

(b) What is the pressure of the tank is psia?

(c) What is the molecular weight of the mixture?

(d) What is the specific gravity of the mixture"

1.3 What are the molecular weight and specific gravity of a gas that is one third eachof methane, ethane, and propane by volume?

1.4 A 10 lb block of dry ice is placed in a 50 Cu ft tank that contains air at atmo-spheric pressure 14 7 psia and 75°F. What will be the final pressure of the sealedtank when all the dry ice has evaporated and cooled the gas to 45°F?

1.5 A welding apparatus for a drilling rig uses acetylene (C2H2), which is purchasedin steel cylinders containing 20 lb of gas, costs $4.50 exclusive of the cylinder.If a welder is using 200 cu ft per day measured at 16 oz gauge and 85°F, whatis the daily cost of acetylene? What is the cost per MCF at 14.7 psia and 60°F'

1.6 (a) A 55,000 bbl (nominal) pipeline tank has a diameter of 110 ft and a heightof 35 ft. It contains 25 ft of oil at the time suction is taken on the oil withpumps that handle 20,000 bbl per day. The breather and safety valves havebecome clogged so that a vacuum is drawn on the tank. If the roof is ratedto withstand 3/4 oz per sq in pressure, how long will it be before the roofcollapses" Barometric pressure is 29 l in. Neglect the fact that the roof ispeaked and that there may be some leaks.

(b) Calculate the total force on the roof at the time of collapse

(c) If the tank had contained more oil, would the collapse time have beengreater or less?

1.7 (a) What percentage of methane by weight does a gas of 0.65 specific gravitycontain if it is composed only of methane and ethane? What percentage byvolume?

(b) Explain why the percentage by volume is greater than the perccntage byweight.

1.8 A 50 cu ft tank contains gas at 50 psia and 50°} It is connected to another tankthat contains gas at 25 psia and 50°l- When the valve between the two is opened.the pressure equalizes at 35 psia at 50°F. What is the volume of the second tank"

1.9 Gas was contracted at $1 10 per MCF at contract conditions of 14 4 psia and80°F What is the equivalent price at a legal temperature of 60°F and pressureof 15 025 psia?

1.10 A cylinder is fitted with a Icakproof piston and calibrated so that the volumewithin the cylinder can be read from a scale for any position of the piston. Thecylinder is immersed in a constant temperature bath maintained at i60°F, whichis the reservoir temperature of the Sabine Gas Field Forty-five thousand cc ofthe gas, measured at 14.7 psia and 60°F, is charged into the cylinder. The volumeis decreased in the steps indicated below, and the corresponding pressures areread with a dead weight tester after temperature equilibrium is reached.

V. cc 2529 964 453 265 180 156 5 142.2p. psia 300 750 1500 2500 4000 5000 (i000

Introduction to Reservoir Engineering

(a) Calculate and place in tabular form the gas deviation factors and the idealvolumes that the initial 45,000 cc occupies at 160°F and at each pressure.

(b) Calculate the gas volume factors at each pressure, in units of ft'/SCF.

(c) Plot the deviation factor and the gas volume factors calculated in part (b)versus pressure on the same graph.

1.11 (a) If the Sabine Field gas gravity is 0 65, calculate the deviation factors fromzero to 6000 psia at 160°F, in 1000 lb increments, using the gas gravitycorrelation of Fig 1.4.

(b) Using the critical pressures and temperatures in Tablc 1 1 ,calculatc and plotthe deviation factors for the Sabinc gas at several pressures and 160°F. Thegas analysis is as

Component C1 C2 C, iC4 nC4 iC5Mole Fraction 0.875 0.083 0.021 0.006 0.008 0.003

Component nC',Mole Fraction 0 002 0.001 0 001

Use the molecular weight and cntical temperature and prcssurc of octanefor the hcptancs-plus. Plot the data of Prob. 1 10(a) and Prob. 1.11(a) on thesame graph for comparison.

(c) Below what pressure at 160°F may the ideal gas law be used for the gas ofthe Sabine Field if errors are to be kept within 2%"

(d) Will a reservoir contain more SCF of a real or of an ideal gas at similarconditions? Explain.

1.12 A high-pressure cell has a volume of 0 330cu ft and contains gas at 2500 psia and130°F, at which conditions its deviation factor is 0.75. When 43.6 SCF measuredat 14.7 psia and 60°F were bled from the cell through a wet test meter. thepressure dropped to 1000 psia, the temperature remaining at 130°F. What is thegas deviation factor at 1000 psia and 130°F?

1.13 A dry gas reservoir is initially at an average pressure of 6000 psia and tem-perature of 160°F. The gas has a specific gravity of 065. What will the averagereservoir pressure be when one-half of the original gas (in SCF) has beenproduced? Assume the volume occupied by the gas in the reservoir remainsconstant. If the reservoir originally contained 1 MM ft' of reservoir gas, howmuch gas has been produced at a final reservoir pressure of 500 psia"

1.14 A reservoir gas has the following gas deviation factors at 150°F?

p, psia 0 500 1000 2000 3000 4000 5000

2 100 092 086 080 0.82 089 100

Plot z versus p and graphically determine the slopes at 1000 psia, 2200 psia,and 4000 psia Then, using Eq (119), find the gas compressibility at thesepressures

1.15 Using Figs 1 4 and 1 5, find the compressibility of a 70% specific gravity gas at5000 psia and 203°F.

1.16 Using Eq. (1 22) and the generali7ed chart for gas deviation factors, Fig. I 5,

Problems 51

find the pseudoreduced compressibility of a gas at a pseudoreduced temperatureof 1.30 and a pseudoreduced pressure of 4 00. Check this value on Fig. 1 7

1.17 Estimate the viscosity of a gas condensate fluid at 7000 psia and 220°F It has aspecific gravity of 090 and contains 2% nitrogen, 4% carbon dioxide, and 6%hydrogen sulfide

1.18 Experiments were made on a bottom-hole sample of the reservoir liquid takenfrom the LaSaile Oil Field to determine the solution gas and the formationvolume factor as functions of pressure. The initial bottom-hole pressure of thereservoir was 3600 psia, and bottom-hole temperature was 160°F'; so all measure-ments in the laboratory were made at 160°F. The following data, converted topractical units, were obtained from the measurements.

Solution gas Formation VolumePressure SCF/STB at Factor,

psia 14 7 psia and 60°F bbl/STB

3600 567 1 3103200 567 1.3172800 567 1.3252500 567 1 3332400 554 1.3101800 436 1 2631200 337 1.210600 223 1.140200 143 1.070

(a) What factors affect the solubility of gas in crude oil"(b) Plot the gas in solution versus pressure

(C) Was the reservoir initially saturated or undersaturated? Explain

(d) Does the reservoir have an initial gas cap'(e) In the region of 200 to 2500 psia, determine the solubililty of the gas from

your graph in SCF/ST B/psi

(1') Suppose 1000 SCF of gas had accumulated with each stock tank barrel of oilin this reservoir instead of 567 SCF. Estimate how much gas would havebeen in solution at 360() psia Would the reservoir oil then be called satur-ated or

1.19 From the bottom-hole sample in Prob 118(a) Plot the formation volume factor versus pressure

(b) Explain the brcak in the curve(c) Why is the slope above the bubble-point pressure negative and smaller than

the positive slope below the bubble-point pressure'

(d) If the reservoir contains 250 MM reservoir barrels of oil initially, what is theinitial number of SIB in place'

(el Whdt is the initial volume of dissolved gas in the reservoir?

(1) What will be the formation volume factor of the oil when the bottom-hole


pressure is essentially atmospheric if the coefficient of expansion of the stocktank oil is 0 1)006 per °F'

1.20 If the gravity of the stock tank oil of thc Big Sandy reservoir is 30°API and thegravity of the solution gas is 0.80. estimate the solution gas-oil ratio and thesingle-phase formation volume factor at 2500 psia and 165°F The bubble-pointprcssure is 2800 psia.

1.21 A 1000 Cu ft tank contains 85 STB of crude and 20,000 SCF of gas. all at 120°F.When equilibrium is established, (i.e.. when as much gas has dissolved in the oilas will), the pressure in the tank is 500 psia. If the solubility of the gas in thecrude is 0 25 SCF/STB/psi and the deviation factor for the gas at 500 psia and120°f- is 0 90. find the liquid formation volume factor at 500 psia and 120°F.

1.22 A crude oil has a compressibility of 20 x 10 6psi 'and a bubble point of 3200psia. Calculate the relative volume factor at 4400 (i c volume relativeto its volume at the bubble point), assuming constant compressibility

1.23 (a) bstimatc the viscosity of an oil at 3000 psia and 130°F It has a stock tankgravity of 35°AP1 at 60°F and contains an estimated 750 SCFISTB of solutiongas at the initial bubble point of 3000 psia

(b) Estimate the viscosity at the initial reservoir pressure of 4500 psia.

(c) Estimate the viscosity at 1000 psiaif there is an estimated 300 SCF1STR ofsolution gas at that pressure.

1.24 Given the following laboratory data.

Cell Oil Volume Gas Volume CellPressure in cell in cell Temperature

(psia) (cc) (cc) (°F)

2000 650 0 195

150 195

195

500 44,500 60

Evaluate R,0. B0. and B at the stated pressures The gas deviation factors at 1000psia and 500 psia have been evaluated as 0 91 and 0 95, respectively.

1.25 (a) Find the compressibility for a connate water that contains 20.000 parts pcrmillion of total solids at a reservoir pressure of 4000 psia and temperatureof 150°F

(b) Find the formation volume factor of the formation water of part (a)

1.26 (a) What is the approximate viscosity of pure water at room temperature andatmosphenc pressure?

(b) Wbdt is the approximate viscosity of pure watcr at 200°F?

1.27 A container has a volume of 500 cc and is full of pure water at 180°F and 6000psia.

(a) I low much water would be expelled if the pressure were reduced to 1000psia9

References 53

(b) What would he the volume of water expelled if the salinity were 20,000 ppmand there werc no gas in solution"

(c) Rework part (b) assuming that the water is initially saturated with gas andthat all the gas is evolved during the pressure change.

(d) Estimate the viscosity of the water.

REFERENCES

1 K C. Sclater and B. R. Stephenson. "Measurements of Onginal Pressure, Tem-perature and Gas-Oil Ratio in Oil Sands," Trans AIME (1928—29), 82, 119.

V. Millikan and Carrol V Sidwell, "Bottom-hole Pressures in Oil Wells," Trans.AIME (1931). 92, 194

'G 11. Fanchcr. J A. Lewis, and K B. Barnes, "Some Physical Charactenstics of OilSands," The Pennsylvania State College, Bull 12 (1933), p. 65

4R D. Wyckoff, H 0 Botset, M Muskat. and 1). W Reed, "Measurement ofPermeability of Porous Media," Bull. AAPG (1934), 18, No. 2, p. 161.

R. U. Wyckoff and 11. 0. Botset. "The Flow of Gas-Liquid Mixtures ThroughUnconsolidated Sands," Physics (1936), 7, 325

6M C. Lcverett and W B. Lewis, "Steady Flow of Oil-Gas-Water Mixtures ThroughUnconsolidated Sands," Trans. AIME (1941), 142, 107.

J Schilthuis, "Technique of Securing and Examining Sub-Surface Samples ofOil and Gas," Drilling and Production Praclice, API (1935), pp 120—126

Howard C. Pyle and P. H. Jones, "Quantitative Determination of the Connate WaterContent of Oil Sands," Drilling and Production Practice, API (1936), pp. 171—180

i. Schilthuis, "Connate Water in Oil and Gas Sands," Trans AIME (1938),127, 199—214.

'°C. V Millikan, "Temperature Surveys in Oil Wells," irans AIME (1941), 142, 15.

Ralph J Schilthuis, "Active Oil and Reservoir Energy," Trans. AIME (1936),118, 33

Coleman, H 0 Wilde, Jr , and Thomas W. Moore, "Quantitathe Effectsof Gas-Oil Ratios on Decline of Average Rock Pressure," Trans. AIME (1930), 86,174.

"A S Odeh and D Ilavlena, "The Materidl Balance As an Equation of a StraightLine," Journal of Petroleum Technology (July 1963), 896—900

'4W. Hurst, "Water influx into a Reservoir and Its Application to the Equation ofVolumetric Balance," irans AIME (1943), 151, 57

15 A F. van Everdingen and W. Hurst, "Application of the LaPlace Transformation toFlow Problems in Reservoirs,' Trans AIME (1949), 186, 305.

'6M J. Fetkovich, "A Simplified Approach to Water Influx Calculations—FiniteAquifer Systems," Jour of Petroleum Technology (July J971), 814-828.

" J. Tamer, "How Different Size Gas Caps and Pressure Maintenance ProgramsAffect Amount of Recoverable Oil " Oil Weekly. June 12, 1944, 144, No 2, 32—44.


18S. E Buckley, and M C Leverctt, "Mechanism of Fluid Displacement in Sands,"Trans. AIME (1942), 146, 107—117.

"M. Muskat, "The Petroleum Histories of Oil Producing (ias-Dnve Reservoirs,"Jour. of Applied Physics (1945), 16, 147.WA Odeh, "Reservoir Simulation—What Is It" Jour, of Petroleum Technology(Nov 1969), 1383-1388

K. 11 Coats, "Use and Misuse ofReservoir Simulation Models," Jour, of PetroleumTechnology (Nov. 1969), 1391—1398

22K. H Coats, "Reservoir Simulation: State of the An," Jour. of Perrolewn Tech-nology (Aug 1982), 1633—1642.

UT. V Moore, "Reservoir Engineering Begins Second 25 Years," Oil and Gas Jour(1955). 54, No. 29, 148.

24Norman J Clark and Arthur J Wessely, "Coordination of' Geology and Reservoir

Engineering—A Growing Need for Management Decisions." Presented before API,

Division of Production, March, 1957.

23Principles of Petroleum Conservation, Engineering Committee, Interstate Oil Com-pact Commission (1955), p. 2.

Reserves Definitions," Jour, of Petroleum Technology (Nov. 1981),

2113—2114.

27B J. Dotson, R. L. Slobod, P. N. McCreery, and James W Spurlock, "Porosity-Measurement Comparisons by Five Laboratories," Trans AIME (1951), 192, 344.

W. van der Knaap, "Non-linear Elastic Behavior of Porous Media," Presentedbefore Society of Petroleum Engineers of A1ME, October 1958, Houston, TX.

290. H. Newman, "Pore-Volume Compressibility of Consolidated, Fnable, and Un-consolidated Reservoir Rocks Under Hydrostatic Loading," Jour, of Petroleum Tech-nology (Feb. 1973), 129—134.

30J. Gecrtsma, "The Effect of Fluid Pressure Decline on Volumetric Changes ofPorous Rocks," Jour of Petroleum Technology (1957), 11, No 12, 332.

31Henry I (iruy and Jack A. Crichton. "A Critical Review of Methods Used in theEstimation of Natural Gas Reserves," Trans. AIME (1949), 179, 249—263

'2R P. Sutton, "Compressibility Factors for High-Molecular-Weight ReservoirGases" Paper SPE 14265 presented at the 1985 SPE Annual Technical Meeting andExhibition, Las Vegas, NV, Sept. 22—25.

33Marsball B. Standing and Donald L. Katt, 'Dcnsity of Natural Gases," Trans.AIME (1942), 146, 144.

P. M. Dranchuk and 3. H Abou-Kassem, "Calculation of Z Factors for NaturalGases Using Equations of State," Jour. of Petroleum Technology (July—Sept. 1975),34—36.

R. W. Ilornbeck. Numerical Methods New York: Quantum Publishers, 1975

'6C. Kenneth ElIcits and Others, Phase Relations of Gas-Condensate Fluids, U.SBureau of Mines Monograph 10, Vol. 1 (New York: Aniencan Gas Association, 1957),427—434.

References 55

"E. Wichert and K Aziz, "Calculate Z's for Sour Gases," Ifyd Proc. (May 1972),119—122.

A. S Trube, "Compressibility of Natural Ciases," Trans. AIME (1957), 210, 61

'9L Mattar, G. S. Brar, and K. Aziz, "Compressibility of Natural Gases," JCPT(Oct.—Dec. 1975), 77—80

A. Blasingame, J. L Johnston, and R. D Poe, Jr: "Properties of Rcservoirfluids," Texas A&M University (1989).

N. L. Carr, R Kobayashi and D. 13. Burrows. "Viscosity of Hydrocarbon GasesUnder Pressure," Trans. AIME (1954), 201, 264-272.

A. L. Lee, M. H Gonzalez and 13. Eakin, "The Viscosity of Natural Gases,"Jour of Petroleum Technology (Aug. 1966), 997-1000; Trans, AIME (1966), 237.

'3M. B. Standing, "A Pressure-Volume-Temperature Correlation for Mixtures ofCalifornia Oils and Gases," Drill, and Prod Prac, API (1947), 275—287.

'4H 1) Beggs, "Oil System Correlations," Petroleum Engineering Ilandboolc, H C.Bradley (ed.), SPE, Richardson, TX (1987) 1, Chap. 22.

.1 Villena-Lanzi, "A Correlation for the Coefficient of Isothermal Com-pressibility of Black Oil at Pressures Below the Bubble Point," M S. Thesis, TexasA&M University, College Station, TX (1985)

Vasqucz and 11.1) Beggs, "Correlations for Fluid Physical Property Prediction,"Jour of Petroleum Engineering (June 1980), 968—970.

'7E 0 "An Improved Temperature-Viscosity Correlation for Crude OilSystems," Paper 83-34-32 presented at the 1983 Annual Technical Meeting of thePetroleum Society of CIM, Banff, Alberta, May 10—13, 1983.

D. Beggs and J R. Robinson, "Estimating the Viscosity of Crude Oil Systems,"Jour, of Petroleum Technology (Sept 1975) 1140—41.

"W D. McCairv, Jr. The Properties of Petroleum Fluids, 2nd ed. Tulsa, OK: PennWcllPublishing Co., 1988.

50W. D. McCain, Jr "Reservoir Water Property Correlationsr-'-State of The Art,"Paper SPE 18573 submitted for publication (1988)

T. L. Osif, "The Effects of Salt, Gas, Temperature, and Pressure on the Com-pressibility of Water," Paper SPE 13174 presented at the 1984 SPE Annual TechnicalConference and Exhibition, houston, TX, Sept. 16—19.

Chapter 2

The General MaterialBalance Equation

1. INTRODUCTION

A general material balance equation that can be applied to all reservoir typesis developed in this chapter. From this general equation, each of the individualequations for the reservoir types, defined in Chapter 1 and discussed in subse-quent chapters, can easily be derived.

The general material balance equation was first developed by Schilthuisin 1936.* 'Since that time, the use of the computer and sophisticated multidi-mensional mathematical models has replaced the zero dimensional Schilthuisequation in many applications.2 However, the Schilthuis equation, if fully un-derstood, can provide great insight for the practicing reservoir engineer. Fol-lowing the derivation of the general material balance equation, a method ofusing the equation, discussed in the literature by Haviena and Odeh, ispresented

2. DERIVATION OF MATERIAL BALANCE EQUATION

When an oil and gas reservoir is tapped with wells, oil and gas. and frequentlysome water, are produced, thereby reducing the reservoir pressure and caus-ing the remaining oil and gas to expand to fiJI the space vacated by the fluids

Refercnces throughout the text arc givun at the end of each chapter

2 Denvation of Material Balance Equation 57

removed. When the oil- and gas-bearing strata are hydraulically connectedwith water-bearing strata, or aquifers, water encroaches into the reservoir asthe pressure drops owing to production, as illustrated in Fig. 2.J. This waterencroachment decreases the extent to which the remaining oil and gas expandand accordingly retards the decline in reservoir pressure. Inasmuch as thetemperature in oil and gas rescrvoirs remains substantially constant during thecourse of production, the degree to which the remaining oil and gas expanddepends only on the pressure. By taking bottom-hole samples of the reservoirfluids under pressure and measuring their relative volumes in the laboratoryat reservoir temperature and under various pressures, it is possible to predicthow these fluids behave in the reservoir as reservoir pressure declines.

!n a subsequent chapter it is shown that, although the connate water andformation compressibilities arc quite small, they are, relative to the com-pressibility of reservoir fluids above their bubble points, significant, and theyaccount for an appreciable fraction of the production above the bubble point.Table 2.1 gives a range of values for formation and fluid compressibilities fromwhich it may be concluded that water and formation compressibilities are lesssignificant in gas and gas cap reservoirs and in undersaturated reservoirs belowthe bubble point where there is appreciable gas saturation. Because of this andthe complications they would introduce in already fairly complex equations,water and formation compressibilities are generally neglected. except in un-

II H

Fig. 2.1. Cross section of a combination drive rcscrvoir. (After Woody and Moscnp,5Trans. AIME)

OIL, GAS CAP GAS OIL AND OIL, SOLUTION GASAND SOLUTION GAS SOLUTION GAS AND WATER

58 The General Material Balance Equation

TABLE 2.1Range of compressibilities

Formation rock 3—l0X10Water 2—4xlO6psiUndersaturatcd oil 5— 100 x 10-6 psiGasatl000psi 900—1300x10Gas at 5000 psi 50—200 x psi

dersaturated reservoirs producing above the bubble point A term accountingfor the change in water and formation volumes owing to their compressibiliticsis included in the material balance derivation; the engineer can choose toeliminate them for particular applications. The gas in solution in the formationwater is neglected, and in many instances the volume of the produced wateris not known with sufficient accuracy to justify the use of a formation volumefactor with the produced water.

The general material balance equation is simply a volumetric balance,which states that since the volume of a reservoir (as defined by its initial limits)is a constant, the algebraic sum of the volume changes of the oil, free gas,water, and rock volumes in the reservoir must be zero. For example, if boththe oil and gas reservoir volumes decrease, the sum of these two decreasesmust be balanced by changes of equal magnitude in the water and rockvolumes. If the assumption is made that compicte equilibrium is attained at alltimes in the reservoir between the oil and its solution gas, it is possible to wntea generalized material balance expression relating the quantities of oil, gas,and water produced, the average reservoir prcssure, the quantity of water thatmay have encroached from the aquifer, and finally the initial oil and gascontent of the reservoir. In making these calculations the following produc.tion, reservoir, and laboratory data are involved:

1. The initial reservoir pressure and the average reservoir pressure at succes-sive intervals after the start of production.

2. The stock tank barrels of oil produced, measured at I atm and 6(Y'F, at anytime or during any production interval.

3. The total standard cubic feet of gas produced. When gas is injected into thereservoir, this will be the difference between the total gas produced andthat returned to the reservoir.

4. The ratio of the initial gas cap volume to the initial oil volume, symbol m.

— Initial reservoir free gas volumem

— Initial reservoir oil volume

If this value can be determined with reasonable precision, there is only oneunknown (N) in the material balance on volumetric gas cap reservoirs, andtwo (N and W,) in water-drive reservoirs. The value of m is determinedfrom log and core data and from well completion data, which frequently

2. Derivation of Material Balance Equation 59

helps to locate the gas-oil and water-oil contacts. The ratio m is known inmany instances much more accurately than the absolute values of the gascap and oil zone volumes For example, when the rock in the gas cap andthat in the oil zone appear to be essentially the same, it may be taken as theratio of the net or even the gross volumes, without knowing the averageconnate water or average porosity, or when gross volumes are used, thefactors for reducing gross to net productive volumes.

5. The gas and oil volume factors and the solution gas-oil ratios. These areobtained as functions of pressure by laboratory measurements on bottom-hole samples by the differential and flash liberation methods.

6. The quantity of water that has been produced.7. The quantity of water that has been encroached into the reservoir from the

aquifer.

For simplicity, the derivation is divided into the changes in the oil, gas,water, and rock volumes that occur between the start of production and anytime t. The change in the rock volume is expressed as a change in the voidspace volume, which is simply the negative of the change in the rock volume.In the development of the general material balance equation, the followingterms are used:

N Initial reservoir oil, SThInitial oil formation volume factor, bbl/STB

Cumulative produced oil, STB

oil formation volume factor, bbl/STI3

0 Initial reservoir gas, SCFBgs Initial gas formation volume factor, bbl/SCF

0, Amount of free gas in the reservoir, SCFR,,, Initial solution gas-oil ratio, SCF/STB

Cumulative produced gas-oil ratio, SCF/STB

R50 Solution gas-oil ratio, SCF/STI3

Gas formation volume factor, bhl/SCF

W Initial reservoir water, bblCumulative produced water, STBWater formation volume factor, bbl/STB

We Water influx into reservoir, bbl

c,,, Water isothermal compressibility, psi -'

Change in average reservoir pressure, psia

S,., Initial water saturationVf Initial void space, bbl

c1 Formation isothermal compressibility, psi


Change in the Oil Volume

Initial reservoir oil volume = NB0,

Oil volume at time t and pressure p = (N —

Change in oil volume = NBth — (N — (2.1)

Change in h'ee Gas Volume:

[Ratio of initial free — —

[ to initial oil volume i— m

— NB1,,

Initial free gas volume = GB,, = NmB0,

SCF free 1 — I SCF initial gas. 1 — [ SCF gas — [scF remaininggas at t i [free and dissolvedj Lproduced I in solution

G1=[ThnBoi

+ NR50, ] — I — [(N —

_NRso, - - (N - Np)Rso] Bg

[Change ;n free]= NmB0,

—+ — — (N — Np)Rso]B,

(2.2)

Change in Water Volume.

Initial reservoir water volume = WCumulative water produced at t = W,

Reservoir volume of cumulative produced water = W,,

Volume of water encroached at t =

W — (W + We - + wc,4/3) = - We +

(2.3)

Change in the Void Space VolumeS

Initial void space volume =

[Change in voidi — —

I. space volume J = — IV, — V1c,,ApJ =

2. Derivation of Material Balance Equation 51

Or, becausethe change in void space volume is the negative of the change inrock volume:

F Change in I —I = — (2.4)[rock volumcj

Combining the changes in water and rock volumes into a single term, yieldsthe following:

=

NB01+NmB,Recognizing that W = V,S.,1 and that Vf

= 1 —and substituting, the

following is obtained:

=

or

(2.5)

Equating the changes in the oil and free gas volumes to the negative ofthe changes in the water and rock volumes and expanding all terms

N B01 - N B0 + N R,qBg +

C S •+c+ NBgR50 — = W, — + (1 + m)N

Now adding and subtracting the term

N N B0 + NIB,, + Nm80, —{NmBO1B1]

— N + +

cS -'c— NpBgR30 + NpBgRzo, — NpJigRsoi = — B,1 + (1 + m) N B,,,

'IThen grouping terms:

NB,,, ÷ — NLR,, + (R30, — + N,[B0 + (R10, — Rg,)BgJ

+ ——

= VI, — B,, + (1 + m) N


Now writing Be,, = B,, and [B0 +(R30, — R50)B,] = B,, where B, is the two phaseformation volume factor, as defined by Eq. 1 28

— B,) + + — Roi)Bgj + NmBta(1—

C S +cl= We — + (1 + m) N B,,

—'s,,, 'j(2.6)

This is the general volumetric material balance equation. It can be rearrangedinto the following form that is useful for discussion purposes.

NmB- cS.+cIN(B,—Bd)+- B• (8, B,j+(l+m) NB,, +W,

WI

= (Rp — R301)B,J + (2.7)

Each term on the left-hand side of Eq. (2.7) accounts for a method offluid production, and each term on the right-hand side represents an amountof hydrocarbon or water production. The first two terms on the left-hand sideaccount for the expansion of any oil and/or gas zones that might be present.The third term accounts for the change in void space volume, which is theexpansion of the formation and connate water. The fourth term is the amountof water influx that has occurred into the reservoir. On the right-hand side, thefirst term represents the production of oil and gas and the second term repre-sents the water production.

Equation (2 7) can be arranged to apply to any of the different reservoirtypes discussed in Chapter 1. Without eliminating any terms, Eq. (2.7)is usedfor the case of a saturated oil reservoir with an associated gas cap. Thesereservoirs are discussed in Chapter 6. When there is no original free gas, suchas in an undersaturated oil reservoir (discussed in Chapter 5), m =0 andEq. (2 7) reduces to:

CW w,+CjN(BI — B,,) 4- N BoLl + 14',

= f — + (2.8)

For gas reservoirs, Eq. (2.7) can be modified by recognizing that =and that = GB,i and substituting these terms into Eq. (2.7):

N(B, — B,,) + G (BR — + (N B,i + GB,,) + W,

= NAB, + — .VR,,,,) B! + (2.9)

2. Derivation of Matenal Balance Equation 63

When working with gas reservoirs, there is no initial oil amount; therefore, Nand arc equal to zero. The general material balancc equation for a gasreservoir can then be obtained

G (B8 + GBgr ÷ We = GpBg + (2 10)

This equation is discussed in conjunction with gas and gas-condensate reser-voirs in Chapters 3 and 4.

In the study of reservoirs that arc produced simultaneously by the threemajor mechanisms of depletion drive, gas cap drive, and water drive, it is ofpractical intere5t to determine the relative magnitude of each of these mech-anisms that contribute to the production. Pirson rearranged the materialbalance Equation (2.7) as follows to obtain three fractions, whose sum is one,which he called the depletion drive mdcx (DDI), the segregation (gas cap)index (SD!), and the water-drive index (WDI) 6

When all three drive mechanisms are contributing to the production ofoil and gas from the reservoir, the compressibility term in Eq (2.7)is negli-gible and can be ignored. Moving the water production term to the left-handsidc of the equation, the following is obtained:

N(B, — B,1) + NlflBti(8— Be,) + — NPLIJ, + —

Dividing through by the term on the hand side of the equation:

+

Np[8t+(RpR301)Bg)

(W—B W)N,,IB,+ — RQ,)B:J —

The numerators of these three fractions that result on the left-hand side of Eq.(2.11) are the expansion of the initial oil the expansion of the initial gaszone, and the net water influx, respectively. The common denominator is thereservoir volume of the cumulative gas and oil production expressed at thelower pressure, which evidently equals the sum of the gas and oil zone expan-sions plus the net water influx. Then, using Pirson's abbreviations:

ODI + SDI + WDI = 1

Calculations are performed in Chapter 6 to illustrate how these dnve indexesare used,

64 The General Matenal Balance Equation

3. USES AND LIMITATIONS OF THE MATERIALBALANCE METHOD

The material balance equation derived in the previous section has been ingeneral use for many years mainly for the following:

1. Determining the initial hydrocarbon in place.2. Calculating water influx.3. Predicting reservoir pressures.

Although in some cases it is possible to solve simultaneously to find the initialhydrocarbon and the water influx, generally one or the other must be knownfrom data or methods that do not depend on the material balance calculationsOne of the most important uses of the equations is predicting the effect ofproduction rate and/or injection rates (gas or water) on reservoir pressure;therefore, it is very desirable to know in advance the initial oil and the ratiom from good core and log data. The presence of a water drive is usuallyindicated by geologic evidence; however, the material balance may be used todetect the existence of water drives by calculating the value of the initialhydrocarbon at successive production periods, assuming zero water influxUnless other complicating factors are present, the constancy in the calculatedvalue of N and/or G indicates a volumetric reservoir, and continually changingvalues of N and ( indicate a water drive.

The precision of the calculated values depends on the accuracy of thedata available to substitute in the equation and on the several assumptions thatunderlie the equations. One such assumption is the attainment of thermo-dynamic equilibrium in the reservoir, mainly between the oil and its solutiongas Wicland and Kennedy have found a tendency for the liquid phase toremain supersaturated with gas as the pressure declines.7 Values in the rangeof 19 psi for East lexas Field fluid and cores and 25 psi for the Slaughter Fieldfluid and cores were measured. The effect of supersaturation causes the reser-voir pressures to be lower than they would be had equilibnum been attained.

It is also implicitly assumed that the PVT data used in the matenalbalances are obtained using gas liberation processes that closely duplicate thegas liberation processes in the reservoir, in the well, and in separators on thesurface. This matter is discussed in detail in Chapter 6, and it is only slatedhere that in some cases the PVT data are based on gas liberation processes thatvary widely from those obtaining for the reservoir fluid, ar.d therefore produceresults in the material balances that arc in considerable error

Another source of error is introduced in the determination of averagereservoir pressure at the end of any production interval. Aside from instru-ment errors and those introduced by difficulties in obtaining true static or finalbuild-up pressures (see Chapter 7), there is often the prcblem of correctlyweighting or averaging the individual well pressures. For thicker formations ofhigher permeabilities and oils of lower viscosities, when final build-up pres-

3. Uses and Limitations of the Matonal Balance Method 65

sures are readily and accurately obtained, and when there are only smallpressure differences across the reservoir, reliable values of average reservoirpressure are easily obtained. On the other hand, for thinner formations oflower permeability and oils of higher viscosity, difficulties are met in obtainingaccurate final build-up pressures, and there are often large pressure variationsthroughout the reservoir. These are commonly averaged by preparing isobancmaps superimposed on isopachous maps This method usually provides re-liable results unless the measured well pressures arc erratic and thereforecannot be accurately contoured These differences may be due to variations information thickness and permeability and in well production and producingrates Also, difficulties are encountered when producing formations arc com-prised of two or more zones or strata of different permeabilities. in this case,the pressures are generally higher in the strata of low permeability, and,because the measured pressures are nearer to those in the zones of highpermeability, the measured static pressures tend to bc lower, and the reservoirbehaves as if it contained less oil. Schilthuis explained this phenomenon byreferring to the oil in the more permeable zones as active oil and by observingthat the calculated active oil usually increases with time because the oil and gasin the zones of lower permeability slowly expand to help offset the pressuredecline. This is also true of fields that arc not fully developed, because theavcrage pressure can be that of the developed portion only whereas the pres-sure is higher in the undeveloped portions.

The effect of pressure errors on calculated values of initial oil or waterinflux depends on the size of the errors in relation to the reservoir pressuredecline This is true because pressure enters the material balance equationmainly as differences (B0 — (R,, — Rj, and — Be,) Because waterinflux and gas cap expansion tend to offset pressure decline, the pressureerrors are more serious than for the undcrsaturated depletion reservoirs. Inthe case of very active water drives and gas caps that are large compared withthe associated oil zone, the material balance is useless to determine the initialoil in place because of the very small pressure decline. Hutchinson emphasiiedthe importance of obtaining accurate values of static well pressures, in hisquantitative study of the effect of data errors on the values of initial gas or ofinitial oil in volumetric gas or undersaturated oil reservoirs, respectively.8

Uncertainties in the ratio of the initial free gas volume to the initialreservoir oil volume also affect the calculations. The error introduced in thecalculated values of initial oil, water influx, or pressure increases with the sizeof this ratio because, as explained in the previous paragraph, larger gas capsreduce the effect of pressure decline For quite large caps relative to the oilzone, the material balance approaches a gas balance modified slightly byproduction from the oil zone. The value of m is obtained from core and logdata used to determine the net productive bulk gas and oil volumes and theiraverage porosities and interstitial water. Because there is frequently oil satu-ration in the gas cap, the oil zone must include this oil, which correspondingly

66 The General Matenat Balance Equation

diminishes the initial free gas volume. Well tests are often useful in locatinggas-oil and water-oil contacts in the determination of m. In some cases, thesecontacts are not horizontal planes but are tilted owing to water movement inthe aquifer, or owing to the effect of capillarity in the less perme-able boundary rocks of volumetnc reservoirs.

Whereas the cumulative oil production is generally known quite pre-cisely, the corresponding gas and water production is usually much less accu-rate, and therefore introduces additional sources of errors. This is particularlytrue when the gas and watcr production is not directly measured but must beinferred from periodic tests to determine the gas-oil ratios and water cuts ofthe individual wells. When two or more wells completed in different reservoirsare producing to common storage, unless there arc individual meters on thewells, only the aggregate production is known and not the individual oilproduction from each reservoir. Under the circumstances that exist in manyfields, it is doubtful that the cumulative gas and water production is known towithin 10% and in some instances the errors may be larger. With the growingimportance of natural gas and because more of the gas associated with the oilis being sold, better values of gas production are becoming available.

4. THE HAVLENA AND ODEH METHOD OF APPLYINGTHE MATERIAL BALANCE EQUATION

As early as 1953, van Everdingen, Timmcrinan, and McMahon recognized amethod of applying the material balance equation as a straight line.9 But itwasn't until Havlena and Odeh published their work that the method becamefully exploited" Normally, when using the material balance equation, anengineer considers each pressure and the corresponding production data asbeing separate points from other pressure values. From each separate point,a calculation for a dependent variable is made. The results of the calculationsare sometimes averaged. The Havlena-Odeh method uses all the data points,with the further requirement that these points must yield solutions to thematerial balance equation that behave linearly to obtain values of the indepen-dent variable.

The straight-line method begins with the material balance equation writ-ten as:

+ (R, — + — W, —

(2.12)

The terms W,, cumulative water injection, G,, cumulative gas injection, andB,4, formation volume factor of the injected gas have been added to Eq. (2.7).in Havlena and Odeh's original development, they chose to neglect the effect

References 67

of the compressibilities of the formation and connate water in the gas capportion of thc reservoir That is, in their development the compressibility termis multiplied by N and not by N(J + m). In Eq. (2.12), the compressibilityterm is multiplied by N(I + m) for completeness. You may choose to ignorethe (1 + m) multiplier in particular applications. Havlena and Odeb definedthe following terms and rewrote Eq. (2.12) as:

F = + — + — GiBig

E0 = — B,,

=

= Bg — Bgi

(2.13)

In Eq. (2.13) F represents the net production from the reservoir. E0, andE1 represent the expansion of oil, formation and water, and gas, respectively.Haviena and Odeb examined several cases of varying reservoir types with thisequation and found that the equation can be rearranged into the form of astraight line. For instance, consider the case of no original gas cap, no waterinflux, and negligible formation and water compressibilities. With thesesumptions, Eq. (2.13) reduces to:

F=NE0 (2.14)

This would suggest that a plot of F as the y coordinate and E,, as the xcoordinate would yield a straight line with slope N and intercept equal to zero.Additional cases can be derived as will be shown in Chapter 6.

Once a linear relationship has been obtained, the plot can be used as apredictive tool for estimating future production. Examples are shown in subse..quent chapters to illustrate the application of the Havclena-Odeh method.

REFERENCES

Ralph 3. Schilthuis, "Active Oil and Reservoir Energy," Trans AIME (1936),118, 33.

2L. P. Dake, Fundamentals of Reservoir Engineering, 1st ed Amsterdam: Elsevier,1978, pp 73—102.

3D. Jiaviena and A S. Odch, "The Material Balance as an Equation of a StraightLine," Part I. Jour, of Petroleum Technology (Aug. 1963), 896-900.

4D. Haviena and A. S. Odch, "The Material Balance as an Equation of a StraightLine Part Il—Field Cascs," Jour of Petroleum Technology (July 1964), 815—822.


L D Woody. Jr. and Robert Moscrip 111. "Performance Calculations for Combina-tion Dnve Reservoirs," Trans AIME (1956), 207, 129.

6Sylvain J. Pirson, Elements of Oil Reservoir Engineering, 2nd ed. New York:McGraw-Hill, 1958, pp. 635—693.

7Denton R. Wieland and Harvey T. Kennedy. "Measurements of Bubble Frequencyin Cores," Trans. AIME (1957), 210, 125.8 A Hutchinson, "Effect of Data Errors on Typical Engineering Calcu-lations." Presented at the Oklahoma City Meeting, Petroleum Branch, AIME, 1951.

'A. F. van Everdingen. E. H. Timmerman, and 3. .1. McMahon, "Application of theMatenal Balance Equation to a Partial Water-Dnve Reservoir," Trans. AIME (1953),198, 51.

Chapter 3

SinglePhase Gas Reservoirs

1. INTRODUCTION

This chapter discusses single-phase gas reservoirs (refer to Fig. 1.1). lhc reser-voir fluid, usually called a natural gas, in a single-phase gas reservoir remainsas nonassociatcd gas during the entire producing life of the reservoir. Thistype of reservoir is frequently referred to as a dry gas reservoir because nocondensate is formed in the reservoir during the life of production. However,this does not preclude the production of condensate, because the temperatureand pressure in the producing well and at the surface can be significantlydifferent from the reservoir temperature and pressure. This change in condi-tions can cause some components in the producing gas phase to condense andbe produced as liquid. The amount of condensation is a function of not onlythe pressure and temperature but also the composition of the natural gas,which typically consists primarily of methane and ethanc. The tendency forcondensate to form on the surface increases as the concentration of heaviercomponents increases in the reservoir fluid.

Wc begin with a consideration of volumetric calculations for initial gasin place.

69

70 Single-Phase Gas Reservoirs

2. CALCULATING GAS iN PLACE BY THE VOLUMETRICMETHOD

The standard cubic feet of gas in a reservoir with a gas pore volume of cuft is simply Vg/Bg, where is expressed in units of cubic feet per standardcubic foot. As the gas volume factor changes with pressure (see Eq 1 17,the gas in place also changes as the pressure declines. The gas pore volumemay also bc changing, owing to water influx into the reservoir. The gas porevolume is related to the bulk, or total, reservoir volume by the averageporosity 4' and the average connate water The bulk reservoir volume Vb iscommonly expressed in acre-fcei, and the standard cubic feet of gas in place,G, is given by:

43,560

a

The arcal extent of the Bell Field gas reservoir was 1500 acres. averagethickness was 40 ft, so that the initial bulk volume was 60,000 ac-ft. Averageporosity was 22%, and average connate water was 23%. at the initialreservoir pressure of 3250 psia was calculated to be 0.00533 cu ft/SCF. There-fore the initial gas in place was

G = 43,560 x 60,000 x 0.22 x (1 —0.23) — 0.00533

= 83.1 MMM SCF

Because the gas volume factor was calculated using 14.7 psia and 60°F asstandard conditions, the initial gas in place is also expressed at these con-ditions.

The volumetric method uses subsurface and isopachous maps based onthe data from electric logs, cores, and drill-stem and production tests. * 12 A

subsurface contour map shows lines connecting points of equal elevations onthe top of a marker bed and therefore shows geologic structure. A net iso-pac/zous map shows lines connecting points of equal net formation thickness;and the individual lines connecting points of equal thickness are called isopachlines. The reservoir engineer uses these maps to determine the bulk produc-tive volume of the reservoir. The contour map is used in preparing the iso-pachous maps when there is an oil-water, gas-water, or gas-oil contact. Thecontact line is the zero isopach line. The volume is obtained by planimeteringthe areas between the isopach lines of the entire reservoir or of the individualunits under consideration. The principal problems in preparing a map of thistype are the proper interpretation of net sand thickness from the well logs andthe outlining of the productive area of the field as defined by the fluid con-tacts, faults, or permeability barriers on the subsurface contour map.

Ret ercncec throughout thc text arc given at the end of each chapter

2. Calculating Gas in Place by the Volumetnc Method 71

Two equations are commonly used to determine the approximate vol.wne of the productive zone from the planimeter readings. The volume of thefrustum of a pyramid is given by:

+ (32)

where is the bulk volume in acre-feet, A,, is the area enclosed by the lowerisopach line in acres, A,,,1 is the area enclosed by the upper isopach line inacres, and h is the interval between the isopach lines in feet. This equation isused to determine the volume between successive isopach lines, and the totalvolume is the sum of these separate voiumes. The volume of a trapezoid is:

+ A,,+1)

or for a series of successive trapezoids:

Vb = + 2A1 + 2A2. . . 2A,,_1 + A,,) + (3.3)

A0 is the area enclosed by the zero isopach line in acres; A1, A2. . . A,, are theareas enclosed by successive isopach lines in acres; avg is the average thicknessabove the top or maximum thickness isopach line in feet; and Ii is the isopachinterval.

For best accuracy the pyramidal formula should be used. Because of itssimpler form, however, the trapezoidal formula is commonly used, but itintroduces an error of 2% when the ratio of successive areas is 0.50. There-fore, a commonly adopted rule in unitization programs is that wherever theratio of the areas of any two successive isopach lines is smaller than 0.5, thepyramidal formula is applied. Whenever the ratio of the areas of any twosuccessivc isopach lines is found to be larger than 0.5, the trapeLoidal formulais applied. Example 3.1 shows the method of calculating the volume of a gasreservoir from an isopachous map. Fig. 3.1 The volume between areas A4 and

by the trapezoidal equation is 570 ac-fl, compared with the more accuratefigure of 558 ac'ft by the pyramidal equation. When the formation is ratheruniformly developed and thcre is good well control, the error in the net bulkreservoir volume should not exceed a few percentagc points.

Example 3.1. Calculating the net volume of an ideaii7ed reservoir fromthe isopachous map.

Given; The planimctcred areas in Fig. 3.1 within each isopach line. A0.A1, A2. etc and the planimeter constant


Fig. 3.1. Cross section and isopachous map of an idealized reservoir

SOLW ION:

PlanimelerProductive

AreaArea4sq in

AreaAcres

Ratio ofAreas

intervalh. feel Equation

V

ac-ft

A0 1964 450

—

A1 1634 375 0.83 5 Trap 206313 19 303 0.80 5 Trap 16951005 231 0.76 5 Trap 1335

A4 6 69 154 0.67 5 Trap 96Y322 74 048 5 Pyr 558b

'lb 000 0 000 4

.

Pyr. c

6713 ac-ft

I in —l000ft,J sq in —2296acS

(231 + 154) —963 ac-ft

hAy + 74 + V154 x 74) 558 ac-ft

ac-ft

FT SOPACHINTERVAL

0 1000 2000 3000

SCALE IN FEET

Well average pressure =

Areal average pressure =

2. Calculating Gas in Place by the Volumetric Method 73

The water in the oil- and gas-bearing parts of a petroleum reservoirabove the transition zone is called connate, or interstitial, water. The two termsarc used more or less interchangeably. Connate water is important primarilybecause it reduces the amount of pore space available to oil and gas and it alsoaffects their recovery. It is generally not uniformly distnhuted throughout thereservoir but varies with the permeability and lithology as shown in Fig. 3.2and with the height above the free water table as shown in Fig. 3.3.

Another problem in any volumetric or material balance calculation isthat of obtaining thc average reservoir pressure at any time after initial pro-duction. Figure 3.4 is a static reservoir pressure survey of the Jones sand in theSchuler Field.4 Because of the large reservoir pressure gradient from east towest, some averaging technique must be used to obtain an average reservoirpressure This can be calculated either as an average well pressure, averageareal pressure, or average volumetric pressure as follows:

(34)

(35)

10 20 30 40 50 60PER CENT CONNATE WATER

Fig. 3.2. Connate water versus permeability. (After Bruce and Wclge! Bruce TransA1ME)

74

Fig. 3.3. Typical capillary pressure curve.

Single-Phase Gas Reservoirs

In

8.I.'

InU,

4-I-a

Fig. 3.4. Reservoir pressure survey showing isobaric lines drawn from the measuredbottom-hole prcssures. (After Kavcler,4 Trans. AIME.)

CONNATE

25%

-

JoIL

14c__.

.!a

w -- — 34% WATER SATURATION —

*1 5 FT ABOVE FREE

0

2

—•010020 40 60 80

WATER SATURATION, PERCENT

2. Calculating Gas in Place by the Volumetric Method 75

Volumctnc average pressure = - (3.6)

0

where ii is the number of wells in Eq. (3.4) and the number of reservoir unitsin Eqs. (3 5) and (3.6). Because we are interested in obtaining the averagepressure of the hydrocarbon contents, the volumetric average, Eq. (3.6),should be used in the volumetric and material balance calculations. Where thepressure gradients in the reservoir are small, the average pressures obtainedwith Eqs. (3.4) and (3.5) will be very close to the volumetric average. Wherethe gradients arc large, there may be considerable differences. Eor example,the average volumetric pressure of the Jones sand survey in Fig. 3.4 is 1658psia as compared with 1598 psia for the well average pressure.

The calculations in Table (3.1) show how the average pressures arcobtained. The figures in Col. (3) are the estimated drainage areas of the wells,which in some cases vary from the well spacing because of the reservoir limits.Owing to the much smaller gradients, the three averages are much closertogether than in the case of the Jones sand.

Most engineers prefer to prepare an isobanc map as illustrated in Fig.(3.5) and to planimeter the areas between the isobaric lines and the isopachlines as shown in Fig. 3.5. Table 3.2, using data taken from Fig 3.5, illustratesthe method of obtaining the average volumetric pressure from this type ofmap.

TABLE 3.1.Calculation of average reservoir pressure

WellNo.

Pressurepsia

Drainage Es:. SdArea Acres p x A Thick. p x A x h Axh

1

23

4

275026H028402700

10,970

160 440,000 20 8,800.000125 335,000 25 8,375,000190 539.600 26 14,029.600145 391.500 31 12,136.500620 1,706,100 43,341.100

3200312549404495

15,760

10,970Well average pressure = = 2743 psia

Arcal average pressure = = 2752 psia

Vo43.341,100

lumetric average pressure=

2750 psia


Inwz.J

C,

4g2eoo

Fig. 3.5. Section of an isobaric and isopadious map

TABLE 3.2.Volumetric calculations of reservoir pressure

(1)

Area

(2)

Acre?

(3)Pressure

psia

(4)

hfi

(5)

Axh

(6)

pxAxhA 25 5 2750 25 637.5 175,313.0001) 15 1 2750 15 226.5 62,288,000C 50 5 2850 25 1262.5 359,813,000D 302 2850 15 453.0

2579.5129,105,000726,519.000

Planimctcred areas of rig 3 5

726,519.000Average pressure on a volume basis =

2579 5

= 2817 psia

3. CALCULATION OF UNIT RECOVERY FROMVOLUMETRIC GAS RESERVOiRS

In many gas reservoirs, particularly during the development period, the bulkvolume is not known. ln this case it is better to place the reservoir calculationson a unit basis, usually 1 ac-ft of bulk reservoir rock. Then one unit, or 1 ac-ft.of bulk reservoir rock contains:

Connate water 43,560 x 4) x cu ft

Reservoir gas volume: 43,560 x 4, (1 — cu ft

Reservoir pore volume: 43,560 x 4, cu ft

3. Calculating of Unit Recovery from Volumetric Gas Reservoirs 77

The initial standard cubic feet of gas in place in the unit is:

G43,560 (c1)(1 — S1

SCFIac-fr (3 7)

G is in standard cubic feet when the gas volume factor B8, is in cubic feet perstandard cubic foot Eq. (1.17). The standard conditions are those used in thecalculation of the gas volume factor, and they may be changed to any otherstandard by means of the ideal gas law. The porosity ch is expressed as afraction of the bulk volume, and the initial connate water as a fraction ofthe pore volume. For a reservoir under volumetric control, there is no changein the interstitial water, so the reservoir gas volume remains the same. Ifis the gas volume factor at the abandonment pressure. then the standard cubicfeet of gas remaining at abandonment is:

Ga = 43,560 (4)(I —SCF/ac-ft (3.8)

Unit recovery is the difference between the initial gas in place and that re-maining at abandonment pressure (i.e , that produced at abandonmentpressure), or

Unit recovery = 43,560(4)(1 — Sw,)—

i—] SCF/ac-fi (3.9)

The unit recovery is also called the initial unit reserve, which is generally lowerthan the initial unit gas. The remaining reserve at any stage of de-pletion is the difference between this initial reserve and the unit production atthat stage of depletion. The fractional recovery or recovery factor expressedin a percentage of the initial in-place gas is

100100(G — Ga) B8, Bga

Recoveryfactor=—1

% (3.10)

Bg,

or

IRecovery factor = 10011 — -s—-

Experience with volumetric gas reservoirs indicates that the recoveries willrange from 80 to 90% Some gas pipeline companies use an abandonmentpressure of 100 psi per 1000 ft of depth.


The gas volume factor in the Bell Gas Field at initial reservoir pressureisO 00533 cu ft/SCF and at 500 psia it isO 03623. The initial unit reserve or unitrecovery based on volumetric performance at an abandonment pressure of 500psia is

Unit recovery = 43,560 X 0 22 X (1 — 0.23))(— 0.03623]

1180M SCF/ac-ft

F 000533Recovery factor 100 [1

— 0.03623

-= 85%

These recovery calculations are valid provided the unit neither drains nor isdrained by adjacent units.

4. CALCULATION OF UNIT RECOVERY FROM GASRESERVOIRS UNDER WATER DRIVE

Under initial conditions one unit (I ac-ft) of bulk reservoir rock contains

Connate water: 43,560 X 4, X SW, cu ft

Reservoir gas volume: 43,560 x 4, x (I — Sw,) cu ft

Surface units of gas: 43,560 x 4, x (1 — — SCF

In many reservoirs under water drive, the pressure suffers an initial decline,after which water enters the reservoir at a rate to equal the production, andthe pressure stabilizes. In this case the stabilized pressure is the abandonmentpressure If is the gas volume factor at the abandonment pressure and S,,is the residual gas saturation, expressed as a fraction of the pore volume, afterwater invades the unit, then under abandonment conditions a unit (1 ac-ft) ofthe reservoir rock contains

Water volume: 43,560 X de X (1 — Se,) CU ft

Reservoir gas volume- 43,560 x x cu ft

Surface units of gas: 43.560 x 4, x -- SCF

Unit recovery is the difference between the initial and the residual surfaceunits of gas, or:

Unit recovery in SCF/ac-ft = 43.560[1

—

(3.11)

4. Calculating of Unit Recovery from Gas Reservoirs Under Water Drive 19

The recovery factor expressed in a percentage of the initial gas in place is

100 -Recovery factor=—— (3.12)

Suppose the Beil Gas Fieid is produced under a water drive such that thepressure stabilizes at 1500 psia. If the residual gas saturation is 24% and thegas volume factor at 1500 psia is 0.01122 cu ft/SCF, then the initial unit reserveor unit recovery is

1(1—0.23) 0.24Unit recovcry = 43,560 x 0.22 x

— 0.0112

1180 M SCF/ac-ft

The recovery factor under these conditions is

[1—0.23 0.241001000533 0.0112

Recovery —= 85%

1.0.005331

Under these particular conditions, the recovery by water drive is the same asthe recovery by volumetric depletion, illustrated in Sect. 3. If the water dnveis very active so that there is essentially no dccline in reservoir pressure, unitrecovery and the recovery factor become

Unit recovery 43,560 x cb x (1 — Se,, — Se,) — SCF/ac-ft (3.13)

100(1 — — S2,)Recovcryfactor=

(1—S.)-% (3.14)

For the Bell Gas Field, assuming a residual gas saturation of 24%:

Unit recovery = 43,560 x 0.22 x (1 —0 23 — 0.24) --0 00533

953 M SCF/ac-ft

100(1—0.23—024)Recovery factor

(I 0.23)

-=69%

Because the residual gas saturation is independent of the pressure, the recov-ery will be greater for the lower stabilization pressure.


The residual gas saturation can he measured in the laboratory on repre-sentative core samples. Table 3.3 gives the residual gas saturations that weremeasured on core samples from a number of producing horizons and on somesynthetic laboratory samples. The values, which range from 16 to 50% andaverage near 30%, help to explain the disappointing recoveries obtained insome water-drive reservoirs. For example. a gas reservoir with an initial watersaturation of 30% and a residual gas saturation of 35% has a recovery factorof only 50% if produced under an active water drive (i e., where the reservoirpressure stabilizes near the initial pressure). When the reservoir permeabilityis uniform, this recovery factor should be representative, except for a correc-tion to allow for the efficiency of the drainage pattern and water coning orcusping. When there are well-defined continuous beds of higher and lowerpermeability, the waler will advance more rapidly through the more perme-able beds so that when a gas well is abandoned owing to excessive waterproduction, considerable unrecovered gas remains in the less permeable beds.Because of these factors, it may be concluded that generally gas recoveries bywater drive arc lower than by volumetric depletion; however, the same conclu-sion does not apply to oil recovery, which is discussed separately. Water-drivegas reservoirs do have the advantage of maintaining higher flowing weliheadpressures and higher well rates compared with depiction gas reservoirs. Thisis due, of course, to the maintenance of higher reservoir pressure as a resultof the water influx

TABLE 3.3.Residual gas saturation after water flood as measured on core plugs(After Getfen, Parish. Haynes, and Morse5.)

Residual Gas Saturation,Porous Material Formation Percentage of Pore Space Remarks

Unconsolidated sand 16 (13-ft Column)Slightly consolidated

sand (synthetic) 21 (1 Core)Synthetic consolidated

materials Selas Porcelain 17 (1 Core)Norton Alundum 24 (1 Core)

ConsolidatedWilcox 25 (3 Cores)1-rio 30 (1 Core)Nellie Bly 30-36 (12 Cores)Frontier 31—34 (3 Cores)Springer 33 (3 Cores)Frio 30—38 (14 Corcs)

(Average 34 6)lbrpcdo 34—37 (6 Cores)ienslccp 40—50 (4 Cores)

Limestone Canyon Reef 50 (2 Cores)

4. Calculating of Unit Recovery from Gas Reservoirs Under Water Drive 81

In calculating the gas reserve of a particular lease or unit, the gas that canbe recovered by the well(s) on the lease is important rather than the totalrecoverable gas initially underlying the lease, some of which may be recoveredby adjacent wells. In volumetric reservoirs where the recoverable gas beneatheach lease (well) is the same, the recoveries will be the same only if all wellsare produced at the same rate. On the other hand, if wells arc atequal rates when the gas beneath the leases (wells) varies, as from variableformation thickness, the initial gas reserve of the lease where the formation isthicker will be less than the initiaJ recoverable gas underlying the Jease.

In water-drive gas reservoirs, when the pressure stabilizes near the initialreservoir pressure, the lowest well on structure will divide its initial recov-erable gas with all updip wells in line with it. For example, if three wells in linealong the dip are drilled at the updip edge of their units, which are presumedequal, and if they all produce at the same rate, the lowest well on structure willrecover approximately one-third of the gas initially underlying it. If the wellis drilled further downstructure near the center of the unit, it will recover stillless. If the pressure stabilizes at some pressure below the initial reservoirpressure, the recovery factor will be improved for the wells low on structure.Example 3.2 shows the calculation of the initial gas reserve of a 160-acre unitby volumetric depletion, partial water drive, and complete water drive.

Example 3.2. Calculating the initial gas reserve of a 160-acre unit of theBell Gas Field by volumetric depletion and under partial and complete waterdrive.

Given:

Average porosity = 22%

Connate water =23%

Residual gas saturation after water displacement =34%

= 0.00533 cu ftISCF at p.=3250 psia

= 0.00667 cu f1ISCF at 2500 psia

= 0.03623 cu ftISCF at 500 psia

Area = 160 acres

Net productive thickness =40 ft

SOLUTION:

Pore volume=43,560x0.22x160X40=61.33X lO6cu ft

Initial gas in place:

G1=6133x 106x(l—0.23)—0.00533—8860MM SCF


Gas in place after volumetric depletion to 2500 psia:

Gas in place after volumetric depletion to 500 psia:

x 106x (1 —0.23)—.0.03623= 1303MM SCF

Gas in place after water invasion at 32.50 psia'

(;4 = 61.33 x x 0.34 — 0.00533 = 3912 MM SCF

Gas in place after water invasion at 2500 psia.

G5=61 33x106x0,34÷0.00667=3126 MM SCF

initial reserve by depletion to 500 psia:

— G3 = (8860 — 1303) x 106 = 7557 MM SCF

initial reserve by water drive at 3250 psia:

G1—G4=(8860—3912)x 106=4948MM SCI

initial reserve by water drive at 2500 psia:

(G1—G,)=(8860—3126)x 1065734MM SCF

If there is one undip well, the initial reserve by water drive at 3250psia is

1(Gi 106=2474MM SCF

The recovery factors calculate to be 85%, 65%, and 56% for the casesof no water drive, partial water drive, and full water dnve, respectively. Theserecoveries are fairly typical and can be explained in the following way. Aswater invades the reservoir, the reservoir pressure is maintained at a higherlevel than if there were no water encroachment. This leads to higher abandon-ment pressures for water-drive reservoirs. Because the main mechanism ofproduction in a gas reservoir is that of depletion, or gas expansion, recoveriesare lower, as shown in Ex. 3.2.

Agarwal, Al-Hussainy. and Ramey conducted a theoretical study andshowed that gas recoveries increased with increasing production rates fromwater-drive reservoirs This technique of "outrunning" the water has been

5. Material Balance 83

attempted in the field and has been found to be successful Matthes, Jackson,Schuler, and Marudiak showed that ultimate recovery increased from 69 to74% by increasing the field production rate from 50 to 75 MM SCFID in theBierwang Field in West Germany.7 Lutes, Chiang, Brady, and Rosscn re-ported an 8.5% increase in ultimate recovery with an increased productionrate in a strong water-drive Gulf Coast gas reservoir.8

A second technique used in the field is the coproduction techniquediscussed by Arcaro and Bassiouni.9 The coproduction tcchniquc is defined asthe simultaneous production of gas and water. In the coproduction process, asdowndip wells begin to be watered out, they are converted to high-rate waterproduction welJs, while the updip wells are maintained on gas productionThis technique enhances the production of gas by several methods First, thehigh-rate downdip water wells act as a pressure sink for the water because thewater is drawn to these wells. This retards the invasion of water into pro-ductive gas zones in the reservoir, therefore prolonging the useful productivelife to these zones. Second, the high-rate production of water lowers theaverage pressure in the reservoir, allowing for more gas expansion and there-fore more gas production. Third, when the average reservoir pressure islowered, immobile gas in the water-swept portion of the reservoir could be-come mobile. The coproduction technique performs best before the reservoiris totally invaded by water. Arcaro and Bassiouni reported the improvementof gas production from 62% to 83% from the Louisiana Gulf Coast EugeneIsland Block 305 Reservoir by using the coproduction technique instead of theconventional production approach. Water-drive reservoirs are discussed inmuch more detail in Chapter 8.

5. MATERIAL BALANCE

In the previous sections, the initial gas in place was calculated on a unit basisof 1 ac-ft of bulk productive rock from a knowledge of the porosity andconnate water To calculate the initial gas in place on any particular portionof a reservoir, it is necessary to know, in addition, the bulk volume of thatportion of the reservoir, if the porosity, connate water, and/or the bulk vol-umes are not known with any reasonable precision, the methods describedcannot be used. In this case, the material balance method may be used tocalculate the initial gas in place; however, this method is applicable only to thereservoir as a whole, because of the migration of gas from one portion of thereservoir to another in both volumetric and water-drive reservoirs.

The general material balance equation for a gas reservoir was derived inChapter 2.

G(88 — B81) + GB8, ÷ W, GpB8 + (2.10)


Equation (2.10) could have been derived by applying the law of conservationof mass to the reservoir and associated production.

For most gas reservoirs, the gas compressibility term is much greaterthan the formation and water compressibilities, and the second term on theleft-hand side of Eq. (2.10) becomes negligible

G(BgBRI)+ W,=GpBg+ (3.15)

When reservoir pressures arc abnormally high, this term is not negligible andshould not be ignored. This situation is discussed in a later section of thischapter.

When there is neither water encroachment into nor water productionfrom a reservoir of interest, the reservoir is said to be volumetric. For avolumetric gas reservoir, Eq. (3.15) reduces to:

G(B, — B111) = GpBg (3.16)

Using Eq. (1.16) and substituting expressions for B, and into Eq. (3.16),the following is obtained:

G — G(PscziE)

(3.17)7ep

Noting that production is essentially an isothermal process (i.e., the reservoirtemperature remains constant), then Eq. (3.17) is reduced to:

G (!) - G =

Rearranging:

(3.18)

Because P1. and G are constants for a given reservoir, Eq. (3.1K) suggeststhat a plot of p/z as the ordinate versus as the abscissa would yield a straightline with:

slope = —Ay interecept =

This plot is shown in Fig. 3 6

5. Matenal Balance 85

—j— —j-—---- -—

4000

-NA.

0U'

I IdVOLUMETRIC

a'z- -

- VOLUMETRIC "VS P

WRONGEXTRAPOLATION -z"\

Z

3 4CUMULATIVE PRODUCTION, MMMSCF

Fig. 3.6. Comparison of theoretical values of p and plz plotted versus cumulativeproduction from a volumetnc gas reservoir.

If plz is set equal to zero, which would represent the production of allthe gas from a reservoir, then the corresponding equals G, the initial gasin place The plot could also be extrapolated to any abandonment plz to findthe initial reserve. Usually this extrapolation requires at least three years ofaccurate pressure depletion and gas production data.

Figure 3.6 also contains a plot of cumulative gas production G,, versuspressure. As indicated by Eq. (3.18), this is not linear, and extrapolationsfrom the pressure-production data may be in considerable error. Because theminimum value of the gas deviation factor generally occurs near 2500 psia, theextrapolations will be low for pressures above 2500 psia and high for pressuresbelow 2500 psia. Equation (3.18) may be uscd graphically as shown in Fig. 3.6to find the initial gas in place or the reserves at any pressure for any selectedabandonment pressure For example, at 1000 psia (or p/z = 1220) abandon-ment pressure, the initial reserve is 4.85 MMM SCF. At 2500 psia (or p/z =3130), the (remaining) reserve is 4.85 less 2.20—that is, 2.65 MMM SCF.

In water-drive reservoirs, the relation between (3,, and p/z is not linear,as can he seen by an inspection of Eq. (3.15) and (3.18). Because of the waterinflux, the pressure drops less rapidly with production than under volumetriccontrol, as shown in the upper curve of Fig 3 6. Consequently, the extrapo-lation technique described for volumetric reservoirs is not applicable. Also,where there is water influx, the initial gas in place calculated at successivestages of depletion, assuming no water influx, takes on successively highervalues, whereas with volumetric reservoirs the calculated values of the initialgas should remain substantially constant


Equation (3 15) may be expressed in terms of the initial porc volume,by recognizing that = GB1, and using Eq. (116) for and Bg,:

I]..ZITPiCGP+BWWP1 fir

For volumetric reservoirs, this equation can be reduced and rearranged togive:

(320)i; z,T z1T

The following three example problems illustrate the use of the various equa-tions that we have described in gas reservoir calculations.

Example 3.3. Calculating the initial gas in place and the initial reserveof a gas reservoir from pressure-production data for a volumetric reservoirNote that the base pressure is 15.025 psia.

;iven:

Initial pressure = 3250 psia

keservoir temperature = 213°F

Standard pressure = 15.025 psia

Standard temperature = 60°F

Cumulative production = 1.00 x SCF

Average reservoir pressure = 2864 psia

Gas deviation factor at 3250 psia 0.9 10

Gas deviation factor at 2864 psia 0.888

Gas deviation factor at 500 psia = 0.951

SoLLrrloN. Solve Eq. (3.20) for the reservoir gas pore volume

1' — 2864 V1

520 0.910x 673 0.888x673

V,= 56.17MM cu ft

5. Matenal Balance 87

The initial gas in place by the real gas law

106x52z,T Psc 0.910)(673x15.025

= 10.32 MMM SCF

The gas remaining at 500 psia abandonment pressure is:

GazTPsc 0.951x673x15.025

= 1.52 MMM SCF

The initial gas reserve based on a 500 psia abandonment pressure is the dif-ference between the initial gas in place and the gas remaining at 500 psia, or:

G,=G —G.=(10.32.— L52)x

= 8.80 MMM SCF

Example 3.4 illustrates the use of equations to calculate the water influxwhen the initial gas in place is known. It also shows the method of estimatingthe residual gas saturation of the portion of the reservoir invaded by water, atwhich time a reliable estimate of the invaded volume can be made. This iscalculated from the isopachous map, the invaded volume being dilineated bythose wells that have gone to water production. The residual gas saturationcalculated in Ex. 3 4 includes that portion of the lower permeability rockwithin the invaded area that actually may not have been invaded at all, thewells having been "drowned" by water production from the more permeablebeds of the formation. Nevertheless, it is still interpreted as the average re-sidual gas saturation, which may be applied to the uninvaded portion of thereservoir.

Example 3.4 Calculating water influx and residual gas saturation inwater-drive gas reservoirs

Bulk reservoir volume, initial = 415.3 MM cu ft

Average porosity = 0.172

Average connate water = 0.25

Initial pressure = 3200 psia


=0 005262 Cu ft/SCF, 14.7 psia and 60°F

Final pressure = 2925 psia

=0005700 cu ft/SCF, 14.7 psia and 60°F

Cumulative water production = 15:200 bbl (surface)

= 1.03 bhl/surface bbl

= 935.4 MM SCF at 14.7 psia and 60°F

Bulk volume invaded by water at 2925 psia = 13.04 MM cu ft

SOLUTION'

415.3x 106X0.172X(1 —025)Initial gas in place = G

0 005262 -

10,180 MM SCF at 14 7 psia and 60°F

Substitute in Eq. (3.15) to find W,

W,=935.4x 106x0.005700—10,180x 106

(0.005700— 0.005262) + 15,200 x 1.03 x 5.615

= 960,400 Cu ft

This much water has invaded 13.04 MM cu ft of bulk rock that initially con-tained 25% connate water. Then the final water saturation of the floodedportion of the reservoir is

— Connate water + Water influx — Produced water— Pore space

_(13.04 x 106x0,172 xO.25)+960,400— 15.200x 1.03

13.04 lOb xO 172

= 0.67 or 67%

Then the residual gas saturation is 33%.

Example 3.5. Using the plz plot to estimate cumulative gas productionA dry gas reservoir contains gas of the following composition:

Mole FractionMethane 0.75Ethane 0.20n-Hexanc 0.05

5. Material Balance 89

The initial reservoir pressurc was 4200 psia, with a temperature of 180°F. Thereservoir has been producing for some time. Two pressure surveys have beenmade at different times.

plz (psia) G, (MMM SCF)4600 03700 1

2800 2

(a) What will be the cumulative gas produced when the average reservoirpressure has dropped to 2000 psia?

(b) Assuming the reservoir rock has a porosity of 12%, the water saturationis 30%, and the reservoir thickness is 15 ft, how many acres does thereservoir cover?

SOLUTION:

Pc. YPe

Methane 0.75 673.1 343.2 504.8 257.4Ethane 0.20 708.3 504.8 141.7 110.0n-Hexanc 0.05 440.1 914.2 22.0 45.7

Totals

(a) To get G, at 2000 psia, calculate z and the p/z. Use pseudocriticalproperties.

2000

r 640—1.

z =0.8

2000p/z = =2500

A linear regression of the data plotted in Fig. 3.7 yields the following equationfor the best straight line through the data:

p/z = — + 4600

Substituting a value of p/z = 2500 in this equation yields:

2500 = — + 4600

= 2.33 SCF


-

N

2000 -. ——

__-.

05 10 15 2.0 2.5

mmm scf

Fig. 3.7. plz versus G,, for Ex. 3.5

(b) Substituting a valuc of p/z =0 into the straight-line equation, would yicldthe amount of produced gas if all of the initial gas wcrc produced; there-fore, the at this p/z is equal to the initial gas in place.

0= — 9(10) + 4600

11(10)9SCF or 511 MMMSCF

RecogniLing that V, = and that = 0.02829 (z1/pjT:

Vi = GBg1 = 20.1(10)6 CU ft

also

20 1(10)6

= 15(0.12)(11595(10)6 or 366 acres

6. The Gas Equivalent of Produced Condensate and Water

6. THE GAS EQUIVALENT OF PRODUCED CONDENSATEAND WATER

In the study of gas reservoirs in the preceding section, it was implicitly as-sumed that the fluid in the reservoir at all pressures as well as on the surfacewas in a single (gas) phase. Most gas reservoirs, however, produce somehydrocarbon liquid, commonly called condensate, in the range of a few to ahundred or more barrels per million standard cubic feet. So long as thereservoir fluid remains in a single (gas) phase, the calculations of the previoussections may be used, provided the cumulative gas production is modifiedto include the condensate liquid production. On the other hand, if a hydro-carbon liquid phase develops in the reservoir, the methods of the previoussections are not applicable, and these retrograde, gas-condensate reservoirsmust be treated specially, as descnbed in Chapter 4.

The reservoir gas production used in the previous sections mustinclude the separator gas production, the stock tank gas production, and thestock tank liquid production converted to its gas equivalent, symbol GE.Figure 3.8 illustrates two common separation schemes. Figure 3.8(a) shows athree-stage separation system with a primary separator, a secondary separa-tor, and a stock tank. The well fluid is introduced into the primary separatorwhere most of the produced gas is obtained. The liquid from the primaryseparator is then sent to the secondary separator where an additional amountof gas is obtained. The liquid from the secondary separator is then flashed intothe stock tank. The liquid from the stock tank is and any gas from the stocktank iS added to the pnmary and sccondary gas to arrive at the total producedsurface gas, Figure 3.8(b) shows a two-stage separation process similarto the one shown in Fig 3.8(a) without the secondary separator.

I

Fig. 3.8. Schematic reprcscntation of surface separation systems

(a) Three-stagc separation system

(b) Two-stage separation system


The produced hydrocarbon liquid is converted to its gas equivalent,assuming it behaves as an ideal gas when vaporized in the produced gas.Taking 14 7 psia and 60°F as standard conditions, the gas equivalcnt of onestock tank barrel of condensate liquid is:

GE ——

133 ()()()YO 321—

— — — '

The gas equivalent of one barrel of condensate of specific gravity of 0.780(water = 1.00) and molecular weight 138 is 752 SCF. The specific gravity maybe calculated from the API gravity. If the molecular weight of the condensateis not measured, as by the freezing point depression method, it may be esti-mated using Eq (3.22).

= = 42.43'yo (3.22)

The total gas equivalent for N,, STB of condensate production is Thetotal reservoir gas production, G,,, is given by Eq (3 23) for a three-stageseparation system and by Eq. (3.24) for a two-stage separation system:

= + = + Ga + + GE(N,,) (3.23)

= + = + + (3.24)

If the gas volumes from the low-pressure separators and the stock tank are notmeasured, then the correlations found in Figs 3.9 and 3.10 can be used toestimate the vapor equivalent of these gas volumes plus the gas equivalent ofthe liquid condensate. Eigure 3 9 is for a three-stage separator system, andFig. 3.10 is for a two-stage separator system. The correlations arc based onproduction parameters that are routinely measured (i.e.. the primary sepa-rator pressure. temperature, and gas gravity), the secondary separator tem-perature, and the stock tank liquid gravity. The total reservoir gas production,

is given by Eq. (3.25) when using the correlations of Figs. 3.9 and 3.10.

(325)

When water is produced on the surface as a condensate from the gasphase in the reservoir, it is fresh water and should be converted to a gasequivalent and added to the gas production. Since the specific gravity of wateris 1.00 and its molecular weight is 18, its gas equivalent is

350.5x 1.00 10.73 x520X—18 14.7

7390 SCF/surface barrel


*ro jIvLtN? 1V1 S

Fig. 3.10. Vapor-Equivalent Noniograph for Two-Stage Separation Systems (AfterGold. McCain, and Jcnningc 10)

cubic feet As rcseroir pressure declines, the water content increases to asmuch as three barrels per million standard cubic feet. Since this additionalcontent has come from vaporization of the connate water, it would appear thatany fresh water produced in excess of the initial content should be treated asproduced water and taken care of in the W,, term rather than the G term. Ifthe water is saline, it definitely is produced water, however, it includes thefraction of a barrel per million cubic feet obtained from the gas phase. if theproduced gas is based on the dehydrated gas volume, the gas volume should

7 Gas Reservoirs as Storage Reservoirs

be increased by the gas equivalent of the water content at the initial reservoirpressure and temperature regardless of the subsequent decline in reservoirpressure, and the water production should be diminished by the water con-tent. This amounts to about a 0.05% increase in the produced gas volumes.

7. GAS RESERVOIRS AS STORAGE RESERVOIRS

The demand for natural gas is seasonal. During winter months, there is a muchgreater demand for natural gas than during the warmer summer months. Tomeet this vanable demand, several means of storing natural gas are used in theindustry. One of the best methods of storing natural gas is with the use ofdepleted gas reservoirs. Gas is injccted during the warm summer months whenthere is an overabundance and produced during the winter months when thereis a shortage of supply. Katz and Tek have presented a good overview of thissubject.12

Katz and Tek listed three primary objectives in the design and operationof a gas storage reservoir: (1) verification of inventory; (2) retention againstmigration; and (3) assurance of deliverability Verification of inventory simplymeans knowing the storage capacity of thc reservoir as a function of pressure.This suggests that a p/z plot or some other measure of material balance beknown for the reservoir of interest. Retention against migration refers to amonitoring system capable of ascertaining if the injected gas remains in thestorage reservoir. Obviously, leaks in casing and so on would be detrimentalto the storage process. The operator needs to be assured that the reservoir canbe produced during peak demand times in order to provide the proper deliv-erability. A major concern with the deliverability is that water encroachmentnot interfere with the gas production. With these design considerations inmind, it is apparent that a good candidate for a storage reservoir would be adepleted volumetric gas reservoir With a depleted volumetric reservoir, thep/z versus curve is usually known and water influx is not a problem.

Ikoku defines three types of gas involved in a gas storage reservoir.13 Thefirst is the base gas. or cushion gas, that remains when the base pressure isreached. The base pressure is the pressure at which production is stopped andinjection is begun The second type of gas is the working gas, or workingstorage, that is produced and injected during the cycle process. The third typeis the unused gas that essentially is the unused capacity of the reservoir. Figure3.11 defines these three types of gas on a p/z plot.

The base pressure, and therefore the amount of base gas, is defined bydeliverability needs. Sufficient pressure must be maintained in the reservoirfor reservoir gas to be delivered to transporting pipelines Economics dictatesthe pressure at which injection of gas during the summer months ends. Com-pression costs must be balanced with the projected supply and demand of thewinter months. In theory, for a volumetric reservoir, the cycles of injection


Fig. 3.11. plz plot showing different types of gas in a gas storage reservoir

and production simply run up and down the p/z versus curve between thepressure limits just discussed.

In certain applications, the use of the delta pressure concept may beadvantageous 12 The delta pressure is defined as the pressure at maximumstorage minus the initial reservoir pressure. Under the right conditions, anamount of gas larger than the initial gas in place can be achieved. This againis dictated by the economics of the given situation

Hollis presented an interesting case history of the considerations in-volved in changing the Rough Gas Field in the North Sea over to a storagereservoir)4 Considerations in the design of storage and deliverability ratesincluded the probability of a severe winter occurring in the demand area. Asevere winter was given a probability of 1 in 50of happening Hollis concludedthat the differences between offshore and onshore storage facilities are owingmainly to economics and the integrated planning that must take place inoffshore development.

Storage is a useful application of gas reservoirs We encourage you topursue the references for more detailed information if it becomes necessary.

-

0

8 Abnormaliy Pressured Gas Reservoirs 97

8. ABNORMALLY PRESSURED GAS RESERVOiRS

Normal pressure gradients obscrvcd in gas reservoirs are in the range of 0 4to 0.5 psia per foot of depth. Reservoirs with abnormal pressures may havegradients as high as 0.7 to 1 0 pisa per foot of depth.'5 16 17 Bernard hasreported that over 300 gas reservoirs have been discovered in the offshore GulfCoast alone with initial gradients in excess of 0.65 psia per foot of depth informations Over 10,000 fcet deep.'8

When the water and formation compressibility term in the materialbalance equation can be ignored, the normal p/z behavior for a volumetric gasreservoir plots a straight line versus cumulative gas produced (Fig 3.6, Sect.5). This is not the case for an abnormally pressured gas reservoir, as can beseen in Figure 3.12, which illustrates the plz behavior for this type of reservoir.

For an abnormally pressured volumetric reservoir, the plz plot is astraight line during the early life of production, but then it usually curvesdownward during the later stages of production. If the early data are used toextrapolate for G or for an abandonment the extrapolation can yieldsignificant errors.

To explain the curvature in the p/z plot for abnormally pressured reser-voirs, Harville and Hawkins postulated a "rock collapse" theory that used ahigh rock compressibility at abnormally high pressures and a reduced rockcompressibility at normal reservoir pressures.'5 Howcver, working with rock

N

Fig. 3.12. piz plot illustrating nonlinear behavior of abnormally pressured reservoir


samples taken from abnormally pressured reservoirs, Jogi, Gray, Ashman,and Thompson and Sinha, Holland, Borshccl, and Schatz reported rock corn-pressibilitics measured at high pressures in the order of 2 to 5 (1O) 20

These values are representative of typical values at low pressures and suggestthat rock compressibilities do not change with pressure. Rarnagost and Far-shad showed that in some cases the p/z data could be adjusted to yield straight-line behavior by including the water and formation compressibility term.2'Bourgoyne, Hawkins, Lavaquial, and Wickenhauser suggested that the non-linear behavior could be due to water influx from

Bernard has proposed a method of analyzing the p/z curve for abnor-mally pressured reservoirs to determine initial gas in place and gas reserve asa function of abandonment plz. 18 method uses two approaches. The firstinvolves the early production life when the piz plot exhibits linear behavior.Bernard developed a correlation for actual gas in place as a function ofapparent gas in place as shown in Fig. 3.13.

The apparent gas in place is obtained by extrapolating the early, linearp/z data. Then, by entering Figure 3.13 with the apparent value of the gas inplace, the ratio of actual gas in place to apparent gas in place can he obtained.The correlation appears to be reasonably accurate for the reservoirs thatBernard studied For later production times, when the p/z data exhibit non-linear behavior, Bernard defined a constant C' according to Eq. (3.26)

LU

a.z

U)4Wi.-<LU

0o4

Fig. 3.13. Correction for initial gas in place (After Bernard

(3.26)

APPARENT GAS IN-PLACE, BCF

9 Limitations of Equations and Errors 99

where

C' = constant

tip = total pressure drop in the reservoir, P — p

in Eq. (3.26), the only unknowns arc C' and G. Bernard suggested that theycan he found by the following procedure:

First, calculate A' and B' from'

andtip(p/z) Ap(p/z)

If there are n data points forp/z and then C' and G can be calculated fromthe following equations:

G— B'/n — Z (B'2)

B'In

C' = ZA 'In + 8'/n

9. LIMITATIONS OF EQUATIONS AND ERRORS

The precision of the reserve calculations by the volumetric method depends onthe accuracy of the data that enter the computations. The precision of theinitial gas in place depends on the probable errors in the averages of theporosity, connate water, pressure, and gas deviation factor, and in the errorin the determination of the bulk productive volume. With the best of core andlog data in rather uniform reservoirs, it is doubtful that the initial gas in placecan be calculated more accurately than about 5%, and the figure will rangeupward to 100% or higher depending on the uniformity of the reservoir andthe quantity and quality of the data available

The reserve is the product of the gas in place and the recovery factor. Forvolumetric reservoirs, the reserve of the reservoir as a whole, for any selectedabandonment pressure, should be known to about the same precision as theinitial gas in place. Water-dnve reservoirs require, in addition, the estimate ofthe volume of the reservoir invaded at abandonment and the average residualgas saturation. When the reservoir exhibits permeability stratification, thedifficulties are increased, and the accuracy is therefore reduced. In general,reserve calculations arc more accurate for volumetric than for water-drivereservoirs. When the reserves are placed on a well or lease basis, the accuracymay be reduced further because of lease drainage, which occurs in bothvolumetric and water-drive reservoirs.

The use of the material balance equation to calculate gas in place in-volves the terms of the gas volume factor. l'he precision of the calculations is,


of course, a function of the probable error in these terms. The error in gasproduction arises from error in gas metering, in the estimate of lease useand leakage, and in the estimate of the low-pressure separator or stock tankgases. Sometimes underground leakage occurs—from the failure in casingcementing, from casing corrosion, or, in the case of dual completions, fromleakage between the two zones. When gas is commingled from two reservoirsat the surface prior to metering, the division of the total between the tworeservoirs depends on periodic well tests, which may introduce additionalinaccuracies. Meters are usually calibrated to an accuracy of 1%, and there-fore it is doubtful that the gas production under the best of circumstances isknown closer than 2%. Average accuracies arc in the range of a few to severalpercentage points.

Pressure errors arc a result of gauge errors and the difficulties in aver-aging, particularly when there are large pressure differences throughout thereservoir. When reservoir pressures arc estimated from measured wcllheadpressures, the errors of this technique enter the calculations. When the fieldis not fully developed, the average pressure is, of course, of the developedportion, which is lower than that of the reservoir as a whole. Water productionwith gas wells is frequently unreported when the amount is small; when it isappreciable, it is often estimated from periodic well tests.

Under the best of circumstances, the material balance estimates of thegas in place are seldom more accurate than 5% and may range much higher.The estimate of reserves is, of course, one step removed.

PROBLEMS

3.1 A volumetric gas field has an initial pressure of 4200 psia, a porosity of 17 2%,and connatc water of 23%. The gas volume factor at 4200 psia is 0.003425 cuItISCF and at 750 psia is 001852 cu ft/SCF.

(a) Calculate the initial in-place gas in standard cubic feet on a unit basis.

(b) Calculate the initial gas reserve in standard cubic feet on a unit basis,assuming an abandonment pressure of 750 psia.

(c) Explain why the calculated initial rcscrvc depends on the abandonmentpressure selected

(d) Calculate the initial reserve of a 640-acre unit whose average net productiveformation thickness is 34 ft. assuming an abandonment pressure of 750 psia

(e) Calculate the recovery factor based on an abandonment pressure of 750 psia.

3.2 The discovery well No I and wells No 2 and No. 4 produce gas in the 7500-ftreservoir of the Echo Lake Field (Fig 3.14). Wells Nos 3 and 7 were dry in the7500-ft reservoir; however, together with their electric logs and the one fromwell No. 1. the fault that seals the northeast side of the reservoir was establishedThe logs of wells Nos. 1, 2, 4. 5, and 6 were used to construct the map of Fig3.14, which was used to locate the gas-water contact and to determine the

Problems 101

average net sand thickness. The reservoir had been producing for 18 monthswhen well No. 6 was drilled at the gas-water contact. The static wcllhead pres-sures of the production wells showed virtually no decline during the 18-monthperiod before dnlling well No.6 and averaged near 3400 psia The following datawere available from electric logs, core analysis, and the like.

Average well depth = 7500 ft

Average static weilbead pressure = 3400 psia


Gas specific gravity =0700

Average porosity =27%

Average connate water =22%

Standard conditions = 14.7 psia and 60°F

Bulk volume of productive reservoir rock at the time No 6 wasdrilled = 22,500 ac-ft

(a) Calculate the reservoir pressure

(b) Estimate the gas deviation factor and the gas volume factor

(c) Calculate the reserve at the time well No. 6 was dnlled, assuming a residualgas saturation of 30%

(d) Discuss the location of well No I with regard to the overall gas(e) Discuss the effect of sand uniformity on overall recovery—for example, a

uniform permeable sand versus a sand in two beds of equal thickness, oneof which has a permeability of 500 md and the other, 100 md.

3.3 The M Sand is a small gas reservoir with an initial bottom-hole pressure of 3200psia and bottom-bole temperature of 220°F It is desired to inventory the gas inplace at three production intervals The pressure-production history and gasvolume factors are as follows:

Fig. 3.14. Echo Lake Field, subsurface map, 7500 ft reservoir.


Cumulative Gas Gas VoiwnePressure Production Factor

psia MM SCF cu f:1SCF

3200 0 0 00526222925 79 0 00570042525 221 000653112125 452 0.9077360

(a) Calculate the initial gas in place using production data at the end of each ofthe production intervals, assuming volumctnc behavior

(b) Explain why the calculations of part (a) indicate a water drive

(c) Show that a water drive exists by plotting the cumulative production versusP/2.

(d) Based on electric log and core data. volumetric calculations On the M Sandshowed that the initial volume of gas in place is 1018 MM SCF. If the sandis under a partial water drive, what is the volume of water encroached at theend of each of the periods9 There was no appreciable water production.

3.4 When the Sabine Gas Field was brought in, it had a reservoir pressure of 1700psia, and a temperature of 160°F After 5.00 MMM SCF was produced, the pres-sure had fallen to 1550 psia. If the reservoir is assumed to be under volumetriccontrol, using the deviation factors of Prob. 1 10, calculate the following

(a) Thc hydrocarbon pore volume of the reservoir.(b) The SCFproduccd when the pressure falisto 1550, 1400, 1100,500, and 200

psia Plot cumulative recovery in SCF versus plz(c) The SCF of gas initially in place.(d) From your graph, find how much gas can he obtained without the usc of

compressors for delivery into a pipeline operating at 750 psia.

(e) What is the approximate pressure drop per MMM SCF of production?

(1) Calculate the minimum value of the initial reserve if the produced gasmeasurement is accurate to 5% and if the average pressures are accurateto ± 12 psi when 5 00 MMM SCF have been produced and the reservoirpressure has dropped to 1550 psia.

3.5 If, however, during the production of 5.00 MMM SCF of gas in the precedingproblem, 4.00 MM bbl of water had encroached into the reservoir and still thepressure had dropped to 1550 psia. calculate the initial in-place gas. How doesthis compare with Prob. 3 4(c)?

3.6 (a) The gas cap of the St. John Oil Field had a bulk volume of 17,000 ac-ft whenthe reservoir pressure had declined to 634 psig Core analysis shows anaverage porosity of 18% and an average interstitial water of 24% It isdesired to increase the recovery of oil from the field by repressuring the gascap to 1100 psig Assuming that no additional gas dissolves in the oil duringrepressuring, calculate the scr required. The deviation factors for both the

Problems 103

reservoir gas and the injected gas are 0 86 at 634 psig and 0 78 at 1100 psig.both at 130°F.

(b) If the injected gas has a deviation factor of 094 at 634 psig and 0.88 at 1100psig, and the reservoir gas deviation factors are as above, recalculate theinjected gas required

(c) Is the assumption that no additional solution gas will enter the reservoir oila valid one?

(d) Considcnng the possibility of some additional solution gas and the produc-tion of oil during the time of injection, will the figure of part (a) be maximumor minimum? Explain

(e) Explain why the gas deviation factors are higher (closer to unity) for theinjected gas in pan (b) than for the reservoir gas.

3.7 The following production data are available from a gas reservoir produced undervolumetric control

Pressure Cumulative(psia) Gas Production (P4MM SCF)

5000 2004000 420

The initial reservoir temperature was 237°F and the reservoir gas gravity is 0.7.

(a) What will be the cumulative gas production at 2500 psia"(b) What fraction of the initial reservoir gas will be produced at 2500 psia"(c) What was the initial reservoir pressure?

3.8 (a) A well drilled into a gas cap for gas recycling purposes is found to be in anisolated fault block. After 50 MM SCF was injected, the pressure increasedfrom 2500 to 3500 psia Deviation factors for the gas are 090 at 3500 and0.80 at 2500 psia and the bottom-hole temperature is 160°F What is thecubic feet of gas storage space in the fault block?

(b) If the average porosity is 16%, average connate water is 24%, and averagesand thickness is 12 ft, what is the areal extent of the fault block"

3.9 The initial volume of gas in place in the P Sand reservoir of the Holden Field iscalculated from electric log and core data to be 200 MMM SCF underlying 2250productive acres, at an initial pressure of 3500 psia and 140°F. The pressure-production history is

Production Gas DeviationPressure, psia MMM SCF Factor

3500 (initial) 0 0 0.852500 75 0 0.82

(a) What is the initial volume of gas in place as calculated from the pressure-production history, assuming no water influx?


(b) Assuming uniform sand thickness, porosity, and connate water, if the vol-ume of gas in place from pressure-production data is believed to be correct,how many acres of cxtcnsion to the present limits of the P Sand arc pre-dicted?

(c) If, on the other hand, the gas in place calculated from the log and core datais believed to be correct, how much water influx must have occurred duringthe 75 MMM SCF of production to make the two figures agree?

3.10 Explain why initial calculations of gas in place are likely to be in greater errorduring the early life of depletion reservoirs. Will these factors make the predic-tions high or low? Explain.

3.11 A gas reservoir under partial water dnve produced 12 0 MMM SCF when theaverage reservoir pressure had dropped from 3000 psia to 2200 psia Dunng thesame interval, an estimated 5.20 MM bbl of water entered the reservoir basedon the volume of the invaded area If the gas deviation factor at 3000 psia andbottom-hole temperature of 170°F is 0.88 and at 2200 psia is 0.78, what is theinitial volume of gas in place measured at 14.7 psia and 60°F?

3.12 A gas-producing formation has uniform thickness of 32 ft. a porosity of 19%,and connate water saturation of 26% The gas deviation factor is 0.83 at theinitial reservoir pressure of 4450 psia and reservoir temperature of 175°F.

(a) Calculate the initial in-place gas per acre-foot of bulk reservoir rock.

(b) How many years will it take a well to deplete by 50% a 640 ac unit at the rateof 3 MM SCF/day?

(c) If the reservoir is under an active water drive so that the decline in reservoirpressure is and during the production of 50 4 MMM SCF of gaswater invades 1280 acres, what is the percentage of recovery by water drive?

(d) What is the gas saturation as a percentage of total pore space in the portionof the reservoir invaded by water'

3.13 Fifty billion standard cubic feet of gas has been produced from a dry gas reser-voir since its discovery The reservoir pressure during this production hasdropped to 3600 psia. Your company, which operates the field, has contractedto use the reservoir as a gas storage reservoir. A gas with a gravity of 0.75 is tobe injected until the average pressure reaches 4800 psia Assume that the reser-voir behaves volumetrically, and determine the amount of SCF of gas that mustbe injected to raise the reservoir pressure from 3600 to 4800 psia. The initialpressure and temperature of the reservoir were 6200 psia and 280°F. respec-tively, and the specific gravity of the reservoir gas is 0.75.

3.14 The production data for a gas field arc given below. Assume volumetric behaviorand calculate the fouowing:

(a) Determine the initial gas in place.

(b) What percentage of the initial gas in place will he recovered at a p/z of 1000?

(c) The field is to be used as a gas storage reservoir into which gas is injectedduring summer months and produced during the peak demand months ofthe winter. What is the minimum plz value that the reservoir needs to bebrought back up to if a supply of 50 MMM SCF of gas is required and theabandonment plz is 1000?

References 105

p/z (psia) - 0,, (MMM SC))

6553 0.3936468 1.6426393 3.2266329 4.2606246 5.5046136 7.5386080 8.749

3.15 Calculate the daily gas production including the condensate and water gas equiv-alents for a reservoir with the following daily production:

Separator gas production =6 MM SCF

Condensate production 100 STB

Stock tank gas production 21 M SCF

Fresh water production = 10 hbl

Initial reservoir pressure = 6000 psia

Current reservoir pressure — 2000 psia


Water vapor content of 6000 psia and 225°F =0 86 bbIIMM SCF

Condensate gravity = 50°API

REFERENCES

'Harold Vance, Elements of Petroleum Subsurface Engineering St Louis, MO Educa-tional Publishers, 1950.

2L. W. LeRoy, Subsurface Geologic Methods, 2nd ed Golden, CO: Colorado Schoolof Mines, 1950.

3W A Bruce and H. J Wclgc, "The Restored-State Method for Determination of Oilin Place and Connate Water," Drilling and Production Praclice, API (1947), 166—173

4H. H Kaveler, "Engineering Features of the Schuler Field and Unit Operation."Trans. AJME (1944), 155, 73.

5T. M. (icifen, D. R Parrish, G. W. Hayncs, and R. A. Morse, "Efficiency of GasDisplacement from Porous Media by Liquid Flooding." Trans. AIME (1952), 195, 37.

6R Agarwal, R Al-Ilussainy, and H J Ramcy, "The Importance of Water Influx inGas Reservoirs," Jour of Petroleum Technology (Nov 1965), 1336.

Matthcs, R F. Jackson, S Schuler, and 0. P. Marudiak, "Reservoir Evaluationand Deliverability Study, Bierwang Field, West Germany." Jour of Petrokum Tech-nology (Jan. 1973), 23

8T. L. Lutes, C. P. Chiang, M M. Brady, and R H Roscen, "Accelerated Blowdownof a Strong Water Dnve Gas Reservoir." SPE 6166 Presented at the 51st Annual FallMeeting of' the Society of Petroleum Engineers of AIME, New Orleans, Oct 3—6,1976


9D P. Arcano and Z Bassiouni, "The Technical and Economic 1-easibility of En-hanced Gas Recovery in the Eugene Island Field by Use of the Coproduction Tech-ruque," Jour, of Petroleum Technology (May 1987), 585.101). K Gold, W. D McCain, J W. Jennings, "An Improved Method for the Deter-mination of the Reservoir Gas Specific Gravity for Retrograde Gases," Jour, of Petro-leum Technology (July 1989)

"Eugene L. McCarthy, William L Boyd. and Lawrence S. Reid, "The Water VaporContent of Essentially Nitrogen-Free Natural Gas Saturated at Various Conditions ofTemperature and Pressure," Trans. AIME (1950), 189, 241—242

12D 1. Kau and M R. Tek, "Overview on Underground Storage of Natural Gas."Jour, of Petroleum Technology (June 1981), 943

13Chi U. Ikoku. Natural Gas Reservoir Engineering. New York: Wiley, 1984.

'4A P. Hollis, "Some Petroleum Engineering Considerations in the Changeover of theRough Gas Field to the Storage Mode," Jour of Petroleum Technology (May 1984),797

15D W. Harville and M F Hawkins, Jr, "Rock Compressibility and Failure asReservoir Mechanisms in Geopressured Gas Reservoirs." Jour. of Petroleum Tech-nology (Dec. 1969), 1528.

I. Fatt. "Compressibility of Sandstones at Low to Moderate Pressures," Bulletin,AAPC'J (1954), No 8, 1924

"J 0. Duggan, "The Anderson 'U—An Abnormally Pressured Gas Reservoir inSouth Texas," Jour, of Petroleum Technology (Feb 1972), 132.

'6W. 3. Bernard. "Reserves kstimation and Performance Prediction For GeopressuredReservoirs," Jour of Petroleum Science and Engineering (J987), 1, 15.

'9P N. Jogi, K. E. Gray, T. R. Ashman, and T. W. Thompson, "Compaction Mea-surements of Cores from the Pleasant Bayou Wells." Proc. 5th ConI. Geopressured-Geothermal Energy, Oct 13—15, 1981, Baton Rouge. LA

20K P. Sinha, M T Holland, T F Borschel, and J. P. Schatz, "Mechanical andGeological Characteristics of Rock Samples from Sweezy No. 1 Well at ParcperdueGeopressured/Geothermal Site "Report of Terra Tek Inc to Dow Chemical Co., U.SDepartment of Energy, 1981.11 B. P. Ramagost and F F. Farshad, "PIZ Abnormally Pressured Gas Reservoirs"Paper SPE 10125 presented at the SPE Fall Meeting, San Antonio, Oct. 5—7, 1981.

T. Bourgoyne. M. F. Hawkins, F. P. Lavaquial. and T. L. Wickenhauser, "ShaleWater as a Pressure Support Mechanism in Superpressure Reservoirs." Paper SPE3851, 1981.

Chapter 4

Gas-Condensate Reservoirs

1. iNTRODUCTION

Gas-condensate production may be thought of as being intermediate betweenoil and gas Oil reservoirs have a dissolved gas content in the range of zero(dead oil) to a few thousand cubic feet per barrel, whereas in gas reservoirsI bbl of liquid (condensate) is vaporized in 100,000 SCF of gas or more, andfrom which, therefore, a small or negligible amount of hydrocarbon liquid isobtained in surface separators Gas-condensate production is predominantlygas from which more or less liquid is condensed in the surface separators.hence the name gas-condensate the liquid is sometimes called by an oldername, dis:illaze, and also sometimes simply oil because it is an oil Gas-condensate reservoirs may be approximately defined as those that producelight-colored or colorless stock tank liquids with gravities above 45° API atgas-oil ratios in the range of 5000 to 100,000 SCF/bbl. Allen has pointed outthe inadequacy of classifying wells and the reservoirs from which they produceentirely on the basis of surface gas-oil ratios, for the classification of reser-voirs, as discussed in Chapter 1, properly depends on (a) the composition ofthe hydrocarbon accumulation and (b) the temperature and pressure of theaccumulation in the earth.*J

* Rcferenccs throughout the text are given at the end uf cach chapter

107


Table 4.1 presents the mole compositions and some additional propertiesof five single-phase reservoir fluids. The volatile oil is intermediate betweenthe gas-condensate and the black, or heavy, oil types. Production with gas-oilratios greater than 100,000 SCFfbbl is commonly called lean or dry gas, al-though there is no generally recognized dividing line between the two catc-gories. In some legal work, statutory gas wells are those with gas-oil ratios inexcess of 100,000 SCFIbbI (i.e., 10 bbllM MCF). The term we: gas is some-times used as being more or less equivalent to gas-condensate In the gas-oilratios, general trends are noticeable in the methane and heptanes-plus contentof the fluids and the color of the tank liquids. Although there is good corre-lation between the molecular weight of the heptanes-plus and the gravity ofthe stock tank liquid, there is virtually no correlation between the gas-oilratios and the gravities of the stock tank liquids, except that most black oilreservoirs have gas-oil ratios below 1000 SCFIbbl and stock tank liquid grav-ities below 450 API. The gas-oil ratios are a good indication of the overallcomposition of the fluid, high gas-oil ratios being associated with low concen-trations of pentanes and heavier, and vice versa.

The gas-oil ratios given in Table 4 1 are for the initial production of theone-phase reservoir fluids producing through one or more surface separatorsoperating at various temperatures and pressures, which may vary considerablyamong the several types of production. The gas-oil ratios and consequently theAPi gravity of the produced liquid vary with the number, pressures, andtemperatures of the separators so that one operator may report a somewhatdifferent gas-oil ratio from another, although both produce the same reservoirfluid. Also, as pressure declines in the black oil, volatile oil, and some gas-

TABLE 4.1.Mole composition and other properties of typical single-phase reservoir fluids

Black Volatile Gas-

Component Oil Oil - Condensate Dry Gas (Jas

C1 48 83 64.36 87.07 95.85 86.67C2 2 75 7 52 4.39 2.67 7.77

C3 1 93 4 74 2.29 0.34 2.95

C4 1.60 412 1.74 052 1.73115 297 0.83 008 0.88159 138 060 012

C; 4215 - 14.91 380 0.42

100.00 00 100 00 100.00 100.00Mol wt c 225 181 112 157

OOR, SCFIbbI 625 2000 18,200 105,000 In!

Tank gravity, °API 34 3 50 1 60.8 54.7Liquid color Greenish Medium light Water

black orange straw whitc

1. Introduction 109

condensate reservoirs, there is generally a considerable increase in the gas-oilratio owing to the reservoir mechanisms that control the relative flow of oiland gas to the well bores. The separator efficiencies also generally decline asflowing welihead pressures decline, which also contributes to increased gas-oilratios.

What has bccn said previously applies to reservoirs initially in a singlephase. The initial gas-oil ratios of production from wells completed either inthe gas cap or in the oil zone of two-phase reservoirs depends, as discussedpreviously, on the compositions of the gas cap hydrocarbons and the oil zonehydrocarbons as well as the reservoir temperature and pressure. The gas capmay contain gas-condensate or dry gas, whereas the oil zone may containblack oil or volatile oil. Naturally, if a well is completed in both the gas andoil zones, the production will be a mixture of the two Sometimes this isunavoidable, as when the gas and oil zones (columns) are only a few feet inthickness. Even when a well is completed in the oil zone only, the downwardconing of gas from the overlying gas cap may occur to increase the gas-oil ratioof the production.

When deeper dnlling began in many areas, the trend to discoveries wastoward reservoirs of the gas and gas-condensate types. Figure 4 1 based onwell test data reported in Ira Rinehart's Yearbooks shows the discovery trendfor 17 parishes in southwest Louisiana for 1952—1956 inclusive 2The reservoirswere separated into oil and gas or gas-condensate types on the basis of well test

Fig. 4.1. Discovery frequency of oil and gas or gas-condensate reservoirs versus depth.For 17 parishes in southwest Louisiana, 1952—1956. inclusive (Data from Ira Rjne-

Yearbooks.

NUMBER OF DISCOVERIESPER 1000 FT INTERVAL

110 Gas-Condensate Reservoirs

'C'I- 0

2 to /0JO — 0

1000I I CONDENSATEto ANOGAS

I \'5

1"0

tOO 1Q00 10,000 100.000GAS OIL RATIO, Cu FT/aBL

Fig. 4.2. Plot showing trend of increase of gas-oil ratio versus depth For 17 parishesin southwest Louisiana during 1955. (Data from Ira Rinehart's Yearbooks 2)

gas-oil ratios and the API gravity of the produced liquid. Oil discoveriespredominated at depths less than 8000 ft, but gas and gas-condensate discov-eries predominated below 10,000 ft. The decline in discoveries below 12,000ft was due to the fewer number of wells drilled below that depth rather thanto a drop in the occurrence of hydrocarbons. Figure 4.2 shows the same datafor the year 1955 with the gas-oil ratio plotted versus depth. The dashed linemarked "oil" indicates the general trend to increased solution gas in oil withincreasing pressure (depth), and the envelop to the lower right encloses thosediscoveries that were of the gas or gas-condensate types.

Muskat, Standing, Thornton, and Eilerts, have discussed the propertiesand behavior of gas-condensate reservoirs.3'6 Table 4.2, taken from Eilerts,shows the distribution of the gas-oil ratio and the API gravity among 172 gasand gas-condensate fields in Texas, Louisiana, and Mississippi.6 These authorsfound no correlation between the gas-oil ratio and the API gravity of the tankliquid for these fields.

2. CALCULATION OF INITIAL GAS AND OIL

The initial gas and oil (condensate) for gas-condensate reservoirs, both retro-grade and nonretrograde, may be calculated from generally available fielddata by recombining the produced gas and oil in the correct ratio to find the

2. Calculation of Initial Gas and Oil 111

TABLE 4.2.Range of gas-oil ratios and tank oil gravities for 172 gas and gas-condensatefields in Texas, Louisiana, and Mississippi

Range ofLiquid-Gas

Ratio, GPM'

Range ofGas-Oil Ratio- M SCF/bbl Texas

Number of I-ields

Louisiana Mississippi lbralPer Centof Total

<0.4 >105 38 12 7 57 331

04to0.8 52.5to105 33 18 4 55 3200.8 to 1.235.0 to 52.5 12 15 5 32 18.61.2tol.6 262to35.0 1 8 1 10 581.6to20 21.0to26.2 1 3 1 5 2.9

>2.0 <21.0 287

561

6

24

13

172

7.6

100.0

Range of StockTank Gravity,

"API

<4040—45

45—50

50—55

55—60

60—65

>65

24

12

23

2419

3

87

1

212

17

13

8

1

54

0007

12

3

2

24

- 36

2447

4930

6

165

1.83.6

14.628.5

29.718.2

3.6

100.0

'Gallons per 1000 SCF

average specific gravity (air = 1.00) of the total well fluid, which is presumablybeing produced initially from a one-phase reservoir. Consider the two-stageseparation system shown in Fig. 3.8. The average specific gravity of the totalwell fluid is given by Eq. (4.1):

where,

R1y1 + 4602•Yo + R3'13'Yw =

R1 + 133 6YoR3

(4.1)

R1, R3—producing gas-oil ratios from the separator (1) and stock tank (3)'yr—specific gravities of separator and stock tank gases'y0—specific gravity of the stock tank oil (water = 1.00). This is given by:


141.5

=(4.2)

weight of the stock tank oil that is given by Eq. (3.22):

= - = 42.43y0(3.22)

1•°°8'Yo

Example 4.1 shows the usc of Eq. (4.1) to calculate the initial gas and oil inplace per acre-foot of a gas-condensate reservoir from the usual productiondata. The three example problems in this chapter represent the type of calcu-lations that an engineer would perform on data generated from laboratorytests on reservoir fluid samples from gas-condensate systems. Sample reportscontaining additional example calculations may be obtained from commerciallaboratories which conduct PVT studies. The engineer dealing with gas-condensate reservoirs should obtain these sample reports to supplement thematerial in this chapter. The gas deviation factor at initial reservoir tempera-ture and pressure is estimated from the gas gravity of the recombined oil andgas as shown in Chapter 1. From the estimated gas deviation factor and thereservoir temperature, pressure, porosity, and connate water, the moles ofhydrocarbons per acre-foot can be calculated, and from this thc initial gas andoil in place.

Example 4.1. Calculate the initial oil and gas in place per acre-foot fora gas-condensate reservoir.

Given:

Initial pressure 2740 psia

Reservoir temperature 21 5°F

Average porosity 25%

Average connate water 30%

Daily tank oil 242 STB

Oil gravity, 60°F 48.0°APJ

Daily separator gas 3100 MCF

Separator gas gravity 0.650

Daily tank gas 120 MCF

Tank gas gravity 1.20

2. Calculation of Initial Gas and Oil 113

SoluTioN:

141.5=0.788

48.0 + 131.5

5954 5954= =

Po.AP1881148.0_8.811151 9

3,100,000R=242

= 12,810

R2= 496

12,180(0.650) + 4602.(0.788) + 496(1.20)= 0.896= 133,316(0.788)

+ 49612,810+151.9

FromFig. 1.4, 7 Then 1= 1.60andp,= 4.30, fromwhich, using Fig. 1.5, the gas deviation factor is 0.825 at the initial conditions.Then the total initial gas in place per acre-foot of bulk reservoir is

G = 379.4 pV = 379.4(2740)(43560)(0.25)(1 — 0.30)= 1326 MCFIac-ft

zR 'T 0.825(10.73)(675)

Because the volume fraction equals the mole fraction in the gas state, thefraction of the total produced on the surface as gas is

R1 R3

3794+3794R1

(4.3)

3794496 350(0.788)0.951

379.4 151.9


Then

Jnitial gas in place = 0.951(1326) = 1261 MCFIac-f:

Initial oil in place= 12,810 + 496

94 8 STBIac-fz

Because the gas production is 95.1% of the total moles produced, the totaldaily gas-condensate production in MCF is

AG,,—0951 0951

The total daily reservoir voidage by the gas law is

=3,386,000 = 19,450 cuftlday

The gas deviation factor of the total well fluid at reservoir temperatureand pressure can also be calculated from its composition. The composition ofthe total well fluid is calculated from the analyses of the produced gas(es) andliquid by recombining them in the ratio in which they are produced. When thecomposition of the stock tank liquid is known, a unit of this liquid must becombined with the proper amounts of gas(es) from the separator(s) and thestock tank, each of which has its own composition. When the compositions ofthe gas and liquid in the first or high-pressure separator are known, theshrinkage the separator liquid undergoes in passing to the stock tank must bemeasured or calculated in order to know the proper proportions in which theseparator gas and liquid must he combined. For example, if the volume factorof the separator liquid is 1.20 separator bbl per stock tank barrel and themeasured gas-oil ratio is 20,000 SCF c'f high-pressure gas per bbl of stock tankliquid, then the separator gas and liquid samples should be recombined in theproportions of 20,000 SCF of gas to 1.20 bbl of separator liquid, since 1.20 bblof separator liquid shrinks to 1.00 bbl in the stock tank.

Example 4 2 shows the calculation of initial gas and oil in place for agas-condensate reservoir from the analyses of the high pressure gas and liquid,assuming the well fluid to be the same as the reservoir fluid The calculationis the same as that shown in Ex. 4.1 except that the gas deviation factor of thereservoir fluid is found from the pseudoreduced temperature and pressure,

2. Calculation of Initial Gas and Cii 115

1400

1300

U.1200

1000

1100 -

500-J

I-0.400 -

300

200100 120 140 160 100 200 220 240

MOLECULAR WEIGHT

Fig. 4.3. Correlation charts for estimation of the pseudocritical temperature and pres-sure of hcptancs plus fractions from molecular weight and specific gravity (AfterMathews, Roland, and Katz, Proc NGAA.)

which arc determined from the composition of the total well fluid rather thanfrom its specific gravity. Figure 4 3 presents charts for estimating the pseudo-critical temperature and pressure of the hcptanes-plus fraction from its molec-ular weight and specific gravity.

Example 4.2. To calculate the initial gas and oil in place from the com-positions of the gas and liquid from the high-pressure separator.

Given:

Reservoir pressure 4350 psiaReservoir temperatureHydrocarbon porosity 17.4 per cent


15.025 psia, 60°F842,600 SCF/day

31.1 STB/day185.0

0.83430.7675

1.235 bbl at 880 psia/STB,both at 60°FCols. (2) and (3), Table 4.3

371.2 cu ft/mole.

Sot.LrrIoN: (column numbers refer to Table 4.3):1. Calculate the mole proportions in which to recombine the separator

gas and liquid. Multiply the molc fraction of each component in the liquid.Col. (3), by its molecular weight, Col. (4), and enter the products in Col. (5).The sum of Col. (5) is the molecular weight of the separator liquid, 127 48.Because the specific gravity of the separator liquid is 0.7675 at 880 psig and60°F, the moles per barrel is

0.7675 x 350 lb/bbl127.48 lb/mole

= 2.107 moles/bbl for the separator liquid

TABLE 4.3.Calculations for Example 4.2 on gas-condensate fluid

(1) (4) (5) (6) (7) (8)Moles

Mole (3) X (6) of Eachcomposition bbl of Componentof Separator Liquid Each in 59 Il

Fluids bbl/mole Component Moles of(2) (3) MoL (3) x (4,1 for Each per Mole of GasGas Liquid wi lb/mole Component sep liq. (2)x 59.1!

CO2 0 0120 0.0000 0 709C1 0.9404 02024 1604 3247 0.1317 002666 55.587

0 0305 0.0484 30.07 1 455 0.1771 0 00857 1.803

C3 00095 0.0312 4409 J.376 0.2480 0.00774 0562i—C4 00024 0.OJ 13 58.12 0.657 0 2948 0.00333 0 142fl—(.4 00023 0.0196 58 12 1139 0.2840 000557 0 136i—Cs 0.0006 0.0159 72.15 1147 0.3298 000524 0.035

0.0003 0.0170 72.15 1 227 0.3264 0.00555 0.018Ct, 00013 00384 86.17 3.309 03706 0.01423 0.077CC 0.0007 06158 1850 113.923 06336 039017 0.041 -

1 0000 10000 127480 046706 59 110

1185 lb/mole — (0 8343 X350 lb/bbl) = 0.6336 bbl/molc

Std. condSeparator gasStock tank oilMol. wt C4 in separator liquidSp. gr. in separator liquidSp. gr. separator liquid at 880 psig and 60°FSeparator liquid volume factor

Compositions of high-pressure gas and liquidMolar volume at 15 025 psia and 60°F

2 Calculation of Initial Gas and Oil 117

The separator liquid rate is 31.1 STB/day x 1.235 sep. bbl/STB so that theseparator gas-oil ratio is

842,60031.1

SCF sep. gas/bbl sep. liquid

Because the 21,940 SCF is 21,940/371.2, or 59.11 moles, the separator gas andliquid must be recombined in the ratio of 59.11 moles of gas to 2.107 molesof liquid.

If the specific gravity of the separator liquid is not available, the mole perbarrel figure may be calculated as follows Multiply the mole fraction of eachcomponent in the liquid, Col. (3), by its barrel per mole figure, Col. (6),obtained from data in Table 1.1 and enter the product in Cal. (7). The sum ofCol. (7), 0.46706 is the number of barrels of separator liquid per mole ofseparator liquid, and the reciprocal is 2.141 moleslbbl (versus 2.107 mea-sured).

2. Recombine 59.11 moles of gas and 2.107 moles of liquid. Multiply themole fraction of each component in the gas, Cot. (2), by 59.11 moles, andenter in Col. (8). Multiply the mole fraction of each component in the liquid,Cal. (3), by 2.107 moles, and enter in Col. (9). Enter the sum of the moles of

(9) (/0) (11) (/2) (13) (14) (15)Moles Moles

of Each of EachComponent Component Mole Partial

in 2.107 in 61 217 Composition Critical PartialMoles of Moles of Gas of Total Critical Pressure, Critical CriticalLiquid and Liquid Well Fluid Pressure, psia Temp. Temp. °R

(3) x2 107 (8) t- (9) (10) psia (II) X (12) °R (11) X (14)

0.0000 07090 0.0116 1070 1241 548 6360.4265 56.0135 09150 673 615.80 343 313850.1020 1.9050 0.0311 708 22.02 550 17.11

0.0657 0.6277 00102 617 629 666 6.790.0238 01658 00027 529 143 735 1.98

00413 01773 00029 550 160 766 2.220.0335 00685 00011 484 053 830 0.910.0358 0 0538 0 0009 490 044 846 0.760.0809 0 1579 0.0026 440 114 914 2.381.2975 1 3385 -

- 2.1070 61.2170

- 0.0219

1.0000

300' 6.57

66823

1227b 26.87

37923

bFrom Fig 4 3. after Mathews, Roland. and Katz for Mo! wt 185 and spgr


each component in the gas and liquid, Col. (8) plus Col. (9), in Cal. (10)Divide each figure in Cal (10) by the sum of Col. (10). 61.217, and enter thequotients in Cal. (11), which is the mole composition of the total well fluidCalculate the pseudocritical temperature, 379.23°R and pressure 668.23 psiafrom the composition. From the pseudocnticals, find the pseudoreduced crit-icals, and then the deviation factor at 4350 psia and 2 17°F, which is 0.963.

3. Find the gas and oil (condensate) in place per acre-foot of net reser-voir rock. From the gas law, the initial moles per acre-foot at 17 4% hydro-carbon porosity is:

pV 4350 x (43,560 x 0.174)zRT O.963x I0.73x677 =4713 moles/ac-ft

59.11Gas mole fraction

= 1070.966

0.966 x 4713 x 371.2Initial gas in place

= 1000= 1690 MCF/ac-ft

(1—0 966) x 4713Initial oil in place = =61 6 STB/ac-ft

2.lOlx 1.235

Because the high-pressure gas is 96.6% of the total mole production, the dailygas-condensate production expressed in standard cubic feet is

AG — Daily hp gas — 842,600— 872 200 SCF/d

0.966 -— 0.966' ay

The daily reservoir voidage at 4350 psia is

= 872,200 X x 0.963 = 3777 cu ft/day

3. THE PERFORMANCE OF VOLUMETRIC RESERVOIRS

rhe behavior of single-phase gas and gas-condensate reservoirs has beentreated in Chapter 3. Since no liquid phase develops within the reservoir,where the temperature is above the cricondentherni, the calculations arcsimplified When the reservoir temperature is below the cricondentherm,however, a liquid phase develops within the reservoir when pressure declinesbelow the dew point owing to retrograde condensation, and the treatment isconsiderably more complex, even for volumetric reservoirs.

One solution is to duplicate closely the reservoir depletion by Laboratorystudies on a representative sample of the initial, single-phase reservoir fluid.The sample is placed in a high-pressure cell at reservoir temperature and

3 The Performance of Volumetric Reservoirs 119

initial reservoir pressure. During the depletion, the volume of the cell is heldconstant to duplicate a volumetric reservoir, and care is taken to remove onlygas-phase hydrocarbons from the cell because for most reservoirs, the retro-grade condensate liquid that forms is trapped as an immobile liquid phasewithin the pore spaces of the reservoir.

Laboratory experiments have shown that with most rocks the oil phaseis essentially immobile until it builds up to a saturation in the range of 10 to20% of the pore space, depending on the nature of the rock pore spaces andthe connate water. Because the liquid saturations for most retrograde fluidsseldom exceed 10%, this is a reasonable assumption for most retrogradecondensate reservoirs. In this same connection, it should be pointed out thatin the vicinity of the well bore retrograde liquid saturations often build up tohigher values so that there is two-phase flow, both gas and retrograde liquid.This buildup of liquid occurs as the one-phase gas suffers a pressure drop asit approaches the well bore. Continued flow increases the retrograde liquidsaturation until there is liquid flow. Although this phenomenon does not affectthe overall performance seriously or enter into the present performancepredictions, it can (a) reduce, sometimes seriously, the flow rate of gas-condensate wells and (b) affect the accuracy of well samples taken assumingone-phase flow into the well bore.

The continuous depletion of the gas phase (only) of the cell at constantvolume can be closely duplicated by the following more convenient technique.The content of the cell is expanded from the initial volume to a larger volumeat a pressure a few hundred psi below the initial pressure by withdrawingmercury from the bottom of the cell, or otherwise increasing the volume. Timeis allowed for equilibrium to be established between the gas phase and theretrograde liquid phase that has formed, and for the liquid to drain to thebottom of the cell so that only gas-phase hydrocarbons are produced fromthe top of the cell. Mercury is injected into the bottom of the cell and gas isremoved at the top at such a rate as to maintain constant pressure in the cell.Thus the volume of gas removed, measured at this lower pressure and cell(reservoir) temperature. equals the volume of mercury injected when thehydrocarbon volume, now two-phase, is returned to the initial cell volume.The volume of retrograde liquid is measured, and the cycle—expansion to anext lower pressure followed by the removal of a second increment of gas—isrepeated down to any selected abandonment pressure. Each increment of gasremoved is analyzed to find its composition, and the volume of each incrementof produced gas is measured at subatmospheric pressure to determine thestandard volume, using the ideal gas law. From this, the gas deviation factorat cell pressure and temperature may be calculated using the real gas law.Alternatively, the gas deviation factor at cell pressure and temperature maybe calculated from the composition of the increment.

Figure 4.4 and Table 4.4 give the composition of a retrograde gas-condensate reservoir fluid at initial pressure and the composition of the gas


0.4

z0I-

0.2

Ii.w-J0

0.1

1.080.06

0.04

removed from a PVT cell in each of five increments, as previously described.Table 4.4 also gives the volume of retrograde liquid in the cell at each pressureand the gas deviation factor and volume of the produced gas increments at cellpressure and temperature.

The liquid recovery from the gas increments produced from the cell maybe measured by passing the gas through small-scale separators, or it may becalculated from the composition for usual field separation methods, or forgasoline plant methods 89 10 recovery of the pentanes-plus is somewhatgreater in gasoline plants than is obtained by field separation, and muchgreater for the propanes and butanes, commonly called liquified petroleumgas (LPG). For simplicity, the liquid recovery from the gas increments ofTable 4.4 is calculated in Ex. 4.3 assuming 25% of the butanes, 50% of thepentancs, 75% of the hexane, and 100% of the heptanes-plus is recovered asliquid

Example 4.3. Calculating the volumetric depletion performance of aretrograde gas-condensate reservoir based on the laboratory tests given inTable 4 4.

1008

0.6

. Ct

IIC4

C?.

.010 500 1000 1500 2000PRESSURE, PSIA

3000

Fig. 4.4 Variations in the composition of the produced gas phase material of a retro-grade gas-condensate fluid with pressure decline (Data from Table 4.4.)

TA

BLE

4.4

.V

olum

e, c

ompo

sitio

n, a

nd g

as d

evia

tion

fact

ors

for

a re

trog

rade

con

dens

ate

fluid

(1)

(4)

('6k)

(8)

(11)

(12)

Pro

duce

dR

etro

grad

eR

etro

grad

eC

ompo

sitio

n of

Pro

duce

dIn

crem

ents

,G

as, C

u cm

Liqu

idV

olum

e,G

as D

evia

tion

Mol

e F

ract

ion

at 1

95 0

fV

olum

e, c

u cm

Per

Cen

t of

Fac

tor

atP

ress

ure

and

Cel

lC

ell V

olum

e,H

ydro

carb

on19

5°F

and

-c2

C3

c4c6

c;P

ress

ure

947.

5 cu

cm

Vol

ume

cell

Pre

ssur

e

2960

0.75

20.

077

0.04

40.

03!

0.02

20.

022

0.05

20.

00.

00

00.

771

2500

0.78

30.

077

0.04

30.

028

0.01

90.

016

0.03

417

5.3

62.5

6.6

0.79

420

000.

795

0078

0042

0.02

70.

017

0.01

40.

027

227.

077

.782

0805

1500

0.79

80.

079

0.04

20.

027

0.01

30.

025

340.

475

.07.

90.

835

1000

0.79

30.

080

0 04

30.

028

0.01

70.

013

0.02

654

4.7

67.2

7.1

0 87

550

00.

768

0.08

200

480.

033

0.02

10.0

150.

033

1080

.756

.96.

009

45

C,,

CD -U CD 0 3 CD 2 a C 3 CD C)

CD


Given

Initial pressure (dew point) 2960 psiaAbandonment pressure 500 psiaReservoir temperature 195°FConnate water 30%Porosity 25%Standard conditions 14.7 psia and 60°FInitial cell volume 947.5 cu cmMol. wt. of C in Initial Fluid 114 lb/lb-moleSp. gr of in Initial Fluid 0.755 at 60°F

Compositions, volumes, and deviation factors given in Table 4.4.

Assume: The same mol. wt. and sp. gr. for the C4 content for all producedgas.

Assume: Liquid recovery from the gas is 25% of the butanes, 50% of thepentane, 75% of the hexane, and 100% of the heptanes and heavier.

SoLuTioN: (column numbers refer to Table 4.5):

1. Calculate the increments of gross production in M SCF per ac-ft ofnet, bulk reservoir rock. Enter in Col. (2):

= 43,560 x 0.25 x (1 —0.30) = 7623 cu ft/ac-ft

For the increment produced from 2960 to 2500 psia, for example,

= 7623 xCU

= 1410 cii ft/ac-ft at 2500 psia and 195°F947.5 cu cm

379.4x2500x 1410 —

— 1000 zRT 1000 x 0.794 x 10.73 x 655— 240.1M SCF

Find the cumulative gross gas production, = Z and enter in Col. (3).

2. Calculate the M SCF of residue gas and the barrels of liquid obtainedfrom each increment of gross gas production. Enter in Col. (4) and Col. (6).Assume that 0.25 C4, 0.50 C5, 0.75 C6, and all C is recovered as stock tankliquid For example, in the 240.1 M SCF produced from 2960 to 2500 psia, themole fraction recovered as liquid is

ian, =025 X0028+0.50x0019+0.75 xO.OJ.6+0.034

= 0.0070 + 0.0095 + 0.0120 + 0.034 = 0.0625 mole fraction

— — UI 0 UI C UI C 00000

C,-'

increments of Gross Gas m 00 00' C Production, M SCF

a C,

Cumulative C,ross Gus 0 1%J tIi 00 'C 0 Ui 00 ... Production, M SCF, )..(2) !.

(I)

CD

UI Ui i,j Residue Gas in Each CD

b increment, M SCF

CD

a) 14

o UI Cumulative Residue Gas

— Production, M SCF

Liquid in Lach increment, bbl

m 'C CD

UI — Cumulative Liquid UI 00 Ui 0 0' Production, bhl

CD

Average Gas-Oil Ratio of '° Each increment, SCF

Ui Ui UJ 0 0 Residue Gas per bbl, 000000 (4) — (6)

Cumulative Gross 000' t.J — 1 0 '4 0 UI 0 Gas Recovery, percentage

(3)xJ001J580

Cumulative Residue I ui o Gas Recovery, percentage

(5)xJOOfJ44J

UI - — Cumulative Liquid

I o 0° o Recovery, percentage (7)x10011432

siIoAJaseH 3!4GWflIOA Jo aqj


As the mole fraction also equals the volume fraction in gas, the M SCErecovered as liquid from 240.1 M SCF is

= 0.0070 x 240 1 + 0.0095 x 240.1 ÷ 0.0120 x 240.1 + 0.034 x 240.1

= 1.681 + 2.281 4 2.881 + 8.163 = 15 006M SCF

The gas volume can be converted to gallons of liquid using the gal/M SCFfigures of Tablc 1.1 for C4, C3 and C6. The average of the iso and normalcompounds is used for C4 and C5.

For

114 lb/lb-mole0.3794M SCF/lb-molex8.337 lb/gal x 0.755

= 47.71 gal/M SC!-

0.3794 is the molar volume at standard conditions of 14.7 psia and 60°F.Then the total liquid recovered from 240.IM SCF is 1.681 x 32.04 + 2.281 x36.32 + 2.881 x 41.03 + 8 163 x 47.71 = 53.9 + 82.8 + 118.2 + 389.5 = 644.4gal 15.3 bbl The residue gas recovered from the 240.1 M SCF is 240.1 x(1 — 0.0625) 225.1 M SCF. Calculate the cumulative residue gas and stocktank liquid recoveries from Cols (4) and (6) and enter in Cols. (5) and (7),respectively.

3. Calculate the gas-oil ratio for each increment of gross production inunits of residue gas per barrel of liquid. Enter in Col. (8) For example, thegas-oil ratio of the incrcment produced from 2960 to 2500 psia is

1000= 14,700 SCF/bbl

4. Calculate the cumulative recovery percentages of gross gas, residuegas, and liquid. Enter in Cols. (9), (10), and (11) The initial gross gas inplace is:

379.4pV 379.4 x2960x76231000 zRT = 1000 x 0.771 x 10.73 x 655

= 1580 M SCF/ac-ft

Of this. the liquid mole fraction is 0.088 and the total liquid recovery is 3.808gal/M SCF of gross gas, which are calculated from the initial composition inthe same manner as shown in Part 2. Then

G (1 — 0.088) x 1580 = 1441 M SCF residue gaslac-ft

3.808x 1580N

42= 143.2 bbllac-ft

3 The Performance of Volumetric Reservoirs 125

At 2500 psia, then

bOx 240.1Grossgasrecovery= =15.2%

100x2251Residue gas recovery

== 15.6%

bOx 153Liquid recovery

= 143.2= 10.7%

The results of the laboratory tests and calculations of Ex 4.3 are plottedversus pressure in Fig. 4.5. The gas-oil ratio rises sharply from 10,060 SCFIbbIto about 19,000 SCFIbb1 near 1600 psia. Maximum retrograde liquid andmaximum gas-oil ratios do not occur at the same pressure because, as pointedout prcviously, the retrograde liquid volume is much larger than its equivalentobtainable stock tank volume, and there is more stock tank liquid in 6.0%retrograde liquid volume at 500 psia than in 7 9% at 1500 psia. Revaporizationbelow 1600 psia helps to reduce the gas-oil ratio; however, there is some doubtthat revaporization equilibrium is reached in the reservoir, the field gas-oilratios generally remain higher than that predicted Part of this is probably aresult of the lower separator efficiency of liquid recovery at the lower pressureand higher separator temperatures. Lower separator temperatures occur athigher welihead pressures owing to the greater cooling of the gas by freeexpansion in flowing through the choke. Although the overall recovery at 500psia abandonment pressure is 80.4%, the liquid recovery is only 50.0%, owingto retrograde condensation. The cumulative production plots are slightlycurved because of the vanation in the gas deviation factor with pressure andwith the composition of the reservoir fluid.

The volumetric depletion performance of a retrograde condensate fluid,such as given in Example 4.3, may also be calculated from the initial com-position of the single-phase reservoir fluid, using equilibrium ratios. An equi-librium ratio (K) is the ratio of the mole fraction (y) of any component in thevapor phase to the mole fraction (x) of the same component in the liquidphase, or K = y/x. These ratios depend on the temperature and pressure, and,unfortunately, on the composition of the system. If a set of equilibrium ratioscan be found that are applicable to a given condensate system, then it ispossible to calculate the mole distribution between the liquid and vapor phasesat any pressure, and reservoir temperature, and also, the composition of theseparate vapor and liquid phases, as shown in Fig. 4.4. From the compositionand total moles in each phase, it is also possible to calculate with reasonabieaccuracy the liquid and vapor volumes at any pressure.

The prediction of retrograde condensate performance using equilibriumratios is a specialized technique. Standing and Rodgers, Harrison, and Regiergive methods for adjusting published equilibrium ratio data for condensate


Fig. 4.5. Gas-oil ratios, retrograde liquid volumes, and recovenes for the depletionperformance of a retrograde gas-condensate reservoir (Data from Tables 4.4 and 4.5.)

systems to apply to systems of different compoiitions.4 1112.13.14 They alsogive step-by-step calculation methods for volumetric performance that consistof starting with a unit volume of the initial reservoir vapor of known com-position. An increment of vapor phase material is assumed to be removedfrom the initial volume at constant pressure, and the remaining fluid is ex-panded to the initial volume. The final pressure, the division in volume be-tween the vapor and retrograde liquid phases, and the individual compositionsof the vapor and liquid phases are then calculated using the adjusted equilib-rium ratios. A second increment of vapor is removed at the lower pressure,and the pressure, volumes, and compositions are calculated as before. Anaccount is kept of the produced moles of each component so that the totalmoles of any component remaining at any pressure is known by subtractionfrom the initial amount. This calculation may be continued down to anyabandonment pressure, just as in the laboratory technique.

Standing points out that the prediction of condensate reservoir per-formance from equilibrium ratios alone is likely to be in considerable errorand that some laboratory data should be available to check the accuracy of theadjusted equilibrium ratios.' Actually, the equilibrium ratios are changingbecause the composition of the system remaining in the reservoir or cellchanges. The changes in the composition of the heptanes-plus particularly

PRESSURE, PS1A

4. Use of Material Balance 127

affect the calculations. Rodgers, Hamson, and Regier point out the need forimproved procedures for developing the equilibrium ratios for the heavierhydrocarbons to improve the overall accuracy of the calculation.a

Jacoby, Koeller, and Berry studied the phase behavior of eight mixturesof separator oil and gas from a lean gas-condensate reservoir at recombinedratios in the range of 2000 to 25,000 SCFIbbI and at several temperatures inthe range of 100 to 250°F.'5 The results arc useful in predicting the depletionperformance of gas-condensate reservoirs for which laboratory studies are notavailable. They show that there is a gradual change in the surface productionperformance from the volatile oil to the nch gas-condensate type of reservoirand that a laboratory examination is necessary to distinguish between thedew-point and bubble-point reservoirs in the range of 2000 to 6000 SCFIbb1gas-oil ratios.

4. USE OF MATER1AL BALANCE

The laboratory test on the retrograde condensate fluid in Ex. 4.3 is itself amaterial balance study of the volumetric performance of the reservoir fromwhich the sample was taken. The application of the basic data and the calcu-lated data of Ex 4.3 to a volumetric reservoir is straightforward. For example,suppose the reservoir had produced 12.05 MMM SCF of gross well fluid whenthe average reservoir pressure declined from 2960 psia initial to 2500 psia.According to Table 4.5 the recovery at 2500 psia under volumetric depletionis 15.2% of the initial gross gas in place, and therefore the initial gross gas inplace is

G = 12.05x= 79.28MMM SCF

Because Table 4.5 shows a recovery of 80.4% down to an abandonmentpressure of 500 psia, the initial recoverable gross gas or the initial reserve is

Initial reserve = 79.28 x x 0.804 63.74 MMM SCF

Since 12.O5MMM SCF has already been recovered, the reserve at 2500 psia is

Reserve at 2500 psia = 63.74— 12.05 = 51.69 MMM SCF

The accuracy of these calculations depends, among other things, on the sam-pling accuracy and the degree of which the laboratory test represents thevolumetric performance. Generally there are pressure gradients throughout areservoir that indicate that the various portions of the reservoir are in varyingstages of depletion. This is due to greater withdrawals in some portions and/orto lower reserves in some portions, because of lower porosities and/or lower


net productive thicknesses. As a consequence the gas-oil ratios of the wellsdiffer, and the average composition of the total reservoir production at anyprevailing average reservoir pressure does not exactly equal the compositionof the total cell production at the same pressure.

Although the gross gas production history of a volumetric reservoirfollows the Jaboratory tests more or less closely, the division of the productioninto residue gas and liquid follows with less accuracy. This is due to thedifferences in the stage of depletion of various portions of the reservoir, asexpiained in the preceding paragraph, and also to the differences between thecalculated liquid recoveries in the laboratory tests and the actual efficiency ofseparators in recovering liquid from the fluid in the field.

The previous remarks apply only to volumetric single-phase gas-con-densate reservoirs. Unfortunately, most retrograde gas-condensate reservoirsthat have been discovered are initially at their dew-point pressures, ratherthan above it. This indicates the presence of an oil zone in contact with thegas-condensate cap. The oil zone may be negligibly small or commensuratewith the siLe of the cap; or it may be much larger. The presence of a small oilzone affects the accuracy of the calculations based on the single-phase studyand is more serious for a larger oil zone. When the oil zone is of a size at allcommensurate with the gas cap. the two must be treated together as a two-phase reservoir, as explained in Chapter 6

Many gas-condensate reservoirs are produced under a partial or totalwater drive When the reservoir pressure stabilizes or stops declining, asoccurs in many reservoirs, recovery depends on the value of the pressure atstabilization and the efficiency with which the invading water displaces the gasphase from the rock. The liquid recovery is lower for the greater retrogradecondensation, because the retrograde liquid is generally immobile and istrapped together with some gas behind the invading water front This situationis aggravated by permeability variations because the wells become "drowned"and are forced off production before the less permeable strata are depleted.In many cases, the recovery by water drive is less than by volumetric per-formance, as explained in Chapter 3, Sect. 4.

When an oil zone is absent or negligible, the material balance Eq. (3.15)may be applied to retrograde reservoirs under both volumetric and water-drive performance. just as for the single-phase (nonretrogradc) gas reservoirsfor which it was developed:

(3.15)

This equation may be used to find either the water influx W, or the initial gasin place G The equation contains the gas deviation factor z at the lowerpressure. It is included in the gas volume factor Bg in Eq. 3.15. Because thisdeviation factor applies to the gas-condensate fluid remaining in the reservoir,when the pressure is below the dew-point pressure in retrograde reservoirs, It

4 Use of Material Balance 129

is a two-phase gas deviation factor. The actual volume in Eq. (1.6) includes thevolume of both the gas and liquid phases, and the ideal volume is calculatedfrom the total moles of gas and liquid, assuming ideal gas behavior. For volu-metric performance, this two-phase deviation factor may be obtained fromsuch laboratory data as obtained in Example 4.3. For example, from the dataof Table 4.6 the cumulative gross gas production down to 2000 psia is 485.3MSCF/ac-ft out of an initial content of 1580.OM SCFIac-ft. Since the initialhydrocarbon pore volume is 7623 cu ft/ac-ft, Ex. 4.3, the two-phase volumefactor for the fluid remaining in the reservoir at 2000 psia and 195°F as calcu-lated using the gas law is

— 379.4 x pV — 379.4 x 2000 x 7623 —Z

— (G — G,)R'T (1580.0—485.3)103x 10.73 x752

Table 4.6 gives the two-phase gas deviation factors for the fluid remaining inthe reservoir at pressures down to 500 psia, calculated as before for the gas-condensate fluid of Ex. 4.3. These data arc not strictly applicable when thereis some water influx because they are based on cell performance in whichvapor equilibrium is maintained between all the gas and liquid remaining inthe cell, whereas in the reservoir some of the gas and retrograde liquid areenveloped by the invading water and are prevented from entering into equi-librium with the hydrocarbons in the rest of the reservoir The deviationfactors of Col (4), Table 4.6, may be used with volumetric reservoirs and, withsome reduction in accuracy, with water-drive reservoirs.

When laboratory data such as given in Ex. 4.3 have not been obtained,the gas deviation factors of the initial reservoir gas may be used to approxi-

TABLE 4.6.Two-phase and single-phase gas deviation factors for the retrogradegas-condensate fluid of Example 43

0) — (4) (6)Gas Deviation Faciors

Pressure,pski M SCI-Iac-fr

(G —

M SCI-iac-ft Two-phase"InitialGa? -

ProducedGasa

2960 00 1580.0 0771 0780 0.7712500 240.1 l339.9 0 768 0.755 0 7942000 485.3 1094.7 0 752 0.755 0.8051500 751.3 828.7 0745 0.790 0.8351000 1022 1 557.9 0 738 0.845 0.875500 - 1270.8 309.2 0 666 0.920 0.945

$ Data from Table 4 5 and Example 4 3

bCalculated from the data of rable 4 5 and Example 4 3

rCalculatcd from initial gas composition using correlation charts


mate those of the remaining reservoir fluid. These arc best measured in thelaboratory but may be estimated from the initial gas gravity or well-streamcomposition using the pseudorcduccd correlations. Although the measureddeviation factors for the initial gas of Ex. 4 3 are not available, it is believedthat they arc closer to the two-phase factors in Cal. (4) than those given in Cal.(5) of Table 4.6, which arc calculated using the pseudoreduced correlations,since the latter method presumes single-phase gases. The deviation factors ofthe produced gas phase arc given in Cal. (6) for comparison.

5. COMPARISON BETWEEN THE PREDiCTEDAND ACTUAL PRODUCTION HISTORIES OFVOLUMETRIC RESERVOIRS

Allen and Roe have reported the performance of a retrograde condensatereservoir that produces from the Bacon Lime Zone of a field located in EastTexas.13 The production history of this reservoir is shown in Figs. 4.6 and 4.7.The reservoir occurs in the lower Glen Rose Formation of Cretaceous age ata depth of 7600 ft (7200 ft subsea) and comprises some 3100 acres. It iscomposed of approximately 50 ft of dense, crystalline, fossiliferous dolomitewith an average permeability of 30 to 40 millidarcys in the more permeablestringers, and an estimated average porosity of about 10%. Interstitial water

Iii

W

z

Lu0

Lu

C,dO

—i ——ER OF-0---

WELLS

—

— — — — —— -.— —— —-- —— - —

— ——-

GASRATE

— --——

—

a PRESSURE

m11j

z

I-

942 1943 1944 1945.YEARS

1946 947 1948 1949

Fig. 4.6. Production history of thc Bacon Lime Zone of an eastern Tcxas gas-condensate reservoir. (After Allcn and Roe.'3 Trans AIME)

5 Comparison Between the Predicted and Actual Production Histones 131

a.

U0IA.'doI-4w4

2 Id>4

Fig. 4.7. Calculated and measured pressure and p/z values versus cumulative gross gasrecovery from the Bacon Lime Zone of an eastern Texas gas-condensate reservoir.(After Allen and Roe,13 Trans AIME.)

is approximately 30%. The reservoir temperature is 220°F, and the initialpressure was 3691 psia at 7200 ft subsea. Because the reservoir was veryheterogeneous regarding porosity and permeability, and, because very poorcommunication between wells was observed, cycling (Sect. 4.6) was not con-sidered feasible. The reservoir was therefore produced by pressure depletion.using three-stage separation to recover the condensate. The recovery at 600psia was 20.500 MM SCF and 830.000 bbl of condensate, or a cumulative(average) gas-oil ratio of 24,700 SCF/bbl, or 1.70 GPM (gallons per MCF).Since the initial gas-oil ratios were about 12,000 SCF/bbl (3.50 GPM), thecondensate recovery of 600 psia was 100 x 1 7/3.5, or 48.6% of the liquidoriginally contained in the produced gas Theoretical calculations based onequilibrium ratios predicted a recovery of only 1.54 GPM (27,300 gas-oilratio), or 44% recovery, which is about 10% lower.

The difference between the actual and predicted recoveries may havebeen due to sampling errors. The initial well samples may have been deficientin the heavier hydrocarbons owing to retrograde condensation of liquid fromthe flowing fluid as it approached the well bore (Sect. 4.3). Another possibilitysuggested by Allen and Roe is the omission of nitrogen as a constituent of thegas from the calculations. A small amount of nitrogen, always below I mole

4 6 8 10 2 4 16 18

CUMULATIVE WET GAS RECOVERY-MILLION STANDARD MCF


was found in several of the samples during the life of the reservoir. Finally,they suggested the possibility of retrograde liquid flow in the reservoir toaccount for a liquid recovery higher than that predicted by their theoreticalcalculations, which presume the immobility of the retrograde liquid phase.Considering the many variables that influence both the calculated recoveryusing equilibrium ratios and the field performance, the agreement betweenthe two appears good.

Figure 4.8 shows good general agreement between the butanes-pluscontent calculated from the composition of the production from two wells andthe content calculated from the study based on equilibrium ratios. The liquidcontent expressed in butanes-plus is higher than the stock tank GPM (Fig. 4.7)because not all the butanes, or, for that matter, all the pentanes-plus arerecovered in the field separators The higher actual butanes-plus content downto 1600 psia is undoubtedly a result of the same causes given in the precedingparagraph to explain why the actual overall recovery of stock tank liquidexceeded the recovery based on equilibrium ratios. The stock tank GPM inFig. 4.7 shows no revaporization; however, the weil-stream compositions be-low 1600 psia in Fig. 4 8 clearly show revaporization of the butanes-plus, andtherefore certainly of the pentanes-plus, which make up the majority of theseparator liquid. The revaporization of the retrograde liquid in the reservoirbelow 1600 psia is evidently just about offset by the decrease in separatorefficiency at lower pressures.

Figure 4.8 also shows a comparison between the calculated reservoirbehavior based on the differential process and the flash process. In the differ-ential process, only the gas is produced and is therefore removed from contact

Fig. 4.8. Calculated and measured butanes-plus in the well streams of the Bacon LimeZone of an eastern Texas gas-condensate reservoir (After Allen and Roe,13 TransAIME)

5 Comparison Between the Predicted and Actual Production Histones 133

with the liquid phase in the rcscrvoir. In the flash process, all the gas remainsin contact with the retrograde liquid, and for this to be so the volume of thesystem must increase as the pressure declines. Thus the differential process isone of constant volume and changing composition, and the flash process is oneof constant composition and changing volume. Laboratory work and calcu-lations based on equilibrium ratios are simpler with the flash process. wherethe overall composition of the system remains constant; however, the reservoirmechanism for the volumetric depletion of retrograde condensate reservoirsis essentially a differential process. The laboratory work and the use of equi-librium ratios discussed in Sect. 4.3 approaches the differential process by aseries of step-by-step flash processes. Figure 4.8 shows the close agreementbetween the flash and differential calcuiations down to 1600 psia. Below 1600psia, the well performance is closer to the differential calculations because thereservoir mechanism largely follows the differential process, provided thatonly gas phase materials arc produccd from the reservoir (i.e., the retrogradeliquid is immobile).

Figure 4.9 shows the good agreement between the reservoir field dataand the laboratory data for a small (one well), noncommercial, gas-condensateaccumulation in the Paradox limestone formation at a depth of 5775 ft in San

-jmm

Ida.II.U

0

4

(I,4U

Fig. 4.9. Comparison of field and laboratory data for a Paradox limestone gas-condensate reservoir in Utah (After Rodgers, 1 Iarrison, and Regier,8 courtesyAIME.)

PRESSURE — PS1A


Juan County, Utah. This afforded Rodgers. Harrison, and Regicr a uniqueopportunity to compare laboratory PVT studics and studies based on equi-librium ratios with actual field depiction under closely controlled and ob-served conditions 8 In the laboratory. a 4000 cu cm cell was charged withrepresentative well samples at reservoir temperature and initial reservoir pres-sure. The cell was pressure depleted so that only the gas phase was removed,and the produced gas was passed through miniature three-stage separators,which were operated at optimum field pressures and temperatures. The calcu-lated performance was also obtained, as explained previously, from equationsinvolving equilibrium ratios, assuming the differential process. Rodgers et ai.concluded that the model laboratory study can adequately reproduce andpredict the behavior of condensate reservoirs. Also, they found that the per-formance can be calculated from the composition of the initial reservoir fluid,provided representative equilibrium ratios are available.

Table 4.7 shows a comparison between the initial compositions of theBacon Lime and Paradox limestone formation fluids. The lower gas-oil ratiosfor the Bacon Lime are consistent with the much larger concentration of thepentanes and heavier in the Bacon Lime fluid.

6. LEAN GAS CYCLING AND WATER DRIVE

Because the liquid content of many condensate reservoirs is a valuable andimportant part of the accumulation, and because through retrograde conden-sation a large fraction of this liquid may be left in the reservoir at abandon-ment, the practice of lean gas cycling has been adopted in many condensatereservoirs. In gas cycling, the condensate liquid is removed from the produced

TABLE 4.7.Comparison of the compositions of the initial fluids in the Bacon Lime andParadox formations

Bacon 1.i,ne Paradox Formation

Nitrogen 17 0.0099Carbon dioxide 0.0135 0 0000Methane 0 7690 0 7741

Ethanc 0 0770 0 1148Propane 0 0335 0.0531Butane 0.0350 0.0230Pentane 0.0210 0.0097Hexane 0.0150 0.0054Heptane-plus 0.0360

1.00000.0100J.0000

Mol wt C 130 116.4Sp. gr. C.,' (60°!) 0.7615 0.7443

6. Lean Gas Cycling and Water Drive 135

(wet) gas, usually in a gasoline plant, and the residue, or dry gas, is returnedto the reservoir through injection wells. The injected gas maintains reservoirpressure and retards retrograde condensation. At the same time, it drives thewct gas toward the producing wells. Because the removed liquids representpart of the wet gas volume, unless additional dry (make-up) gas is injected,reservoir pressure will decline slowly. At the conclusion of cycling (i.e., whenthe producing wells have been invaded by the dry gas), the reservoir is thenpressure depleted (blown down) to recover the gas plus some of the remainingliquids from the portions not swept.

Although lean gas cycling appears to be an ideal solution to the retro-grade condensate problem, a number of practical considerations make it lessattractive First, there is the deferred income from the sale of the gas, whichmay not be produced for 10 to 20 years. Second, cycling requires additionalexpenditures, usually some more wells, a gas compression and distributionsystem to the injection wells, and a liquid recovery plant Third, it must berealized that even when reservoir pressure is maintained above the dew point,the liquid recovery by cycling may be considerably less than 100%

Cycling recoveries may be broken down into three separate recoveryfactors, or efficiencies. When dry gas displaces wet gas within the pores of thereservoir rock, the microscopic displacement efficiency is in the range of 70 to90%. Then, owing to the location and flow rates of the production and injec-tion wells, there are areas of the reservoir that are not swept by dry gas at thetime the producing wells have been invaded by dry gas, resulting in sweepefficiencies in the range of 50 to 90% (i.e., 50 to 90% of the initial pore volumeis invaded by dry gas). Finally, many reservoirs are stratified in such a way thatsome stringers are much more permeable than others, so that the dry gassweeps through them quite rapidly. Although much wet gas remains in thelower permeability (tightcr) stringers, dry gas will have entered the producingwells in the more permeable stringers, eventually reducing the liquid contentof the gas to the plant to an unprofitable level.

Now suppose a particular gas-condensate reservoir has a displacementefficiency of 80%, a sweep efficiency of 80%, and a permeability stratificationfactor of 80%. The product of these separate factors given an overall con-densate recovery by cycling of 51 .2%. Under these conditions, cycling may notbe particularly attrative, because retrograde condensate losses by depictionperformance seldom exceed 50%. However, during pressure depletion (blowdown) of the reservoir following cycling, some additional liquid may be recov-ered from both the swept and unswept portions of the reservoir. Also, liquidrecovenes of propane and butane in gasoline plants are much higher thanthose from stage separation of low-temperature separation, which would beused if cycling is not adopted. From what has been said, it is evident thatwhether to cycle is a problem whose many aspects must be carefully studiedbefore a proper decision can be reached.

Cycling is also adopted in nonretrograde gas caps overlying oil zones,


particularly when the oil is itself underlain by an active body of water. If thegas cap is produced concurrently with the oil, as the water drives thc oil zoneinto the shrinking gas cap Lone, unrecoverable oil remains not only in theoriginal oil zone but also in that portion of the gas cap invaded by the oil. Onthe othcr hand, if the gas cap is cycled at essentially initial pressure, the activewater drive displaces the oil into the producing oil wells with maximum recov-ery. In the meantime, some of the valuable liquids from the gas cap may berecovered by cycling Additional benefit will accrue, of course, if the gas capis retrograde. Even when water drive is absent, the concurrent depletion of thegas cap and the oil zone results in lowered oil recoveries; and increased oilrecovery is had by depleting the oil zone first and allowing the gas cap toexpand and sweep through the oil zone.

When gas-condensate reservoirs are produced under an active waterdrive such that reservoir pressure declines very little below the initial pressure,there is little or no retrograde condensation and the gas-oil ratio of the produc-tion remains substantially constant. The recovery is the same as in nonretro-grade gas reservoirs under the same conditions and depends on (a) the initialconnate water, S,,1, (b) the residual gas saturation, S,,, in the portion of thereservoir invaded by water, and (c) the fraction, F, of the initial reservoirvolume invaded by water. The gas volume factor Dgj Ifl cu ft/SCF remainssubstantially constant because reservoir pressure does not decline, so thefractional recovery is

Recover = — S,. — = (1 — Sw, — S1,)F(44)" (l—S,.,)

where V1 is the initial gross reservoir volume, is the residual gas saturationin thc flooded area, S,. is the initial connate water saturation, and F is thefraction of the total volume invaded. Table 3.3 shows that residual gas satu-rations lie in range of 20 to 50% following water displacement The fractionof the total volume invaded at any time or at abandonment depends primarilyon well location and the effect of permeability stratification in edge-waterdrives, and primarily on well spacing and the degree of water coning inbottom-water drives.

Table 4.8 shows the recovery factors calculated from Eq. (4.4) assuminga reasonable range of values for the connate water, residual gas saturation,and the fractional invasion by water at abandonment. The recovery factorsapply equally to gas and gas-condensate reservoirs because under active waterdrive there is no retrograde loss

Table 4.9 shows a comparison of gas-condensate recovery for the reser-voir of Ex. 4.3 by (a) volumetric depletion, (b) water drive at initial pressureof 2960 psia, and (c) partial water drive where the pressure stabilizes at 2000psia. Thc initial gross fluid, gas, and condensate, and the recoveries by de-pletion performance at an assumed abandonment pressure of 500 psia are

6. Lean Gas Cycling and Water Drive 137

TABLE 4.8.Recovery factors for complete water-drive reservoirs based on Eq (44)

S,., F=40 F=60 F=80 F=90 F=100

20 10203040

31.130028.626.7

46.745.042.840.0

62 260.057153.4

70.0675643600

77.875071.466.7

30 10203040

26.725.022.820.0

40.037.534.3300

53.450.045.740.0

60 056 351.4450

66.762.557.150.0

40 10

203040

22.220 017.113.3

33.330.025.720.0

44.440.034226.6

50.045.038530.0

55.650 042.833.3

50 10203040

17.715.011.46.7

26622.517.110.0

35.530.022.813.6 -—

40033.825.715.0

44437 528.516.7

TABLE 4.9.Comparison of gas-condensate recovery by volumetric performance.complete water drive, and partial water drive (based on the data of Tables 44and 45 and Ex 4.3. =30%, F =80%)

CondensalRecovery

Mechanism bbllac-fr

e RecoveryPer-

cenzage

Gas RecoveryPer-

MCFI ac-ft centage

Gross R

MCF/ac-fz

ecoveryPer-

cenzage

Initial in-place - 143 2 100.0 1441 100 9 !580 — 1000

Depletion to 500 psia 716 500 1200 833 1211 804

Water drive at 2960 psia 81.8 57 1 823 57 1 902 57 1

(a) Depletion to 2000 psia 28 4 198 457 31 7 485 30 7(b) Water drive at 2000 psia 31 2 21.8 553 38 4 584 37.0fotal by partial water

drive, (a) +(b) - 59 6 41.6 1010 70 1 1069 677

obtained from Ex. 4.3 and Tables 4.4 and 4.5 Under complete water drive,the recovery is 57.1% for a residual gas saturation of 20%, a connate water of30%, and a fractional invasion of 80% at abandonment, as may be found byEq. (4.4) or Table 4.8. Because there is no retrograde loss, this figure appliesequally to the gross gas, gas, and condensate recovery.

When a partial water drive exists and the reservoir pressure stabilizes atsome pressure, here 2000 psia, the recovery is approximately the sum of therecovery by pressure depletion down to the stabilization pressure, plus the


recovery of the remaining fluid by complete water drive at the stabilizationpressure. Because the retrograde liquid at the stabilization pressure is immo-bile, it is enveloped by the invading water, and the residual hydrocarbonsaturation (gas plus retrograde liquid) is about the same as for gas alone, or20% for this example. The recovery figures of Table 4.9 by depletion down to2000 psia are obtained from 'lable 4.5. The additional recovery by water driveat 2000 psia may be explained using the figures of Table 4.10. At 2000 psia theretrograde condensate volume is 625 cu ft/ac-ft, or 8.2% of the initial hydro-carbon pore volume of 7623 cu ft/ac-ft, 8 2% being found from the PVT datagiven in Table 4.4. If the residual hydrocarbon (both gas and condensate)saturation after water invasion is assumed to be 20%, as previously assumedfor the residual gas saturation by complete water drive, the water volume afterwater drive is 80% of 10,890, or 8712 cu fl/ac-ft. Of the remainder (i.e., 2178cu ftlac-ft) sinCC the pressure is assumed to stabilize at 2000 psia, 8.2%, or 625cu ft/ac-ft. will be condensate Jiquid and 1553 free gas. The reservoir vapor at2000 psia prior to water drive is

2000 x 6998 x 379.4l000xO.805x 1073 x655 9385 M SCF/ac-ft

The fractional recovery of this vapor phase by complcte water drive at 2000psia is

6998—15530778 or 77.8%

If V = 0.80-—or only 80% of each acre-foot, on the average, is invaded bywater at abandonment—the overall recovery reduces to 0.80 X 0 778, or62.2% of the vapor content at 2000 psia, or 584M SCF/ac-ft. Table 4.5 indi-cates that at 2000 psia the ratio of gross gas to residue gas after separation is

TABLE 4.10.Volumes of water, gas, and condensate in one acre-foot of bulk rock for thereservoir of Example 43

Initial Volumes After - Volumes AfterReservoir Depletion to Water DriveVolwnes, 2000 psia at 2000 psiacu ft/ac-ft cu ft/ac-ft cii ft/ac-ft

Water 3267

-

3267 8712Gas 7623 6998 1553Condensate .. 625 625Total 10,890 10,890 10,890

1 Use of Nitrogen for Pressure Maintenance 139

245.2 to 232 3, and that the gas-oil ratio on a residue gas basis is 17.730SCF/bbl. Then 584 M SCF of gross gas contains residue gas in the amount of

584 X = 553 M SCF/ac-ft

and tank or surface condensate liquid in the amount of

553x 1000

17 730=31.2 bbl/ac-ft

Table 4.9 indicates that for the gas-condensate reservoir of Example 4.3,using the assumed values for F and S,,, best overall recovery is obtained bystraight depletion performance. Best condensate recovery is by active waterdrive because no retrograde liquid forms 'Ilie value of the products obtaineddepends, of course, on the relative unit prices at which the gas and condensateare sold.

7. USE OF NITROGEN FOR PRESSURE MAINTENANCE

One of the major disadvantages associated with the use of lean gas in gascycling applications is that the income that would be derived from the sale ofthe lean gas is deferred for several years. For this reason, the use of nitrogenhas been suggested as a replacement for the lean gas.'6 However, one mightexpect the phase behavior of nitrogen and a wet gas to exhibit differentcharacteristics from that of lean gas and the same wet gas. Researchers havefound that mixing nitrogen and a typical wet gas causes the dew point of theresulting mixture to bc higher than the dew point of the original wet gas IS

This is also true for lean gas, but the dew point is raised higher with nitrogenIf, in a reservoir situation, the reservoir pressure is not maintained higher thanthis new dew point, then retrograde condensation will occur. This condensa-tion may be as much or more than what would occur if the reservoir was notcycled with gas. Studies have shown, however, that very little mixing occursbetween an injected gas and the reservoir gas in the reservoir.17 18 Mixingoccurs as a result of molecular diffusion and dispersion forces, and the re-sulting mixing zone width is usually only a few feet.'9 20 dew point may heraised in this local area of mixing, but this will be a very small volume and, asa result, only a small amount of condensate may drop out. Vogel and Yar-borough have also shown that, under certain conditions, nitrogen revaporizesthe condensate.'8 The conclusion from these studies indicates that nitrogencan be used as a replacement for lean gas in cycling operations with thepotential for some condensate formation that should be minimal in mostapplications.


Kicinsteiber, Wendscblag, and Calvin conducted a study to determinethe optimum plan of depletion for the Anschutz Ranch East Unit, which islocated in Summit County, Utah and Uinta County, Wyoming.2' The AnschutzRanch East Ficld, discovered in 1979, is one of the largest hydrocarbonaccumulations found in the Western Ovcrthrust Belt. Tests have indicated thatthe original in-place hydrocarbon content was over 800 million bbl of oilequivalent. Laboratory experiments conducted on several surface-recombinedsamples indicated that the reservoir fluid was a rich gas condensate The fluidhad a dew point only 150 to 300 psia below the original reservoir pressure of5310 psia. The dew-point pressure was a function of the structural position inthe reservoir, with fluid near the water-oil contact having a dew point about300 psia lower than the original pressure and field near the crest having a dewpoint only about 150 psia lower. The liquid saturation, observed in constantcomposition expansion tests, accumulated very rapidly below the dew point,suggesting that depletion of the reservoir and the subsequent drop in reservoirpressure could cause the loss of significant amounts of condensate. Because ofthis potential loss of valuable hydrocarbons, a project was undertaken todetermine the optimum method of production.2'

To begin the study, a modified Redlich-Kwong equation of state wascalibrated with the laboratory phase behavior data that had beenThe equation of state was then used in a compositional reservoir simulator.Several depletion schemes were considered, including primary depletion andpartial or full pressure maintenance. Wet hydrocarbon gas, dry hydrocarbongas, carbon dioxide, combustion flue gas, and nitrogen were all considered aspotential gases to inject. Carbon dioxide and flue gas were eliminated due tolack of availability and high cost. The results of the study led to the conclusionthat full pressure maintenance should be used. Liquid recoveries were foundto be better with dry hydrocarbon gas than with nitrogen. However, whennitrogen injection was preceded by a 10 to 20% buffer of dry hydrocarbon gas,the liquid recoveries were nearly the same When an economic analysis wascoupled with the simulation study, the decision was to conduct a full-pressuremaintenance program with nitrogen as the injected gas. A 10% pore volumebuffer, consisting of 35% nitrogen and 65% wet hydrocarbon gas, was to beinjected before the nitrogen to improve the recovery of liquid condensate.

The approach taken in the study by Kleinsteiber, Wendschlag, andCalvin would he appropriate for the evaluation of any gas-condensate reser-voir. The conclusions regarding which injected material is best or whether abuffer would be necessary may be different for a reservoir gas of differentcomposition.

PROBLEMS

4.J A gas-condensate reservoir initially contains 130DM SCF of residue (dry or salesgas) per acre-foot and 115 STB of condensate. Gas recovery is calculated to be85% and condensate recovery 58% by depletion performance. Calculate the

Problems 141

value of the initial gas and condensate reserves per acre-foot if the condensatesells for $20 00/bbl, and the gas sells for $1.90 per 1000 std cu ft.

4.2 A well produces 45 3 STB ofcondensate and 742 M SCF of sales gas daily Thecondensate has a molecular weight of 121.2 and a gravity of 52.O°API at 60°c

(a) What is the gas-oil ratio on a dry gas basis'(b) What is the liquid content expressed in barrels per million standard cubic

feet on a dry gas basis?

(c) What is the liquid content expressed in GPM on a dry gas basis'(d) Repeat parts (a), (b), and (c) expressing the figures on a wet, or gross, gas

basis.

4.3 The initial daily production from a gas-condensate reservoir is 186 STh ofcondensate, 3750M SCF of high-pressure gas, and 95 M SCF of stock tank gasThe tank oil has a gravity of 51.2°API at 60°F The specific gravity of theseparator gas is 0.712 and of the stock tank gas. 1.30. The initial reservoirpressure is 3480 psia, and reservoir temperature is 220°F. Average hydrocarbonporosity is 17.2%. Assume standard conditions of 14.7 psia and 60°F.

(a) What is the average gravity of the produced gases?(b) What is the initial gas-oil ratio?

(C) Estimate the molecular weight of the condensate.

(d) Calculate the specific gravity (air = 1.00) of the total well production.

(e) Calculate the gas deviation factor of the initial reservoir fluid (vapor) atinitial reservoir pressure.

(1) Calculate the initial moles in place per acre-foot(g) Calculate the mole fraction that is gas in the initial reservoir fluid(h) Calculate the initial (sales) gas and condensate in place per acre-foot.

4.4 (a) Calculate the gas deviation factor for the gas-condensate fluid the com-position of which is given in Table 4.1 at 5820 psia and 265°F. Use the criticalvalues of C8 for the C74 fraction.

(b) If half the butanes and all the pentanes and heavier gases are recovered asliquids, calculate the gas-oil ratio of the initial production Compare with themeasured gas-oil ratio.

4.5 Calculate the composition of the reservoir retrograde liquid at 2500 psia for thedata of Tables 4.4 and 4.5 and Ex. 4 3. Assume the molecular weight of theheptanes-plus fraction to be the same as for the initial reservoir fluid.

4.6 Estimate the gas and condensate recovery for the reservoir of Ex. 4.3 underpartial water drive if reservoir pressure stabilizes at 2500 psia Assume a residualhydrocarbon saturation of 20% and F = 52.5%.

4.7 Calculate the recovery factor by cycling in a condensate reservoir if the displace-ment efficiency is 85%, the sweep efficiency is 65%, and the permeabilitystratification factor is 60%

4.8 The following data are taken from a study on a recombined sample of separatorgas and separator condensate in a PVT cell with an initial hydrocarbon volumeof 3958.14 cu cm The wet gas GPM and the residue gas-oil ratios were calcu-lated using equilibrium ratios for production through a separator operating at


300 psia and 70°F. The initial reservoir pressure was 4000 psia. which was alsoclose to the dew-point pressure, and reservoir temperature was 186°F

(a) On the basis of an initial reservoir content of 1 00 MM SCF of wet gas,calculate the wet gas, residue gas, and condensate recovery by pressuredepletion for each pressure interval

(b) Calculate the dry gas and condensate initially in place in 1 00 MM SCI- ofwet gas.

(c) Calculate the cumulative recovery and the percentage of recovery of wet gas,residue gas, and condensate by depletion performance at each pressure.

(ci) Place the recoveries at an abandonment pressure of 605 psia on an acre-footbasis for a porosity of 10% and a connate water of 20%.

CornPressure. psia

position iis Mok Pc4000

rceniages3500 2900 2100 1300 605

('02 018 018 018 018 019 021N, 013 013 014 0.15 0.15 014C1 6772 6310 6521 69.79 7077 6659C, 1410 1427 1410 14.12 1463 1606(.3 837 825 8.10 7.57 773 911i-C4 098 09! 0.95 0.81 079 1.01

n-C4 345 340 316 271 259 3.31

i-C, 0.91 086 084 067 055 0.68n-Cs 152 140 139 097 081 102C6 179 160 152 103 073 0.80C7. 685 590 441 200 106 1.07

Mel wt.C,. 143 138 128 116 lii 110

Qas deviation factor for wet gas at 186°F 0 867 0 799 0 748 0 762 0.819 0 902

Wet gas production. cu cm at ccliP and T 0 224 0 474.0 1303 2600 5198

Wet gas 6PM (calculated) 5 254 4 578 3.347 1 553 0 835 0 895

Residue gas-oil ratio 7.127 8.283 11.621 26.051 49,312 45.872

Retrograde liquid, percentage ofcellvolume 0 332 1936 2391 22.46 1807

4.9 If the retrograde liquid for the reservoir of Prob. 4.8 becomes mobile at 15%retrograde liquid saturation, what effect will this have on the condensaterecovery9

4.10 if the initial pressure of the reservoir of Prob. 4.8 had been 5713 psia with thedew point at 4000 psia. calculate the additional recovery of wet gas, residue gas,and condensate per acre-foot. The gas deviation fattor at 5713 psia is 1 107, andthe (JPM and GOR between 5713 and 4000 psia are the same as at 4000 psia

4.11 Calculate the value of the products by each mechanism in Table 4 9 assuming (a)$17 50 per STB for condensate and Si 50 per M SCF for gas; (b) $20 00 per STBand SI .90 per M SCF; and (c) $2000 per STB and $2.50 per M SCF.

4. U In a PVT study of a gas-condensate fluid, 17.5 cu cm of wet gas (vapor),mcasurcd at cell pressure of 2500 psia and temperature of 195°F, was displaced

Problems 143

into an evacuated low-pressure receiver of 5000 cu cm volume that was maintain-ed at 250°1 to ensure that no liquid phase developed in the expansion If thepressure of the receiver rises to 620 mm Hg. what will be the deviation factor ofthe gas in the cell at 2500 psia and 195°F, assuming the gas in the receiverbehaves ideally?

4.13 Using the assumptions of Ex. 4.3 and the data of Table 4.4, show that thecondensate recovery between 2000 and 1500 psia is 13.3 STB/ac-ft and theresidue gas-oil ratio is 19,010 SCFIbb1.

4.14 A stock tank barrel of condensate has a gravity of 55°API. Estimate the volumein ft' occupied by this condensate as a single-phase gas in a reservoir at 2740 psiaand 215°F. The reservoir wet gas has a gravity of 0.76.

4.15 A gas-condensate reservoir has an areal extent of 200 acres, an average thicknessof 15 ft. an average porosity of 0.18, and an initial water saturation of 0.23 APV'F cell is used to simulate the production from the reservoir, and the followingdata are collected:

Condensate

Pressure(psia)

Wet (lasProduced

(cc)z

Wet Gas

Produced fromSeparator(moles)

4000 (dew point) 0 0.75 03700 400 0.77 0.00033300 450 0.81 0.0002

The initial cell volume was 1850 cc, and the initial gas contained 0.002 moles ofcondensate. The initial pressure is 4000 psia, and the reservoir temperature is200°F. Calculate the amount of dry gas (SCF) and condensate (STB) recoveredat 3300 psia from the reservoir. The molecular weight and specific gravity of thecondensate are 145 and 08, respectively.

4.16 Production from a gas-condensate reservoir is listed bclow. The molecularweight and the specific gravity of the condensate are 150 and 0.8, respectivelyThe initial wet gas in place was 35 MMM SCF, and the initial condensate was2MM STB. Assume a volumetric reservoir and that the recoveries of condensateand water are identical, and determine the following:(a) What is the percentage of recovery of residue gas at 3300 psia?(b) Can a PVT cell experiment be used to simulate the production from this

reservoir9 Why or why not?

Pressurc. psia 4000 3500 3300

Compressibility of wet gas, z 0.85 080 083Wet gas produced dunng pressureincrement, SCF 0 2 4 MMM 2 2 MMM

Liquid condensate producedduring pressure increment, STB 0 80,000 70.000

Water produced during pressureincrement. S-rB 0 5000 4375


4.17 A PVT cell is used to simulate a gas-condensate reservoir. The initial cell volumeis 1500 cc, and the initial reservoir temperature is 175°F. Show by calculationsthat the PVT cell will or will not adequately simulate the reservoir behavior. Thedata generated by the PVT expenments as well as the actual production historyare as follows:

Pressure1 psia 4000 3600 3000

Wet gas produced in pressure increment, cc 0 300 700

Compressibility of produced gas 0.70 0.73 0.77

Actual Production history, M SCF 0 1000 2300

REFERENCES

'3. C. Allen, "Factors Affecting the Classification of Oil and Gas Wells," API Drillingand Production Practice (1952), p. 118.

2Ira Rinehart.c Yearbooks. Dallas, TX: Rinehart Oil News, 1953—1957, Vol. No. 2.

3M. Muskat, Physical Principles of Oil Production, (New York: McGraw-Hill, BookCompany, mc, 1949), Chap. 15

4M. B. Standing, Volumetric and Phase Behavior of Oil Field Hydrocarbon Systems.New York: Reinhold Publishing, 1952, Chap. 6.

F. Thornton, "Gas-Condensate Reservoirs—a Review," API Drilling andProduc-tion Practice (1946), p. 150

6C. K. Eiierts, Phase Relations of Gas-Condensate Fluids, Monograph 10, Bureau ofMines. New York: American Gas Association, 1957, Vol. 1.

7T. A. Mathews, C. H. Roland, and 0. L Katz, "High Pressure Gas Measurement,"Proc. NGAA (1942), p. 41.

8J. K. Rodgers, N. H. Harrison, and S Regier, "Comparison Between the Predictedand Actual Production History of a Condensate Reservoir." Paper No. 883—0presented before the AIME Meeting, Dallas, TX, Oct. 1957.

9W. E. Portman and J. M. Campbell, "Effect of Pressure, Temperature, and Well-stream Composition on the Quantity of Stabilized Separator Fluid," Trans. AIME(1956), 207, 308.

'°R. L. Huntington, Natural Gas and Natural Gasoline. New York: McGraw-l-hll,1950, Chap. 7

"Natural Gasoline Supply Men's Association Engineering Data Book, 7th ed. Tulsa,Natural Gasoline Supply Men's Association, 1957), p. 161.

2A E. Hoffmann, 1. S. Crump, and C. R. llocott, "Equilibrium Constants for aGas-Condensate System," Trans. AJME (1953), 198, 1

"F. H. Allen and R. P. Roe, "Performance Characteristics of a Volumetric Conden-sate Reservoir," Trans. AIME (1950), 189, 83.

14J. E. Berryman, "The Predicted Performance of a Gas-Condensate System, Wash-ington Field, Louisiana," Trans. AIME (1957), 210, 102.

"R. 11. Jacoby, R. C. Koeller, and V 1. Berry, Jr. "Effect of Composition and

References 145

Temperature on Phasc Behavior and Depletion Performance of Gas-Condensate Sys-tems.' Paper presented before SI'E of AIME, Houston, TX, Oct 5—8, 1958.

'6C. W. Donohoe and R D. Buchanan. "Economic Evaluation of Cycling Gas-Condensate Reservoirs with Nitrogen," Jour, of Petroleum Technology (Feb. 1981),263

"P. L. Moses and K. Wilson, "Phase Equilibnum Considerations in Using Nitrogenfor Improved Recovery From Retrograde Condensate Reservoirs," Jour. of PetroleumTechnology (Feb. 1981), 256

L. Vogel and L. Yarborough, "The Effect of Nitrogen on the Phase Behavior andPhysical Properties of Reservoir Fluids." Paper SPE 8815 presented at the First JointSPE/DOE Symposium on Enhanced Oil Recovery, Tulsa, OK, April, 1980.19p, M Sigmund, 'Prediction of Molecular Diffusion at Reservoir Conditions. PartT—Measurement and Prediction of Binary Dense Gas Diffusion Coefficients," Jour, ofCanadian Petroleum Technology (Apnl—iune 1976), 48.

20P. M. Sigmund, "Prediction of Molecular Diffusion of Reservoir Conditions Part11—Estimating the Effects of Molecular Diffusion and Convective Mixing in Multi-component Systems," Jour of Canadian Petroleum Technology (JuIy—Scpt 1976). 53.

W Kleinstciber, D. D. Wendschlag, and i. W. Calvin, "A Study for Developmentof a Plan of Depletion in a Rich Gas Condensate Reservoir Anschutz Ranch EastUnit, Summit County, Utah, Uinta County, Wyoming." Paper SPE 12042 presentedat 58th Annual Conference of SPE of AIME, San Francisco. Oct. 1983.22L. Yarborough, "Application of a Generalized Equation of State to PetroleumReservoir Fluids," in Equations of Stale in Engineering, Advances in Chemical Series,K. C. Cbao and R. L. Robinson, ed.. Amencan Chemical Society, 1979, 385

Chapter 5

Undersaturated OilReservoirs

1. INTRODUCTION

Oil reservoir fluids are mainly complex mixtures of the hydrocarbon com-pounds, which frequently contain impurities such as nitrogen, carbon dioxide,and hydrogen sulfide. The composition in mole percentages of several typicalreservoir liquids is given in Table 5.1, together with the tank gravity of thecrude oil, the gas-oil ratio of the reservoir mixture, and other characteristicsof the fluids.* 1 The composition of the tank oils obtained from the reservoirfluids are quite different from the composition of the reservoir fluids, owingmainly to the release of most of the methane and ethane from solution and thevaporization of sizeable fractions of the propane. butanes. and pentanes aspressure is reduced in passing from the reservoir to the stock tank. The tableshows a good correlation between the -gas-oil ratios of the fluids and thepercentages of methane and ethane they contain over a range of gas-oil ratiosfrom only 22 SCF/STB up to 4053 SCF/STB

Several methods arc available for collecting samples of reservoir fluidsThe samples may he taken with subsurface sampling equipment lowered intothe well on a wire line, or samples of the gas and oil may be collected at the

• Rcfcrenccs throughout the text are given at thc end of cach chapter

146

1. Introduction 147

TABLE 5.1.Reservoir fluid compositions and properties (After Kennery,' courtesy CoreLaboratories, Inc.)

South North West SouihComponent or Property ('alif. Wyo Texas Texas Texas La

Methane 2262 1.08 48.04 25.63 2863 6501kthane 1.69 2.41 336 5.26 1075 784Propane 0.81 2.86 1 94 10.36 9 95 642iso-Butane 0.51 0.86 0 43 1.84 4.36 2 14n-Butane 0.38 283 075 5.67 4.16 291iso-Pentane 0.19 1 68 0.78 3.14 2.03 1 65n-Pentane 0.19 2 17 0 73 1.91 3.83 0 83Hcxancs 0.62 4 51 2 79 4.26 2 35 119Heptanes-plus 72.99 81 60 4118 41.93 33.94 12 01Density heptanes-

plus. g/cc 0.957 0.920 0.860 0 843 0.792 0814Mol wt hcptancs-plus 360 289 198 231 177 177

Sampling depth, feet 2980 3160 8010 4520 12,400 10,600Reservoir temperature.

op 141 108 210 140 202 241Saturation pressure. psig 1217 95 3660 1205 1822 4730GOR, SCF/STB 105 22 750 480 895 4053Formation volume

factor, bbIISTB 1 065 1.031 1 428 1 305 1.659 3 610Tank oil °API 16.3 25 1 34.8 40 6 50 8 43.5Gas gravity (air= 1.00) 0669 . . 0 715 1.032 1151 0.880

surface and later recombined in proportion to the gas-oil ratio measured atthe time of sampling. Samples should be obtained as early as possible in thelife of the reservoir, preferably at the completion of the discovery well, so thatthe sample approaches as nearly as possible the original reservoir fluid. Thetype of fluid collected in a sampler is dependent on the well history priorto sampling. Unless the well has been properly conditioned before sampling,it is impossible to collect representative samples of the reservoir fluid. Acomplete well-conditioning procedure has been described by Kennerlyand Reudelhuber) 2 The information obtained from the usual fluid sampleanalysis includes the following properties:

1. Solution and evolved gas-oil ratios and liquid phase volumes.2. Formation volume factors, tank oil gravities, and separator and stock tank

gas-oil ratios for various separator pressures3. Bubble-point pressure of the reservoir fluid.4. Compressibility of the saturated reservoir oil

Undersaturated Oil Reservoirs

5. Viscosity of the reservoir oil as a function of pressure.6. Fractional analysis of a casing head gas sample and of the saturated reser-

voir fluid.

if laboratory data are not available, satisfactory estimations for a preliminaryanalysis can often be made from empirical correlations based on data that areusually available. These data include the gravity of the tank oil, the specificgravity of the produced gas, the initial producing gas-oil ratio, the viscosity ofthe tank oil, reservoir temperature, and initial reservoir pressure.

In most reservoirs, the variations in the reservoir fluid properties amongsamples taken from different portions of the reservoir are not large, and theylie within the variations inherent in the techniques of fluid sampling andanalysis. In some reservoirs, on the other hand, particularly those with largeclosures, there are large variations in the fluid properties For example, in theElk Basin Field, Wyoming and Montana. under initial reservoir conditionsthere was 490 SCF of gas in solution per barrel of oil in a sample taken nearthe crest of the structure but only 134 SCF/bbl in a sample taken on the flanksof the field, 1762 ft lower in elevation.3 This is a solution gas gradient of 20SCF/bbl per 100 ft of elevation. Because the quantity of solution gas has alarge effect on the other fluid properties. large variations also occur in the fluidviscosity, the formation volume factor, and the like. Similar variations havebeen reported for the Weber sandstone reservoir of the Rangely Field, Colo-rado, and the Scurry Reef Field, Texas, where the solution gas gradients were25 and 46 SCFIbbL per 100 ft of elevation, respectively.' These variations influid properties may be explained by a combination of (a) temperature gra-dients, (b) gravitational segregation, and (c) the lack of equilibrium betweenthe oil and the solution gas Cook, Spencer, Bobrowski, and Chin, andMcCord have presented methods for handling calculations when there aresignificant variations in the fluid properties.'

2. CALCULATION OF INITIAL OIL IN PLACE BY THEVOLUMETRIC METHOD AND ESTIMATiON OF OILRECOVERIES

One of the important functions of the reservoir engineer is the periodic calcu-lation of the reservoir oil (and gas) in place and the recovery anticipated underthe prevailing reservoir mechanism(s). In some companies, this work is doneby a group that periodically renders an accounting of the company's reservestogether with the rates at which they can be recovered in the future The com-pany's financial position depends primarily on its reserves, the rate at whichit increases or loses them, and the rates at which they can be recovered. Aknowledge of the reserves and rates of recovery is also important in the saleor exchange of oil properties. The calculation of reserves of new discoveriesis particularly important because it serves as a guide to sound developmentprograms. Likewise, an accurate knowledge of the initial contents of reser-

2. Calculation of Initial Oil in Place 149

voirs is invaluable to the reservoir engineer who studies the reservoir behaviorwith the aim of calculating and/or improving primary recoveries, for it elimi-nates one of the unknown quantities in equations.

Oil reserves are usually obtained by applying recovery factors to the oilin place. They are also estimated from decline curve studies and by applyingappropriate barrel-per-acre-foot recovery figures obtained from experience orstatistical studies. The oil in place is calculated either (a) by the volumetricmethod or (b) by material balance studies, both of which were presented forgas reservoirs in Chapter 3 and which will be given for oil reservoirs in this andfollowing chapters. The recovery factors are determined from (a) displace-ment efficiency studies and (b) correlations based on statistical studies ofparticular types of reservoir mechanisms.

The volumetric method for estimating oil in place is based on log and coreanalysis data to determine the bulk volume, the porosity, and the fluid satu-rations, and on fluid analysis to determine the oil volume factor. Under initialconditions 1 ac-ft of bulk oil productive rock contains

Interstitial water 7758 x 4) x S,,,

Reservoir oil 7758 X 4, X (1 —

7758 barrels is the equivalent of I ac-ft. 4, is the porosity as a fractionof the bulk volume, S,,, is the interstitial water as a fraction of thc pore volume,and B0, is the initial formation volume factor of the reservoir oil. Usingsomewhat average values, 4, 0.20, SW = 0.20, and = 1.24, the initial stocktank oil in place per acre-foot is on the order of 1000 or

7758 x 0.20 x (1 —0.20)Stock tank oil =

= 1000 STB/ac-ft

For oil reservoirs under volumetric control, there is no water influx toreplace the produced oil, so it must be replaced by gas the saturation of whichincreases as the oil saturation decreases. If is the gas saturation and theoil volume factor at abandonment, then at abandonment conditions 1 ac-ft ofbulk rock contains

Interstitial water 7758 x 4, X S,,

Reservoir gas 7758 x4, X S1

Reservoir oil 7758 X 4, X (1 — Sw — S1)

7758x4,x(lStock tank oil

U0

150 Undersaturated OH Reservoirs

Then the recovery in stock tank barrels per acre-foot is

Recovery = 7758 )(— 13

and the fractional recovery in terms of stock tank barrels is

Recovery = 1 — X (5.2)

The total free gas saturation to he expected at abandonment can bc estimatedfrom the oil and water saturations as reported in core analysis.7 This expecta-tion is based on the assumption that, while being removed from the well, thecore is subjected to fluid removal by the expansion of the gas liberated fromthe residual oil, and that this process is somewhat similar to the depletionprocess in the reservoir, in a study of the well spacing problem, Craze andBuckley collected a large amount of statistical data on 103 oil reservoirs, 27 ofwhich were considered to be producing under volumetric control.8 The finalgas saturation in most of these reservoirs ranged from 20 to 40% of the porespace, with an average saturation of 30 4%. Recoveries may also he calculatedfor depletion performance from a knowledge of the properties of the reservoirrock and fluids.

lii the case of reservoirs under hydraulic control, where there is no ap-preciable decline in reservoir pressure. water influx is either inward and par-allel to the bedding planes as found in thin, relatively steep dipping beds(edge-water drive), or upward where the producing oil zone (column) is un-derlain by water (bottom-water drive) The oil remaining at abandonment inthose portions of the reservoir invaded by water, in barrels per is:

Reservoir oil 7758 x d x S0.

Stock tank oil

where S0. is the residual oil saturation remaining after water displacement.Since it was assumed that the reservoir pressure was maintained at its initialvalue by the water influx, no free gas saturation develops in the oil zone andthe oil volume factor at abandonment remains B01. The recovery by activewater drive then is

7758 x 4,( 1 — — S0,)Recovery

= 130,STB/ac-ft (5.3)

and the recovery factor is

(1Recovery factor

= (1 5) (5.4)

2. Calculation of Initial Oil in Place 151

It is generally believed that the oil Content of cores, reported from theanalysis of cores taken with a water-base drilling fluid, is a reasonable esti-mation of the unrecoverable oil, because the core has been subjected to apartial water displacement (by the mud filtrate) during coring and to displace-ment by the expansion of the solution gas as the pressure on the core isreduced to atmospheric pressure.'° If this figure is used for the resident oilsaturation in Eqs. (5.3) and (5.4), it should he increased by the formationvolume factor. For example, a residual oil saturation of 20% from core analy-sis indicates a residual reservoir saturation of 30% for an oil volume factor of1 50 bbl/STB. The residual oil saturation may also be estimated using the dataof Table 3 3, which should be applicable to residual oil saturations as well asgas saturations (i.e., in the range of 25 to 40% for the consolidated sandstonesstudied).

In the reservoir analysis made by Craze and Buckley, some 70 of the 103fields analyLed produced wholly or partially under water-drive conditions, andthe residual oil saturations ranged from 17.9 to 60 9% of the pore space 8According to Arps, the data apparently relate according to the reservoir oilviscosity and permeability.7 The average correlation between oil viscosity andresidual oil saturation, both under reservoir conditions, is shown in Table 5 2Also included in Table 5.2 is the deviation of this trend against averageformation permeability. For example, the residual oil saturation under reset-

TABLE 5.2.Correlation between reservoir oil viscosity, average reservoir permeability,and residual oil saturation (After Craze and Buckle? and Arps7)

Reservoir Oil Viscosity(in cp)

Residual Oil Saturation(percentage of pore space) -

0.2 300.5 321.0 3452.0 375.0 40.5

10.0 43.5-- 200 64.5

Average Reservoir PermeabilityDeviation of Residual OilSaturation from Viscosity

(in md) Trend (percentage of pore space)

50 +12100 +9200 4-6500 +2

1000 -12000 —455000 — 8.5

152 Undersaturated Oil Reservoirs

voir conditions for a formation containing 2 cp oil and having an averagepermeability of 500 md is estimated at 37+ 2, or 39% of the pore space.

Because Craze and Buckley's data were arrivcd at by comparing recov-eries from the reservoir as a whole with the estimated initial content, theresidual oil calculated by this method includes a sweep efficiency as well as theresidual oil saturation; that is, the figures are higher than the residual oilsaturations in those portions of the reservoir invaded by water at abandon-ment. This sweep efficiency reflects the effect of well location, the bypassingof some of the oil in the less permeable strata, and the abandonment of someleases before thc flooding action in all zones is complete, owing to excessivewater-oil ratios, both in edge-water and bottom-waler drives.

in a statistical study of Craze and Buckley's water-drive recovery data,Guthrie and Greenberger, using multiple correlation analysis methods, foundthe following correlation between water-drive recovery and five variables thataffect recovery in sandstone reservoirs."

RFO.114+0.272 log k +0256 log

— I 53&b—0.00035 h (5.5)

For k = 1000 md, = 0.25, 20 cp, 4) 0.20, and h = 10 ft.

RF=0.114+0.272x1og l000+0.256x025—0.136

log 2— l.538x0.20—0.00035 x 10

0.642 or 64 2% (of initial stock tank oil)

where,

RE = recovery factor

A test of the equation showed that 50% of the fields had recoveries within±6.2 recovery % of that predicted by Eq (5.5), 75% were within ±9 0recovery %. and 100% were within ±19 0 recovery %. For instance, it is 75%probable that the recovery from the foregoing example is 64.2 ± 9.0%.

Although it is usually possible to determine a reasonably accurate re-covery factor for a reservoir as a whole, the figure may be wholly unrealisticwhen applied to a particular lease or portion of a reservoir, owing to theproblem of fluid migration in the reservoir, also referred to as lease drainage.F'or example, a flank lease in a water-drive reservoir may have 50,000 STB ofrecoverable stock tank oil in place but will divide its reserve with all updipwells in line with it. 'fhe degree to which migration may affect the ultimaterecoveries from various leases is Illustrated in Fig. 5 1 12 If the wells are locatedon 40-acre units, if each well has the same daily allowable, if there is uniformpermeability, and ii the reservoir is under an active water drive so that the

3 Matenal Balance in Undersaturated Reservoirs 153

Fig. 5.1. Effect of water drive on oil migration. (After Buckley,'2 ATME.)

water advances along a horizontal surface, then the recovery from lease A isonly one-seventh of the recoverable oil in place, whereas lease G recoversone-seventh of the recoverable oil under lease A, one-sixth under lease B,one-fifth under lease C, and so on Lease drainage is generally less severe withother reservoir mechanisms, but it occurs to some extent in all reservoirs.

3. MATERIAL BALANCE IN UNDERSATURATEDRESERVOIRS

The material balance equation for undersaturatcd reservoirs was developed inChapter 2 and is

N(B,- B,,) + + (Re— R,01) + BwWp (2.8)

Neglecting the change in porosity of rocks with the change of internal fluidpressure, which is treated later, reservoirs with zero or negligible water influxare constant volume or volumetric reservoirs. If the reservoir oil is initiallyundersaturated, then initially it contains only connate water and oil, with thcirsolution gas. The solubility of gas in reservoir waters is generally quite low andis considered negligible for the present discussion. Because the water produc-tion from volumetric reservoirs is generally small or negligible, it will beconsidered as zero. From initial reservoir pressure down to the bubble point,then, the reservoir oil volume remains a constant, and oil, is produced byliquid expansion. Incorporating these assumptions into Eq. (2.8), we get

N(81— 4 (5.6)


While the reservoir pressure is maintained above the bubble-point pressureand the oil remains undersaturated, only liquid will exist in thc reservoir. Anygas that is produced on the surface will be gas coming out of solution as theoil moves up through the weilbore and through the surface facilities. All thisgas will be gas that was in solution at reservoir conditions. Therefore, duringthis period, will equal R50 and will equal since the solution gas-oilratio remains constant (see Chapter 1). The material balance equationbecomes

N(B,— = NAB, (5.7)

This can be rearranged to yield fractional recovery, RF, as

(5.8)

The fractional recovery is generally expressed as a fraction of the initial stocktank oil in place. The PVT data for the 3—A —2 reservoir of a field is given inFig. 5 2.

The formation volume factor plotted in Fig. 5.2 is the single-phaseformation volume factor, The material balance equation has been derivedusing the two-phase formation volume factor, B,. B0 and are related byEq. (1.28)

B, = + Bg(Rsoi (1.28)

It should be apparent that B, = B0 above the bubble-point pressure because R,0is constant and equal to R,01.

The reservoir fluid has an oil volume factor of 1 572 bbl/STB at the initialpressure 4400 psia and 1.600 bbliSTB at the bubble-point pressure of 3550psia. Then by volumetric depletion, the fractional recovery of the stock tankoil at 3550 psia by Eq. (5.8) is

1.600—1 572RF= 0.0175or1.75%

If the reservoir produced 680,000 STB when the pressure dropped at 3550psia. then the initial oil in place by Eq. (5.7) is

— 1.600 xMM STBN— 1.600—1572

88

Below 3550 psia a free gas phase develops; and for a volumetric, under-

PRODUC1 IONSIB AND

N0 SCF

7 IN-ND) STBI

156 Uridersaturated Oil Reservoirs

if the free gas remains distributed throughout the reservoir as isolated bubbles.Equation (5.6) can he rearranged to solve for N and the fractional recovery,RF, for any undersaturated reservoir below the bubble point

N — 4 (Re—'5 10

— (8,-B,,) (.(B,!3,,)

511— N - (

The nez cumulative produced gas-oil ratio is the quotient of all the gasproduced from the reservoir Ga,, and all the oil produced In some reser-voirs, some of the produced gas is returned to the same reservoir, so that thenet produced gas is only that which is not returned to the reservoir. When allthe produced gas is returned to the reservoir, is zero.

An inspection of Eq. (5.11) indicates that all the terms except the pro-duccd gas-oil ratio, are functions of pressure wily and are the propertiesof the reservoir fluid. Because the nature of the fluid is fixed, it follows thatthe fractional recovery RF is fixed by the PVT properties of the reservoir fluidand the produced gas-oil ratio. Since the produced gas-oil ratio occurs in thedenominator of Eq. (5.11), large gas-oil ratios give low recoveries and viceversa.

Example 5.1. Calculations to show the effect of the produced gas-oilratio R,, on fractional recovery in volumetric, undersaturated reservoirs.

Give,r

The PVT data for the 3—A—2 reservoir, Fig 5.1.

Cumulative GOR at 2800 psia 3300 SCF/STB

Reservoir temperature = 190°F = 650°R.

Standard conditions = 14.7 psia and 60°F.

Sol.uTJoN: = 1100 SCF/STB; B,,, = 1.572 bbL/STB; R,,, at 2800 psia900 SCF/STB; 80 at 2800 psia = 1.520 bbl/STB; = 3300 SCF/STB; andat 280() psia calculated as

_znR'T_0.870X10.73x650_0 0 b C— 5.615 p 5.615 x 379.4 x 2800

.001 2 bl/S F

B, = B,, +

B, = 1.520 + 0.00102(1100— 900) = 1 724 bbl/STB

3. Material Balance in Undersaturated Reservoirs

Then at 2800 psia

— 1.724—1.572RF

— 1.724 + 0.00102(3300 — 1100)

= 0.0383, or 3.83%

If two-thirds of the produced gas had been returned to the reservoir, at thesame pressure (i.e., 2800 psia), the fractional recovery would have been

— 1.724—1.572RF

— 1.724 + 0.00102(1100 — 1100)

= 0.088, or 8.8%

Equation (5.10) may be used to find the initial oil in place. For example, if1.486 MM STB had been produced down to 2800 psia, for = 3300 SCF/STB, the initial oil in place is

— 1.486 x 1061 1.724+0 00102(3300 — 1100))N—

1.724—1.572

=38.8MM STB

The calculations of Ex 5.1 for the 3—A —2 reservoir show that for = 3300SCF/STB the recovery at 2800 psia is 3.83%, and that if had been only 1100SCF/STB, the recovery would have becn 8.80%. Neglecting in each case the1.75% recovery by liquid expansion down to the bubblc-point pressure, theeffect of reducing the gas-oil ratio by one-third is approximately to triplethe recovery. The produced gas-oil ratio can be controlled by working overhigh gas-oil ratio wells, shutting in or reducing the producing rates of highratio wells, and/or by returning some or all of the produced gas to the reser-voir. If gravitational segregation occurs during production so that a gas capforms, as shown in Fig. 5.3 and if the producing wells are completed low in theformation, their gas-oil ratios will be lower and recovery will be improved.Simply from the material balance point of view, by returning all produced gasto the reservoir it is possible to obtain 100% recoveries. From the point of viewof flow dynamics, however, a practical limit is reached when the reservoir gassaturation rises to values in the range of 10 to 40% because the reservoirbecomes so permeable to gas that the returned gas moves rapidly from theinjection wells to the production wells, displacing with it only a small quantityof oil. Thus although gas-oil ratio control is important in solution gas-drivereservoirs, recoveries are inherently low because the gas is produced fasterthan the oil Outside of the energy stored up in the liquid above thc bubblepoint, the energy for producing the oil is stored up in the solution gas. When


this gas has been produced, the only remaining natural source of energy isgravity drainage, and there may be a considerable period in which the oildrains downward to the wells from which it is pumped to the surface.

In the next section, we present a method that allows the material balanceequation to be used as a predictive tool. The method was used by engineersperforming calculations on the Canyon Reef Reservoir in the Kelly-SnyderField.

4. KELLY.SNYDER FIELD, CANYON REEF RESERVOIR

The Canyon Reef reservoir of the Kelly-Snyder Field, Texas, was discoveredin 1948. During the early years of production, there was much concern aboutthe very rapid decline in reservoir pressure; however, reservoir engineers wereable to show that this was to be expected of a volumetric undersaturatedreservoir with an initial pressure of 3112 psig and a bubble-point pressure ofonly 1725 psig, both at a datum of 4300 ft subsea calculations furthershowed that when the bubble-point pressure is reached, the pressure declineshould be much less rapid, and that the reservoir could be produced withoutpressure maintenance for many years thereafter without prejudice to thepressure maintenance program eventually adopted. in the meantime, withadditional pressure drop and production, further reservoir studies could eval-uate the potentialities of water influx, gravity drainage, and intrareservoircommunication. These, together with laboratory studies on cores to determinerecovery efficiencies of oil by depletion and by gas and water displacement,should enable the operators to make a more prudent selection of the pressuremaintenance program to be used, or demonstrate that a pressure maintenanceprogram would not be successful.

Althougji additional and revised data have become available in subse-quent years, the following calculations, which were made in 1950 by reservoirengineers, are based on data available in 1950. Table 5 3 gives the basicreservoir data for the Canyon Reef reservoir. Geologic and other evidenceindicated that the reservoir was volumetric (i.e., that there would be negligiblewater influx), so the calculations were based on volumetric behavior. If anywater entry should occur, the effect would be to make the calculations moreoptimistic, that is, there would be more recovery at any reservoir pressure.The reservoir was undersaturated, so the recovery from initial pressure tobubble-point pressure is by liquid expansion, and the fractional recovery at thebubble point is

Bt—Bft 14509—1.4235

145090.0189 or 1.89%

Based on an initial content of 1.4235 reservoir barrels or 1.00 STB, this isrecovery of 0.0189 STB. Because the solution gas remains at 885 SCF/STB

4. Kelly-Snyder Field Canyon Reef Reservoir 159

TABLE 53.Reservoir rock and fluid properties for the Canyon Reef Reservoir of theKelly-Snyder Field, Texas (Courtesy. The Oil and Gas Journal 14)

Initial reservoir pressure 3112 psig (at 4300 ft subsea)Bubble-point pressure 1725 psig (at 4300 ft subsea)Average reservoir temperature 125°FAverage porosity 7.7%Average connate water 20%Critical gas saturation (estimated) 10%

Differential Liberation Analyses of a Bottom-hole Sample from the Standard Oil Com-pany of Texas No. 2—1, i W Brown, at 125°F

Pressurepsig

- B0 -bbl/STB bbl/SCF -

Solution GORSCFISTB

B,bbliSTB

3112 1.4235 ... 885 1.42352800 1.4290 ... 885 1.42902400 1.4370 . 885 1.43702000 1.4446 . 885 1.44461725 1.4509 ... 885 1.45091700 1.4468 0.00141 876 1.45951600 14303 000151 842 1.49521500 1 4139 0.00162 807 1 54031400 1 3978 0.00174 772 1.5944

down to 1725 p51g. the producing gas-oil ratio and the cumulative producedgas-oil ratio should remain near 885 SCF/STB during this pressure decline

Below 1725 psig, a free gas phase develops in the reservoir. As long asthis gas phase remains immobile, it can neither flow to the well bores normigrate upward to develop a gas cap but must remain distributed throughoutthe reservoir, increasing in size as the pressure declines. Because pressurechanges much less rapidly with reservoir voidage for gases than for liquids, thereservoir pressure declines at a much lower rate below the bubble point. It wasestimated that the gas in the Canyon Reef reservoir would remain immobileuntil the gas saturation reached a value near 10% of the pore volume. Whenthe free gas begins to flow, the calculations become quite complex (see Chap-ter 9); but as long as the free gas is immobile, calculations may be madeassuming that the producing gas-oil ratio R at any pressure will equal thesolution gas-oil ratio R30 at the pressure, since the only gas that reaches thewell bore is that in solution, the free gas being immobile. Then the averageproducing (daily) gas-oil ratio between any two pressures Pt and P2 is

approximately

(5.12)

160 Iindersaturated Oil Reservoirs

and the cumulative gas-oil ratio at any pressure is

=______

= X R,01 +(N,3— Npb)Ravgl + — Npi)Ravgz t etc.5 13

NPb+(NP1—NPb)+(NP2—-NPI)+etc.

On the basis of 1.00 STB of initial oil, the production at bubble-point pressureis 0.0189 STB. The average producing gas-oil ratio between 1725 and 1600

psig will be

== 864 SCF/STB

The cumulative recovery at 1600 psig is unknown: however the cumulativegas-oil ratio may be expressed by Eq. (5.13) as

— 0.0189 x 885 + — 0.0189)864

This value of may be placed in Eq (5.11) together with the PVT values at1600 psig as.

1.4952 — 1.4235

= 1.4952+000151[0M189x885 — 0.0189)864

— 885]

0 0486 STB at 1600 psig

In a similar manner the recovery at 1400 psig may be calculated, the resultsbeing valid only if the gas saturation remains below the critical gas saturation,assumed to be 10% for the present calculations.

When stock tank barrels of oil have bccn produced from a volumetricundersaturated reservoir, and the average reservoir pressure is p, the volumeof the remaining oil is (N — Since the initial pore volume of the reser-voir %i, is

(5.14)

and since the oil saturation is the oil volume divided by the pore volume,

s 515NB01 (.

4 Kelly-Snyder Field, Canyon Reef Reservoir 161

On the basis of N = 1.00 STB initially, N,, is the fractional recovery RF, orNe/N, and Eq. (5.15) may be written as

f B0= (1— RF)( 1— (5.16)

where S,, is the connate water, which is assumed to remain constant forvolumetric reservoirs. Then at 1600 psig the oil saturation is

/1.4303\S0 = (1 — 0.0486)(1 —0.20)

=0.765

The gas saturation is (1 — S0 — or

Sg= 10.7650.200=0.035

Figure 5 4 shows the calculated performance of the Kelly-Snyder Field downto a pressure of 1400 psig. Calculations were not continued beyond this pointbecause the free gas saturation had reached approximately 10%, the esti-mated critical gas saturation for the reservoir. The graph shows the rapidpressure decline above the bubble point and the predicted flattening below the

Fig. 5.4. Material balance calculations and performance. Canyon Reef reservoir.Kelly-Snyder Field

RECOVERY IN PER CENT


bubble point. The predictions are in good agreement with the field perform-ance, which is calculated in Table 5 4 using field pressures and productiondata, and a value of 2.25 MMM STh for thc initial oil in place. The producinggas-oil ratio, Col (2), increases instead of decreasing, as predicted by theprevious theory. This is due to the more rapid depletion of some portions ofthe reservoir—for example, those drilled first, those of low net productivethickness, and those in the vicinity of the well bores. For the present predic-tions, it is pointed out that the previous calculations would not be alteredgreatly if a constant producing gas-oil ratio of 885 SCF/STB (i.e., the initialdissolved ratio) had bccn assumed throughout the entire calculation.

The initial oil under a 40 acre unit of the Canyon Reef reservoir for a netformation thickness of 200 feet is

— 7758 x4Ox 200 x 0077 x (1 — 0.20)N—

1.4235

= 2.69MM STh

Then at thc average daily well rate of 92 BOPD in 1950, the time to produceII 35% of the initial oil (i.e., at 1400 psig when the gas saturation is calculatedto be near 10%) is

0. 1135 x2 69 x 10692x365

By means of this calculation, the reservoir engineers were able to show thatthere was no immediate need for a curtailment of production and that therewas plenty of time in which to make further reservoir studies and carefully

TABLE 5.4.Recovery from Kelly-Snyder Canyon Reef Reservoir based on productiondata and measured average reservoir pressures, and assuming an initial oilcontent of 2.25 MMM STB

(1) (2) (3) (4) (5)Percentage

Pressure Ave. Producing Incremental Cumulative Recoveryinterval, Gas-Oil Ratio, Oil Production, Oil Production, (N 2.25 MMM

psig SCFISTR MM SiB MMSTB - STB)

3312w 1771 896 60.421 60421 2691771 to1713 934 11.958 72379 3221713w 1662 971 13.320 85699 381

1662 to 1570 1023 20.009 lOS 708 4.70

1570to 1561 1045 11.864 117572 523

5. The Gloyd-Mitchell Zone of the Rodessa Field 163

considered plans for the optimum pressure maintenance program. Followingcomprehensive and exhaustive studies by engineers, the field was unitized inMarch, 1953, and placed under the management of an operating committee.This group proceeded to put into operation a pressure maintenance programconsisting of (a) water injection into wells located along the longitudinal axisof the field, and (b) shutting in the high gas-oil ratio wells and transferringtheir allowables to low gas-oil ratio wells. The high-ratio wells were shut in assoon as the field was unitized, and water injection was started in 1954. Theoperation has gone as planned and approximately 50% of the initial oil inplace has been recovered, in contrast to approximately 25% by primary deple-tion, an increase of approximately 600 MM STB of recoverable 011.15

5. THE GLOYD-MITCHELL ZONE OF THE RODESSAFIELD

Many reservoirs arc of the volumetric undersaturatcd type and their produc-tion, therefore, is controlled largely by the solution gas-drive mechanism. Inmany cases, the mechanism is altered to a greater or lesser extent by gravita-tional segregation of the gas and oil, by small water drives, and by pressuremaintenance, all of which improve recovery. The important characteristics ofthis type of production may be summarized as follows and observed in thegraph of Fig. 5.5 for the Gloyd-Mitchell zone of the Rodessa Field. Above thebubble point, the reservoir is produced by liquid-expansion, and there is arapid decline in reservoir pressure that accompanies the recovery of a fractionof 1% to a few percentage points of the initial oil in place. The gas-oil ratiosremain low and generally near the value of the initial solution gas-oil ratio.Below the bubble point, a gas phase develops that in most cases is immobileuntil the gas saturation reaches the critical gas saturation in the range of a fewpercentage points to 20%. During this period, the reservoir produces by gasexpansion, which is characterized by a much slower decline in pressure andgas-oil ratios near, or in some cascs even below the initial solution gas-oilratio. After the critical gas saturation is reached, free gas begins to flow Thisreduces the oil flow rate and depletes the reservoir of its main source ofenergy. By the time the gas saturation reaches a value usually in the range of15 to 30%, the flow of oil is small compared with the gas (high gas-oil ratios),and the reservoir gas is rapidly depleted. At abandonment the recoveries areusually in the range of 10 to 25% by the solution gas-drive mechanism alone,but they may be improved by gravitational segregation and the control of highgas-oil ratio wells.

The production of the Gloyd-Mitchell zone of the Rodessa Field, Louisi-ana, is a good example of a reservoir which produced during the major portionof its life by the dissolved gas-drive mechanism.'6 Reasonably accurate data onthis reservoir relating to oil and gas production, reservoir pressure decline,


and sand thickness, and the number of producing wells provide an cxcellentexample of the theoretical features of the dissolved gas-dnve mechanism. TheGloyd-Mitchcll zone is practically flat and produced an oil of 42.8°API gravitywhich, under the original bottom-hole pressure of 2700 psig, had a solutiongas-oil ratio of 627 SCF/STB. There was no free gas originally present, andthere is no evidence of an active water drive The wells were produced at highrates and had a rapid decline in production. The behavior of the gas-oil ratios,reservoir pressures, and oil production, had the characteristics expected of adissolved gas drive, although there is some evidence that there was a modifica-tion of the recovery mechanism in the later stages of depletion. The ultimaterecovery was estimated at 20% of the initial oil in place.

Many unsuccessful attempts were made to decrease the gas-oil ratios byshutting in the wells, by blanking off upper portions of the formation inproducing wells, and by perforating only the lowest sand members. The failureto reduce the gas-oil ratios is typical of the dissolved gas-drive mechanism,because when the critical gas saturation is reached, the gas-oil ratio is a func-tion of the decline in reservoir pressure or depiction and is not materiallychanged by production rate or completion methods. Evidently there wasnegligible gravitational segregation by which an artificial gas cap develops andcauses abnormally high gas-oil ratios in wells completed high on the structureor in the upper portion of the formation.

Table 5.5 gives the number of producing wells, average daily production,average gas-oil ratio, and average pressure for the Gloyd-Mitchell zone. Thedaily oil production per well, monthly oil production, cumulative oil produc-tion, monthly gas production, cumulative gas production, and cumulativegas-oil ratios have been calculated from these figures. The source of data is ofinterest. The number of producing wells at the end of any period is obtainedeither from the operators in the field, from the completion records as filedwith the state regulatory body, or from the periodic potential tests. The aver-age daily oil production is available from the monthly production reports filedwith the state regulatory commission. Accurate values for the average dailygas-oil ratios can be obtained only when all the produced gas is metered.Alternativcly, this information is obtained from the potential tests. To obtainthe average daily gas-oil ratio from potential tests during any month, thegas-oil ratio for each well is multiplied by the daily oil allowable or dailyproduction rate for the same well, giving the total daily gas production. Theaverage daily gas-oil ratio for any month is the total daily gas production fromall producing wells divided by the total daily oil production from all the wellsinvolved. For example, if the gas-oil ratio of well A is 1000 SCF/STB and thedaily rate is 100 hbl/day, and the ratio of well B is 4000 SCF/STB and the dailyrate is 50 bbl/day, then the average daily gas-oil ratio R of the two wells is

R =100+4000 x 50=

2000 SCF/STB

5. The Gloyd-Mitchell Zone of the Rodessa Field 165

This figure is lower than the arithmetic average ratio of 2500 SCF/STB. Theaverage gas-oil ratio of a largenumber of weLls, then, can be expressed by

(5.17)

where R and q0 are the individual gas-oil ratios and stock tank oil productionrates.

Figure 5.5 shows, plottcd in block diagram, the number of producingwells, the daily gas-oil ratio, and the daily oil production pcr well. Also in asmooth curve, pressure is plotted against time. The initial increase in daily oilproduction is due to the increase in the number of producing wells, and notto the improvement in individual well rates. If all the wells had been com-pleted and put on production at the same time, the daily production ratewould have been a plateau during the time all the wells could make theirallowables. followed by an exponential decline, which is shown beginning at16 months after the start of production. Since the daily oil allowable and dailyproduction of a well are dependent on the bottom-hole pressure and gas-oilratio, the oil recovery is larger for wells completed early in the life of a field.Because the controlling factor in this type of mechanism is gas flow in thereservoir, the rate of production has no material effect on the ultimate recov-ery, unless some gravity drainage occurs. Likewise, well spacing has no proveneffect on recovery; however, well spacing and production rate directly affectthe economic return.

-Liih_

—

-- -- - —

4 8 1Z IS 28 32 36 40MONTHS AFTER INITiAL PRODUCTION

Fig. 5.5. Development, production. and reservoir pressure curves for the Gloyd.Mitchell Zone, Rodessa rield, Louisiana.

- - U,

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625

21

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750

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11,4

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694

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233

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8048

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18,6

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54

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640

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626,

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1181

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1443

1255

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213

355,

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1752

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310,

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2.72

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554M

1880

1679

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434

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769,

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The rapid increase in gas-oil ratios in the Rodessa Field led to the enact-ment of a gas-conservation order. In this order, oil and gas production wereallocated partly on a volumetric basis to restrict production from wells withhigh gas-oil ratios. The basic ratio for oil wells was set at 2000 SCFIbb1. Forleases on which the wells produced more than 2000 SCF/STI3, the allowablein barrels per day per well, based on acreage and pressure, was multiplied by2000 and divided by the gas-oil ratio of the well. This cut in productionproduced a double hump in the daily production curve.

In addition to a graph showing the production history versus time, it isusually desirable to have a graph that shows the production history plottedversus the cumulative produced oil. Figure 5.6 is such a plot for the Gloyd-Mitchell zone data and is also obtained from Table 5.5. This graph shows somefeatures that do not appear in the time graph. For example, a study of thereservoir pressure curve shows the Gloyd-Mitchell zone was producing byliquid expansion until approximately 200,000 bbl were produced. This wasfollowed by a period of production by gas expansion with a limited amount offree gas flow. When approximately 3 million bbl had been produced, the gasbegan to flow much more rapidly than the oil, resulting in a rapid increase inthe gas-oil ratio. In the course of this trend, the gas-oil ratio curve reached amaximum, then declined as the gas was depleted and the reservoir pressureapproached zero. The decline in gas-oil ratio beginning after approximately4.5 million bbl were produced was due mainly to the expansion of the flowingreservoir gas as pressure declined Thus the same gas-oil ratio in standardcubic feet per day gives approximately twice the reservoir flow rate at 400 psig

Fig. 5.6. 1 listory of the Gloyd-Mitchell Zone of the Rodessa Field plotted versuscumulative recovery.

5 The Gloyd.MitcheU Zone of the Rodessa Field 169

as at 800 psig; hence, the surface gas-oil ratio may decline and yct the ratio ofthe rate of flow of gas to the rate of flow of oil under reservoir conditionscontinucs to increase. It may also be reduced by the occurrence of somegravitational segregation, and also, from a quite practical point of view. by thefailure of operators to measure or report gas production on wells producingfairiy low volumes of low-pressure gas.

The results of a differential gas-liberation test on a bottom-hole samplefrom the Oioyd zone show that the solution gas-oil ratio was 624 SCF/STB,which is in excellent agreement with the initial producing gas-oil ratio of 625SCF/STB.'7 In the absence of gas-liberation tests on a bottom-hole sample, theinitial gas-oil ratio of a properly completed well in either a dissolved gas drive,gas cap drive, or water-drive reservoir, is usually a reliable value to use for theinitial solution gas-oil ratio of the reservoir. The extrapolations of the pres-sure, oil rate, and producing gas-oil ratio curves on the cumulative oil plot allindicate an ultimate recovery of about 7 million bbl. However, no such extrap-olation can be made on the time plot It is also of interest that whereas thedaily producing rate is exponential on the time plot, it is close to a straight lineon the cumulative oil plot.

The average gas-oil ratio during any production interval and the cumu-lative gas-oil ratio may be indicated by integrals and shaded areas on a typicaldaily gas-oil ratio versus cumulative stock tank oil production curve as shownin Fig. 5.7. If R represents the daily gas-oil ratio at any time, and thecumulative stock tank production at the same time, then the production dur-ing a short interval of time is dN,, and the total volume of gas produced dunngthat production interval is R The gas produced over a longer period whenthe gas-oil ratio is changing is given by

(5.18).vpI

The shaded area between and is proportional to the gas producedduring the interval The average daily gas-oil ratio during the production

dNp

CUMULATIVE STB PRODUCED

Fig. 5.7. Typical daily gas-oil ratio curve for a dissolved gas drive reservoir


interval equals the area under the gas-oil ratio curve between and N,,2 inunits given by the coordinate scales, divided by the oil produced in the interval

and

JRdN,,

Ray8 = (iv;,—(5.19)

The cumulative gas-oil ratio, is the total net gas produced up to any perioddivided by the total oil produced up to that period, or

JRdN,,

N,,(520)

The cumulative produced gas-oil ratio was calculated in this manner inCol. (11) of Table 5.5. For exampie, at the end of the third period,

625x 12,160+750x15,200+875x21,280 F/TB12,J60+15,200+21,280

h73 SC S

6. CALCULATIONS INCLUDING FORMATION ANDWATER COMPRESSIBIUTIES

In Chapter 1, it was shown that both formation and water compressibilities arefunctions of pressure. This suggests thai there are in fact no volumetricreservoirs—that is, those in which the hydrocarbon pore volume of the reser-voir remains constant. Hall showed the magnitude of the effect of formationcompressibility on volumetric reservoir calculations.'8 The term volumetric,however, is retained to indicate those reservoirs in which there is no waterinflux but in which volumes change slightly with pressure due to the effects justmentioned.

The effect of compressibilities above the bubble point on calculations forN are examined first. Equation (2.8), with R,, = R,0, above the bubble pointbecomes

N B,, Aji + = + LW,, (5.21)

This equation may be rearranged to solve for N

w Wi —

1— *+.1

6. Calculations Including Formation and Water Compressibilities 171

Although this equation is entirely satisfactory, often an oil compressibility, c0,is introduced with the following defining relationship

— _B0—B0,C0

— v01(p1 —

and

B0 = B0, + (5.23)

The definition of c0 uses the single-phase formation volume factor, but itshould be apparent that as long as the calculations are being conducted abovethe bubble point, = B,. If Eq. (5.23) is substituted into thc first term inEq. (5.21), thc result is

N(B, — B,) + N B, = — W, + LW,, (5.24)

Multiplying both the numerator and the denominator of the term containingc0 by S0 and realizing that above the bubble point there is no gas saturation,S0=1—S,,,,, Eq. (5.24) becomes

NB,, + N = NAB, — +

or

NB,,[c +cwS1i + C1]

= N,,B, — W, + B,, (5.25)

The expression in brackets of Eq. (5.25) is called the effective fluid com-pressibility, c,, which includes the compressibilitics of the oil, the connatewater, and the formation, or

526C, ( . )

Finally, Eq. (5.26) may be written as

= NIB, — W, + (5.27)

For volumetric reservoirs, W, =0 and W,, is generally negligible, and Eq (5.27)can be rearranged to solve for N.

N IB,\(5.28)c4p


Finally, if the formation and water compressibilities, c, and both equal zero,then c, is simply c0, and Eq. (5 28) reduces to Eq. (5.8) derived in Sect 3, forproduction above the bubble point.

- -N B,,

Example 5 3 shows the use of Eq. (5.22) and (5.28) to find the initial oil inplace from the pressure-production data of a reservoir that all geologic evi-dence indicates is volumetric (i.e., it is bounded on all sides by impermeablerocks). Because the equations are basically identical, they give the samecalculation of initial oil, 51.73 MM STB. A calculation is also included to showthat an error of 61 % is introduced by neglecting the formation and watercompressibilities.

Example 5.3. Calculation of initial oil in place in a volumetric, under-saturated reservoir.

Given

B,, = 1 35469 bhl/STB

B, at 3600 psig = 1.37500 bbIISTB

Connate water = 0.20

Li',. at 3600 psig = 1.04 hbl/STB

c1= 5.0 (10)_b

5000 psig

N,, = 1.25 MM SIBat 3600 psig = 1400 psi

= 32.00() SIBwe=0

SOLUTION Substituting into Eq (5.23)

— 1,250,000(1.37500) + 32,000(1.04)N

— 1 37500—1.35469 +

= 51.73MM STB

The average compressibility of the reservoir oil is

B0—!?0, 1375—135469 6C0 =

- B04p = 1.35469 (5000 — 3600)= 10 71 (10) psi

6. Calculations Including Formation and Water Compressibilities 173

and the effective fluid compressibility by Eq (5.26) is

6psi 1

Then the initial oil in place by Eq. (5.28) is

N — 1,250,000(1.37500) + 32,000(1.04)—51 73 MM—

. STB

If the water and formation compressibilitics are neglected, c, -= c0, and theinitial oil in place is calculated to be

N1,250,000(1.37500) + 32,000(1.04)

—86 25 MM STB—

As can he seen from the example calculations, the inclusion of the compressi-bility terms significantly affects the value of N. This is true above the bubblepoint where the oil-producing mechanism is depletion, or the swelling ofreservoir fluids. After the bubble point is reached, the compressihilitics havea much smaller effect on the calculations.

When Eq (2.8) is rearranged and solved for N, we get the following:

N - + (Rp R10JBg] - + Bwl4j

C —— B1, + B1,

s,,,,

This is the general material balance equation written fur an undersaturatedreservoir below the bubble point. The effect of water and formation compres-sibilities are accounted for in this equation. Example Problem 5.4 comparesthe calculations for recovery factor, for an undersaturated reservoirwith and without including the effects of the water and formation comprcssi-bilities.

Example 5.4 Calculation of for an undersaturated reservoir withno water production and neglible water influx The calculation is performedwith and without including the effect of comprcssibilities. Assume that thecritical gas saturation is not reached until after the reservoir pressure dropsbelow 2200 psia

Given.c =3x 10 6psr'

Pb=2SOO psia c,=5 X

174 Undersaturated OH Reservoirs

Pressurepsia

H1,,

SCF/STB8R

bbl/SCFB,

bbl/SCF

4000 1000 0.00083 1 30002500 1000 0.00133 1 32002300 920 0.00144 1 39522250 900 0 00148 1 41802200 880 000151 1.4412

SoLuTIoN: The calculations are performed first by including the cffcctof compressibilities. Equation (5.22) is used to calculate the recovery at thebubble point.

IcSNBIBI+Bht[ i-Sn,

NL32 -130+1.30 1500

= 0.0276

Below the bubble point, Eqs. (5.29) and (5.13) arc used to calculate therecovery.

IC,.Sw1 + c11

N

and

R— -

-

During the pressure increment 2500 — 2300 psia, the calculations yield

N — 1.3952 (Re— 1000)0 00144

R— 0.0276(1000) + (iVy/N —

p Ne/N

where equals the average value of the solution GOR during the pressureincrement

1000 + 920Rawi=

2=960

6 Calculations Including Formation and Water Compressibilities 175

Solving these three equations for N,,/N yields

= 0.08391

Repeating the calculations for the pressure increment 2300 — 2250 psia, theis found to be

'=0.11754

Now, the calculations are performed by assuming that the effect of includingthe compressibility terms is negligible. For this case, at the bubble point, therecovery can be calculated by using Eq. (5.8)

N,, 1.32— L30_001515N 1.32

Below the bubble point, Eq. (5.11) and (5.13) are used to calculate N,,/N

N B:+(RpR301)Bg

and

R - —' -For the pressure increment 2500 —2300 psia

N,,_ 13952—1.30N 1 3952 + — 1000)0.00144

R — 0.01515(1000)p Na/N

where is given by

1000+920= 2

= 960

Solving these three equations yields

=0 07051


Repeating the calculations for the pressure increment 2300 — 2250 psia:

= 0.08707

For the pressure increment 2250 — 2200 psia

=010377

Figure 5.8 is a plot of the results for the two different cases—that is, with andwithout considering the compressibility term.

The calculations suggest that there is a very significant difference in theresults of the two cases down to the bubble point. The difference is the resuitof the fact that the rock and water comprcssibilities are on the same order ofmagnitude as the oil compressibility. By including them, the fractional recov-ery has been significantly affected. The case that used the rock and watercompressibilitics comes closer to simulating real production above the bubblepoint from this type of reservoir. This is because the actual mechanism of oil

a,

a,a,a,

5000

4000

2000

Fig. 5.8. Pressure versus fractional recovery for the calculations of Example Problem5.4.

F;actional Recovery

Problems 177

production is the expansion of the oil, water, and rock phases; there is no freegas phase.

Below the bubble point, the magnitude of the fractional recoveries calcu-lated by the two schemes still differ by about what the difference was at thebubble point, suggesting that bclow the bubble point, the compressibility ofthe gas phase is so large that the water and rock comprcssibilities do notcontribute significantly to the calculated fractional recoveries. This corre-sponds to the actual mechanism of oil production below the bubble pointwhere gas is coming out of solution and free gas is expanding as the reservoirpressure declines.

The results of the calculations of Ex. 5.4 are meant to help you tounderstand the fundamental production mechanisms that occur in under-saturated reservoirs. lucy are not meant to suggest that the calculations canbe made easier by ignoring terms in equations for particular reservoir situ-ations. The calculations are relatively easy to perform, whether or not allterms are included. Since nearly all calculations are conducted with the use ofa computer, there is no need to neglect terms from the equations.

PROBLEMS

5.1 Using the letter symbols for reservoir engineering, write expressions for thefollowing terms for a volumetnc, undersaturated reservoir:

(a) Thc initial reservoir oil in place in stock tank barrels.(b) The fractional recovery after producing N,, STB

(c) The volume occupied by the remaining oil (liquid) after producing N,, STB.

(d) The SCF of gas produced.

(e) The SCF of initial gas.(1) The SCF of gas in solution in the remaining oil

(g) By difference, the SCF of escaped or free gas in the reservoir after producingSTB

(h) The volume occupied by the escaped, or free, gas5.2 The physical characteristics of the 3—A —2 reservoir are given in Fig 5 2

(a) Calculate the percentage of recovery, assuming this reservoir could be pro-duced at a constant cumulative produced gas-oil ratio of 1100 SCF/STB.when the pressure falls to 3550. 2800, 2000, 1200, and 800 psia Plot thepercentage of recovery versus pressure

(b) To demonstrate the effect of increased GOR on recovery. reca'culate therecoveries, assuming that the cumulative produced (.}OR is 3300 SCF/STRPlot the percentage of recovery versus pressure on the same graph used forthe previous problem.


(c) To a first approximation, what does tripling the produced 00k do to thepercentage of recovery?

(d) Does this make it appear reasonable that to improve recovery high-ratio(OCR) wells should be worked over or shut in when feasible?

5.3 If 1 million STB of oil have been produced from the 3—A—2 reservoir at acumulative produced 00k of 2700 SCF/STB, causing the reservoir pressure todrop from the initial reservoir pressure of' 4400 psia to 2800 psia, what is theinitial stock tank oil in place?

5.4 The following data are taken from an oil field that had no original gas cap andno water drive

Oil pore volume of reservoir =75 MM cu ftSolubility of gas in crude = 0.42 SCF/STB/psiInitial bottom-hole pressure =3500 psiaBottom-hole temperature = 140°FBubble-point pressure of the reservoir = 2400 psiaFormation volume factor at 3500 psia = 1 333 bbIISTBCompressibility factor of the gas at 1500 psia and 140°F =0 95Oil produced when pressure is 1500 psia = 1.0 MM STENet cumulative produced CiOR = 2800 SCF/STB

(a) Calculate the initial STB of oil in the reservoir.(b) Calculate the initial SCF of gas in the reservoir

(C) Calculate the initial dissolved GOR of' the reservoir

(d) Calculate the SCF of gas remaining in the reservoir at 1500 psia.(e) Calculate the SCF of free gas in the reservoir at 1500 psia(f) Calculate the gas volume factor of the escaped gas at 1500 psia at standard

conditions of 14.7 psia and 60°F.

(g) Calculate the reservoir volume of the free gas at 1500 psia

(h) Calculate the total reservoir 00k at 1500 psia(I) Calculate the dissolved 00k at 1500 psia

(j) Calculate the liquid volume factor of the oil at 1500 psia

(k) Calculate the total, or two-phase, oil volume factor of the oil and its initialcomplement of dissolved gas at 1500 psia

5.5 (a) Continuing the calculations of the Kelly-Snyder Field, calculate the fractional recovery and the gas saturation at 1400 psig.

(b) What is the deviation factor for the gas at 1600 psig and BHT 125°F?

5.6 1 he R Sand is a volumetric oil whose PVT properties are shown in Fig5 9 When the reservoir pressure dropped from an initial pressure of 2500 psiato an average pressure of 1600 psia. a total of' 26.0 MM STB of oil had beenproduced The cumulative (JOR at 1600 psia was 954 SCF/STB, and the current00k was 2250 SCF/STB The average porosity for the field is 18% and average

Problems 179

connate watcr 18%. No appreciable amount of water was produced. and stan-dard conditions were J4 7 psia and 60°F

(a) Calculate the initial oil in place

(b) Calculate the SCF of evolved gas remaining in the reservoir at 1600 psia

(c) Calculate the gas saturation in thc reservoir at 1600 psia

(d) Calculate the barrels of oil that would have been recovered at 1600 psia ifall the produced gas had been returned to the reservoir

(e) Calculate the two-phase volume factor at 1600 psia

(f) Assuming no free gas flow, calculate the recovery expected by depletiondrive performance down to 2000 psia

(g) Calculate the initial SCF of free gas in the reservoir at 2500 psia

5.7 If the reservoir of Prob 5 6 had been a reservoir, in which 25 x 10bbbl of water had encroached into the reservoir when the pressure had fallen to1600 psia. calculate the initial oil in place Use the same current and cumulativeGORs, the same PVT data. and assume no water production

5.8 The following production and gas injection data pertain to a

(a) Calculate the average producing (JOR during the production interval from6 MM to 8 MM STB

(b) What is the cumulative produced GOR when 8 MM STB has been pro-duced'

(C) Calculate the net average producing GOR during the production intervalfrom 6 MM to 8 MM STB.

(d) Calculate the net cumulative produced GOR when 8 MM STh has beenproduced'

PRESSURE, PSIA

Fig. 5.9. PVT data for the R Sand Reservoir at 150°F


(e) Plot on the same graph the average daily the cumulative producedgas, the net cumulative produced gas, and the cumulative injected gas versuscumulative oil production

Cumulative OilProduction,

MM STB

Average Daily Gas-Oil Ratio, R

cumulative GasInjected,MM SCF

0 300 0 -

1 280 02 280 03 340 04 560 05 850 06 1120 5207 1420 9308 1640 14409 1700 2104

10 1640 2743

5.9 An undersaturatcd reservoir producing above the buhhlc point had an initialpressure of 5000 psia. at which pressure the oil volume factor was 1.510 bbl/STBWhen the pressure dropped to 4600 psia, owing to the production of 100,000S1'B of oil, the oil volume factor was 1 520 bbl/STB. The connate wateration was 25%, water compressibility 3 2 x 10 ',and, based on an averageporosity of 16%, the rock compressibility was 4 0 X 10 6 psi ' The averagecompressibility of the oil between 5000 and 4600 psia relative to the volume at5000 psia was 17 00 x 10-6 psi

(a) Geologic evidence and the absence of water production indicated a volumet-nc reservoir Assuming this was so. what was the calculated initial oil inplace"

(b) It was desired to inventoTy the initial stock tank barrels in place at a secondproduction interval. When the pressure had dropped to 4200 psia, formationvolume factor 1 531 bbl/STB, 205 M STB had been produced. If the averageoil compressibility was 17.65 X 10-6 what was the initial oil in place?

(C) When all cores and logs had been analyzed, the volumetric estimate of theinitial oil in place was 7 5 MM STB. If this figure is correct, how much waterentered the reservoir when the pressure declined to 4600 psia?

5.10 Estimate the fraction recovery from a sandstone reservoir by water drive if thepermeability is 1500 md, the connate water is 20%, the reservoir oil viscosity is1 5 cp. the porosity is 25%, and the average formation thickness is 50 ft.

5.11 The following PVT data are available for a reservoir, which from volumetricreserve estimation is considered to have 275 MM STB of oil initially in placeThe original pressure was 3600 psia. The curTent pressure is 3400 psia, and732,800 STB have been produced. How much oil will have been produced by thetime the reservoir pressure is 2700 psia?

Problems 181

Pressure(psia)

SolutionGas Oil Ratio

(SCF/STB)

FormationVolume Factor

(bbIISTB)

3600 567 1 3103200 567 1 317

2800 567 1 325

2500 567 1 333

2400 554 I 310

1800 434 1 263

1200 337 1 210

600 223 1140200 143 - 1070

5.12 Production data, along with reservoir and fluid data. for an undersaturatcdreservoir follow There was no measurable water produced. and it can be as-sumed that there was no free gas flow in the reservoir. Determine the

(a) Saturations of oil, gas, and water at a reservoir pressure of 2258.

(b) has water encroachment occurred and, if so. what is the volumc'

Gas gravity = 0.78Reservoir temperature = 160°FInitial water saturation = 25%Onginal oil in place 180 MM STBBubble-point pressure - 2819 psia

The following expressions for B0 and as functions of pressure were deter-mined from laboratory data:

l.0O+O.00015p, bbl/STBR,0 =50+ O.42p. SCF/STB

Cumulative Cumulative InstantaneousPressure Oil Produced Gas Produced GOR

(psia) (5TH) -- - ISCFISTB)

2819 0 0 10002742 4 38 MM 4.38 MM 12802639 10 16MM 1036MM 1480

2506 20 09 MM 21 295 MM 20002403 27 02 MM 30 26 MM 25002258 34 29 MM 41.15 MM 3300

5.13 The following table provides fluid property data for an initially undersaturatedlense type of' oil reservoir. The initial connate water saturation was 25%. Initialreservoir temperature and pressure were 97°F and 2110 psia. respectively Thebubble-point pressure was 1700 psia Average compressibility factors betweenthe initial and bubble-point pressures were 4 0(10) psia' and 3 1(10) 6psiafor the formation and water, respectively The initial oil formation volume factor


was 1.256 bhl/STB The critical gas saturation is estimated to he 10% Deter-mine the recovery versus pressure curve for this reservoir

Pressure(psia)

Oil FormationVolume Factor

(bb1ISTB)

SolutionGOR

(SCF1STB)

Gas FormationVolume Factor

(fr3ISCt)

1700 1265 540 00074121500 1.241 490 0.0084231300 1 214 440 00098261100 1.191 387 0.011792900 1.161 334 0.014711700 1.147 278 0.019316500 1117 220 - 0.027794

5.14 The Wildcat reservoir was discovered in 1970. The reservoir had an initialpressure of 3000 psia, and laboratory data indicated a bubble point pressure of2500 psia The connate water saturation was 22%. Calculate the fractionalrecovery, Ne/N, from initial conditions down to a pressure of 2300 psia Stateany assumptions which you make relative to the calculations.

Porosity 0.16.5Formation compressibility 2 5 (10) 6 psiaReservoir temperature = 150°F

ViscosityPressure

(psia)B0

(bb1ISTB) (SCFISTB) iB1

(bbIISCF)Ratiolao/P.R

3000 1.315 650 0 745 0.000726 53.912500 1.325 650 0680 0.000796 56 602300 1.311 618 0663 0.000843 6146

REFERENCES

'T L Kennerly, "Oil Reservoir Fluids (Sampling, Analysis, and Application ofData)," presented before the Delia Section of AIME, January. 1953. (Available fromCore Laboratories, Inc ,Dallas)2 0 Reudelbuber, "Sampling Procedures for Oil Reservoir Fluids," Jour. ofPetroleum Technology (December 1957) 9, 15—18.

Ralph H. kspach and Joseph Fry, "Variable Characteristics of the Oil in the TensleepSandstone Reservoir, Elk Basin Field. Wyoming and Montana." Trans. AIME (l951),192, 75

4Cccil 0 Cupps, Philip 11. Upstate. Jr , and Joseph Fry, "Variance in Characteristicsin the Oil in the Weber Sandstone Reservoir, Rangcly Field, Cob ," Bureau of MinesR. 1. 4761, U S.D.1 , April 1951, also World Oil. December, 1957, 133, No 7, 192

'A B. Cook. G. B. Spencer, F P Bobrowski, and Tim Chin, "A New Method ofDetermining Variations in PhysicalProperties of Oil in a Reservoir, with Application

References 183

to the Scurry Reef Field, Scurry County. Tex." U S Bureau of Mines R 1 5106,U.S.D.I. February, 1955, pp 12—23.

6D. R. McCord, "Performance Predictions Incorporating Gravity Drainage and GasCap Pressure Maintenance—LL-370 Area, Bolivar Coastal Field," Trans. AIME(1953). 198, 232.

7J. .1. Arps, "Estimation of Primary Oil Reserves," Trans AIME (1956), 207,183—186.

'R. C. Craze and S. E. Buckley, "A Factual Analysis of the Effect of Well Spacing onOil Recovery," Drilling and Production Practice, API (1945). pp 144—155

9J J. Arps and T. 0. Roberts, "The Effect of Relative Permeability Ratio, the OilGravity, and the Solution Gas-Oil Ratio on the Primary Recovery from a DepictionType Reservoir," Trans. AIME (1955), 204, 120-126.

'°H. 0. Botset and M Muskat, "Effect of Pressure Reduction upon Core Saturation,"Trans AIME (1939), 132, 172—183

"R K. Guthric and Martin K. Greenberger, "Tbe Use of Multiple CorrelationAnalyses for Interpreting Pctroleum-Enginccnng Data," Drilling and ProductionPractice, API (1955), pp. 135—137

'2Stewart E Buckley, "Petroleum Conservation New York: American Institute ofMining and Metallurgical Engineers, 1951, p 239.13K. B Barnes and R. F. Carison, "Scurry Analysis," Oil and Gas Journal (1950), 48,No. 51, 64.

'4"Matenal-Balancc Calculations, North Snyder Field Canyon Reef Reservoir," Oiland Gas Journal (1950), 49, No. 1, 85.

"R. M. Dicharry, T L. Pcrryman. and J. D. Ronquille, "Evaluation and Design ofa CO2 Miscible Flood Projcct—SACROC Unit, Kelly-Snyder Field," Jour, of Petro-leum Technology, November 1973, 1309—1318

'6Petroleum Reservoir Efficiency and Well Spacing (New Jersey: Standard Oil Devel-opment Company, 1943), p. 22

"ii B Hill and R K. Guthne, "Engineenng Study of the Rodessa Oil Field inLouisiana, Texas, and Arkansas," U.S. Bureau of Mines R.I 3715 (1943), p. 87'8Howard N. Hall, "Compressibility of Reservoir Rocks," Trans AIME (1953), 198,309.

Chapter 6

Saturated Oil Reservoirs

1. INTRODUCTION

The material balance equations discussed in Chapter 5 apply to volumetric andwater-drive rcscrvoirs in which there is no initial gas cap (I e., they are initiallyundersaturated). However, the equations apply to reservoirs in which anartificial gas cap forms owing either to gravitational segregation of the oil andfree gas phases below the bubble point, or to the injection of gas, usually inthe higher structural portions of the reservoir. When there is an initial gas cap(i.e.. the oil is initially saturated), there is negligible liquid expansion energy.However, the energy stored in the dissolved gas is supplemented by that in thecap, and it is not surprising that recoveries from gas cap reservoirs arc gen-erally higher than from those without caps, other things remaining equal. Ingas cap drives, as production proceeds and reservoir pressure declines, theexpansion of the gas displaces oil downward toward the wells. This phenom-enon is observed in the increase of the gas-oil ratios in successively lower wells.At the same time, by virtue of its expansion. the gas cap retards pressuredecline and therefore the liberation of solution gas within the oil zone, thusimproving recovery by reducing the producing gas-oil ratios of the wells. Thismechanism is most effective in those reservoirs of marked structural relief,which introduces a vertical component Qf fluid flow whereby gravitational

I 84

1. Introduction 185

segregation of the oil and free gas in the sand may occur. * 1 The recoveriesfrom volumetric gas cap reservoirs could range from the recoveries for under-saturated reservoirs up to 70 to 80% of the initial stock tank oil in place andwill be higher for large gas caps, continuous uniform formations, and goodgravitational segregation characteristics.

Large gas caps

The size of the gas cap is usually expressed relative to the size of the oil zoneby the ratio rn as defined in Chapter 2.

Continuous uniform formations

Continuous uniform formations reduce the channeling of the expanding gascap ahead of the oil and the bypassing of oil in the less permeable portions

Good gravitational segregation characteristics

These characteristics include primarily (a) pronounced structure, (b) low oilviscosity, (C) high permeability, and (d) low oil velocities.

Water drive and hydraulic control are terms used in designating a mech-anism that involves the movement of water into the reservoir as gas and oil areproduced. Water influx into a reservoir may be edge water or bottom water,the latter indicating that the oil is underlain by a water zone of sufficientthickness so that the water movement is essentially vertical. The most commonsource of water drive is a result of expansion of the water and the com-pressibility of the rock in the aquifer; however, it may result from artesianflow. The important charactenstics of a water-drive recovery process are thefollowing'

1. The volume of the reservoir is constantly reduced by the water influx. Thisinflux is a source of energy in addition to the energy of liquid expansionabove the bubble point and the energy stored in the solution gas and in thefree, or cap, gas.

2. The bottom-hole pressure is related to the ratio of water influx to voidage.When the voidage only slightly exceeds the influx, there is only a slightpressure decline. When the voidage considerably exceeds the influx, thepressure decline is pronounced and approaches that for gas cap or dissolvedgas-drive reservoirs, as the case may be.

3. For edge-water drives, regional migration is pronounced in the direction ofthe higher structural areas.

4. As the water encroaches in both edge-water and bottom-water drives, thereis an increasing volume of water produced, and eventually water is pro-duced by all wells.

'Refcrcnccs throughout the text are given at the end of each chapter

186 Saturated Oil Reservoirs

5. Under favorable conditions, the oil recoveries are high and range from 60to 80% of the oil in place.

2. MATERIAL BALANCE IN SATURATED RESERVOIRS

The general Schilthuis material balance equation was developed in Chapter 2and is as follows:

N(B, — B11)+

N B,, +

= + — + (2.7)

Equation (2.7) can be rearranged and solved for N, the initial oil in place:

— + — +

B, — B,, + — Bgi) + (1 + [c.I;s:i; 1]

If the expansion term due to the compressibilities of the formation and con-nate water can be neglected, as they usually arc in a saturated reservoir, thenEq. (6.1) becomes

— +N— (6.2)

Example 6.1 shows the application of Eq. (6.2) to the calculation of initial oilin place for a water-drive reservoir with an initial gas cap. The calculations aredone once by converting all barrel units to cubic feet units and then convertingall cubic feet units to barrel units. It does not matter which set of units is used,only that each term in the equation is consistent. Problems sometimes arisebecause gas formation volume factors are either reported in cu ItISCF or inbhl/SCF. Usually when applying the material balance equation for a liquidreservoir, gas formation volume factors are reported in bhl/SCF. Use care inmaking sure that the units are correct.

Example 6.1 To calculate the stock tank barrels of oil initially in placein a combination drive reservoir

(;iven.

Volume of bulk oil zone 112,000 ac-ftVolume of bulk gas ione = 19,600 ac-ft

2 Matenal Balance in Saturated Reservoirs 187

Initial reservoir pressure 2710 psiaInitial FVF = 1.340 bbl/STBInitial gas volume factor 0 006266 cu ft/SCFInitial dissolved GOR = 562 SCF/STBOil produced during the interval = 20 MM STBReservoir pressure at the end of the interval = 2000 psiaAverage produced GOR = 700 SCF/STBTwo-phase FVF at 2000 psia = 1.4954 bbl/STBVolume of water encroached = 11.58MM bblVolume of water produced = 1.05MM STI3FVF of the water =1 028 bbl/STBGas volume factor at 2000 psia 0.008479 cu ftISCF

SoI.uTuoN: In the use of Eq. (6.2):

B,, 1.3400X56157.5241 cuft/STBB, = 1.4954 x 5.615 = 8.3967 cu fIISTB

MM cuft05x1 028X5.615=6.O6MMrescuft

Assuming the same porosity and connate water for the oil and gas zones:

19,600m = 112,0000.175

Substituting in Eq. (6.2):

N=20 x 106 [8.3967 + (700 — 562)0.0084891 — (65.02 — 6.06) x I

8.3967 — 7 5241 + 0.175 (0.008489 — 0.006266)

=98.97MM STB

If B, is in barrels per stock tank barrel, then Bg must be in barrels per standardcubic foot and W, and in barrles, and the substitution is as follows.

N = 20 x 106 [1.4954 4 (700— 562)0.001510J — (11.58 — 1 05 x 1.028) x

1.4954— 1.3400+0.175 (0.001510—0001116)

= 98.97 MM STI3

In Chapter 2. the concept of dnve indexes, first introduced to the reser-voir engineering literature by Pirson, was developed.2 To illustrate the use ofthese drive indexes, calculations are performed on the Conroc Field, Texas.


Figure 6.1 shows the pressure and production history of the Conroe Field, andFig. 6.2 gives the gas and two-phase oil formation volume factor for the

reservoir fluids. Table 6 1 contains other reservoir and production data andthe calculations in column form for three different periods.

q) I

w

U)C,)'U

a.

TWO-PHASE OIL VOLUME FACTOR, CU FT PER ST!314 82 90 98 06

- - - —

2000——-- —

-

1400 — 'c -—

120('S

——

006 008 010 012

GAS VOLUME FACTOR, CU FT PER SCF

Fig. 6.2. Pressure volume relations for Conroc Field oil and original complement of

dissolved gas. (After Schilthuis,' Trans AIME.)

— ——

II RESCRYQIR'I 's.1 AT 4000 FEET SUISUS -

I ASS-OIL RATIO

-..-J' L——— —

OAsLT PADDUCTIOII — —

z CIUIILATIVI *OOUCTIQUS

-,

6

500

-uTIME, rEARS

Fig. 6.1. Reservoir pressure and production data, Conroe Held. (After Schilthuis3,Trans AIME.)

TA

BLE

6.1

.M

ater

ial b

alan

ce c

alcu

latio

n of

wat

er in

flux

or o

il in

pla

ce fo

r oi

l res

ervo

irs b

elow

the

bubb

le-p

oint

pre

ssur

e

For

Con

roe

Fie

ld. 8

= 7

.37

cuft/STB

mB

,,1B

,, =

259

SC

F/S

TB

= 0

.006

37C

Uft/

SC

FR

,0, =

600

SCF/

STB

(14.

4 ps

ia a

nd 6

0°F)

181,

225

ac-f

t =0.

224

810.

000

ac-f

t

Line

No.

Qua

niuy

Mon

ths

afte

r S

tart

of P

rodu

ctio

n

-U

nits

1218

2430

36

IM

MST

B90

7022

.34

3203

40.1

848

.24

2A

,,SC

F/ST

B16

3011

8010

7010

2599

53

pps

ig21

4321

0820

9820

8720

914

BR

cu f

t/SC

F00

0676

0.00

687

0.00

691

0.00

694

0.00

693

5B

,cu

ft/S

TB

7.46

7.51

7.51

7.53

7.52

6N

,,R,.

MM

SC

F14

,800

34,4

0048

.100

7SC

F/ST

B10

3047

039

58

(RpR

cu.)

Bg

CU

ft/S

'lB695

3.24

2.74

9(5

)+(8

)cu

ft/S

TB

1441

10.75

1026

10A

IR,,

CU

ft/S

CF

0.00

039

0 00

054

0.00

056

11(1

0) x

CU

ft/S

TB

0 10

10.

137

0.14

512

B, —

B,,

CU

ft/S

TB

0.09

0.14

0 15

13(1

1)i-

(12)

cuft/

ST

B0.

191

0.27

702

9514

(1)x

(9)

MM

cuft

131

345

495

15

W,—

W,,

MM

cuft

51.5

178

320

16

(14)—(15)

MMcuft

795

167

175

17

N=

(16)

/(13

)MMSTB

415

602

594

18

DD1

Fraction

0285

0244

0180

19SD

IFr

actio

n0.

320

0.23

90.

174

20W

DI

Frac

tion

0395

0516

0646

(1)

C m CD a. CD

Cl)

CD 0


The use of such tabular forms is common in many calculations of reser-voir engineering in the interest of standardizing and summarizing calculationsthat may not be reviewed or repeated for intervals of months or sometimeslonger. They also enable an engineer to take over the work of a predecessorwith a minimum of briefing and study. Tabular forms also have the advantageof providing at a glance the component parts of a calculation, many of whichhave significance themselves. The more important factors can be readily dis-tinguished from the less important ones, and trends in some of the componentparts often provide insight into the reservoir behavior. For example, the valuesof line 11 in Table 6.1 show the expansion of the gas cap of the Conroe Fieldas the pressure declines. Line 17 shows the values of the initial oil in placecalculated at three production intervals. These values and othcrs calculatedelsewhere are plotted versus cumulative production in Fig 6.3, which alsoincludes the recovery at each period, expressed as the percentage the cumu-lative oil is of the initial oil in place as calculated at that period. The increasingvalues of the initial oil during the early life of the field may be explained bysome of the limitations of the material balance equation discussed in Chap-ter 2, particularly the average reservoir pressure. Lower values of the averagereservoir pressure in the more permeable and in the developed portion of thereservoir cause the calculated values of the initial oil to be low through theeffect on the oil and gas volume factors. The indications of Fig. 6.3 are thatthe reservoir contains approximately 600 MM STB of initial oil and thatreliable values of the initial oil are not obtained until about 5% of the oil hasbeen produced. This is not a universal figure but depends on a number offactors, particularly the amount of pressure decline. For Conroe, the driveindexes have been calculated at each of three periods, as given in lines 18, 19,and 20 of Table 6.1. For example, at the end of 12 months the calculated initial

—I

IdU4-J0.

z-j

0w0a0

"a.I-zIiiU

'aa.A

/D,nr

,,— — — — — — — S

-j4

z10 20 30 40

PRQDUCTION,.MM BBL

Fig. 6.3. Activc oil, Conroc Field. (After Schilthuis3, Trans AIME.)

3. Matenal Balance as a Straight Line I

oil in place is 415 MM STB, and the value of + — given inline 14 is 131 MM cu ft. Then

DDJ_415 x 106(7.46 7.37)_

0 285131x6415 xlO6xO.224x7.37

SDI=- 131x 106 -

-=0.320

These figures indicate that during the first 12 months 39.5% of the productionwas by water drive, 32.0% by gas cap expansion, and 28.5% by depletiondrive. At the end of 36 months, as the prcssure stabilized, the current mech-anism was essentially 100% water drive and the cumulative mechanism in-creascd to 64.6% by water drive. If figures for recovery by each of the threemechanisms could be obtained, the overall recovery could be estimated usingthe drive indexes. An increase in the depletion drive and gas-drive indexeswould be reflected by declining pressures and increasing gas-oil ratios, andmight indicate the need for water injection to suppJement the natural waterinflux and to turn the recovery mechanism more toward water drive.

3. MATERIAL BALANCE AS A STRAIGHT LINE

In Chaptcr 2 Sect. 4, the method developed by Ilaviena and Odch of applyingthe genereal material balance equation was presented.4 This approach de-fines several new variables (see Chapter 2) and rewrites the material balanceequation as Eq. (2.13):

F = NE0 + N(1 ++ [NmBu]

E1 + W, (2.13)

This equation is then reduced for a particular application and arranged into aform of a straight line. When this is done, the slope and intercept often yieldvaluable assistance in determining such parameters as N and m. The use-fulness of this approach is illustrated by applying the method to the data fromthe Conroe Field example discussed in the last section.

For the case of a saturated reservoir with an initial gas cap, such as theConroe Field, and neglecting the compressibility term, Eq. (2.13)becomes

NmB1,FNL0+ (6.3)lJgj


If N is factored out of the first two terms on the right-hand side and both sidesof the equation arc divided by the expression remaining after factoring, we get

F W,=N+ (6.4)

For the example of the Conroc Field in the previous section, the water produc-lion values were not known. For this reason, two dummy parameters arcdefined as F' = F — and = W, — Equation (6.4) then becomes

F' =N+ w:(6.5)

Equation (6.5) is now in the desired form. If a plot of F'f[E,, + asthe ordinate and WYI E,, + mB,,Eg/Bg,t as the abscissa is constructed, a straightline with slope equal to I and intercept equal to N is obtained. Table 6.2contains the calculated values of the ordinate, lineS, and abscissa, line 7, usingthe Conroe Field data from Table 6.1. Figure 6.4 is a plot of these values.

if a least squares regression analysis is done on all three data pointscalculated in Table 6.2, the result is the solid line shown in Fig. 6.4. The linehas a slope of 1.21 and an intercept, of N, of 396 MM STB. This slope issignificantly larger than 1, which is what we should have obtained from theHavlena-Odeh method. If we now ignore the first data point, which representsthe earliest production, and determine the slope and intercept of a line drawnthrough the remaining two points (the dashed line in Fig. 6.4), we get 1.00 for

TABLE 6.2.Tabulated values from the Conroe Field for use in the Havlena-Odeh method

lineNo. Quantity Units

MMcuft

Months after Sian of Production

12 24 - 36

I F 131 345 4952 cuft/STB 0.09 0.14 0.153 ER cu ft/SCF 0.00039 000054 0 00056

4 cu ft/STB 0 191 0.280 0.295

5 (1)/(4) MM STB 686 1232 16786 W MMcuft 515 178 3207 (6)1(4) MM STB 270 636 1085

3. Material Balance as a Straight Line 193

F-U)

1000IL?

800

400

- mB,,E04—E9 .MMSTB

Fig. 6.4. Haviena-Odeb plot for the Conroe Field. Sobd line represents line drawnthrough all the data points Dashed line represents line drawn through data points fromthe later production periods

a slope and 600 MM STB for N, the intercept. This value of the slope meetsthe requirement for the I-Javlena-Odeh method for this case. We should nowraise the question: Can we justify ignoring the first point? If we realize that theproduction rcprcscnts less than 5% of the initial oil in place and the fact thatwe have met the requirement for the slope of I for this case, then there isjustification for not including the first point in our analysis. We conclude fromour analysis that the initial oil in place is 600 MM STB for the Conroe Field.

You may take issue with the fact that an analsysis was done on only twopoints. Clearly, it would have been better to use more data points, but nonewere available in this particular example As more production data are col-lected, then the plot in Fig. 6.4 can be updated and the calculation for Nreviewed The important point to remember is that if the Havlena-Odehmethod is used, the condition of the slope and/or intercept must bc met for theparticular case you are working with. This imposes another restriction onthe data and can be used to justify the exclusion of some data, as was done inthe case of the Conroc Field example.

1800

1600

1400

1200

2000 200 400 600 800 1000 1200


4. FLASH AND DIFFERENTIAL GAS LIBERATION

For heavy crudes whose dissolved gases are almost entirely methane andethane, the manner of separation is relatively unimportant. For lighter crudesand heavier gases (i.e., for reservoir fluids with larger fractions of the inter-mediate hydrocarbons—mainly propane, butanes, and pentanes), the mannerof separation raises some important questions. The nature of the difficulty liesmainly with these intermediate hydrocarbons which are, relatively speaking,intermediate between true gases and true liquids. They are therefore dividedbetween the gas and liquid phases in proportions that are affected by themanner of separation. The situation may be explained with reference to twowell-defined, isothermal, gas-liberation processes commonly used in labora-tory PVT studies. In the flash liberation process, all the gas evolved during areduction in pressure remains in contact and presumably in equilibrium withthe liquid phase from which it is liberated. In the differential process, on theother hand, the gas evolved during a pressure reduction is removed fromcontact with the liquid phase as rapidly as it is liberated. Figure 6.5 showsthe variation of solution gas with pressure for the differential process andthe specific gravity of the gas that is being liberated at any pressure. Since thespecific gravity of the gas is quite constant down to about 800 psia, it can beinferred that very close to the same quantity of gas would have been liberatedby the flash process, down to 800 psia and at the same temperature. Below 800psia, the vaporization of the intermediate hydrocarbons begins to be appre-ciable for the fluid of Fig. 6.5. In more volatile crudes, it begins at higher

Fig. 6.5. Gas solubility and gas gravity by the differential libcration process on asubsurface sample from the Magnolia Field, Arkansas (After Carpenter. Schroeder.and Cook,6 U.S. Bureau of Mines.)

RESERVOIR PRESSURE. P514

4 Flash and Differential Gas Liberation

pressures, and vice versa. The vaporization is indicted by the rise in the gasgravity and by the increasing rate of gas liberation, indicated by the steepeningof the slope dR30/dp If all the gas liberated down to, say, 400 psia remains incontact with the liquid phase, as in the flash process, more gas is Liberatedbecause the intermediate hydrocarbons in the liquid phase vaporize into theentire gas space in contact with the liquid until equilibrium is reached Be-cause gas is removed as rapidly as it is formed in the differential process, lessvaporization of the intermediates occurs. The release of solution gas at lowerpressures by the flash process, a: the same temperature, is further acceleratedbecause the loss of more of the intermediate hydrocarbons reduces the gassolubility. In some flash processes, the temperature is reduced at some pres-sure during the gas liberation process, whereas differential liberations arcgenerally run at reservoir temperature. Because of the increased gas solubilityand the lower volatility of the intermediates at lower temperatures, the quan-tity of gas released by the flash process is less at the lower temperatures andis commonly less than that by the differential process at rcservoir temperature.

Table 6.3 gives the PVT data obtained from a laboratory study of areservoir fluid sampie at reservoir temperature of 220°F The volumes givenin Col. (2) are the result of the flash gas-liberation process and arc shownplotted in Fig. 6.6. Below the bubble-point pressure at 2695 psig, the volumesinclude the volume of the liberated gas and are therefore two-phase volumefactors. Since the stock tank oil remaining at atmospheric pressure depends onthe pressure, temperature, and separation stages by which the gas is liberatedat lower pressures, the volumes are reported relative to the volume at thebubble-point pressure, To relate the reservoir volumes to the stock tank oilvolumes, additional tests are performed on other samples using small-scaleseparators, which are operated in the range of pressures and temperaturesused in the field separation of the gas and oil. Table 6.4 shows the results offour laboratory tests at separator pressures of 0, 50, 100, and 200 psig, andseparator temperatures from 74 to 77°F. The temperatures are lower for thelower separator pressures because of the greater cooling effect of the gasexpansion and the greater vaporization of the intermediate hydrocarbon com-ponents at the lower pressures. At 100 psig and 76°F, the tcsts indicate that 505SCF arc liberated at the separator and 49 SCF in the stock tank, or a total of554 SCF/STB by two-stage separation if the stock tank is called a stage.Otherwise, it is a single stage. Then the initial solution gas-oil ratio, R,0g, is 554SCF/STB. The tests also shows that under these separation conditions 1.335bbl of fluid at the bubble-point pressure yield 1.000 STB of oil. Hence, theformation volume factor at the bubble-point pressure is 1.335 bbl/STB and thetwo-phase flash FVF at 1773 psig is

1.335 x 1.1814= 1.577 bbl/STB

The data of Table 6.4 indicate that both oil gravity and recovery can be


TABLE 6.3.Reservoir fluid sample tabular data (After Kennerly, CoreLaboratories, Inc.)

Differential Liberation a: 220°FFlash Liberation at

220°F. Relative Volume Liberated Gas-Oil Solution Gas-Oil RelativePressure of Oil and Gas, Ratio, SCF/bbl Ratio, SCFIbhI Oil Volume

p:g V1Vb of Residual Oil of Residual Oil V/V,

5000 0 9739 1 3554700 0.9768 1 359

4400 0.9799 1 363

4100 0.9829 1 361

3800 0.9862 1 372

3600 0.9886 1 375

3400 0 9909 1 378

3200 0 9934 1 3823000 0 9960 1 385

2900 09972 1.387

2800 0 9985 I 389

2695 1 0000 0 638 1 391

2663 1 00382607 1 0101

2512 42 596 1 373

2503 1 0233

2358 1.04-17

2300 89 549 1 351

2197 1 07272008 150 488 1 323

2000 11160 152 486 1.3221773 118141702 213 425 1 295

1550 1 2691

1351 1 3792

1315 290 348 1.260

1180 1 5117

1010 351 287 1.232

992 1 7108

711 22404

705 412 226 1 205

540 28606410 37149

405 474 164 1175289 5 1788

150 539 99 1141

0 638 0 1.066

Residual volume at 60°F = I 000Residual oil gravity = 28 8°APISp gr of liberated gas = 1 0626

V — at givcn pressure

Vb = volume at bubble-point pressure

V. = residual oil volume at 14 7 psia and 60°F by diffcrential gas liberation at 220°F

4. Flash and Differential Gas Liberation 197

Fig. 6.6. Flash liberation PVT data for a reservoir fluid at 220°FCore Laboratoncs, mc)

(After Kennerly.

improved by using an optimum separator pressure of 100 psig, by reducingthe loss of liquid components, particularly the intermediate hydrocarbons, tothe separated gas. In reference to material balance calculations, they alsoindicate that the volume factors and solution gas-oil ratios depend on how thegas and oil are separated at the surface. When diffenng separation practicesarc used in the various wells owing to operator prefcrence or to limitations ofthe flowing wellhead pressures, further complications are introduced. Figure6.7 shows the variation in oil shrinkage with separator pressure for a West

TABLE 6.4.Separator tests of reservoir fluid sample (After Kennerly, CoreLaboratories, Inc)

Separator Stock Tank Formation SpecificSeparator Separator Gas-Oil Gas.Oil Stock Tank Volume Gravity ofPressure lemperarure Razio° Ratio° Gravity Factora Flash Gas

psig °F SCFISTB SCFISTB °,1 Pt a: 60°F VbN, Air =100

0 74 620 0 299 1.3142 0972550 75 539 23 31 5 1.340

100 76 505 49 319 1335200 77 459 98 318 1337

Standard conditions 14 7 psia and 60°Fb VbIV, — barrels of oil at the bubble-point pressure 2695 psig and 220°F per stuck tank

barrel at 14 7 psia and 60°F


Central Texas and a South Louisiana Field. Each crude oil has an optimumseparator pressure at which the shrinkage is a minimum and stock tank oilgravity a maximum. For example, in the case of the West Central Texas reser-voir oil, there is an increased recovery of 7% when the operating separatorpressure is increased from atmospheric pressure to 70 psig The effect of usingtwo stages of separation with the South Louisiana reservoir oil is shown by thetnangle.

The effect of changes in separator pressures and temperatures on gas-oilratios, oil gravities, and shrinkage in reservoir oil was determined for theScurry Reef Field by Cook, Spencer, Bobrowski, and Chin.7 The data ob-tained from field and laboratory tests showed that the amount of gas liberatedfrom the oil produced was affected materially by changes in both separatortemperatures and pressures. For example, when the separator temperaturewas reduced to 62.5°F, the gas-oil ratio decreased from 1068 to 844 SCF/STI3and the production increased from 125 to 135 STB/day. This was a decreasein gas-oil ratio of 21% and a production increase of 8%. Therefore, to yield

'Iia.

'Ii-

35'U>

U)-w

ZO4'-U)-I

790

780 — — — ;; —71C — —

— —

2 ) 4

WEST

E) 80 ICOEPARATOR PRESSURE, PSIG

TEXAS RESERVOiR OIL500-50-0 STAGING

Ia.

u,2-J'U

4m

)

7 z —

645 —

640 —200 400 600 600 1000 200

SEPARATOR PRESSURE, PSIGSOUTH LOUISIANA RESERVOIR OIL

Fig. 6.7. Variation in stock tank recovery with separator pressure (After Kennerly,Core I.aboratorics, Inc.)

4 Flash and Differential Gas Uberation 199

the same volume of stock tank oil, the production of 8% more reservoir fluidwas needed when the separator was operating at the higher temperature.

Table 6.3 also gives the solution gas and oil volume factors for the samereservoir fluid by differential liberation at 220°F all the way down to atmo-spheric pressure, whereas the flash tests were stopped at 289 psig, owing tolimitations of the volume of the PVT cell. Figure 6.8 shows a plot of the oil(liquid) volume factor and the liberated gas-oil ratios relative to a barrel ofresidual oil (i.e., the oil remaining at I atm and 60°F after a differentialliberation down to I atm at 220°F) The volume change from 1.066 at 220°Fto 1.000 at 60°F is a measure of the coefficient of thermal expansion of theresidual oil. In some cases, a barrel of residual oil by the differential processis close to a stock tank barrel by a particular flash process, and the two arctaken as equivalent. In the present case, the volume factor at the bubble-pointpressure is 1.335 bbl per stock tank barrel by the flash process, using sepa-ration at 100 psig and 76°F versus 1.391 bbl per residual barrel by the differen-tial process. The initial solution gas-oil ratios are 554 SCF/STB versus 638SCF/rcsidual barrel, respectively.

Fig. 6.8. PVT data for the differential gas liberation of a reservoir fluid at 220°F.(After Kennerly, Core Laboratories, Inc.)


In addition to the volumetric data of Table 6.3, PVT studies usuallyobtain values for (a) the specific volume of the bubble-point oil, (b) thc therrnmal expansion of the saturated oil, and (c) the compressibility of the reservoirfluid at or above the bubble point. For the fluid of Table 6.3, the specificvolume of the fluid at 220°F and 2695 psig is 0.02163 cu ft/lb. and the thermalexpansion is 1 07741 volumes at 220°F and 5000 psia per volume at 74°F and5000 psia, or a coefficient of 0.00053 per The compressibility of the under-saturated liquid has been discussed and calculated from the data of Table 6 3in Chapter 1, Sect. 6, as 10.27 X 10 1 between 5000 psia and 4100 psia at220°F.

The deviation factor of the gas released by the differential liberationprocess may be measured, or it may he estimated from the measured specificgravity. Alternatively, the gas composition may be calculated using a set ofvalid equilibrium constants and the composition of the reservoir fluid, and thegas deviation factor may be calculated from the gas composition.

5. THE CALCULATION OF FORMATION VOLUMEFACTOR AND SOLUTION GAS-OIL RATIO FROMDIFFERENTiAL VAPORIZATION AND SEPARATOR TESTS

The data in Tables 6 3 and 6.4 can be combined to yield values for the oilformation volume factor and the solution gas-oil ratio. The formation volumefactor is calculated from Eq. (6.6) or (6 7), depending on whether the pressureis above or below the bubble-point pressure:For p > bubble-point pressure

= R V. (Beib) (6 6)

For p <bubble-point pressure

(6.7)

where

R. V. —Relative volume from the flash liberation test listed in labte 6.3as VIV,,

volume factor from separator tests listed in Table 6.4as Vb/V,

B,,d—Formation volume factor from differential liberation test listed in Table6 3 as V/V,

6 Volatile Oil Reservoirs 201

volume factor at the bubble point from differential liberationtest

The solution gas-oil ratio can be calculated using Eq. (6.8).

= RS0Ih — — (6.8)

where

RhOfb—Sum of separator gas and the stock tank gas from separator tests listedin 'lable 6.4

gas-oil ratio from differential liberation test listed in Ta-ble 6.3

R30db—Thc value of at the bubble point

For example, at a pressure of 5000 psia and separator conditions of 200 psigand 77°F

= 0.9739(1.337) = 1.302 bbVSTB

and

R50 =459+98=557 SCF/STB

(Recall that = R30b for pressures above the bubble point.)At a pressure of 2512 psig, which is below the bubble-point pressure, andR30 become

= 1.373 = 1.320 bbIISTB

and

557— [(638— = 516 SCF/STB

6. VOLATILE OIL RESERVOIRS

If all gas in reservoirs were methane and all oil were decane and heavier, thePVT properties of the reservoir fluids would be quite simple because thequantities of oil and gas obtained from a mixture of the two would be almostindependent of the temperatures, pressures, and type of the gas liberationprocess by which the two are separated. Low volatility crudes approach this


behavior, which is approximately indicated by reservoir temperatures below150°F, solution gas-oil ratios below 500 SCFIbb1, and stock tank gravitiesbelow 35°API. Because the propane, butane, and pcntane content of thesefluids is low, the volatility is low.

For conditions above, but not too far above the approximate limits of lowvolatility fluids, satisfactory PVT data for material balance use are obtainedby combining separator tests at appropriate temperatures and pressures withthe flash and differential tests according to the procedure discussed in theprevious section. Although this procedure is satisfactory for fluids of moderatevolatility, it becomes less satisfactory as the volatility increases; and more com-plicated, extensive, and precise laboratory tests are necessary to provide PVTdata that are realistic in the application, particularly to reservoirs of thedepletion typc.

With present-day deeper drilling, many reservoirs of higher volatility arebeing discovered that include the gas-condensate reservoirs discussed in Chap-ter 4. The volatility is higher because of the higher reservoir temperatures atdepth, approaching 500°F in some cases, and also because of the compositionof the fluids, which are high in propane through decanc. The volatile oilreservoir is recognized as a type intermediate in volatility between the mod-erately volatile reservoir and the gas-condensate reservoir. Jacoby and Berryhave approximately defined the volatilc type of reservoir as one containingrelatively large proportions of ethane through decane at a reservoir tempera-ture near or above 250°F with a high formation volume factor and stock tankoil gravity above 45°API.9 The fluid of the Elk City Field, Oklahoma, is anexample. The reservoir fluid at the initial pressure of 4364 psia and reservoirtemperature of 180°F had a formation volume factor of 2.624 bbl/STB and asolution gas-oil ratio of 2821 SCF/STB, both relative to production through asingle separator operating at 50 psig and 60°F. The stock tank gravity was51 .4°API for these separator conditions. Cook, Spencer, and Bobrowski de-scribed the Elk City Field and a technique for predicting recovery by depletiondrive-performance.'0 Reudelhuber and Hinds, and Jacoby and Berry alsodescribed somewhat similar laboratory techniques and prediction methods forthe depletion drive performance of these volatile oil reservoirs.0'9 Themethods arc similar to those used for gas-condensate reservoirs discussed inChapter 4.

A typical laboratory method of estimating the recovery from volatilereservoirs is as follows. Samples of primary separator gas and liquid areobtained and analyzed for composition. With these compositions and a knowl-edge of separator gas and oil flow rates, the reservoir fluid composition can becalculated. Also, by recombining the separator fluids in the appropriate ratio,a reservoir fluid sample can be obtained. This reservoir fluid sample is placedin a PVT cell and brought to reservoir temperature and pressure. At thispoint, several tests are conducted. A constant composition expansion is per-

7 Maximum Efficient Rate (MER) 203

formcd to determine relative volume data These data are the flash liberationvolume data listed in Table 6.3. On a separate reservoir sample, a constantvolume expansion is performed while the volumes and compositions of theproduced phases are monitored. The produced phases are passed through aseparator system that simulates the surface facilities. By expanding the origi-nal reservoir fluid from the initial reservoir pressure down to an abandonmentpressure, the actual production process from the reservoir is simulated. Usingthe data from the laboratory expansion, the field production can be estimatedwith a procedure similar to the one used in Ex. 4.3 to predict performancefrom a gas-condensate reservoir.

7. MAXiMUM EFFICIENT RATE (MER)

Many studies indicate that the recovery from true solution gas-drive reservoirsby primary depletion is essentially independent of both individual well ratesand total or reservoir production rates. Keller, Tracy, and Roe showed thatthis is true even for reservoirs with severe permeability stratification where thestrata are separated by impermeable barriers and are hydraulically connectedonly at the wells.'2 The Gloyd-Mitchell zone of the Rodessa Field (see Chap-ter 5, Sect. 5) is an example of a solution gas-drive reservoir that is essentiallynot rate sensitive (i.e., the recovery is unrelated to the rate at which thereservoir is produced). The recovery from very permeable, uniform reservoirsunder very active water drives may also be essentially independent of the ratesat which they are produced.

Many reservoirs are clearly rate-sensitive, and there is a maximum effi-cient rate (MER) above which there will be a significant reduction in thepractical ultimate oil recovery 13 "Rate-sensitive reservoirs imply that there issome mechanism(s) at work in the reservoir, which, in a practical period, cansubstantially improve the recovery of the oil in place These mechanismsinclude (a) partial water drive, (b) gravitational segregation, and (c) thoseeffective in reservoirs of heterogeneous permeability.

When initially undersaturated reservoirs are produced under partial wa-ter drive at voidage rates (gas, oil, and water) considerably in excess of thenatural influx rate, they are produced essentially as solution gas-drive reser-voirs modified by a small water influx. Assuming that recovery by waterdisplacement is considerably larger than by solution gas drive, there will be aconsiderable loss in recoverable oil by the high production rate, even whenthe oil zone is eventually entirely invaded by water. The loss is caused by theincrease in the viscosity of the oil and the decrease in the volume factor of theoil at lower pressures, and also to the earlier abandonment of the wells thatmust be produced by artificial lift. Because of the higher oil viscosity at the


lower pressure, producing ratios wilt be higher, and the economiclimit of production rate will be reached at lower oil recoveries. Because of thelower oil volume factor at the lower pressure, at the same residual oil satur-ation in the invaded area, more stock tank oil will be left at low pressure.There are, of course, additional benefits to be realized by producing at sucha rate so as to maintain high reservoir pressure. If there is no appreciablegravitational segregation and the effects of reservoir heterogeneity are small,the MER for a partial water-drive reservoir can be inferred from a study of theeffect of the net reservoir voidage rate on reservoir pressure, and the con-sequent effect of pressure on the gas saturation relative to the critical gassaturation. (i.e., on gas-oil ratios) The MER may also be inferred fromstudies of the drive indexes, Eq. (6.3). The presence of a gas cap in a partialwater-drive field introduces complications in determining the MER, which isaffected by the relative size of the gas cap and the relative efficiencies of oildisplacement by the expanding gas cap and by the encroaching water.

The Gloyd-Mitchell zone of the Rodessa Field was not rate sensitivebecause there was no water influx and because there was essentially no gravi-tational segregation of the free gas released from solution and the oil. If therebad been substantial segregation, the well completion and well workovermeasures, which were taken in an effort to reduce gas-oil ratios, would havebeen effective as they are in many solution gas-drive reservoirs In some casesa gas cap forms in the higher portions of the reservoir, and, when high gas-oilratio wells are penalized or shut in, there may be a substantial improvementin recovery, as indicated by Eq. (5.11) for a reduction in the value of theproduced gas-oil ratio R,,. Under these conditions, the MER is that rate atwhich gravitational segregation is substantial for practical producing rates.

Gravitational segregation is also important in many gas cap reservoirsThe effect of displacement rate on recovery by gas cap expansion when thereis substantial segregation of the oil and gas is discussed in Chapter 9. Thestudies presented on the Mile Six Pool show that at the adopted displacementrate the recovery will be approximately 52.4% If the displacement rate isdoubled, the recovery will be reduced to about 36.0%, and at very high ratesit will drop to 14.4% for negligible gravity segregation

Gravitational segregation also occurs in the displacement of oil by water,and like the gas-oil segregation. it is also dependent on the time factor.Gravity segregation is generally of less relative importance in water drive thanin gas cap drive because of the much higher recoveries usually obtained bywater drive. The MER for water-drive reservoirs is that rate above which therewill be insufficient time for effective segregation and, therefore, a substantialloss of recoverable oil. The rate may be inferred from calculations similar tothose used for gas displacement in Chapter 9 or from laboratory studies. it isinteresting that in the case of gravitational segregation. the reservoir pressureis not the index of the MER. In an active water-drive field, for example, theremay be no appreciable difference in the reservoir pressure decline for a

Problems 205

several-fold change in the production rate, and yet recovery at the lower ratemay be substantially higher if gravity segregation is effective at the lower ratebut not at the higher.

As water invades a reservoir of heterogeneous permeability, the dis-placement is more rapid in the more permeable portions, and considerablequantities of oil may be bypassed if the displacement rate is too high. At lowerrates there is time for water to enter the less permeable portions of the rockand recover a larger portion of the oil As the water level rises, water issometimes imbibed or drawn into the less permeable portions by capillaryaction, and this may also help to recover oil from the less permeable areas.Because water imbihition and the consequent capillary expulsion of oil is farfrom instantaneous, if appreciable additional oil can be recovered by thismechanism, the displacement rate should be lowered when practicable to takeadvantage of it. Although the maximum efficient rate under these circum-stances is more difficult to establish, it may be inferred from the degree of thereservoir heterogeneity and the capillary pressure charactenstics of the reser-voir rocks.

In the present discussion of MER, it is realized that the recovery of oilis also affected by the reservoir mechanisms. fluid injection, gas-oil and water-oil ratio control, and other factors, and that it is difficult to speak ofratc-sensitive mechanisms entirely independently of these other factors, whichin many cases are far more important.

PROBLEMS

6.1 Calculate the values for the second and fourth periods through the fourteenthstep of Table 6.1 for Conroe Ficid.

6.2 Calculate the drive indexes at Conroe for the second and fourth periods

6.3 If the recovery by water drive at Conroe is 70%, by segregation dnve 50%, andby depletion drive 25%, using the drive indexes for the fifth period calculate theultimate oil recovery expected at Conroe.

6.4 Explain why the first material balance calculation at Conroc gives a low "alucfor the initial oil in place

6.5 (a) Calculate the single-phase formation volume factor on a stock tank basis.from the PVT data given in Tables 6.3 and 6.4. at a reservoir pressure of 1702psig, for separator conditions of 100 psig and 76°F.

(b) Calculate the solution CIOR at 1702 psig on a stock tank basis for the sameseparator conditions.

(c) the two-phase formation volume factor by flash separation at 1550psig for separator conditions of 100 psig and 76°F.

6.6 From the core data that follow, calculate the initial volume of oil and free gasin place by the volumetric method Then, using the material balance equation.calculate the cubic feet of water that have encroached into the reservoir at theend of the four periods for which production data are given.


Pressure B,(psia) (bbl/STF3) (fr3/SCF) (STB) (SCFISTB)

- W,,

(STB)

3480 1 4765 0.0048844 0 0 03190 15092 0.0052380 11.17MM 885 224.5M3139 15159 0.0053086 1380MM 884 534.2M3093 1.5223 0.0053747 16.41 MM 884 1005.0 M

3060 1 5270 0.0054237 18 59 MM 896 1554.0 M

Average porosity = 16.8%Connate water saturation = 27%Productive oil zone volume = 346,000 ac-ftProductive gas zone volume 73,700 ac-ft

1 025 bbl/STBReservoir temperature 207°FInitial reservoir pressure = 3480 psia

The following PVT data are for the Anetb Field in Utah.

Pressure(psia) (bb!ISTR) (SCF/STB) (bbl/SCF)

2200 1.383 727 — —1850 1 388 727 0.00130 351600 1.358 654 0.00150 391300 I 321 563 0.00182 471000 1 280 469 0.00250 56700 1.241 374 0.00375 68400 1.199 277 0.00691 85100 1.139 143 002495 13040 1100 78 005430 420

The initial reservoir temperature was 133°F The initial pressure was 2200 psia.and the bubble-point pressure was 1850 psia There was no active water driveFrom 1850 psia to 1300 psia a total of 720 MM STB of oil were produced and5906 MMM SCFof gas

(a) many reservoir barrels of oil were in place at 1850 psia9

(bi The average porosity was 10%. and connate water saturation was 28%. Thefield covered 50.000 acres. What is the average formation thickness in feet9

6.8 You have been asked to review the performance of a combination solution gas.gas-cap drive reservoir Well test and log information show that the reservoirinitially had a gas cap half the size of the initial oil volume. Initial reservoirpressure and solution gas-oil ratio were 2500 psia and 721 SCF/STB, respec-tively Using the volumetric approach, initial oil in place was found to be 56 MMSTB. As you precede with the analysis. you discover that your boss has not givenyou all the data you need to make the analysis. The missing information is thatat some point in the life of the project a pressure maintenance program wasinitiated using gas injection. The time of the gas injection and the total amountof gas injected are not known There was no active water drive or wdter produc-tion PVT and production data are in the following tableS

6.7

Problems 207

Pressure(psia)

B,(bbl/SCF)

B,(bbl/STJ3) (STB)

R,,

(SCFISTB)

2500 0.001048 1.498 0 0

2300 0001155 1.523 3.741MM 7162100 0.001280 1.562 6 849MM 9661900 0.001440 1.620 9 173MM 12971700 0.001634 I 701 10.99MM 16231500 0001884 I 817 12.42MM 19531300 0.002206 1.967 14.39MM 25511100 0.002654 2251 16.14MM 3214900 0.003300 2.597 17 38MM 3765700 0.004315 3.209 18.50MM 4317500 0.006163 4.361 1959MM 4839

(a) At what point (1 e., pressure) did the pressure maintenance program begin?

(b) How much gas in SCF had been injectcd when the reservoir pressure is 500psia? Assume that the reservoir gas and the injected gas have the samecompressibility factor

6.9 An oil reservoir initially contains 4 MM SIB of oil at its bubble-point pressureof 3150 psia with 600 SCF/STB of gas in solution. When the average reservoirpressure has dropped to 2900 psia, the gas in solution is 550 SCF/STB. B01 was1 34 bbl/STB and at a pressure of 2900 psia is 1.32 bbl/STB.

Other data:600 SCF/STB at 2900 psia

S1.,, 0.25B, = 0.0011 bbl/SCF at 2900 psia

volumetnc reservoirno original gas cap

(a) How many STB of oil will be produced when the pressure has decreased to2900 psia9

(b) Calculate the free gas saturation that exists at 2900 psia.

6.10 Given the following data froni laboratory core tests, production data, and log-ging information:

well spacing = 320 acnet pay thickness = 50 ft with the gas/oil contact 10 ft from the topporosity 0 17initial water saturation = 0.26initial gas saturation 0 15

bubble-point pressure 3600 psiainitial reservoir pressure 3000 psiareservoir temperature = 120°F

B,,, 1.26 bbl/STB8,, 1.37 bbl/STB at the bubble-point pressureB,, 119 bbl/STB at 2000 psia

2 0 MM SIB at 2000 psia2.4 MMM SCF at 2000 psia


gas compressibility factor, z = 1.0—0 000lpsolution gas-oil ratio, R,0 = 0.2p

Calculate the amount of water that has influxed and the drive indexes at 20(X)psia.

6.11 From the following information determine:

(a) Cumulative water influx at pressures 3625, 3530, and 3200 psia

(b) Water-drivc index for the pressures in (a).

Pressure(psia) (STB) (SCF)

W,,

(STB) (bhl/SCF) (SCt7STB) (bbl/STB)

3640 0 0 0 0.000892 888 1 4643625 0.06 MM 0 49 MM 0 0.000895 884 1.4663610 036MM 2.31MM 0001 MM 0000899 880 1.4683585 0.79 MM 4.12 MM 008 MM 0 000905 874 1.4693530 1.21MM 568MM 026MM 0000918 860 1.4763460 1.54 MM 7 00 MM 0.41 MM 0 000936 846 1.4823385 208MM 841MM 060MM 0000957 825 1.4913300 2.58 MM 9.71 MM 092 MM 0 000982 804 1.501

3200 3.40MM 1162MM 1.38MM 0001014 779 1.519

6.12 The cumulative oil production, and cumulative gas oil ratio, as functionsof the average reservoir pressure over the first 10 years of production for a gascap reservoir follow Use the Havlcna-Odeh approach to solve for the initial oiland gas (both free and solution) in place

Pressure(psia) (STB)

B0(bb!ISTB)

R,, -(SCFISTR)

Bg

(bbl/SCF,J

3330 0 0 1.2511 510 0000873150 3.295 MM 1050 1 2353 477 0.00092

3000 5 903 MM 1060 1 2222 450 0.000962850 8.852MM 1160 1.2122 425 0.001012700 11.503 MM 1235 1 2022 401 0 001072550 14513 MM 1265 11922 375 0001132400 17730MM 1300 11822 352 000120

6.13 Using the following data, determine the onginal oil in place by the Havlena-Odeh method Assume there is no water influx and no initial gas cap. Thebubble-point pressure is 1800 psia

Pressure B Rb,, Bg

(psia) (STB) (SCF/STR) (bbIISTB) (SG1STR)

1800 0 0 1.268 577 000097

1482 2223MM 634 1.335 491 000119

1367 2981 MM 707 1.372 460 0.001301053 5 787 MM 1034 1 540 375 0.00175

References 209

REFERENCES

Petroleum Reservoir Efficiency and Well Spacing (New Jersey Standard Oil Develop-ment Company, 1934), p. 24.

Sylvain J. Pirson, Ilemenis of Oil Reservoir Engineering, 2nd ed. New York:McGraw-hill, 1958, pp 635—693.

'Ralph J. Schilthuis, "Active Oil and Reservoir Energy," Trans. AIME (1936),118, 33.

'D. Haviena and A. S. Odeh, "The Material Balance as an Equation of a StraightLine," Jour, of Petroleum Technology (August 1963), 896—900.

5D. Haviena and A S. Odeh, "The Material Balance as an Equation of a StraightLine. Part 11—Field Cases," Jour, of Petroleum Technology (July 1964), 815-822.

bCharles B. Carpenter, H J. Shroeder, and Alton B. Cook, "Magnolia Oil Field,Columbia County, Arkansas," U S. Bureau of Mines, R. 1.3720(1943), pp 46.47,82

'Alton B. Cook, 0. B. Spencer, F P. Bobrowski, and Tim Chin, "Changes in Gas-OilRatios with Variations in Separator Pressures and Temperatures," Petroleum Engineer(March 1954), 26, B 77—B 82.

SAlton B. Cook, 0. B. Spencer, F P. Bobrowski, and Tim Chin, "A New Method ofDetermining Variations in Physical Properties in a Reservoir, with Application to theScurry Reef Field, Scurry County, Texas," U.S Bureau of Mines, R. 1.5106, February1955, pp. 10—11.

9R. H. Jacoby and V. J. Berry, Jr., "A Method for Predicting Depletion Performanceof a Reservoir Producing Volatile Crude Oil," Trans. AIME (1957), 201, 27.

'°Alton B. Cook, 0 B. Spencer, and F. P. Bobrowski, "Special Considerations inPredicting Reservoir Performance of Highly Volatile Type Oil Reservoirs," Trans.AIME (1951), 192, 37—46.

"F. 0. Reudelhuber and Richard F. Hinds, "A Compositional Material BalanceMethod for Prediction of Recovery from Volatile Oil Depletion Drive Reservoirs,"Trans. AIME (1957), 201, 19—26.

'2W. 0. Keller, 0. W. Tracy, and R. P Roc, "Effects of Permeability on Recoveryby Gas Displacement," Drilling and Production Practice, API (1949).

p. 218.Kraus, "MER—A History," Drilling and Production Practice, API (1947),

pp. 108—1 10.

"Stewart E Buckley, Petroleum Conservation. New York Amencan Institute ofMining and Metallurgical Engineers, 1951, pp. 151—163.

Chapter 7

Single-Phase Fluid Flow inReservoirs

1. INTRODUCTION

In the previous four chapters, the material balance equations for each of thefour reservoir types defined in Chapter 1 were developed. These matenalbalance equations may be used to calculate the production of oil and/or gas asa function of reservoir pressure. The reservoir engineer, however, would liketo know the production as a function of time. To learn this, it is necessary todevelop a model containing time or some related property such as flow rate.This chapter discusses the flow of fluids in reservoirs and the models thatare used to relate reservoir pressure to flow rate. The discussion in this chapteris limited to single-phase flow. Some multiphase flow considerations arcpresented in Chapters 9 and 10.

2. DARCY'S LAW AND PERMEABILITY

In 1856, as a result of experimental studies on the flow of water throughunconsolidated sand filter beds, Henry Darcy formulated the law that bearshis name. This law has been extended to describe, with some limitations, themovement of other fluids, including two or more immiscible fluids, in consol-

210

2 Darcy's Law and Permeability 211

idated rocks and other porous media. Darcy's law states that the velocity ofa homogeneous fluid in a porous medium is proportional to the driving forceand inversely proportional to the fluid viscosity, or

kldp 1

11= —0001127—11—0.433 'y cos aj (7 1)

where,

u the apparent velocity, bbls/day-ft2

k = permeability, millidarcies (md)

p. = fluid viscosity, cp

p = pressure, psia

s distance along flow path in ft= fluid specific gravity (always relative to water)

a = the angle measured counterclockwise from the downward vertical to thepositive s direction

and the term

— 0.433 y'cos a]

represents the driving force. The driving force may be caused by fluid pressuregradients (dplds) andfor hydraulic (gravitational) gradients (0.433 y'cos a). Inmany cases of practical interest, the hydraulic gradients, although alwayspresent, are small compared with the fluid pressure gradients, and arc fre-quently neglected. In other cases, notably production by pumping from reser-voirs whose pressures have been depleted, and gas cap expansion reservoirswith good gravity drainage characteristics, the hydraulic gradients are im-portant, and must be considered.

The apparent velocity, u, is equal to qB/A, where q is the volumetric flowrate in STB/day, B is the formation volume factor, and A is the apparent ortotal cross-sectional area of the rock in square feet. In other words, A includesthe area of the rock material as well as the area of the pore channels. The fluidpressure gradient, dplds, is taken in the same direction as u and q. Thenegative sign in front of the constant 0.001127 indicates that if the flow is takenas positive in the positive s-direction, then the pressure decreases in thatdirection so that the slope dplds is negative.

Darcy's law applies only in the region of laminar flow; in turbulent flow,which occurs at higher velocities, the pressure gradient increases at a greaterrate than does the flow rate. Fortunately, except for some instances of quitelarge production or injection rates in the vicinity of the well bore, flow in thereservoir and in most laboratory tests is by design, streamlined and Darcy'slaw is valid. Darcy's law does not apply to flow within individual pore channels

212 Single-Phase Fluid Flow in Reservoirs

but to portions of a rock the dimensions of which are reasonably large com-pared with the size of the pore channels—that is, it is a statistical law thataverages the behavior of many pore channels. For this reason, although sam-ples with dimensions of a centimeter or two are satisfactory for permeabilitymeasurements on uniform sandstones, much larger samples arc required forreliable measurements on rocks of the fracture and vugular types.

Owing to the porosity of the rock, the tortuosity of the flow paths, andthe absence of flow in some of the (dead) pore spaccs, the actual fluid velocityvaries from point to point within the rock and maintains an average that ismany times the apparent velocity Because actual velocities are in general notmeasurable, and to keep porosity and permeability separated, apparentveloc-ities form the basis of Darcy's law. This means that the actual average forwardvelocity of a fluid is the apparent velocity divided by the porosity where thefluid completely saturates the rock.

A basic unit of permeability is the darcy (d). A rock of 1 darcy perme-ability is one in which a fluid of I cp viscosity will move at a velocity of 1 cm/secunder a pressure gradient of 1 atm/cm Since this is a fairly large unit for mostproducing rocks, permeabilities are commonly expressed in units one thou-sandth as large, the milidarcy, or 0.001 darcy Throughout this text, the unitof permeability used is the millidarcy (md). Commercial oil and gas sands havepermeabilities varying from a few millidarcies to several thousands. Inter-granular limestone permeabilities may be only a fraction of a millidarcy andyet be commercial if the rock contains additional natural or artificial fracturesor other kinds of openings. Fractured and vugular rocks may have enormouspermcabilities, and some cavernous limestones approach the equivalent ofunderground tanks.

The permeability of a sample as measured in the laboratory may varyconsiderably from the average of the reservoir as a whole or a portion thereof,for there are often wide variations both laterally and vertically, the perme-ability sometimes changing several-fold within an inch in rock that appearsquite uniform. Generally, the permeability measured parallel to the beddingplanes of stratified rocks is larger than the vertical permeability. Also, in somecases the permeability along the bedding plane varies considerably and consis-tently with Core orientation, owing presumably to the oriented deposition ofmore or less elongated particles and/or the subsequent leaching or cementingaction of migrating waters. Some reservoirs show general permeability trendsfrom one portion to another, and many reservoirs are closed on all or part oftheir boundaries by rock of very low permeability, certainly by the overlyingcaprock. The occurrence of one or more strata of consistent permeability overa portion or all of a reservoir is common. In the proper development of reser-voirs, it is customary to core selected wells throughout the productive area,measunng the permeability and porosity on each foot of core recovered. Theresults are frequently handled statistically. I 2

Refcrcnces throughout the text are given at the cnd of each chapter

3. The Classification of Reservoir Flow Systems 213

Hydraulic gradients in reservoirs vary from a maximum near 0.500 psi/ftfor brines to 0.433 psi/ft for fresh water at 60°F, depending on the pressure,temperature, and salinity of the water. Reservoir oils and high-pressure gasand gas-condensate gradients lie in the range of 0.10 to 0.30 psi/fl, dependingon the temperature, pressure, and composition of the fluid. Gases at lowpressure will have very tow gradients (e.g., about 0.002 psi/ft for natural gasat 100 psia). The figures given are the vertical gradients The effective gradientis reduced by the factor cos a. Thus a reservoir oil with a reservoir specificgravity of 0.60 will have a vertical gradient of 0.260 psi/ft; however, if the fluidis constrained to flow along the bedding plane of its stratum, which dips at 150(a = 75°), then the effective hydraulic gradient is only 0.26 cos 750, or 0.067psi/ft. Although these hydraulic gradients are small compared with usualreservoir pressures, the fluid pressure gradients, except in the vicinity of wellbores, are also quite small and in the same range. Fluid pressure gradientswithin a few feet of well bores may be as high as tens of psi per foot but willfall off rapidly away from the well, inversely with the radius.

Measured static well pressures are usually corrected to the top of theproduction (perforated) interval using gradients measured within the wellbore, and thence up or down to a datum level using the gradient of thereservoir fluid. The datum level is selected near the center of gravity of theinitial hydrocarbon accumulation. Isobaric maps prepared from pressures cor-rected to datum may be used to find the average pressure of the hydrocarbonsand also to determine the direction and magnitude of fluid movement fromone portion of a reservoir to another.

Equation (7.1) suggests that the velocity and pressure gradient are re-lated by the mobility. The mobility, given the symbol A, is the ratio of perme-ability to viscosity, k/p.. The mobility appears in all equations describing theflow of single-phase fluids in reservoir rocks When two fluids are flowingsimultaneously—for example, gas and oil to a well bore, it is the ratio of themobility of the gas, to that of the oil, A0, which determines their individualflow rates. The mobility ratio M (sec Chapter 9) is an important factor affect-ing the displacement efficiency of oil by water. When one fluid displacesanother, the standard notation for the mobility ratio is the mobility of thedisplacing fluid to that of the displaced fluid. For water displacing oil, it isAwIA°.

3. THE CLASSIFICATION OF RESERVOIR FLOWSYSTEMS

Reservoir flow systems are usually classed according to (a) the type of fluid,(b) the geometry of the reservoir or portion thereof, and (c) the relative rateat which the flow approaches a steady-state condition following a disturbance

For most engineering purposes, the reservoir fluid may be classed eitheras (a) incompressible, (b) slightly compressible, or (c) compressible. The con-


cept of the incomprcssible fluid, volume of which does not change with pres-sure, simplifies the derivation and the final form of many equations, which arcsufficiently accurate for many purposes. However, the engineer should realizethat there are no true incompressible fluids.

A slightly compressible fluid, .which is the description of nearly allliquids, is sometimes defined as one whose volume change with pressure isquite small and expressible by the equation'

V = (7.2)

where

R = reference conditions

Equation (7.2) may be derived by integrating Eq (1.1), which definescompressibility between limits, assuming an average compressibility c, as

fVdVI —cdp=i —

JpR JVR V

P)_YR

V = P))

But ex may be represented by a series expansion as

Where x is small, the first two terms, 1 ÷ x, suffice, and where the exponentx is C(pR — p). the equation may be written as

V = + c(pR—p)1 (7.3)

A compressible fluid is one in which the volume has a strong dependenceon pressure. All gases are in this catcgory.Jn Chapter 1, the real gas law wasused to describe how gas volumes vary with pressure:

p

Unlike the case of the slightly compressible fluids, the gas isothermal corn-


pressibility, cannot be treated as a constant with varying pressure In fact,the following expression for c1 was developed:

I ldzCg = — — —

p zdp (1.19)

Although fluids are typed mainly by their compressibilities, in addition theremay be single phase or multiphase flow. Many systems are either only gas, oil,or water, and most of the remainder are either gas-oil or oil-water systems. Forthe purposes of this chapter, discussion is restricted to cases where there isonly a single phase flowing.

The two geometries of greatest practical interest are those that give riseto linear and to radial flow. In Jinear flow, as shown in Fig. 7.1, the flow linesare parallel and the cross section exposed to flow is constant. In radial flow,the flow lines are straight and converge in two dimensions toward a commoncenter (i.e., a well). The cross section exposed to flow decreases as the centeris approached. Occasionally, spherical flow is of interest, in which the flowlines are straight and converge toward a common center in three dimensions.Although the actual paths of the fluid particles in rocks are irregular due tothe shape of the pore spaces, the overall or average paths may be representedas straight lines in linear, radial, or sphencal flow.

Actually, none of these geometries is found precisely in petroleum reser-voirs, but for many engineering purposes the actual may often be closelyrepresented by one of the these idealizations. In some types of reservoirstudies (i.e., water flooding and gas cycling), these idealizations are inade-quate, and more sophisticated models are commonly used in their stead.

Flow systems in reservoir rocks are classified, according to their timedependence, as steady-state, transient, late transient, or pseudosteady-state.During the life of a well or reservoir, the type of system can changc severaltimes, which suggests that it is critical to know as much about the flow systemas possible in order to use the appropriate model to describe the relationshipbetween the pressure and the flow rate In steady-state systems, the pressureand fluid saturations at every point throughout the system adjust instanta-neously to a change in pressure or flow rate in any part of the system. Ofcourse, no real system can respond instantaneously, but some reservoir flow

Fig. 7.1. Common flow geometries

LINEAR RADIAL. SPHERICAL

Single-Phase Fluid Flow in Reservoirs

systems do resemble a steady-state process. This resemblance occurs whenany production from a reservoir is replaced with an equal mass of fluid fromsome external source. In Chapter 8, we consider a case of water influx thatcomcs close to meeting this requirement, but in general there are very fewsystems that can be assumed to be at steady-state conditions.

To consider the remaining three classifications of time dependence, wediscuss the movement of pressure that occurs when a change in the flow rateof a well located in the center of a reservoir, as illustrated in Figure 7.2, causesa pressure disturbance in the reservoir. The discussion assumes the following.(I) the flow system is made up of a reservoir of constant thickness and rockproperties; (2) the radius of the circular reservoir is and (3) the flow rate,after the disturbance has occurred, is constant As the flow rate is changed atthe well, the movement of pressure begins to occur away from the well. Themovement of pressure is a diffusion phenomenon and is modeled by thediffusivity equation (see Sect. 5). The pressure moves at a rate proportionalto the formation diffusivity,

k(7.4)

where k is the effective permeability of the flowing phase, 4 is the total effec-tive porosity, p. is the fluid viscosity of the flowing phase, and c, is the totalcompressibility. The total compressibility is obtained by weighting the com-pressibility of each phase by its saturation and adding the formation corn-pressibility, or:

C,C5S,+ (7.5)

T

I

Fig. 7.2. Schematic of a single well in a circular reservoir.


The formation compressibility, c1, should be expressed in change in porevolume per unit pore volume per psi. During the time the pressure is travelingat this rate, the flow is said to be transient. While the pressure is in thistransient region, the outer boundary of the reservoir has no influence on thepressure movement, and the reservoir acts as if it were infinite in size.

The late transient region is the period after the pressure has reached theouter boundary of the reservoir and bcfore the pressure behavior has had timeto stabilize in the reservoir. In this region, the pressure no longer travels at arate proportional to it is very difficult to describe the pressure behaviorduring this period.

The fourth period, the pseudosteady-state, is the period after the pres-sure behavior has stabilized in the reservoir. During this period, the pressureat every point throughout the rescrvoir is changing at a constant rate and asa linear function of time. This period has been incorrectly referred to as thesteady-state period.

An estimation for the time when a flow system of the type shown inFigure 7.2 reaches pseudosteady-state can be made from the followingequation

—7 6

— k (.)where is the time to reach pseudosteady-state expressed in hours.3 For awell producing an oil with a reservoir viscosity of 1.5 cp and a total conipressi-bility of 15 X 10-6 from a circular reservoir of 1000-ft radius with apermeability of 100 md and a total effective porosity of 20%:

1200(0 2)(1.5)(15(l0) 6)(J0002)54 h-

This means that approximately 54 hours, or 2 25 days, is required for the flowin this reservoir to reach pseudosteady-state conditions after a well located inits center is opened to flow, or following a change in the well flow rate. It alsomeans that if the well is shut in, it will take approximately this time for thepressure to equalize throughout the drainage area of the well, so that themeasured subsurface pressure equals the average drainage area pressure ofthe well.

This same criterion may be applied approximately to gas reservoirs butwith less certainty because the gas is more compressible. For a gas viscosity of0.015 cp and a compressibility of 400 x psi ',

— 1200(0.2)(0.015)(400(10)_6)(10002)— 14 4

100.hr

Thus, under somewhat comparable conditions (i.e., the same r, and k), gasreservoirs reach pseudosteady-state conditions more rapidly than do oil reser-


voirs. This is due to the much lower viscosity of gases, which more than offsetsthe increase in fluid compressibility. On the other hand, gas wells are usuallydrilled on wider spacings so that the value of r, generally is larger for gas wellsthan for oil wells, thus increasing the time required to reach pseudosteady-state. Many gas reservoirs, such as those found in the overthrust belt, areassociated with sands of low permeability. If we consider an r, value of 2500ft and a permeability of 1 md, which would represent a tight gas sand, then wecalculate the following value for

— 1200(0.2)(0.015)(400(10)&6)(25002)— 9000 h— r

The calculations suggest that reaching pseudosteady-state conditions in a typ-ical tight gas reservoir takes a very long time compared to a typical oil reser-voir. In general, pscudostcady-state mechanics suffice when the time requiredto reach pseudosteady-state is small compared with the time between substan-tial changes in the flow rate, or, in the case of reservoirs, small compared withthe total producing life (time) of the reservoir.

4. STEADY-STATE FLOW SYSTEMS

Now that Darcy's law has been reviewed and the classification of flow systemshas been discussed, the actual models that relate flow rate to reservoir pres-sure can be developed. The next several sections discuss the steady-statemodels. In this discussion, both linear and radial flow geometries are discussedsince there are many applications for these types of systems. For both thelinear and radial geometries, equations are developed for all three generaltypes of fluids (i.e., incompressible, slightly compressible, and compressible).

4.1. Linear Flow of Incompressible Fluids, Steady State

Figure 7.3 represents linear flow through a body of constant cross section,where both ends are entirely open to flow, and where no flow crosses the sides,top, or bottom if the fluid is incompressible, or essentially so for all en-

7

XLFig. 7.3.

4 Steady-State Flow Systems 219

gineering purposes, then the velocity is the same at all points, as is the totalflow rate across any cross section, so that,

—o

Separating variables and integrating over the length of the porous body;

Bf'dx =

q (7.7)Bp.L

For example, under a pressure differential of 100 psi for a permeability of 250md, a fluid viscosity of 2.5 cp, a formation volume factor of 1.127 bbIISTB, alength of 450 ft, and a cross-sectional area of 45 sq ft, the flow rate is

0 001127(250)(45)(100)

— 1 0 STB/d(1.127)(2.5)(450)

ay

In this intcgi'ation B, q, p., and k have been removed from the intcgral sign,assuming they are invariant with pressure. Actually, for flow above the bubblepoint, the volume, and hence the rate of flow, varies with the pressure asexpressed by Eq. (7.2). The formation volume factor and viscosity also varywith pressure, as explained in Chapter 1. Fatt and Davis have shown a vari-ation in permeability with net overburden pressure for several sandstones.4The net overburden pressure is the gross less the internal fluid pressure;therefore, a variation of permeability with pressure is indicated, particularlyin the shallower reservoirs. Becausc none of thcse effects is serious for a fewhundred psi difference, values at the average pressure may be used for mostpurposes.

4.2. Linear Flow of Slightly Compressible Fluids,Steady State

The equation for flow of slightly compressible fluids is modified from what wasjust dcnved in the previous section since the volume of slightly compressiblefluids increases as pressure decreases. Earlier in this chapter, Eq. (7.3) wasderived which describes the relationship between pressure and volume for aslightly compressible fluid The product of the flow rate, defined in SIB units,and the formation volume factor has a similar dependence on pressure and isgiven by

qB=qRIl+c(pR—p)I (7.8)


wherc is the flow rate at some reference pressure, PR- If Darcy's law iswritten for this case, variables separated, and the rcsulting equation integratedover the length of the porous body, the following is obtained:

= -- dp —-

o p. Pi l+c(PR—p)Ii

pLcIn11

+C(PR—PI)j(7.9)

This integration assumes a constant compressibility over the Cntirc pressuredrop. For example, under a pressure differential of 100 psi for a permeabilityof 250 md, a fluid viscosity of 2.5 cp, a length of 450 ft, a cross-sectional areaof 45 sq it, a constant compressibility of 65(10-6) psi', and choosing Pi as thereference pressure, the flow rate is

— (0.001127)(250)(45) {1+65x101

bbllday

When compared with the flow rate calculation in the preceding section, q1 isfound to be different due to the assumption of a slightly compressible fluid inthe calculation rather than an incompressible fluid. Note also, that the flowrate is not in STh units because the calculation is being done at a referencepressure that is not the standard pressure. If P2 is chosen to be the referencepressure, then the result of the calculation will be q2, and the value of thecalculated flow rate will be different still because of the volume dependenceon the reference pressure:

(0.001 127)(250)(45) { I Iq2(25)(450)(65 x 10 6(_joo)j = 1.131 bbl/day

The calculations show that q1 and q2 are not largeiy different, which confirmswhat was discussed earlier: the fact that volume is not a strong function ofpressure for slightly compressible fluids.

4.3. Linear Flow of Compressible Fluids, Steady State

The rate of flow of gas expressed in standard cubic feet per day is the same atall cross sections in a steady-state, linear system. However, because the gasexpands as the pressure drops, the velocity is greater at the downstream endthan at the upstream end, and consequently the pressure gradient increasestoward the downstream end. The flow at any cross section x of Fig. 7.3 wherethe pressure is p may be expressed in terms of the flow in standard cubic feetper day by substituting the definition of the gas formation volume factor

B-

q—


Substituting in Darcy's law:

qp,,Tz— 0001127

k dp- p.dr

Separating variables and integrating;

qpTzp..2 2

= — Jp1pdp — —P2)

Finally

A LI 2 2P2

7 10— PSCTZLIL

For example, where = 60°F, =45 sq ft, k = 125 md, P' = 1000 psia, P2=500 psia, pS = 14.7 psia, T = 140°F, z = 0.92, L = 450 ft, and = 0.015 cp

— 0.003164(520)(45)(125) (10002 _5002)— 12 Cq—-

14.7(600)(0.92)(450)(0.015)— 6.7 MS F/day

Here again, T, k, and the product p.z have been withdrawn from the integralsas if they were invariant with pressure, and as before, avcrage values may beused in this case. At this point, it is instructive to examine an observation thatWattenbarger and Ramey made about the behavior of the gas deviationfactor—viscosity product as a function of pressure.5 Figure 7.4 is atypical plotof liz versus pressure for a real gas. Note that the product, jiz, is nearlyconstant for pressures less than about 2000 psia. Above 2000 psia, the productp.z/p is nearly constant. Although the shape of the curve varies slightly fordifferent gases at different temperatures, the pressure dependence is repre-sentative of most natural gases of interest. The pressure at which the curvebends varies from about 1500 to 2000 psia for various gases. This variationsuggests that Eq. (7.10) is valid only for pressures less than about 1500 to 2000psia, depending on the properties of the flowing gas. Above this pressurerange, it would be more accurate to assume that the product p.zlp is constant.For the case of p.zlp constant, the following is obtained:

qpj(zpJp) —dx—

j,,1

0 OO6328kTKAC(pl _p2)(7.11)

In applying Eq. (7.11), the product p.z/p should be evaluated at the averagepressure between and P2.


a.U

P1

Fig. 7.4. Isothermal variation of p.z with pressure.

4.4. Permeability Variations in Linear Systems

Consider two or more beds of equal cross section hut of uncqual lengths andpermeabilities (Fig 7.5) in which the same linear flow rate q exists, assumingan incompressible fluid. Obviously the pressure drops are additive, and

P4) = (Pi — P2) + (P2 — p3) ÷ — JJ4)

Substituting the cquivalents of these pressure drops from Equation (7.7),

+ q313p.L3

— 0 001127k2Ari

I— L,—1—-L, I- - L1—1

Fig. 7.5. Scries flow in linear beds

4000 6000 6000

PRESSURE, p, PSIA

4. Steady-State Flow Systems 223

But since the flow rates, the cross sections, the viscosities, and the formationvolume factors (neglecting the change with pressure) are equal in all beds

L, L1 L2 L3

kL, 72avg L1 L2 Li/k, ( .1)

k1+

k2+

k3

The average permeability as defined by Eq. (7.12) is that permeability towhich a number of beds of various geometries and permeabilities could bechanged, and yet the same total flow rate under the same applied pressuredrop could be obtained.

Equation (7.12) was derived using the incompressible fluid equation.Because the permeability is a property of the rock and not of the fluids flowingthrough it, except for gases at low pressure, the average permeability must beequally applicable to gases. This rcquiremcnt may be demonstrated by observ-ing that for pressures below 1500 to 2000 psia:

(pf — = — + — + —

Substituting the equivalents from Eq. (7.10), the same Eq. (7.12) is obtained.The average permeability of 10 md, 50 md, and 1000 md beds, which are

6 ft, 18 ft. and 40 ft in length, respectively, but of equal cross section, whenplaced in series is

6+18+40k,,5

— L,1k1 — 6/10 + 18/50 + 40/1000 —flid

Consider two or more beds of equal length but unequal cross sectionsand permeabilitics flowing the same fluid in linear flow under the same pres-sure drop to P2) as shown in Fig 7.6.Obviously the total flow is the sum of the individual flows, or

q1q3 +q2-I-q3

And

—pi) = + +Bp.L


Cancelling

kavgAr: = k1 + +

=(7.13)

And where all beds are of the same width, so that their areas are proportionalto their thicknesses,

kavg1kM,

(7.14)

Where the parallel beds are homogeneous in permeability and fluid content,the pressure and the pressure gradient are the Same in all beds at equaldistances. Thus there will be no cross-flow between beds, owing to fluidpressure differences. However, when water displaces oil—for example from aset of parallel beds—the rates of advance of the flood fronts will be greater inthe more permeable beds. Because the mobility of the oil (k0/p.0) ahead of theflood front is different from the mobility of water behind the floodfront, the pressure gradients will be different In this instance, there will bepressure differences between two points at the same distance through therock, and cross flow will take place between the beds if they are not separatedby impermeable barriers. Under these circumstances, Eqs (7.13) and (7.14)arc not stnctly applicable, and the average permeability changes with the stageof displacement. Water may also move from the more permeable to the lesspermeable beds by capillary action, which further complicates the study ofparallel flow.

The average permeability of three beds of 10 md, 50 md, and 1000 md,

Fig. 7.6. Parallel flow in linear beds


and 6 ft, 18 ft, and 36 ft respectively in thickness but of equal width, whenplaced in parallel is

Ik,h1 10x6+18x50+36x1000_6+18+36 —616md

4.5. Flow Through Capillaries and Fractures

Although the pore spaces within rocks seldom resemble straight, smooth-walled capillary tubes of constant diameter, it is often convenient and instruc-tive to treat these pore spaces as if they were composed of bundles of parallelcapillary tubes of various diameters. Consider a capillary tube of length L andinside radius r0 which is flowing an incompressible fluid of p. viscosity inlaminar or viscous flow under a pressure difference of (Pi —p�). From fluiddynamics. Poiseuille's law, which describes the total flow rate through thecapillary, can be written as:

q = (7.15)Bp.L

Darcy's law the linear flow of incompressible fluids in permeable beds, Eq.(7.7), and Poiscuille's law for incompressible fluid capillary flow, Eq. (7.15),are quite similar:

q= —Pi) (7.7)Bp.L

Writing for area in Eq. (7.7), and equating it to Eq. (7.15);

k = (7.16)

Thus the permeability of a rock composed of closely packed capillaries, eachhaving a radius of 4.17(JO)6 foot (0.00005 in.), is about 200 md. And if only25% of the rock consists of pore channels (i.e , it has 25% porosity), thepermeability is about one-quarter as large, or about 50 md.

An equation for the viscous flow of incompressible wetting fluids throughsmooth fractures of constant width may be obtained as:

q = (7.17)Bp.L

In Eq (7.17). W is the width of the fracture, is the cross-sectional area ofthe fracture, which equals the product of the width Wand lateral extent of thefracture, and the pressure difference is that which exists between the ends of


the fracture of length L. Equation (7.17) may be combined with Eq. (7.7) toobtain an expression for the permeability of a fracture

a fracture only ft wide (0.001 in.) is53,SOOmd.Fractures and solution channels account for economic production rates

in many dolomite, limestone, and sandstone rocks, which could not be pro-duced economically if such openings did not exist. Consider, for example, arock of very low primary or matrix permcability, say 0.01 md, but whichcontains on the average a fracture 4. 1 ft in lateral extentper square foot of rock. Assuming the fracture is in the direction in which flowis desired, the law of parallel flow, Eq. (7.13) will apply, and

k— — (1)(4.17(1(J)4)j + (7 7(10)12(4.17(I0)_4)2(i(4.17(1O)_4)I

1

kavg=558 md

4.6. Radial Flow of incompressible Fluid, Steady State

Consider radial flow toward a vertical weilbore of radius r,,, in a horizontalstratum of uniform thickness and permeability, as shown in Fig. 7.7. If thefluid is incompressible, the flow across any circumference is a constant.be the pressure maintained in the welibore when the well is flowing q STB/dayand a pressure p. is maintained at the external radius re. Let the pressure atany radius r be p. Then at this radius r

qU qB Icdp— —0.001127—

2trrh p.dr

where positive q is in the positive r direction. Separating variables and inte-grating between any two radii, r1 and r2, where the pressures are p3 andrespectively

— —0.001127 dpPi P'

— () (10708 kh (P2 — P1)

pB In (r2/r3)

The minus sign is usually dispensed with, for where P2 is greater than theflow is known to be negative—that is. in the negative r direction, or toward the

— 0.00708 kh (P2 P1)q

p.R in (r2/r1)


Pe

- r,,

/I

1H

I I

Fig. 7.7.

Frequently the two radii of interest are the wellbore radius r,, and the externalor drainage radius r,. Thcn

0.00708kh (PePw)(7 19)q pB ln (re/rw)

The external radius is usually inferred from the well spacing. For example, acircle of 660 ft radius can be inscribed within a square 40 ac unit; so 660 ft iscommonly uscd for re with 40 ac spacing. Sometimes a radius of 745 ft is used,this being the radius of a circle 40 ac in area The weilbore radius is usuallyassigned from the bit diameter, the casing diameter, or a caliper survey. Inpractice, neither the external radius nor the wellbore radius is generallyknown with precision. Fortunately, they enter the equation as a logarithm, sothat the error in the equation will be much less than the errors in the radii.Since wellbore radii are about 113 ft and 40 ac spacing (r, = 660 ft) is quitecommon, a ratio 2000 is quite commonly used for Since in 2000 is 7.60and ln 3000 is 8.00, a 50% increase in the value of gives only a 5.3%increase in the value of the loganthm.

The external pressure p, used in Eq. (7.19) is generally taken as the staticwell pressure corrected to the middle of the producing interval, and theflowing well pressure Pw is the flowing well pressure also corrected to themiddle of the producing interval during a period of stabilized flow at rate qWhen reservoir pressure stabilizes as under natural water drive or pressuremaintenance, Eq. (7 19) is quite applicable because the pressure is maintainedat the external boundary, and the fluid produced at the well is replaced byfluid crossing the external boundary. The flow, however, may not be strictlyradial.

4.7. Radial Flow of Slightly Compressible Fluids, SteadyState

Equation (7 3) is again used to express the volume dependence on pressure forslightly compressible fluids. If this equation is substituted into the radial formof Darcy's law. the following is obtained:


Bq— 2'rrrh — p.dr

Scparating the variables, assuming a constant compressibility over the entirepressure drop, and integrating ovcr the length of the porous medium,

— 0.00708 kh [1 + C(PR(7.20)

p.c ln(r2/r1) 11+c(pR—p,)

4.8. RadIal Flow of Compressible Fluids, Steady State

The flow of a gas at any radius r of Fig. 7.7, where the pressure isp, may beexpressed in terms of the flow in standard cubic feet per day by:

B — qpjzq

Substituting in the radial form of Darcy's law

5.615T,p(2irrh) p. dr

Separating variables and integrating

qp,lzp. f'2dr (PP2 2

5.615(0.001127)(2ir)7,khJr1 — = —j pdp

qp,Tzp.2 2(r2/ri

Finally

= 0.019881kh (p1—( 21q

(r2/r1)

The product p.z has been assumed to be constant for the derivation of Eq.(7 21). It was pointed out in Sect. 4.3, that this is usually true only forpressures less than about 1500 to 2000 psia. For greater pressures, it was statedthat a better assumption was that the product p.z/p was constant. For this case,the following is obtained:

003976TJch (P1—p2)(722q

(r2Iri)

4. Steady-State Flow Systems 229

in applying Eqs. (7.21) and (7.22), the products p.z and p.z/p should becalculated at the average pressure between Pi and P2.

4.9. Permeability Variations in Radial Flow

Many producing formations are composed of strata or stringers that may varywidely in permeability and thickness, as illustrated in Fig. 7.8. If these strataare producing fluid to a common wellbore under the same drawdown and fromthe same drainage radius, then

0.00708 k.vgh:(Pe — Pw) = 0.00708 k1hi( — Pw) + 0.00708 k2h2(p, Pw) + etcjiB In (Tel,',,) p.B in (r,/,',,) pB In (relrw)

Then cancelling

kavghr=kihi+kih2+

(7.23)

This equation is the same as for parallel flow in linear beds with the same bedwidth. Here, again, average permeability refers to that permeability by whichall beds could be replaced and still obtain the same production rate under thesame drawdown. The product kh is called the capacity ofa bed or stratum, andthe total capacity of the producing formation, is usually expressed inmiiidarcy-feer. Because the rate of flow is directly proportional to the capac-ity, Eq. (7.19), a 10 ft bed of 100 md will have the same production rate as a100 ft bed of 10 md permeability, other things being equal. There are limitsof formation capacity below which production rates arc not economic, just asthere arc limits of net productive formation thicknesses below which wells willnever pay out. Of two formations with the same capacity, the one with the

h3 k3

Hg. 7.8. Radial flow in parallel beds


lower oil viscosity may be economic but the other not; and the availableprcssurc drawdown enters in similarly. Net sand thicknesses of the order of 5ft and capacities of the order of a few hundred millidarcy-feet are likely to beuneconomic, depending on other factors such as available drawdown, vis-cosity, porosity, connate water, depth, and the like. The capacity of the forma-tion together with the viscosity also determines to a large extent whether a wellwill flow or whether artifical lift must be used. The amount of solution gas isan important factor With the discovery of formation fracturing, to be dis-cussed later, the productivity of low capacity wells in many cases has beengreatly improved.

We now consider a radial flow system of constant thickness with a perme-ability of between the drainage radius and some lesser radius Ta, and analtered permeability ka between the radius r and the wellhorc radius as

shown in Fig. 7 9. The pressure drops are additive, and

(Papw)

Then from Eq. (7.19)

_qp.B in In

0.00708 kavgh — 0.00708 keh 0.00708 k0h

Cancelling and solving for kay5

k — kgk, ln (r,/r,,)24

avg — ka in (re/ra) + k. In (Ta/F,,,)

Equation (7.24) may be extended to include three or more zones in series.This equation is important in studying the effect of a decrease or increase ofpermeability in the zone about the welibore on the well productivity.

ELat k

P

Fig. 7.9. Radial flow in beds in sencc

5. Development of the Radial Differential Equation 231

5. DEVELOPMENT OF THE RADIAL DIFFERENTIALEQUATION

The radial differential equation, which is the general differential equationused to model time-dependent flow systems, is now developed. Consider thevolume element shown in Fig. 7.10. The element has a thickness

r distance from the center of the well. Mass is allowed to flow into andout of the volume element during a period The volume element is in areservoir of constant thickness and constant properties. Flow is allowed in onlythe radial direction. The following nomenclature, which is the same nomen-clature defined previously, is used:

q = volume flow rate, STh/day for incompressible and slightly compressiblefluids and SCF/day for compressible fluids

p = density of flowing fluid at reservoir conditions, lb/ft3

r = distance from wellbore, ft

h = formation thickness, ft

u = velocity of flowing fluid, bbl/day-ft2

= hours

= porosity, fraction

k = permeability, md

p. = flowing fluid viscosity, cp

With these assumptions and definitions, a mass balance can be written aroundthe volume element over the time interval In word form, the mass balanceis written as:

Fig. 7.10. Volume clcmcnt used in the development of the radial differential equation.


Mass entering Mass leaving Rate at whichvolume element — volume element = mass accumulates

during interval A: during interval A: dunng interval A:

The mass entering the volume eiement dunng At is given by'

+ (7.25)

The mass leaving the volume element during At is given by:

(qBp), = 2irrh(pv(5.615124)), (7.26)

The rate at which mass accumulates during the interval At is given by:

2rrrArh — (7.27)

Combining Eqs. (7 25), (7.26), and (7.27), as suggested by the wordtion" written above,

2ir(r + Ar)h(pu(5.615/24)),*a, — 2trrh(pu(5.615/24)),=— (èp),]

If both sides of this equation are divided by 2'rrrArh and the limit is taken ineach term as Ar and & approach zero, the following is obtained:

(0.234 pu) +! (0.234 pu) = (4)p)

or

0234a a(7.28)

Equation (7.28) is the continuity equation and is valid for any flow system ofradial geometry 'lb obtain the radial differential equation that will bc thebasis for time-dependent models, pressure must be introduced and 4) elimi-nated from the partial derivative term on the right-hand side of Eq. (7.28). Todo this, Darcy's equation must be introduced to relate the fluid flow rate toreservoir pressure:

ii=

Realizing that the minus sign can be dropped from Darcy's equation because

6. Transient Flow Systems 233

of the sign convention for fluid flow in porous media and substituting Darcy'sequation into Eq. (7.28):

(7.29)

The porosity from the partial derivative term on the right-hand side is elimi-nated by expanding the right-hand side by taking the indicated derivatives:

a ap ad)(7.30)

It can be shown that porosity is related to the formation compressibility by thefollowing

(7.31)

Applying the chain rule of differentiation to a4/at,

ad) — ad) op

Ot OpOt

Substituting Eq. (7.31) into this equation,

Od) Op-

Finally, substituting this equation into Eq. (7.30) and the result into Eq.(7.29),

(7.32)

Equation 7.32 is the general partial differential equation used to describe theflow of any fluid flowing in a radial direction in porous media. In addition tothe initial assumptions, Darcy's equation has been added, which implies thatthe flow is laminar. Otherwise, the equation is not restricted to any type offluid or any particular time region

6. TRANSIENT FLOW SYSTEMS

By applying appropriate boundary and initial conditions, particular solutionsto the differential equation derived in the preceding section can be discussed.The solutions obtained pertain to the transient and pseudosteady-state flow


periods for both slightly compressible and compressible fluids. Since thc in-compressible fluid does not exist, solutions involving this type of fluid are notdiscussed. Only the radial flow geometry is considered because it is the mostuseful and applicable geometry. If you are interested in linear flow, Matthewsand Russell present the necessary equations.6 Also, due to the complex natureof the pressure behavior during the late-transient period, solutions of thedifferential equation for this time region arc not considered. To further justifynot considering flow models from this period, it is true that the most practicalapplications involve the transient and pseudosteady-state periods.

6.1. Radial Flow of Slightly Compressible Fluids,Transient Flow

If Eq. (7.2) is expressed in terms of density, p. which is the inverse of specificvolume, then the following is obtained:

PR) (733)

where PR is some reference pressure and PR IS (he density at that referencepressure. Inherent in this equation is the assumption that the compressibilityof the fluid is constant. This is nearly always a good assumption over thepressure range of a given application. Substituting Eq. (7.33) into Eq. (7.32),

(o.oo1J27 = +

'ro simplify this equation, one must make the assumption that k and p. areconstant over the pressure, time, and distance ranges in our applications. Thisis rarely true about k. However, if k is assumed to be a volumetric averagepermeability over these ranges, then the assumption is good. In addition, ithas been found that viscosities of liquids do not change significantly overtypical pressure ranges of interest. Making this assumption allows k/p. to bebrought outside the derivative. Taking the necessary derivatives and sim-plifying,

Or2 r Or I. Or] 0.0002637k / 01

or

02p + 1 Op + Op ép.C1 Op34

Or2 r ilr COr 0.0002637k a: . )

The last term on the left-hand side of Eq (7.34) causes this equation to benonlinear and very difficult to solve however, it has been found that the term

6. Transient Flow Systems 235

is very small for most applications of fluid flow involving liquids. When thisterm becomes negligible for the case of liquid flow. Eq. (7.34) reduces to:

a2p + 1 ap — 4p.C, ap7 35r r3r 00002637k at

This equation is the diffusivity equation in radial form. The name comes fromits application to the radial flow of the diffusion of heat. Basically, the flow ofheat, the flow of electricity, and the flow of fluids in permeable rocks can bedescribed by the same mathematical forms. The group of terms waspreviously defined to be equal to I/q, where is called the diffusivity constant(see Sect. 3). This same constant was encountered in Eq. (7.6) for the read-justment time.

To obtain a solution to Eq. (7.35), it is necessary first to specify oneinitial and two boundary conditions. The initial condition is simply that at timet =0, the reservoir pressure is equal to the initial reservoir pressure, p,. Thefirst boundary condition is given by Darcy's equation if it is required that therehe a constant rate at the weilbore.

The second boundary condition is given by the fact that the desired solutionis for the transient period. For this period, the reservoir behaves as if it wereinfinite in size. This suggests that at r = x, the reservoir pressure will remaincquai to the initial reservoir pressure,p,. With these conditions, Matthews andRussel gave the following solution:

706qp.B (p(r,:)=p,—kh

(736)

where all variables are consistent with units that have been defined previ-inpsia, qisin STB/day, p.isincp, fl(formation

volume factor) is in bblISTI3, k is in md, h is in ft, is in psi', r is in ft.t is in hr.6 Equation (7.36) is called the line source solution to the diffusivityequation and is used to predict the reservoir pressure as a function of time andposition. The mathematical function, E,, is the exponential integral and isdefined by.

(Zeudu I x x2 x3L1(—x)=

—J u[In X etc.

This integral has been calculated as a function of x and is presented in Table7.1, from which Fig. 7.11 was developed.


x —E,(—x)

0.1 1.822920.2 1.222650.3 0.905680.4 0.702380.5 0559770.6 0.454380.7 0373770.8 0.310600.9 0.260181.0 0.219381.1 0.185991.2 0.158411.3 0 1354514 0.116221 5 0.100021 6 0.086311 7 0.074651 8 0.064711 9 0.056202 0 0.048902.1 0.042612.2 0.037192.3 0.032502.4 0.028442.5 0.024912.6 0.021852.7 0.019182.8 0.016862 9 0.014823.0 0.013053.1 0.011493.2 0.010133 3 0.008943 4 0.007893 5 0.006973.6 0.006163.7 0 005453.8 0.004823.9 0 004274 0 0.003784 1 0.003354 2 0.00297

x —E,(—x)

4.3 0.002634 4 0.02344.5 0.002074.6 0.001844.7 0.001644.8 0.001454.9 0.001295.0 0.001155.1 0.001025.2 0.000915.3 0.000815.4 0000725.5 0 000645.6 0.000575.7 0.000515.8 0000455.9 0000406.0 0.000366.1 0000326.2 0 000296.3 0.000266.4 0 000236.5 0 000206.6 0.000186.7 0 000166.8 00001469 00001370 0.000127 1 0.000107.2 0.000097.3 0.000087.4 0.000077.5 0.000077.6 0.000067.7 0.0000578 00000579 00000480 0000048.1 0000038.2 0.000038.3 0.000038.4 0.00002

TABLE 7.1.Values of —EA—x) as a function of x

x —E,(—x)

8.58687888.99.09.1929.39.49.59.6979.899

10 0

0.000020.000020.000020.000020000010.000010.000010.000010.000010.000010.000010.000010.000010.000010.000000.00000


Equation (7.36) can be used to find the pressure drop (p, — p) that willhave occurred at any radius about a flowing wef I after the well has flowed ata rate, q, for some time, t. For example, consider a reservoir where oil isflowing and = 0.72 cp; = 1.475 bbIISTB; k = 100 md; h = 15 ft;c, = 15 x psi'; 4) =23.4%; P'= 3000 psia. After a well is produced at 200STB/day for 10 days, thc pressure at a radius of 1000 ft will be.

— 70.6(200)(0.72)(1.475)

{E,

(100(15)

Then

0.234(0 72)15(10) 6(1000)20.00105(100)( lO)(24)

p =3000+10.0 E,(—0.10)

From Fig. 7.11, = —1.82. Therefore

p = 3000 + 10.0 (—1.82) = 2981.8 psia

Figure 7 12 shows this pressure plotted on the 10-day curve and shows thepressure distributions at 0.1, 1.0, and 100 days for the same flow conditions.

It has been shown that for values of the function argument less than0.01 the following approximation can be made:

4U)Q.

w

U)U)

a.

—E1(—x) = —ln (x) — 0.5772

Fig. 7.12. Pressure distribution about a well at four time periods after start of produc-tion

RADIUS, FEET

6 Transient Flow Systems 239

This suggests that

<0010.OOlO5kt

By rearranging the equation and solving for t, the time required to make thisapproximation valid for the pressure determination 1000 ft from the producingwell can be found:

0.234(0.72)15( I0)_6(1000)2

O.00105(100)(O.O1)2400 hr= 100 days

To dctcrmine if the approximation to the E, function is valid when calculatingthe pressure at the sandface of a producing well, it is necessary to assume awelibore radius. (0.25 ft) and to calculate the time that would make theapproximation valid. The following is obtained:

0.234(0.72)15(10)_6(0.25)20

0.00105(I00)(0.O1).0002 hours

It is apparent from these calculations that whethcr the approximation can beused is a strong function of the distance from the pressure disturbance to thepoint at which the pressure determination is desired or, in this case, from theproducing well. For all practical purposes, the assumption is valid when con-sidering pressures at the point of the disturbance. Therefore, at the weilboreand wherever the assumption is valid, Eq. (7.36) can be rewritten as:

= —

{— ln

( —________— 0.5772]

Substituting the log base 10 into this equation for the ln term, rearranging andsimplifying, one gets:

1) =— 162.6 qpJ3 [log

— 3.23] (7.37)

Equation (7.37) serves as the basis for a well testing procedure called transientwell testing, a very useful technique that is discussed later in this chapter

6.2. Radial Flow of Compressible Fluids, Transient Flow

In Sect. 5, Eq (7.32)

(0.001127 = + (7.32)


was developed to describe the flow of any fluid flowing in a radial geometryin porous media. To develop a solution to Eq. (7.32) for the compressiblefluid, or gas, case, two additional equations are required: (1) an equation ofstate, usually the real gas law, which is Eq. (1.7), and (2) Eq. (1.19), whichdescribes how the gas isothermal compressibility varies with pressure:

pVznR'T (1.7)

1 ldz(1.19)p zdp

These three equations can be combined to yield

I a ( p api— 4C1p738tar \ p.zarJO.0002637kz a: )

Al-Hussainy, Ramey, and Crawford and Russel, Goodrich, Perry, and Brus-kotter introduced a transformation of variables to obtain a solution to Eq.

The transformation involves the real gas pseudopressure, m(p),which has units of psia7kp in standard field units and is defined as:

IP nm(p)=2J (7.39)

PR ILZ

where PR iS a reference pressure, usually chosen to be 14.7 psia. from whichthe function is evaluated. Since p. and z are only functions of pressure for agiven reservoir system, which we have assumed to be isothennal, Eq. (7.39)can be differentiated and the chain rule of differentiation applied to obtain thefollowing relationships:

(7.40)ap p.z

am(p)am(p)ap741)ar ap är (.

am(p) = am(p) ap7 42)

at ap at

Substituting Eq. (7.40) into Eq. (7.41) and (7.42) yields

app.zam(p)743ar2p ar

app.zam(p)744

a: 2p at (.

6 Transient Flow Systems 241

Combining Eq. (7.43), (7 44), and (7 38) yields

82m(p) 1 am(p) — Om(p)(745)

— ar2 ar 0.0002637k a:

Equation (7.45) is the diffusivity equation for compressible fluids, and it hasa very similar form to Eq. (7.35), which is the diffusivity equation for slightlycompressible fluids. The only difference in the appearance of the two equa-tions is that Eq. (7.45) has the real gas pseudopressurc, m(p), substituted forp. There is another significant difference that is not apparent from looking atthe two equations. This difference is in the assumption concerning the mag-nitude of the c(t9plär)2 term in Eq (7.34) which represents the pressuregradient. To linearize Eq. (7.34), it is necessary to limit the term to a smallvalue so that it results in a negligible quantity, which is normally the case forliquid flow applications. This limitation is not necessary for the gas equation.Since pressure gradients around the gas wells can be very large, the trans-formation of variables has led to a much more practical and useful equationfor gascs.

Equation (7.45) is still a nonlinear differential equation because of thedcpendence of p and c, on pressure or the real gas pseudopressure. Thus,there is no analytical solution for Eq. (7.45). Al-Flussainy and Ramey, how-ever, used finite difference techniques to obtain an approximate solution toEq. (7.45)•9 The result of their studies for pressures at the welibore (i.e.,where thc logarithm approximation to the function can be made) is thefollowing equation:

1637(1O)3qT [ f kt Ilog —3.23j (7.46)L Wf'PiCIITw

where is the flowing pressure at the welibore, p, is the initial rcscrvoirpressure, q is the flow rate in SCFIday at standard conditions of &Y'Fand 14.7psia, T is the reservoir temperature in °R, k is in md, h is in ft, t is in hr, p.,is in cp and is evaluated at the initial pressure, p', c,, is in psi' and is alsoevaluated at p., and is the welibore radius in feet. Equation (7.46) can beused to calculate the flowing pressure at the sandface of a gas well.

The use of Eq. (7.46) requires values of the real gas pseudopressure Theprocedure used to find values of m(p) has been discussed in the literature.'0"The procedure involves determining p. and z for several pressures over thepressure range of interest, using the methods of Chapter 1. Values of plp.z arethen calculated, and a plot of p/p.z versus p is made, as illustrated in Fig 7.13.A numcrical integration scheme such as Simpson's rule is then used to deter-mine the value of the area from the reference pressure up to a pressure ofinterest, p1. The value of m(p,) that corresponds with pressure, p,, is given by:

m(p,) = 2 (area1)


p

pZ

Fig. 7.13. Graphical determination of

where

area1= JPi

PR PZ

The real gas pseudopressure method can be applied at any pressure of interestif the data are available.

7. PSEUDOSTEADY•STATE FLOW SYSTEMS

For the transient flow cases that were considered in the previous section, thewell was assumed to be located in a very large reservoir. This assumption wasmade so that the flow from or to the well would not be affected by boundariesthat would inhibit the flow. Obviously, the time that this assumption can bemade is a finite amount and often is very short in length. As soon as the flowbegins to feel the affect of a boundary1 it is no longer in the transient regime.At this point, it becomes necessary to make a new assumption that will leadto a different solution to the radial diffusivity equation. The following sectionsdiscuss solutions to the radial diffusivity equation that allow calculations dur-ing the pseudosteady-state flow regime.

7.1. RadIal Flow of Slightly Compressible Fluids,Pseudosteady-State Flow

Once the pressure disturbance has been felt throughout the reservoir includingat the boundary, then the reservoir can no longer be considered as beinginfinite in size and the flow is not in the transient regime. This situationnecessitates another solution to Eq. (7.35), using a different boundary condi-tion at the outer boundary. The initial condition remains the same as before(i.e., the reservoir pressure is p throughout the reservoir at time a' = 0). Theflow rate is again treated as constant at the weilbore. The new boundary

PR p

7 Pseudosteady-State Flow Systems 243

condition used to find a solution to the radial diffusivity equation is that theouter boundary of the reservoir is a no-flow boundary. In mathematical terms,

—0 atr=r,

Applying these conditions to Eq. (7.35), the solution for thc pressure at theweilbore becomes

162.6qp.B F 4Aj

02339qBtPwfPs — kh — Ak4c, (7.47)

where A is the drainage area of the well in square feet and CA is a reservoir

shape factor. Values of the shape factor are given in Table 7.2 for several

reservoir types. Eq. (7.47) is valid only for sufficiently long enough times forthe flow to have reached the pseudosteady-state time period

After reaching pseudosteady-state flow, the pressure at every point inthe reservoir is changing at the same rate, which suggests that the averagereservoir pressure is also changing at the same rate. The volumetric averagereservoir pressure, which is usually designated and is the pressure used tocalculate fluid properties in material balance equations, is defined as:

±(7.48)

j—i

where is the average pressure in the jth drainage volume and is the volumeof the jth drainage volume. It is useful to rewrite Eq. (7.47) in terms of theaverage reservoir pressure,

— 1626qJLB I 4A I=

— kh- log

1.781(7.49)

For a well in the center of a circular reservoir with a distance to the outerboundary of Te, Eq. (7 49) reduces to:

— 70.6qp.B Fkh

—1.5

If this equation is rearranged and solved for q,

= 0.00708kh [ p —50q ( . )

MT

AB

LE 7

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Sha

pe fa

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s fo

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1387

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3.31

78

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95

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—0.

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0005


7.2. Radial Flow of Compressible Fluids,Pseudosteady-State Flow

The differential equation for the flow of compressible fluids in terms of thereal gas pscudopressurc was derived in Eq. (7 45). When the appropriateboundary conditions are applied to Eq. (7.45), the pseudosteady-state solu-tion rearranged and solved for q yields Eq. (7 51).

19.88(JO)(751)q

I. ln(r,/r,.,) —0 75

8. PRODUCTIVITY INDEX (P1)

Thc ratio of the rate of production, expressed in STL3/day for liquid flow, tothe pressure drawdown at the midpoint of the producing interval, is called theproductivity index, symbol J.

q(7.52)

P

The P1 is a measure of the well potential, or the ability of the well to produce,and is a commonly measured well property To calculate I from a produc-tion test, it is necessary to flow the well a sufficiently long time to reachpseudostcady-state flow. Only during this flow regime will the differencebetween and be constant. It was pointed out in Section 3 that once thepseudosteady-state period had been reached, then the pressure changes atevery point in the reservoir at the same rate. This is not true for the otherperiods, and a calculation of productivity index during other periods wouldnot bc accurate.

In some wells, the P1 remains constant over a wide variation in flow ratesuch that the flow rate is directly proportional to the bottom-hole pressuredrawdown. In other wells, at higher flow rates the linearity fails, and the P1index declines, as shown in Fig 7 14 The cause of this decline may be (a)turbulence at increased rates of flow, (b) decrease in the permeability to oildue to presence of free gas caused by the drop in pressure at the well bore,(c) the increase in oil viscosity with pressure drop below bubble point, and/or(d) reduction in permeability due to formation compressibility.

In depletion reservoirs, the productivity indexes of the wells decline asdepletion proceeds, owing to the increase in oil viscosity as gas is releasedfrom solution and to the decrease in the permeability of the rock to oil as theoil saturation decreases. Since each of these factors may change from a few toseveral-fold dunng depletion, the Fl may decline to a small fraction of theinitial value Also, as the permeability to oil decreases, there is a correspond-ing increase in the permeability to gas, which results in rising gas-oil ratios.The maximum rate at which a well can produce depends on the productivity

8. Productivity Index (P1) 247

Fig. 7.14. Decline in productivity index at higher flow rates.

a.

LV Q

x0z>-

>I—000a.

index at prevailing reservoir conditions and on the available pressure draw-down. If the producing bottom-hole pressure is maintained near zero bykeeping the well "pumped off," then the available drawdown is the prevailingreservoir pressure, and the maximum rate is x J.

In wells producing water, the P1, which is based on dry oil production,declines as the water cut increases because of the decrease in oil permeability,even though there is no substantial drop in reservoir pressure. In the study ofthese "water wells," it is sometimes useful to place the P1 on the basis of totalflow, including both oil and water, where in some cases the water cut may riseto 99% or more.

The injectzvily index is used with salt water disposal wells and withinjection wells for secondary recovery or pressure maintenance, it is the ratioof the injection rate in STh per day to the excess pressure above reservoirpressure that causes that injection rate, or

qInjectivity index = I = STB/day/psi

P(7.53)

With both productivity index and injectivity index, the pressures referred toare sandfacc pressures, so that frictional pressure drops in the tubing or casingarc not included In the case of injecting or producing at high rates, thesepressure losses may be appreciable.

200 400PRESSURE DRAWDOWN, PSI


In comparing one well with another in a given field, particularly whenthere is a variation in net productive thickness but when the other factorsaffecting the productivity index arc essentially the same, the specific produc-tivity index J5 is sometimes used, which is the productivity index divided by thenet feet of pay, or,

Specific productivity index = =— q

STB/day/psi/ft (7.54)h

8.1. ProductivIty Ratio (PR)

In evaluating well performance, the standard usually referred to is the produc-tivity index of an open hole that completely penetrates a circular formationnormal to the strata, and in which no alteration in permeability has occurredin the vicinity of the weilbore. Substituting Eq. (7.50) into Eq. (7.52) we get

I = 0.00708pJ3(ln(re/rv) — 035)

(7.55)

The PR then is the ratio of the P1 of a well in any condition to the PT of thisstandard well:

(756)

Thus, the productivity ratio may be less than one, greater than one, or equalto one Although the productivity index of the standard well is generally un-known, the relative effect of certain changes in the well system may he evalu-atcd from theoretical considerations, laboratory models, or well tests. Forexample, the theoretical productivity ratio of a well reamed from an 8-in.borehole diameter to 16 in is derived by Eq (7.55):

—0.75PR

— 18 — In (r,/0.667) — 0.75

Assuming 'e 660 ft.

PR — ln (660/0.333) — 0.75— Ill

— ln(660/O.667)—0.75 —

Thus, doubling the borehole diameter should increase the P1 approximately11%. An inspection of Eq. (7.55) indicates that the P1 can be improved byincreasing the average permeability k, decreasing the viscosity p., or increasingthe wellbore radius r,..

9. Superposition 249

9. SUPERPOSITION

Earlougher and others have discussed the application of the principle of super-position to fluid flow in rcservoirs.3 12 13.14 This principle allows the use of theconstant rate, single well equations that have been developed earlier in thischapter and applies them to a variety of other cases. We illustrate the applica-tion by examining the solution to Eq. (7.35), which is a linear, second-orderdifferential equation. The principle of superposition can be stated as' Theaddition of solutions to a linear differential equation results in a new solutionto the originial differential equation. For example, consider the reservoirsystem depicted in Fig. 7.15. In the example shown in Fig 7.15 weiis 1 and 2arc opened up to their respective flow rates, q1 and q2, and the pressure dropthat occurs in the observation well is monitored. The principle of super-position states that the total pressure drop will be the sum of the pressure dropcaused by the flow from well 1 and the pressure drop caused by the flow fromwell 2:

+Ap2

Each of the individual terms is given by Eq. (7.36), or:

A 706qp.B[ .( 4p.c1r2 \— p. p(r,

— kh I. — 'k O.OOlOSktJ

To apply the method of superposition, pressure drops or changes are added.It is not correct simply to add or subtract individual pressure terms. It isobvious that if there are more than two flowing wells in the reservoir system,the procedure is the same, and the total pressure drop is given by the following,

(7.57)

Observation Well

Well 2

Fig. 7.15. Two Ilowing-well rcscrvoir system to illustrate the principle of super-position.

Ti

Well 1r2


where N equals the number of flowing wells in the system. Example 7.1illustrates the calculations involved when more than one well affects the pres-sure of a point in a reservoir.

Example 71. For the well layout shown in Fig. 7.16, calculate the totalpressure drop as measured in the observation well (well 3) caused by the fourflowing wells (wells 1, 2, 4, and 5) after 10 days. The wells were shut in for along time before opening them to flow.

Given: The following data apply to the reservoir system:

B0=l.50 bbl/STBformation thickness =50 ft

= 15 x 10 6 psi_I

SoLuTIoN: The individual pressure drops can be calculated with Eq.(7 36), and the total pressure drop is given by Eq. (7.57). For well 1

Fig. 7.16. Well layout for Ex Prob. 7.1.

oil viscosity = 0.40 cpk 47 mdporosity= 11.2%

WellFlow Rate(STO/day)

Distance toObservation well

(feet)

1

-

265 17002 270 19204 287 18705 260 1690

_70 6(265)(0 40)(1.5) r-(47)(50) [- E1 (

= 4.78

5

(1 12)(0.40)(15 x— 0.001O5(47)(240)

o Flowing well

• Observation welt

9 Superposition 251

From Fig. 7.11

Eg(0.164)= 1.39

Therefore

=478(1.39) =6.6 psi

Similarly, for wells 2, 4, and 5

= 4.87[—E,(--O 209)1 = 5 7 psi

= 5.14[—E1(—O.198)J = 6.4 psi

= 4.69[—E1(--0.162)I = 6.6 psi

Using Eq. (7.57) to find the total pressure drop at the observation well, Well3, the individual pressure drops arc added together to give the total:

= + AP2 + Ap4 +

or

psi

The superposition principle can also be applied in the time dimension,as is illustrated in Fig. 7.17 In this case, one well (which means the positionwhere the prcssure disturbances occur remains constant) has been producedat two flow rates. The change in the flow rate from q1 to q2 occurred at time

Figure 7.17 shows that the total pressure drop is given by the sum ofthe pressure drop caused by the flow rate qj and the pressure drop caused bythe change in flow rate — q1. This new flow rate, q7 — q1, has flowed for time( — tl.

The pressure drop for this flow rate, q2 — q1, is given by

— — _70.6(q2—q1)pii[ Et' 4p,c,r2p—p, p(r,r)—kh 1. 'kO.00105k(z—z1)

As in the case of the multiwell system just described, superposition can alsobc applied to multirate systems as well as the two rate examples depicted inFig. 7.17.

9.1. Superposition In Bounded or Partially BoundedReservoirs

Although Eq. (7.36) applies to infinite reservoirs, it may be used in conjunc-tion with the superposition principle to simulate boundaries of closed orpartially closed reservoirs. The effect of boundaries is always to cause greater


q2q1C2(I

2a-

ti

lime

Pi •— — - . -— ——— —-

}

=

to

Time

Fig. 7.17. Production ratc and pressure history for a well with two flow rates.

pressure drops than those calculated for the infinite reservoirs. The method ofimages is useful in handling the effect of boundaries For example, the pres-sure drop at point X (Fig 7.18), owing to production in a well located adistancc d from a scaling fault, will be the sum of the effects of the producingwell and an image well that is superimposed at a distance d behind the fault.In this case the total pressure drop is given by Eq. (7.57), where the individualpressure drops are again given by Eq (7.36), or for the case shown in Fig7.18:

+ APimage

I (Pi— kh

— L EPimage kh - I 'k—OoolOsk:

10. Introduction to Pressure Transient Testing 253

Well image Well

Fig. 7.18. Method of images used in the solution of boundary problcms.

10. INTRODUCTION TO PRESSURE TRANSIENTTESTING

Pressure transient testing is an important diagnostic tool that can provide valu-able information for the reservoir engineer. A transient test is initiated bycreating a disturbance at a welibore, (i.e., a change in the flow rate) and thenmonitoring the pressure as a function of time. An efficiently conducted testthat yields good data can provide information such as average permeability,drainage volume, welibore damage or stimulation, and reservoir pressure.

A pressure transient test does not always yield a unique solution. Thereare often anomalies associated with the reservoir system that yield pressuredata that could lead to multiple conclusions. In these cases, the strength oftransient testing is realized only when the procedure is used in conjunctionwith other diagnostic tools or other information.

In the next two subsections, the two most popular tests (i.e., the draw-down and buildup tests) are introduced. However, notice that the material isintended to be only an introduction. You must be aware of many otherconsiderations in order to conduct a proper transient test. We refer you tosome excellent books in this area by Earloughcr, Matthews and Russell, andLee.3 6.15

10.1. IntroductIon to Drawdown Testing

The drawdown test consists of flowing a well at a constant rate following ashut-in period. The shut-in period should be sufficiently long for the reservoirpressure to have stabilized. The basis for the drawdown test is found in Eq.(7.37),


p(r, 1) p — 162.6 qp.B— 3.23J (7.37)

which predicts the pressure at any radius, r, as a function of time for a givenreservoir flow system during the transient period If r = then p(r, t) will bethe pressure at the wellbore. For a given reservoir system. p,, q, p., B, k, h,

b, and are constant, and Eq (7.37) can be written as

Pwf =1 + mlog(t) (7.58)

where,

= flowing well pressure in psiab = constant

— time in hrs162.6qp.B

m =- constant =— kh

(7.59)

Equation (7.59) suggests that a plot of versus i on semilog graph paperwould yield a straight line with slope m through the early time data thatcorrespond with the transient time period. This is providing that the assump-tions inherent in the derivation of Eq. (7.37) are met. These assumptions arethe following:

1. Laminar, honzontal flow in a homogeneous reservoir.2. Reservoir and fluid properties, k, 4,, h, c1, p., and B are independent of

pressure.3. Single-phase liquid flow in the transient time region.4. Negligible pressure gradients.

The expression for the slope, Eq. (7.59), can be rearranged to solve for thecapacity. kh, of the drainage area of the flowing well, if the thickness isknown, then the average permeability can be obtained by Eq (7.60):

760)nih

if the drawdown test is conducted long enough for the pressure transients toreach the pseudosteady-state period, then Eq. (7.47) is used to describe thepressure behavior:

162.6qp.B 4A 1 0 2339qBt= — kh

log—

(7.47)


Again grouping together the terms that arc constant for a given reservoirsystem, Eq. (7.47) becomes

(7.61)

where

b' = constant0.2339qB

m = constant =— Ah4,c (7.62)

Now a plot of pressure versus time on regular Cartesian graph paper yields astraight line with slope equal to m' through the late time data that correspondto the pseudosteady-state period. If Eq. (7.62) is rearranged, an expression forthe drainage volume of the test well can be obtained:

= — O.2339qB(7.63)m c,

The drawdown test can also yield information about damage that mayhave occurred around the welibore during the initial drilling or during subse-quent production. We now develop an equation that allows the calculation ofa damage factor, using information from the transient flow region.

A damage zone yields an additional pressure drop because of the re-duced permeability in that zone. Van Everdingen and Hurst developed anexpression for this pressure drop and defined a dimensionless damage factor,S, called the skin factor'6 17,

=(7.64)

or

= —0.87mS (7.65)

From Eq. (7 65), a positive value of S causes a positive pressure drop andtherefore represents a damage situation. A negative value of S causes a nega-tive pressure drop that represents a stimulated condition like a fracture. No-tice that these pressure drops caused by the skin factor are compared to thepressure drop that would normally occur through this affected zone as pre-dicted by Eq. (7.37). Combining Eq. (7.37) and (7.65), the following expres-sion is obtained for the pressure at the weilbore:

Pwf = Pt—

[log —3 23+ O.87S] (7.66)


This equation can be rearranged and solved for the skin factor, S:

ktS — 1.151 162.6qp.B — log 2 + 3.23

k/i

The value of p11 must he obtained from the straight line in the transient flowregion. Usually a time corresponding to 1 hr is used, and the correspondingpressure is given by the designation, Plhr Substituting m into this equation andrecognizing that the denominator of the first term within the brackets isactually —m,

k2+3.23] (767)

m

Eq. (7.67) can be used to calculate a value for S from the slope of the transientflow region and the value of Plhr also taken from the straight line in thetransient period.

A drawdown test is often conducted during the initial production froma well since the reservoir has obviously been stabilized at the initial pressure,p,. The difficult aspect of the test is maintaining a constant flow rate, q. If theflow rate is not kept constant during the length of the test, then the pressurebehavior will reflect this varying flow rate and the correct straight line regionson the semilog and regular cartesian plots may nut be identified. Other factorssuch as weilbore storage (shut-in well) orunloading (flowing well) can inter-fere with the analysis. Weilbore storage and unloading are phenomena thatoccur in every well to a certain degree and cause anomalies in the pressurebehavior. Weilbore storage is caused by fluid flowing into the weilbore aftera well has been shut in on the surface and by the pressure in the weilborechanging as the height of the fluid in the welibore changes. Wellbore unload-ing in a flowing well will lead to more production at the surface than whatactually occurs down hole. 'Lhe effects of wcllbore storage and unloading canbe so dominating that they completely mask the transient time data. If thishappens and if the engineer does not know how to analyze for these effects,the pressure data may be misinterpreted and errors in calculated values ofpermeability, skin, and the like may occur. Wellbore storage and unloadingeffects are discussed in detail by Earlougher These effects should always betaken into consideration when you evaluate pressure transient data.

The following example problem illustrates the analysis of drawdown testdata

Example 7.2. A drawdown test was conducted on a new oil well in alarge reservoir At the time of the test, the well was the only well that had beendeveloped in the reservoir. Analysis of the data indicates that wellbore storagedoes not affect the pressure measurements. Use the data to calculate the


average permeability of the area around the well the skin factor, and thedrainage area of the well.

= 4000 psia formation thickness =20 ft

q=500STB/day c,=30x106psia 1

= 1.5 cp porosity =25%

B0= 1.2 bbl/STB r,, = 0.333 ft

Flowing pressure, p..,,psia

time,hi's

3503 23469 5

3443 103417 203383 503368 753350 1003306 1503282 2(k)3250 300

SoumoN: Figure 7.19 contains a scmilog plot of the pressure data. Theslope of the early time data, which are in the transient time rcgion, was foundto he —86 psi/cycle and the value of PIhr was found to be 3526 psia.Equation (7.60) can now be used to calculate the permeability.

k — 162.6(500)(1.5)(1.2)_—86(20)

85 1 md

The skin factor is found from Eq. (7.67).

[3526—4000 f 85.1S — 1151

—86— log

\(O.25)(J .5)(30+ 3.23

5=1.04

This positive value for the skin factor suggests the well is slightly damagedFrom the slope of a plot of P versus time on regular cartesian graph

paper, shown in Fig. 7.20, and using Eq. (7 63), an estimate for the drainagearea of the well can be obtained. From the semilog plot of pressure versustime, the first six data points fell on the straight line region indicating they

258

4,

Iia.

a,

1!

3500

3400

3300-

3200


0 100 200 300 400

time

Fig. 7.20. Plot of prcssure versus time for the data of Ex 7 2

10 100

Log t

Fig. 7.19. Plot of pressure versus log time for the data of Ex. 7.2

S

.

________

S


were in the transient time period. Therefore, the last two to three points ofthe data are in the pseudosteady-state period and can be used to calculate thedrainage area. The slope of a line drawn through the last three points is—0.373. Therefore,

O.2339(500)(1 .2)2

= (—.373)(30 x 10) 6(20)(0.25) = 2,508,000 ft = 57.6 ac

10.2. introduction to Buildup Testing

The buildup test is the most popular transicnt test used in the industry. It isconducted by shutting in a producing well that has obtained a stabilizedpressure in the pseudosteady-state time region by flowing the well at a con-stant rate for a sufficiently long period The pressure is then monitored duringthe length of the shut-in period. The primary reason for the popularity of thebuildup test is the fact that it is easy to maintain the flow rate constant at zeroduring the length of the test. The main disadvantage of the buildup test overthat of the drawdown test is that there is no production during the test andtherefore no subsequent income.

A pressure buildup test is simulated mathematically by using the pnn-ciple of superposition. Before the shut-in period, a well is flowed at a constantflow rate, q. At the time corresponding to the point of shut-in, a secondwell, superimposed over the location of the first well, is opened to flow at arate equal to —q, while the first well is allowed to continue to flow at rate q.The time that the second well is flowed is given the symbol of At When theeffects of the two wells are added, the result is that a well has been allowedto flow at rate q for time I,, and then shut in for time At. This simulates theactual test procedure, which is shown schematically in Fig. 7.21. The timecorresponding to the point of shut-in, :,,, can be estimated from the followingequation,

(7.68)

where, = cumulative production that has occurred dunng the time beforeshut-in that the well was flowed at the constant flow rate q.

Equations (7.37) and (7.57) can be used to describe the pressure behav-ior of the shut-in well:

— — 162 6(—q)pil kAtPws — kh

ogj — kh

— 3.23

Expanding this equation and cancelling terms,

162.6qp.B 1 (: + Ar)1Pws = Pi

— kh (7.69)


H

time time --

q

Flow o

time

Fig. 7.21. Graphical simulation of pressure buildup test using superposition.

where,

= bottom-hole shut-in pressure

Equation (7.69) is used to calculate the shut-in pressure as a function of theshut-in time and suggests that a plot of this pressure versus the ratio of(ç, + on semilog graph paper will yield a straight line. This plot isreferred to as a Homer plot after the man who introduced it into the petro-leum literature.'8 The slope of the Homer plot is equal to m, or

— 162.6qp.B

kh

This equation can he rearranged to solve for the permeability:

k = —162.6qp.B

(7.70)nih

The skin factor equation for buildup is found by combining Eq. (7.67), writtenfort =0), and Eq. (7.69):

+3.23]m

The shut-in pressure, Pws, can be taken at any on the straight line of thetransient flow period. For convenience, is set equal to 1 hr, and p,,,, is taken


it that point. At a time of At = 1 hr, is much larger than At for most tests,md 1,, + Ar 1,,. With these considerations, the skin factor equation becomes

= 1.151[pwi(At =0) Puuit

— 'r2 + 3.23] (7.7J)

We concludc this section with an example problem illustrating the analy-sis of a buildup test. Notice again that there is much more to this overall areaof pressure transient testing. Pressure transient testing is a very useful diagnos-tic tool for the reservoir engineer if used correctly The intent of this sectionwas simply to introduce these important concepts. You should pursue themdicated references if you need a more thorough coverage of the material.

Example 7.3. Calculation of permeability and skin from a pressurebuildup test.

Given.

flow rate before shut in period = 280 STB/day

during constant rate period before shut in = 2682 STB

Pwf at the time of shut-in = 1123 psia

From the foregoing data and Eq. (7.68), can be calculated

= = = 230 hours

Other given data arc

B0 = 1.31 bbl/STB = 2.0 cph=401t4=010

time aftershut in, Pressure, p,,,, t,. +

hours - psia At

2 2290 116.04 2514 58.58 2584 29.8

12 2612 20.216 2632 15.420 2643 J2524 2650 10 630 2658 87


SOLIrnON The slope of the straight line region (notice the difficulty inidentifying the straight line region) of the Homer plot in Fig. 7.22 is —170psi/cycle From Eq. (7.70)

162.6 (280)(2 0)(1.31)k—(—170)(40)

—17.5md

Again from the Horncr plot, Plhr is 2435 psia and from Eq. (7.71)

[1123—2435 / 17.5S = 1.1511 170

— l)(20)(15 x 1O_6)(O.333)2) + 3.23

S

The difficulty in identifying the straight line of the transient time region resultsfrom the two earliest data points being dominated by weilbore storage effects.The time ratio + decreases as increases. Therefore, the early timedata are on the right of Fig. 7.22 This difficulty in identifying the properstraight line region points out the importance of a thorough understanding ofwellborc storage and other anomalies that could affect the pressure transientdata

2700-

.S

2600- - -- -

1 10 100 1000

log

Fig. 7.22. Plot of pressure versus time ratio for Ex 7.3

Problems 263

PROBLEMS

7.1 Two wells are located 2500 ft apart. The static well pressure at the top ofperforations (9332 ft subsea) in well A is 4365 psia and at the top of perforations(9672 ft subsea) in well B is 4372 psia The reservoir fluid gradient is 0 25 psi/ft.reservoir permeability is 245 md, and reservoir fluid viscosity is 0.63 cp.

(a) Correct the two static pressures to a datum level of 9100 ft subsea(b) In what direction is the fluid flowing between the wells?(c) What is the average effective pressure gradient between the wells'(d) What is the fluid velocity9

(e) Is this the total velocity or only the component of the velocity in the directionbetween the two wells?

(f) Show that the same fluid velocity is obtained using Eq. (7.1).

7.2 A sand body is 1500 ft long, 300 ft wide, and 12 ft thick. It has a uniformpermeability of 345 md to oil at 17% connate water saturation. The porosity is32%. The oil has a reservoir viscosity of 3.2 cp and B0 of 1.25 bbl/STB at thebubble point.(a) If flow takes place above the bubble-point pressure, what pressure drop will

cause 100 reservoir bbLlday to flow through the sand body, assuming thefluid behaves essentially as an incompressible fluid? What for 200 reservoirbbllday?

(b) What is the apparent velocity of the oil in feet per day at the 100 bbl/day flowrate?

(c) What is the actual average velocity?

(d) What time will bc required for complete displacement of the oil from thesand?

(e) What pressure gradient exists in the sand"(f) What will be the effect of raising both the upstream and downstream pres-

sures by, say. 1000 psi"

(g) Considenng the oil as a fluid with a very high compressibility of 65(10)psi how much greater is the flow rate at the downstream end than theupstream end at 100 bbl/day"

(h) What pressure drop will be required to flow 100 bbl/day, measured at theupstream pressure, through the sand lithe compressibility of the oil is65(10) 6 Consider the oil to be a slightly compressible fluid.

(I) What will be the downstream flow rate?(I) What conclusion can he drawn from these calculations concerning the use of

the incompressible flow equation for the flow of slightly compressible liq-uids, even with high compressibilities"

7.3 If the sand body of Prob 7.2 had been a gas reservoir with a bottom-holetemperature of 140°F but with the same connate water and permeability to gas,calculate the following:

(a) With an upstream pressure of 2500 psia, what downstream pressure will


causc 5.00 MM SCF/day to flow through the sand? Assume an average gasviscosity of 0.023 cp and an average gas deviation factor of 088.

(b) What downstream pressure will cause 25 MM SCF/day to flow if the gasviscosity and deviation factors remain the same9

(c) Explain why it takes more than five times the pressure drop to cause flvctimes the gas flow.

(d) What is the pressure at the midpoint of the sand when 25 MM SCF/day isflowing?

(e) What is the mean pressure at 25 MM SCF/ciay?

(f) Why is there a greater pressure drop in the downstream half of the sand bodythan in the upstream half?

(g) From the gas law calculate the rate of flow at the mean pressure p.,,. andshow that the equation in terms of qm is valid by numencal substitution.

7.4 (a) Plot pressure versus distance through the sand of the previous problem atthe 25 MM SCF/day flow rate.

(b) Plot the pressure gradient versus distance through the sand body.7.5 A rectangular sand body is flowing gas at 10 MM SCFfday under a downstream

pressure of 1000 psia. Standard conditions are 14.4 psia and 80°F. The averagedeviation factor is 0 80 The sand body is 1000 ft long, tOO ft wide, and 10 ftthick. Porosity is 22%, and average permeability to gas at 17% connate wateris 125 md. Bottom-hole temperature is 160°F, and gas viscosity is 0.029 cp.

(a) What is the upstream pressure?(b) What is the pressure gradient at the midpoint of' the sand?

(c) What is the average pressure gradient throughout the sand"(d) Whcrc does the mean pressure occur?

7.6 A horizontal pipe 10cm in diameter (1.D.) and 3000cm long is filled with a sandof' 20% porosity. It has a connate water saturation of 30% and, at that watersaturation, a permeability of oil of 200 md. The viscosity of the oil is 0.65 cp andthe water is immobile.

(a) What is the apparent velocity of the oil under a 100 psi pressure differential?

(b) What is the flow rate9(c) Calculate the oil contained in the pipe and the time needed to displace it at

the rate of 0.055 cu cm/sec.

(d) From this actual time and the length of the pipe, calculate the actual averagevelocity.

(e) Calculate the actual average velocity from the apparent velocity, porosity,and connate water.

(1) Which velocity is used to calculate flow rates, and which is used to calculatedisplacement times?

(g) if the oil is displaced with water so that 20% unrecoverable (or residual) oilsaturation is left behind the water flood front, what are the apparent andactual average velocities in the watered zone behind the flood front if the oil

Problems 265

production rate is maintained at 0.055 cu cmisec? Assume piston-like dis-placement of the oil by the water.

(h) What is thc rate of advance of the flood front"(i) How long will it takc to obtain all the recoverable oil, and how much will bc

recovered"

(j) How much pressure drop wil be required to produce oil at the rate of 0 055cu cm/sec when the water flood front is at the midpoint of the pipe9

7.7 (a) Three beds of equal cross section have permeabilities of 50,200, and 500 mdand lengths of 40, 10. and 75 ft. respectively. What is the average perme-ability of the beds placed in series"

(b) What are the ratios of the pressure drops across the individual beds for liquidflow?

(c) For gas flow will the overall pressure drop through beds in series be the samefor flow in either direction? Will the individual pressure drops he the same?

(d) The gas flow constant for a given linear system is 900, so that — = 900Lik If the upstream pressure is 500 psia, calculate the pressure drops in eachof two beds for senes flow in both directions. The one bed is 10 ft long andJ00 md, the second is 70 ft and 900 md.

(e) A producing formation from top to bottom consists of 10 ft of 350 md sand,4 in of 0 5 md shale, 4 ft of 1230 md sand, 2 in. of 2.4 md shale, and 8 ftof 520 md sand. What is the average vertical permeability?

0) If the 8 ft of 520 nid sand is in the lower part of the formation and carrieswater, what well completion technique will you use to keep the water-oilratio low for the well" Discuss the effect of the magnitude of the lateralextent of the shale breaks on the well production

7.8 (a) Three beds of 40, 100, and 800 md, and 4, 6, and 10 ft thick, respectively,are conducting fluid in parallel flow If all are of equal length and width,what is the average permeability"

(b) In what ratio are the separate flows in the three beds?7.9 As project supervisor for an in situ uranium leaching project. you have observed

that to maintain a constant injection rate in well A, the pump pressure has hadto be increased so that p. — p.. has increased by a factor of 20 from the value atstartup An average permeability of 100 md was measured from plugs coredbefore the injection of leachant. You suspect buildup of a calcium carbonateprecipitate has damaged the formation near the injection well If the perme-ability of the damaged section can be assumed to be I md, find the extent of thedamage. The wcllborc radius is 0.5 ft, and the distance to the outer boundaryof the uranium deposit is estimated to be 1000 ft

7.10 A well was given a large fracture treatment, creating a fracture that extends toa radius of about 150 ft. Thc effective permeability of the fracture area wasestimated to be 200 md. The permeability of the area beyond the fracture is 15md. Assume that the flow is steady-state, single-phase, incompressible flow. Theouter boundary at r = 1500 ft has a pressure of 2200 psia and the weliborepressure is 100 psia = 0.5 ft). The reservoir thickness is 20 ft and the porosity


is 18% The flowing fluid has a formation volume factor of 112 bbl/STB and aviscosity of 1 5 cp

(a) Calculate the flow rate in STB/day.

(b) Calculate the pressure in the reservoir at a distance of 300 ft from the centerof the wcllhore

7.11 (a) A limestone formation has a matnx (primary or intcrgranular) permeabilityof less than 1 md. I Towever, it contains 10 solution channels per square foot,each 0 02 in. in diameter. If the channels lie in the direction of fluid flow,what is the permeability of the rock?

(b) If the porosity of the matrix rock is 10%, what percentage of the fluid isstored in the primary pores, and what in the secondary pores (vugs, frac-tures. etc)?

(c) If the secondary pore system is well connected throughout a reservoir, whatconclusions must be drawn concerning the probable result of gas or waterdrive on the recovery of either oil, gas, or gas-condensate" What then arcthe means of recovering the hydrocarbons from the primary pores"

7.12 During a gravel rock operation the 6 in. l.D liner became filled with gravel, anda layer of mill scale and dirt accumulated to a thickness of 1 in on top of thegravel within the pipe If the permeability of the accumulation is 1000 md, whatadditional pressure drop is placed on the system when pumping a 1 cp fluid atthe rate of 100 bbl/hr?

7.13 One hundred capillary tubes of 0.02 in. ID and 50 capillary tubes of 0.04 in. ID,all of equal length, arc placed inside a pipe of 2 in inside diameter. The spacebetween the tubes is filled with wax so that flow is only through the capillarytubes. What is the permeability of this 'rock'?

7.14 Suppose, after cementing, an opening 0.01 in wide is left between the cementand an 8 in. diameter hole. If this circular fracture extends from the producingformation through an impermeable shale 20 ft thick to an underlying water sand,at what rate will water enter the producing formation (well) under a 100 psipressure drawdown9 The water contains 60,000 ppm salt and the bottom-holetemperature is I5ODF

7.15 A high water-oil ratio is being produced from a well. It is thought that the wateris corning from an underlying aquifer 20 ft from the oil producing zone. Inbetween the aquifer and the producing zone is an impermeable shale zoneAssume that the water is coming up through an incomplete cementing job thatleft an opening 0 Olin. wide between the cement and the 8 in. hole The waterhas a viscosity of 0.5 cp Determine the rate at which water is entering the wellat the producing formation level if the pressure in the aquifer is 150 psi greaterthan the pressure in the well at the producing formation level

7.16 Derive the equation for the steady-state. semispherical flow of an incom-pressible fluid

7.17 A well has a shut-in bottom-hole pressure of 2300 psia and flows 215 bbl/day ofoil under a drawdown of 500 psi The well produces from a formation of 36 ftnet productive thickness. Use = 6 in; = 66() feet, 0.88 cp, B0 = 1 32bbl/STB

(a) What is the productivity index of the well"

Problems 267

(b) What is the average permeability of the formation"

(C) What is the capacity of the formation"

7.18 A producing formation consists of two one 15 ft thick and 150 md inpermeability; the other 10 ft thick and 400 md in permeability

(a) What is the average permeability"

(b) What is the capacity of the formation?

(c) if dunng a well workover the 150 md stratum permeability is reduced to 25md out to a radius of 4 ft, and the 400 md stratum is reduced to 40 md outto an 8 ft radius, what is the average permeability after the workover,assuming no cross-flow between beds? Use r, 500 ft and r,. = 0 5 ft.

(d) To what percentage of the original productivity index will the well be re-duced"

(e) What is the capacity of the damaged formation"

7.19 (a) Plot pressure versus radius on both linear and semilog paper at 0 1. 1 0.10, and 100 days for - 2500 psia, q - 300 STB/day; B,, = 1.32 bbl/STI3;p.=044cp;k —25md;h ';cb—0.16

(b) Assuming that a pressure drop of 5 psi can be easily detected with a pressuregauge, how long must the well be flowed to produce this drop in a welllocated 1200 ft away?

(C) Suppose the flowing well is located 200 ft due east of a north-south fault.What pressure drop will occur after 10 days of flow, in a shut-in well located600 ft due north of the flowing well?

(d) What will the pressure drop be in a shut-in well 500 ft from the flowing wellwhen the flowing well has been shut in tor one day following a flow penodof 5 days at 300 STB/day"

7.20 A shut-in well is located 500 ft from one well and 1000 ft from a second well 1 hefirst well flows for 3 days at 250 STB/day, at which time the second well beginsto flow at 400 STh/day What is the pressure drop in the shut-in well when thesecond well has been flowing for 5 days (i.e., the first has been flowing a totalof 8 days)? Use the reservoir constants of Prob. 7.19.

7.21 A well is opened to flow at 200 STh/day for I day. The second day its flow isincreased to 400 SThIday and the third to 600 STB/day What is the pressuredrop caused in a shut-in well 500 ft away after the third day" Use the reservoirconstants of Prob 7.19

7.22 The following data pertain to a volumetric gas reservoir:

Net formation thickness = 15 ft

Hydrocarbon porosity = 20%

initial reservoir pressure 6000 psia

Reservoir temperature =Gas viscosity 0.020 cp

Casing diameter =6 inAverage formation permeability —6 md

(a) Assuming ideal gas behavior and uniform permeability, calculate the per-


centage of recovery from a 640 ac unit for a producing rate of 4.00 MMSCF/day when the flowing well pressure reaches 500 psia.

(b) If the average reservoir permeability had been 60 md instead of 6 md, whatrecovery would be obtained at 4.00MM SCF/day and a flowing well pressureof 500 psia?

(c) Recalculate part (a) for a production rate of 2 00 MM SCF/day.(d) Suppose four wells are dnlled on the 640 ac unit, and each is produced at

4.00 MM SCF/day. For 6 md and 500 psia nunimum flowing well pressure,calculate the rccovcry.

7,23 A sandstone reservoir, producing well above its bubble-point pressure, containsonly one producing welJ, which is flowing only oil at a constant rate of 175SThIday Ten weeks after this well began producing, another well was com-pleted 660ft away in the same formation On the basis of the reservoir properticsthat follow, estimate the initial formation pressure that should be encounteredby the second well at the time of completion

h=3Oft

c0=18(l0) 6psi 1 p.=29cp

6psi' k=35mdc1=43(l0) 6psi 1

p,=4300psia

= 1 25 bbl/STB

7.24 Develop an equation to calculate and then calculate the pressure at well 1,illustrated in Fig. 7.23. if the well has flowed for 5 days at a flow rate of 200STB/day

h 20 ft

c,30(10) "psi ' p.=0.Scp

k =SOmd B0=l.32bb1/STB

p,=4000psia

7.25 A pressure drawdown test was conducted on the discovery well in a new reser-voir to estimate the drainage volume of the reservoir. The well was flowed at aconstant rate of 125 STB/day The bottom-hole pressure data, as well as otherrock and fluid property data, follow What arc the drainage volume of the welland the average permeability of the drainage volume? The initial reservoirpressure was 3900 psia.

13., = 11 bbl/STI3 p.., =0 80 cp

è=20% h=22ftS.,—80%

c.,=10(lO) "psi '1

c,4(10)" psi' ft

Problems 269

Fault 1

Fault 2

Fig. 7.23. Well layout for Prob. 7.24

- Time in Hours pw1, psi

0.5 365710 36391.5 362920 362030 361250 359870 3591

10.0 358320.0 356530.0 355140.0 354850.0 3544600 354170 0 353780 0 353390 0 3529

100.0 35251200 3518150.0 3505

7.26 The initial average reservoir pressure in the vicinity of a new well was 4150 psiaA pressure drawdown test was conducted while the well was flowed at a constantoil flow rate of 550 STB/day. The oil had a viscosity of 3.3 cp and avolume factor of J.55 bbl/STB. Other data, along with the bottom-hole pressuredata recorded during the drawdown test, follow Assume that welibore storageconsiderations may be neglected, and determine the following'

(a) The permeability of the formation around the well

Well 1


(b) Any damage to the well

(C) The drainage voiuTnc of the reservoir communicating to the well.

b=34.3% h=93ft

Time in Hours

1 40252 40063 39994 39966 39938 3990

10 398920 398230 397940 397950 397860 397770 397680 3975

7.27 The first oil well in a new reservoir was flowed at a constant flow rate of 195STB/day until a cumulative volume of 361 STh had been produced After thisproduction period, the well was shut in and the bottom-hole pressure monitoredfor several hours. The flowing pressure just as the well was being shut in was1790 psia. For the data that follow, calculate the formation permeability and theinitial reservoir pressure.

B0—215bb1/STB p0=085cp4=11.5% h=23ftc,=1(10) 5psi

& in Hours Pws,

05 242510 288020 33003.0 331540 33205.0 33246.0 33308.0 3337

100 334312.0 334714.0 335216.0 335318.0 3356

References 271

7.28 A well located in the center of several other wells in a consolidated sandstonereservoir was chosen for a pressure buildup test. The well had been put onproduction at the same time as the other wells and had been produced for 80 hrat a constant oil flow rate of 375 SThfday The wells were dnlled on 80 acspacing. For the pressure buildup data and other rock and fluid property datathat follow, estimate a value for the formation permeability and determine if thewell is damaged. The flowing pressure at shut-in was 3470 psia.

B0='1.31 bbl/STS p.0=0.87cp

4=253% h=22ftS0=80%

c0=17(10)6 psi '

c,4(l0)

At in Hours

0.114 3701

0.201 3705

0.432 3711

0.808 3715

2051 3722

4.000 3726

8.000 3728

17 780 3730

REFERENCES

'W T Cardwell, Jr., and R. L. Parsons, "Average Permeabilities of HeterogeneousOil Sands," Trans. AIME (1945), 160, 34.

2J. Law, "A Statistical Approach to the Interstitial Heterogeneity of Sand Reser-voirs," Trans. AIME (1948), 174, 165.

3Robert C. Earlouglier. Jr., "Advances in W:ll Test Analysis," Monograph Vol. 5,Society of Petroleum Engineers of AIME. Dallas, TX: Millet the Printer, 1977

Fatt and D. H. Davis, "Reduction in Permeability with Overburden Pressure,"Trans. AIME (1952), 195, 329.

'Robert A. Wattenbargcr and H. i Ramey, Jr.. "Gas Well Testing With Turbulence.Damage and Welibore Storage," Jour, of Petroleum Technology (Aug. 1968), 877—887;Trans. A1ME, 243.

6C. 5. Matthews and D. G. Russell, "Pressure Buildup and Flow Tests in Wclls,"Monograph Vol. 1, Society of Petroleum Engineers of AIME. Dallas, 1'X Millet thePrinter, 1967.

7R Al-Hussainy, ii J. Ramey, Jr., and P B. Crawford, "The Flow of Real GasesThrough Porous Media," Trans. AIME (1966), 237, 624


0. Russell, J. H Goodrich, 0. E. Perry, and J. F. Bruskotter, "Methods forPredicting Gas Well Performance," Jour, of Petrolewn Technology (Jan. 1966).99—108; Trans AJME, 237.

9R. Al-Hussainy and H J. Ramey, Jr., "Application of Real Gas flow Thcory to Welll'esting and Deliverability Forecasting." Jour of petroleum Technology (May 1966),637—642; Trans AIME, 237. Also Reprint Series. No 9—Pressure Analysis Methods,Society of Petroleum Engineers of AIME, Dallas (1967) 245—250.

'°Theory and Practice of the Testing of Gas Wells, 3rd ed. Calgary, Canada: knergyResources Conservation Board, 1975.

"L. P Dake, Fundamentals of Reservoir Engineering. Amsterdam. Elsevier, 1978.

'2A. F. van Everdingen and W. Hurst, "The Application of the Laplace Trans-formation to Flow Problems in Reservoirs," Trans AIME (1949), 186, 305—324.

'3D R. Homer, "Pressure Build-Up in Wells," Proc, Third World Pet. Cong., TheHague (1951), Sec. 11, 503—523. Also Reprint Series, No. 9—Pressure Analysis Meth-ods, Society of Petroleum Engineers of AIME, Dallas, TX (1967) 25-43.

Eugenc Collins, Now of Fluids Through Porous Materials New York: Rein-hold, 1961, 108—123

'5W. John Lee, Well Testing, Society of Petroleum Engineers, Dallas, TX (1982).

'6A. F. van Everdingen, "The Skin Effect and Its Influence on the Productive Capacityof a Well," Trans., AIME (1953), 198, 171.

"W. Hurst, "Establishment of the Skin Effect and Its Impcdiment to Fluid Flow intoa Wellborc," Petroleum Engineering (Oct 1953), 25, B—6.

'II) R. Homer, "Pressure Build-Up in Wells," Proc.. Third World Petroleum Con-gress, The Hague (1951), Sec. II, 503. Also Reprint Series, No. 9—Pressure AnalysisMethods, Society of Petroleum Engineers of AIME, Dallas, TX (1961), 25.

Chapter 8

Water Influx

1. iNTRODUCTION

Many reservoirs arc bounded on a portion or all of their peripheries by water-bearing rocks called aquifers (Latin aqua—water, ferrc—to bear). The aqui-fers may be so large compared with the reservoirs they adjoin as to appearinfinite for all practical purposes, and they may range down to those so smallas to be negligible in their effect on reservoir performance. The aquifer itselfmay be entirely bounded by impermeable rock so that the reservoir andaquifer together form a closed, or volumetric, unit (Fig. 8.1). On the otherhand, the reservoir may outcrop at one or more places where it may bereplenished by surface waters (Fig. 8.2). Finally, an aquifer may be essentiallyhonzontal with the reservoir it adjoins, or it may rise, as at the edge ofstructural basins, considerably above the reservoir to provide some artesiankind of flow of water to the reservoir. *

in response to a pressure drop in the reservoir, the aquifer reacts tooffset, or retard, pressure decline by providing a source of water influx orencroachment by (a) cxpansion of the water; (b) expansion of other known orunknown hydrocarbon accumulations in the aquifer rock; (c) compressibility

throughout the text are at cod of cach chapter

273

274 Water Influx

FIg. 8.1. A reservoir analyzer study of five fields completed in a closed aquifer in theEllenburger formation in West Texas. (After Moore and Truby.')

Fig. 8.2. (icologic cross section through the Torchlight Tensleep Reservoir, Wyoming(After Stewart, Callaway, and Gladfclter.2)

SCALE IN

20 IS 10 5 0 5 10 15 20NORTHWEST DiSTANCE FROM TORCHLIGHT-MiLES EAST

2 Steady-State Models 275

of the aquifer rock; and/or (d) artesian flow, which occurs when thc aquiferrises to a level above the reservoir, whether it outcrops or not, and whetheror not the outcrop is replenished by surface water.

To determine the effect that an aquifer has on the production from ahydrocarbon reservoir, it is necessary to be able to calculate the amount ofwater that has influxed into the reservoir from the aquifer. This calculation canbe made using the material balance equation when the initial hydrocarbonamount and the production are known The Havlena and Odeh approach tomaterial balance calculations, presented in Chapter 2. can sometimes be usedto obtain an estimate for both water influx and initial hydrocarbon amount.34For the case of a water-drive reservoir, no original gas cap, and negligiblecoinpressibilitics, Eq (2.13) reduces to the followingS

=NE0+

or

we= N i

L.0

if correct values of W, are placed in this cquation as a function of reservoirpressure, then the equation should plot as a straight line with intercept, N, andslope equal to unity The procedure to solve for both and N in this case in-volves assuming a model for as a function of pressure, calculating 14, mak-ing the plot of F/E0 versus We/Eo, and observing if a straight line is obtained.If a straight line is not obtained, then a new model for is assumed and theprocedure repeated.

Choosing an appropriate model for water influx involves many uncer-tainties. Some of these include the size and shape of the aquifer and aquiferproperties such as porosity and permeability. Normally, little iS known aboutthese parameters largely because the cost to drill into the aquifer to obtain thenecessary data is not often justified.

In this chapter, several models that have been used in reservoir studiesto calculate water influx amounts are considered. These models can be gen-erally categorized by a time dependence (i.e., steady-state or unsteady-state)and whether the aquifer is an edge-water or bottom-water drive.

2. STEADY-STATE MODELS

The simplest model we discuss is the Schilthuis steady-state model, in whichthe rate of water influx, dWeIdI, Is directly proportional to (p, — p), where thepressure, p. is measured at the original oil-water contact.5 This model assumesthat the pressure at the external boundary of the aquifer is maintained at the

276 Water Influx

initial value and that flow to the reservoir is, by L)arcy's Law, proportionalto the pressure differential, assuming the water viscosity, average perme-ability, and aquifer geometry remain constant,

(8.1)

(8.2)

where k • is the water influx constant in barrels per day per pounds per squareinch and (p, — p) is the boundary pressure drop in pounds per square inch. Ifthe value of k' can be found, then the value of the cumulative water influx W,can be found from Eq (8.1), from a knowledge of the pressure history of thereservoir. If during any reasonably long period the rate of production andreservoir pressure remain substantially constant, it is obvious that the volu-metric withdrawal rate, or reservoir voidage rate, must equal the water influxrate, or

dW Rate of active Rate of free Rate of water= oil volumetric + gas volumetric + volumetric

voidage voidage voidage

In terms of single-phase oil volume factors

(8.3)

where dN,,/dt is the daily oil rate in STO/day and (R — is the dailyfree gas rate in SCF/day. The solution gas-oil ratio is subtracted from thenet daily or current gas-oil ratio R because the solution gas R,0 is accounted forin the oil volume factor B0 of the oil voidage term Equation (8.3) may beconverted to an equivalent one using two-phase volume factors by adding andsubtracting the term and grouping as

+ lB0 + (R — R301)B, +

and since LA0 + (R50, — is the two-phase volume factor

-di — 'di 'di Wdt

When dWe/dt has been obtained in terms of the voidage rates by Eqs. (8.3) or

2. Steady-State Models 277

(8.4), then the influx constant k' may be found using (8 2). Although theinflux constant can be obtained in this manner only when the reservoir pres-sure stabilizes, once it has been found it may be applied to both stabilized andchanging reservoir pressures.

Figure 8.3 shows the pressure and production history of the ConrocField, Texas, and Fig 8.4 gives the gas and two-phase oil volume factors forthe reservoir fluids. Between 33 and 39 months after the start of production.the reservoir pressure stabilized near 2090 psig and the production rate wassubstantially constant at 44,100 STB/day with a constant gas-oil ratio of 825SCF/STB. Water production during the period was negligible. Example 8.1shows the calculation of the water influx constant k' for the Conroe Field fromdata for this period of stabilized pressure. If the pressure stabilizes and thewithdrawal rates are not reasonably constant, the water influx for the penodof stabilized pressure may be obtained from the total oil, gas, and watervoidages for the period,

where and are the gas, oil, and water produced during theperiod in surface units. The influx constant is obtained by dividing by theproduct of the days in the interval andthe stabilized pressure drop (p, —

k' =— p,)

1500 oI-4

U,

U

C)

-

'I-

cn2

—

"Jo

SI

L

— —AVERAGE RESERVOIR PRESSURE

AT4000 FEET SUBSEA

GAS OIL RATIO

nl_. I IDAILY PRODUCIION

— — —

r— —4-

FIME YEARS

Fig. 8.3. Reservoir pressure and production data, Conroe Field (After

276 Water Influx

C,

U)a.

U)Cl)wa-

.006 .008 .0l0 .012

GAS VOLUME FACTOR, Cu FT PER SCF

Fig. 8.4. Pressure volume relations for Conroe Field oil and onginal complement ofdissolved gas (After Schilthuis.3)

Example 8.1. Calculating the water influx constant when reservoir pres-sure stabilizes

Given

'rhe PVT data for Conroe Field, Fig. 8.4:

= 2275 psig

= 2090 psig (stabilized pressure)

= 7.520 cu ft/STB at 2090 psig

BR = 0.00693 cu ft/SCF at 2090 psig

= 600 SCF/STB (initial solution gas)

R = 825 SCF/STB, from production data

dNa/dr = 44,100 STB/day, from production data

=0

Sot u'rIoN: At 2090 psig by Eq (8.4) the daily voidage rate is

220C

TWO-PHASE OIL VOLUME FACTOR, CU FT PER STB7.4 8-2 9.0 98 106

gout:.) — —

1800 — — — — — —-

IgIvu —

—

—

—

— — — — ——

—

—

— ———

—

2 Steady-State Models 279

= 7 520 x 44,100 + (825— 600)0.00693 x 44,100 + 0

= 401,000 cu ft/day

Since this must equal the water influx rate at stabilized pressure condi-tions, by Eq. (8.2)

= = 401 000 k '(2275 — 2090)di di

k' = 2170 cu ft/day/psi

A water influx constant of 2170 cu ft/day/psi means that if the reservoirpressure suddenly drops from an initial pressure of 2275 to, say, 2265 psig(i.e., A,., = 10 psi) and remains there for 10 days, during this period the waterinflux will be

=2170x lOx 10=217,000 cu ft

If at the end of 10 days it drops to, say, 2255 (i.e., Ap =20 psi) and remains

there for 20 days, the water influx during this second period will be

AWe2=2170x20x20=868,000cu ft

There is four times the influx in the second period because the influx rate wastwice as great (because the pressure drop was twice as great) and because themterval was twice as long. The cumulative water influx at the end of 30 days,then, is

30

(p1—p)dz=k2(p1—p)Az

= 2170[(2275 — 2265) x 10+ (2275 — 2255) x 20]

=1,085,000cuft

In Fig 8.5 the —p)dt is represented by the area beneath the curveof pressure drop, (p, — p). plotted versus time; or it represents the area above

the curve of pressure versus time. The areas may be found by graphicalintegration.

One of the problems associated with the Schilthuis steady-state model isthat as the water is drained from the aquifer, the distance that the water hasto travel to the reservoir increases. Hurst suggested a modification to theSchilthuis equation by including a logarithmic term to account for this increas-ing distance.6 The Hurst method has met with limited application and isinfrequently used.

280 Water Influx

TIME

Fig. 8.5. Plot of pressure and pressure drop versus time

0 logaz

dW _c'(p,—p)di — logai

where c' is the water influx constant in barrels per day per pounds per squareinch, is the boundary pressure drop in pounds per square inch, anda is a time conversion constant that depends on the units of the time t.

3. UNSTEADY-STATE MODELS

in nearly all applications, the steady-state models discussed in the previoussection arc not adequate in describing the water influx The transient natureof the aquifers suggests that a time-dependent term be included in the calcu-lations for W,. in the next two sections, unsteady-state models for both edge-water and bottom-water drives are presented. An edge-water drive is definedas water influxing the reservoir from its flanks with negligible flow in thevertical direction. in contrast, a bottom-water dnve has significant verticalflow

3.1. The van Everdingen and Hurst Edge-Water DriveModel

Consider a circular reservoir of radius rR, as shown in Fig. 8.6, in a horizontal,circular aquifer of radius Te, which is uniform in thickness, permeability, andporosity, and in rock and water compressibilitics. The radial diffusivity equa-tion, Eq. (7.35), expresses the relationship between pressure, radius, and time

TIME

3. Unsteady-State Models 281

Fig. 8.6. Circular rescrvoir inside a circular aquifer.

for a radial system such as Fig. 8.6, where the driving potential of thc systemis the water expandibility and the rock compressibility:

a2p bl.LC1735

ar 0.0002637k 81 (

This equation was solved in Chapter 7 for what is referred to as the constantterminal rare case. The constant terminal rate case requires a constant flowrate at the inner boundary, which was the welibore for the solutions of Chap-ter 7. This was appropriate for the applications of Chapter 7 since it wasdesirous to know the pressure behavior at various points in the reservoirbecause a constant flow of fluid came into the wellbore from the reservoir.

In this chapter, the diffusivity equation is applied to the aquifer wherethe inner boundary is defined as the interfacc between the reservoir and theaquifer With the interface as the inner boundary, it would be more useful torequire the pressure at the inner boundary to remain constant and observe theflow rate as it crosses the boundary or as it enters the reservoir from theaquifer Mathematically, this condition is stated as

p =constant=pI—Apatr=rR (85)

where TR is a constant and is equal to the outer radius of the rescrvoir (i.c . theoriginal oil-water contact). The pressure p must be determined at this originaloil-water contact. Van Evcrdingcn and Hurst7 solved the diffusivity equationfor this condition, which is referred to as the constant terminal pressure case,and the following initial and outer boundary conditions.

282 Water Influx

Initial condition:

p =p for all values of r

Outer boundary condition:

For an infinite aquifer:

p at r =00

For a finite aquifer:

op =0 at r =

At this point, we rewrite the diffusivity equation in terms of the followingdimensionless parameters:

Dimensionless time: = 0.0002637 2 (8.6)

Dimensionless radius: rD =TR

Dimensionless pressure: Po= PPPi —

where k = average aquifer pcrmcability, md; : = time, hours; è = aquiferporosity, fraction; p. = water viscosity, cp; c, = aquifer compressibility,and TR = reservoir radius, feet. With these dimensionless parameters, the dif-fusivity equation becomes.

(870r20 1D OrD 01D

)

van Everdingen and Hurst converted their solutions to dimensionless, cumu-lative water influx values and made the results available in a convenient formhere given in Tables 8 1 and 8.2 for various ratios of aquifer to reservoir size,expressed by the ratio of their radii, r,/rR Figures 8.7 to 8.10 are plots of someof the tabular values. The data are given in terms of dimensionless time, 'D'and dimensionless water influx, W,D, so that one set of values suffices for allaquifers whose behavior can be by the radial form of the dif-fusivity equation The water influx is then found by using Eq (8.8):

(8.8)

3 Unsteady-State Models 283

where

D'—.l iin.&.I) —

B' is the water influx constant in barrels per pounds per square inch and 0 isthe angle subtended by thc reservoir circumference, (i.e.. for a full circle,0 360° and for a semicircular reservoir against a fault, 0= 180°). c, is in psi'and rR and Ii are in feet.

Example 8.1 shows the use of Eq. (8.8) and the values of Tables 8.1 and8.2 to calculate the cumulative water influx at successive periods for the caseof a constant reservoir boundary pressure. The infinite aquifer values may beused for small time values even though the aquifer is limited in size.

Example 8.1. Calculate the water influx after 100 days, 200 days, 400days, and 800 days into a reservoir the boundary pressure of which is suddenlylowered and held at 2724 psia (p, = 2734 psia).

Given:

cf,=0.20 k=83mdC,8(lO)6p5i_t TR=3000 ft

r, = 30,000 ft p. = 0.62 cp

0=3600 h=40ft

SoLuTioN: From Eq. (8 6):

O.0002637(83)r8(10) 6130002

= 0 00245:

From Eq. (8 9):

B' = 111 = 644 5

At 100 days D 0.00245(100)(24) 5.88 dimensionless time units. From ther,/rR = 10 curve of Fig. 8.8 find corresponding to D = 5.88. Wç0 = 5.07 dimen-sionless influx units. This samc value may also be found by interpolation ofTable 8.1, since below CD 15 the aqwfeT behaves essentially as if it wereinfinite, and no values are given in Table 8.2 Since = 2734 — 2724 = 10 psi,and water influx at 100 days from Eq. (8.8) is

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36


and middle curves represent the pressure distribution in the aquifer in re-sponse to the first signal alone, at times ti and respectively. The upper curvemay also be used to represent the pressure distribution for the second pressuresignal alone at time 12 because in this simplified case lip1 = and =The lower curve, then, is the sum of the upper and middle curves. Mathe-matically, this means that Eq. (8.10) can be used to calculate the cumulativewater influx:

(8.10)

This calculation is illustrated in Ex. Prob. 8.2.

Example 8.2. Suppose in Ex. 8.1 at the end of 100 days the reservoirboundary pressure suddenly drops to 22704 psia (i.e., lip2 = p'— Pi =20psi, not p•— P2 =30 psi). Calculate the water influx at 400 days total time.

The water influx due to the first pressure drop lip1 = 10 psi at 400 dayswas calculated in Example 8.1 to be 89,590 bbl. This will be the same eventhough a second pressure drop occurs at 100 days and continues to 400 daysThis second drop wilt have acted for 300 days, or a dimensionless time of1D 0.0588 x 300 = 17.6. From Fig 8.8 or Table 8.2 for rIrR = 10 WeD = 11.14for tD = 17 6 and the water influx is

= B' X x = 644.5 x 20 x 11.14= 143,600 bbl

= + = B' X Ap1 x We01 + B' X X = B'ThPW,0

=644.5 (lOx 13.90+20x 11.14)

= 89,590 + 143,600 = 233,190 bbl

Example 8.2 illustrates the calculation of water influx when a secondpressure drop occurs 100 days after the first drop in Example 8.1. A con-tinuation of this method may be used to calculate the water influx into reser-voirs for which boundary pressure histories are known, and also for whichsufficient information is known about the aquifer to calculate the constant B'and the dimensionless time z0.

The history of thc reservoir boundary pressure may be approximated asclosely as desired by a series of step-by-step pressure reductions (or increases),as illustrated in Fig. 8.12 The best approximation of the pressure history ismade as shown by making the pressure step at any time equal to half of thedrop in the previous interval of time plus half of the drop in the succeedingperiod of time.8 When reservoir boundary pressures are not known, averagereservoir pressures may be substituted with some reduction in the accuracy ofthe results. In addition, for best accuracy, the average boundary pressureshould always be that at the initial rather than the current oil-water contact;otherwise, among other changes, a decreasing value of rR is unaccounted for.

296 Water Influx

Fig. 8.12. Sketch showing the use of step pressures to approximate the pressure-timecurve

Example 8 3 iUustratcs the calculation of water influx at two successive timevalues for the reservoir shown in Fig. 8.13

Example 8.3. Calculate the water influx at the third and fourth quarteryears of production for the reservoir shown in Fig. 8 13. Use 4) = 0.209;k = 275 md (average reservoir permeability, presumed the same for the aqui-fer); p= 0.25 cp; c, 6 X 10 psi'; h = 19.2 ft; area of rcscrvoir= 1216 ac;estimated area of aquifer 250,000 ac; 0= 1800.

SoLLrr!oN: Since the reservoir is against a fault A = and:

21216x43,560

rR 0.5x3.1416

rR=5807ft

For z = 91.3 days (one-quarter year or one period)-

6.323 x x 275 x 91.30.209 x x x (5807)2 15.0

B = 1.119 x 0.209 x 6 x 10 6 x(5807)2 X 19.2 x(l80°/360°)

= 455 bbUpsi

Since the aquifer is 250,000/1216 = 206 times the area of the reservoir,for a considerable time the infinite aquifer values may be used. Table 8.3

TiME PERIODS


Fig. 8.13. Sketch showing the equivalent radius of a reservoir

TABLE 8.3.Boundary step pressures and values for Ex. 8.3

TimePeriod

Timein Days

Dimensionlessless Time,

p. AvgReservoirPressure,

Pa. Avg.BoundaryPressure,

Ap. SlepPressure,

I I 1D We,, psia psia psi

0 0 0 0.0 3793 3793

-

001 913 15 10.0 3786 3788 25

2 1826 30 16.7 3768 3774 953 2739 45 22.9 3739 3748 2004 365.2 6() 28.7 3699 3709 32 55 456.5 75 34 3 3651 3680 34.06 547.8 90 396 3613 3643 33.0

a Infinite aquifer values from Fig 8 9 or 8.1

shows the values of boundary step pressures and the W,1, values for the first sixperiods. The calculation of the step pressures is illustrated in Fig. 8.12. Forexample,

= — — 3748) = 20 0 psi

Tables 8.4 and 8.5 show the calculation of X at the end of the third

298 Water Influx

TABLE 8.4.Water influx at the end of the third quarter for Example 83

- - W,0 - )( -

45 229 25 57330 167 9.5 158715 100 20.0 200.0

TABLE 8.5.Water influx at the end of the fourth quarter for Example 83

to

60 28.7 2.5 71.845 229 95 217630 167 20.0 334.015 10.0 32.5 325.0

and fourth periods, the values being 416 0 and 948.0, respectively. Then thecorresponding water influx at the end of these periods is

We (3rd quarter) = B 'SAp x W,D = 455 x 416.0 = 189,300 bbl

W, (4th quarter) B'Y.,Ap x 455 x 948.4 = 431,500 bbl

In calculating the water influx in Ex. 8.3 at the end of the third quarter,it should be carefully noted in Table 8.4 that since the first pressurc drop,

=2 5 psi, had been operating for the full three quarters (z0 = 45), it wasmultiplied by W,D = 22.9, which corresponds to t,, = 45. Similarly, for thefourth quarter calculation in Table 8.5, the 2.5 psi was multiplied by WeD28 7, which is the value for tD = 60. Thus the WeD values are inverted so thatthe one corresponding to the longest time is multiplied by the first pressuredrop, and vice versa. Also, in calculating each successive value of x WeD,it is not simply a matter of adding a new x W,D term to the former sum-mation, but a complete recalculation, as shown in Tables 8.4 and 8 5.

You should show that the water influx values at the end of thc fifth andsixth quarters are 773,100 and 1,201,600 bbl, respectively.

From the previous discussion it is evident that it is possible to calculatewater influx independently of material balance calculations from a knowledgeof the history of the reservoir, boundary pressure, and the dimensions andphysical characteristics of the aquifer, as shown by Chatas.9 Although strictlyspeaking, the van Everdingen and Hurst solutions to the diffusivity equationapply only to circular rcservoirs surrounded concentrically by horizontal, cir-cular (or infinite) aquifers of constant thickness, porosity, permeability, and

3 Models 299

effective water compressibility, for many engineering purposes good resultsmay be obtained when the situation is somewhat less than ideal, as it nearlyalways is. The radius of the reservoir may be approximated by using the radiusof a circle equal in area to the area of the reservoir; and where the approxi-mate size of the aquifer is known, the same approximation may be used for theaquifer radius. Where the aquifer is more than approximately 99 times the size(volume) of the reservoir (re/rR = 10), the aquifer behaves essentially as if itwere infinite for a considerable period so that the values of Table 8.1 may beused. There arc to be sure, uncertainties in the permeability, porosity, andthickness of the aquifer which must be estimated from information obtainedfrom wells drilled in the reservoir and whatever wells are drilled in the aquifer.The viscosity of the water can be estimated from the temperature and pressure(Chapter 1, Sect. 7.4) and the water and rock comprcssibilities from the dataof Chapter 1. Sects. 7.3 and 4 2

Because of the many uncertainties in the dimensions and properties ofthe aquifer, the calculation of water influx independently of material balanceappears somewhat unreliable. For instance in Ex. 8.3 it was assumed that thefault against which the reservoir accumulated was of large (actually infinite)extent; and since the permeability of only the reservoir rock was known, it wasassumed that the average permeability of the aquifer was also 275 md. Theremay be variations in the aquifer thickness and porosity, and the aquifer maycontain faults, impermeable areas, and unknown hydrocarbon accumulations,all of which can introduce variations of greater or lesser importance.

As Ex. Probs. 8.1, 8.2, and 8.3 suggest, the calculations for water influxcan become long and tedious. The use of the computer with these calculationsrequires large data files containing the values of W,1,, 1D, and r,IrR from Tables8.1 and 8.2 as well as a table lookup routine. Several authors have attemptedto develop equations to describe the dimensionless water influx as a functionof the dimensionless time and radius ratio. These equations reduce thedata storage required when using the computer in calculations of water influx.

3.2. Bottom-Water Drive

The van Everdingen and Hurst model discussed in the previous section isbased on the radial diffusivity equation written without a term describingvertical flow from the aquifer. in theory, this model should not be used whenthere iS significant movement of water into the reservoir from a batdrive. To account for the flow of water in a vertical direction, Coats and laterAllard and Chcn, added a term to Eq. (7.35) to yield the following'

apat

(8.10)

where is the ratio of vertical to horizontal permeability.'3 14

Water Influx

Using the definitions of dimensionless time, radius, and pressure and in-troducingaseconddimensionlessdistance,

z2D

rR

+ + = (811)TD aTD äZ/, t3tD

Coats solved Eq. (8.11) for the terminal rate case for infinite aquifers. 13 Allardand Chen used a numerical simulator to solve the problem for the terminalpressure case.14 They defined a water influx constant, B', and a dimensionlesswater influx, W,,, analogous to those defined by van Everdingen and Hurstexcept that B' does not include the angle

B' (8.12)

The actual values of W,0 will be different from those of the van Everdingen andHurst model because for the bottom-water drive is a function of thevertical permeability. Because of this functionality, the solutions presented byAllard and Chen, found in Tables 8.6 to 8 10, are functions of two dimen-sionless parameters, rt, and z,.

(813)rR

hrI'2 (8.14)

rRL-

The method of calculating water influx from the dimensionless values ob-tained from these tables follows exactly the method illustratcd in Exs. 8 1 to8.3. The procedure is shown in Ex. 8 4, which is a problem taken from Allardand Chen 14

Example 8.4. Calculate the water influx as a function of time for thereservoir data and boundary pressure data that follow:

Given.

rR2000fth=200ft k=5Omd

4=O.1O1


Average -

Time Boundaryin Days Pressure, psia

(I) (PB)

0 300030 29566() 291790 2877

120 2844150 2811

180 2791210 2773240 2755

SOLUTION:

=

200

== 0.5

0 0002637(50)1D = 0.10(0.395)8(10) -620002 = 0.OJO4i (where us in hours)

B' = = 716 bbl/psi

Averagelimein Days

(t)

DimensionlessTime(Ia) -

BoundaryPressure, psia

(pn)

SiepPressure

(Ap)

WalerInflux, M bbl

(We) --

0 0 0 3000 0 030 75 5.038 2956 220 7960 15.0 8.389 2917 41.5 28290 22.5 11.414 2877 395 572

120 30.0 14.263 2844 36.5 933150 37.5 16.994 2811 33.0 1353180 45.0 19.641 2791 265 1810210 52.5 22.214 2773 19.0 2284240 60.0 24 728 2755 18.0 2782

C,, 0

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4.53

6 x

JO4

6.04

8 )c

6.00

4 x

10'

5.99

9 x

10'

5.92

0 x

(0'

7.43

6 x

10'

7 43

1 x

10'

7.37

6 x

10'

7.28

0 x

10'

8 80

5 x

10'

8.79

8 X

10'

8.73

510

'8.

623

x 10

'1.

016x

10'

1015

x10'

1.O

OB

XIO

'9.

951x

10'

1.15

0x10

'1.

149x

10'

1.14

1x10

'lJ

27x1

0'1.

283

x 10

'1.

282

x 10

'12

73 x

10'

1.25

7x 1

0'1.

415

X 1

0'1.

412X

10'

1 40

4x 1

0'1.

387

x 10

'2.

059

x 10

'2.

060

x 10

'2.

041

x 10

'2.

016

x 10

'2.

695'

clO

'2.

695x

10'

2676

x10'

2.64

4x10

'3.

319x

10'

3.29

6x10

'3.

254x

10'

3.93

7 x

10'

3.93

6 x

10'

3.90

9x

10'

x 10

'10

'10

'x

10'

x 10

'X

10'

8.70

9)c1

0'8.

705x

10'

8.65

0x10

'8.

556x

10'

9.97

2x10

'98

67x1

0'9.

806x

10'

9.69

9x10

'1

103x

106

1.10

2x10

'10

95x1

0'1.

084x

10'

1217

x10'

1217

x10'

1.20

9x10

'1.

196x

10'

1.78

1x10

'1.

771x

10'

l.752

x106

2.33

7 x

106

2.33

6 x

10'

2.32

2 x

10'

2.29

8 x

10'

2 88

4 x

10°

2.88

2 x

106

2.86

6 x

10'

2.83

7 x

10'

3.42

5x10

'34

23x

106

3.40

4x 1

063.

369x

106

4.49

3 x

10°

4 49

1 x

10'

4 46

610

'44

22 x

106

5.54

7x10

°5.

544x

t0'

S.S

l4xl

O'

5.46

0x 1

06

4290

X 1

0'5.

606

x 10

'6.

900

x 10

'8.

178x

10'

9.44

2 x

10'

l.070

x 10

'1.

194

x 10

'1.

317

x 10

'1

918

x 10

'2.

518

X 1

0'3.

101

x 10

'3.

684

x 10

'4.

828

)C

5.95

6)(

10'

7.07

2 x

10'

8.17

7x 1

0'92

73x

10'

1.03

6x it

t1

144X

10°

1.67

8 x

10°

2.20

2 x

10'

2.72

0 X

itt

3.23

2x 1

0'42

44X

10'

5.24

3 )(

10'

600x

10'

7.O

Ox

10'

800x

9.O

Ox

10'

1.O

Ox

108

1.50

x 10

82.

OO

x 10

82.

50x

108

3.O

Ox

108

4.O

Ox

108

5.O

Ox

6.00

x 10

87.

OO

x 10

88.

OO

x 10

°9.

00 x

108

lOO

x1.

SOx

2.O

Ox

2.50

x3.

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x10

°4.

OO

x iU

°5.

OO

x6.

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x7.

OO

x8.

OO

x9.

OO

x 10

°IQ

Ox

10'°

1.SO

x 1Q

102.

OO

x 10

102.

50x

lOb

'03.

OO

x 10

10

6.59

0x10

°7.

624x

106

8.65

1 x

156

7 x

10'

2.05

9 x

jQ7

2.54

6x

10'

3.02

7 x

IC'

3,97

9x 1

0'4.

920

x5.

852

x 10

'6.

777

x tO

'7.

700x

8.60

9x 1

0'9.

518

x jQ

71.

401x

1.84

3 x

10°

2.28

1x 1

0°2.

714

x 10

83.

573

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084.

422

x 10

85.

265

)c 1

086

101

x 10

86.

932

x 10

°7.

760x

8.58

3x 1

081

263

'C J

O9

1 66

6 x

10'

2.06

5 'C

2.45

8 x

10°

6.40

1 x

106

7.40

710

6

8.40

7x 1

069.

400

x 10

61.

039

x 10

'1.

522x

tO'

2.00

4 x

2.47

6x 1

0'2.

947

x 10

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875

x 10

'4.

793

'C 1

0'5.

702

x 10

'6.

605

x 10

'7.

501

x 10

'8.

393

x 10

'9.

281

x 10

'1.

367

x 10

81.

799

x 10

82.

226x

108

2.65

0 x

3.48

9 'C

108

4.32

0 x

ION

5.14

3 x

10°

5.96

1 x

10°

6.77

5 x

10°

7.58

4 'C

8.38

9 x

108

I 23

5x 1

0'1.

630

x 10

°2.

018

x 10

'2.

405

x

6299

x10°

6232

x10°

7.29

0 x

108

7.21

3 x

106

8.27

4 x

106

8.18

8 x

106

9.25

2x 1

069.

156x

10°

1.02

3 x

101

1.01

2 x

10'

1 49

9 x

10'

1.48

310

'1.

974

x 10

'1.

954

x 10

'2.

439

x 10

'2.

415

x 10

'29

04x1

073.

819

x 10

'3.

782

x 10

14.

724x

4679

x 10

'S6

21xW

'55

68x1

0'6.

511

x 10

'6.

450

x 10

'7.

396

x 10

'7.

327

x8.

275

x 10

'8.

199

x 10

'9

151

'C 1

0'9.

066

x 10

'1.

348x

108

1.33

6x17

74x1

081.

758x

10°

2.19

6x10

'2.

177x

IO

N26

15 x

108

2.59

2x 1

083.

443

x 10

°3.

413

x 10

842

63x1

(f42

27x1

085.

077

x 10

°5.

033

x 10

85.

885x

108

5.83

5x 1

086.

688

x 10

°6.

632

x 10

°74

87x

108

7.42

4x 1

088.

283

x 10

88.

214

x 10

81.

219x

10°

1 20

9x10

°1

610x

10'

1.59

6x 1

0'1.

993x

109

1.97

7x 1

0'2.

376

10°

2.35

7 x

10'

6.58

7 x

10°

6.55

1 x

10°

6.48

8 x

106

7.62

0x10

°75

79x

106

7.50

7x10

°8.

647x

106

8.60

0x10

°8.

519x

10°

9.66

6 x

10°

9.61

5 x

10°

9.52

4 x

106

1.06

7 x

10'

1.06

2 x

10'

1.05

2 x

1.56

7 'C

10'

1.55

5 x

10'

1.54

1 x

10'

2.05

9x10

'2.

048x

10'

2.02

9x 1

0'2.

545

x 10

'2.

531

x 10

'2.

507

x 10

'3.

026x

10'

3.O

lOxl

O'

2.98

4x10

'3.

978x

10'

3958

x10'

3.92

3x10

'4.

918

x iO

74.

894

x 10

'4.

851

x 10

'5.

850x

IO'

5.82

1x10

'5.

771x

10'

6.77

4 x

10'

6.74

1 X

10'

6.68

4 x

jQ7

7.69

3x10

'7.

655x

10'

7.59

0x10

'8.

606x

10'

8.56

4x10

'8.

492x

10'

9.51

5 x

10'

9.46

9 x

10'

9.39

0 x

10'

1.40

0x10

81.

394x

10°

1382

x108

1.84

3x10

°l.8

34x1

0°1.

819x

10°

2.28

0x10

'2.

269x

108

2.25

1x10

°2.

713x

10°

2.70

1x10

°2.

680x

108

3.57

2x10

°3.

S56x

10°

3.52

8x 1

084.

421

x 10

°4.

401

x 10

°4.

367

x 10

85.

262x

10'

5.24

0x 1

085.

199x

10°

6.09

8 x

108

6.07

2 x

108

6.02

5 x

10°

6.93

Ox

108

6900

x108

6.84

7x10

°7.

756x

10°

7.72

3x 1

087.

664x

10°

8.57

4 x

8.54

3 x

108

8.47

8 x

10°

1.26

4x10

°1.

257x

10°

1247

x10°

1666

x10°

1659

x10°

1.64

6x10

'2.

063x

10'

2.05

5x10

'2.

038x

10°

2.45

8x

jQ9

2.44

7 X

10°

2.43

0 x

10'

TA

BLE

8.6

.(C

ont'd

)

z6

ii,00

501

0305

0709

10

4 00

x 1

010

3.24

0x

3 23

9 x

3.22

6 x

10"

3.20

3 x

10"

3.17

1 x

10"

3.13

3 X

3 10

8 x

JO"

5.00

x 1

010

4.01

4x

4.01

3 x

10"

3.99

7 x

10"

3.96

8 x

10"

3.92

9 x

10"

3.88

3 X

10"

3.85

2)(

10"

600x

101°

4782

x10"

4.78

1x10

'47

62x1

0"4.

728x

109

4.68

2x10

946

27x1

0'4.

591x

10"

7.00

x5.

546

x 10

"5.

544

x iO

"5.

522

x 10

"5.

483

X10

"5.

430

x 10

"5.

366

x 10

"5

325

x 10

"8.

00)(

6 30

5 X

10"

6303

X6.

278

X 1

0"6.

234

X 1

0"6

174

X 1

0"6.

102

10"

6.05

5 'c

10"

9.00

x7.

060

x 10

"7.

058

x7.

030

x 10

"6.

982

x 10

"6

914

x 10

"6

834

x6

782

x 10

"1.

00 x

7.81

3 x

10"

7.81

0 x

10"

7 78

0 x

10"

7.72

6 x

II?

7.65

2 X

10"

7.56

4 x

10"

7.50

6 x

10"

1.50

x 1

0"11

54 x

101

01.

153

x lO

b1.

149

x1.

141

X10

'°1.

130

X10

"'11

18 X

1010

110

9 x

10"'

2.00

x 1

0"1.

522

x 10

"'1

521)

c 10

"'1

515

x 10

101.

505

)c 1

0'°

1.49

1 X

10"'

1.47

4 x

10"'

1.46

3 x

2.50

x 1

0"1

886

X10

101

885

X 1

020

1 87

8 x

10'°

1.86

6 x

10'°

1.84

9 x

1011

1.82

8)(

101

1.81

4)(

10"

2.24

8 x

10"'

2.24

7 x

2.23

9 x

2.22

4 x

220

4 x

2.17

9x

1011

2.16

3 x

4.00

x 1

0"2.

965

x 10

202.

964

x jO

"2.

953

)c 1

0"'

2.93

4 x

10"'

2.90

7 x

1010

2.87

6 X

102

02.

855

1020

5.00

x 1

011

3.67

7 x

10"'

3.67

5 x

10"'

3.66

2 x

10"'

3.63

8 )

1010

3.60

5 X

10"

3.56

6 X

10"'

3.54

0 x

10"'

600

x 10

"4

383

x 10

104

381

x lO

ll4

365

x 10

104.

337

x 10

"'4.

298

x 10

"'4.

252

X 1

020

4.22

1 x

l0'°

7.00

x 1

0"5.

085

10"'

5.08

2 X

10"

'5.

064

X 1

010

5.03

2 x

iO'°

4.98

7 x

10"'

4.93

3 x

10"'

4.89

810

°8.

00 x

10"

5.78

3 x

10"'

5.78

1 x

1010

5.76

0 x

1010

5.72

3 x

10"'

5673

x 1

010

5.61

2 x

iO'°

5.57

2tO

"'9.

00 x

10"

6.47

8 x

10"'

6.47

6 x

1010

6.45

3 x

6.41

2 x

10"'

6.35

5 x

10"'

6.28

8 x

6.24

3 c

1020

1.00

x 1

0's

7.17

1 x

10'°

7.16

8 x

10"'

7.14

3 x

10"'

7.09

8 x

10"'

7.03

5 x

10"'

6.96

1 X

10"

'6.

912

< 1

0"'

150x

10'2

1060

x101

'10

60x1

0"1.

041x

10"

1.03

0X10

"1.

022)

clO

"2.

OO

xlO

'21.

400x

10"

1.39

9X10

"1.

394x

10'1

1386

x10"

L37

4x10

"1.

350X

10"


TABLE 8.7.Dimensionless Influx, for =4 for bottom-water drive

0.05 01 0.3 0.5 0.7

2 2398 2.389 2.284 2.031 1.824 1.620 1.507

3 3 006 2.993 2.874 2.629 2.390 2.149 2 012

4 3552 3.528 3.404 3.158 2893 2.620 24665 4053 4.017 3.893 3627 3.341 3.045 2.876

6 4.490 4 452 4332 4.047 3 744 3 430 3.2497 4.867 4.829 4715 4.420 4.107 3.778 35878 5.191 5 157 5 043 4.757 4.437 4 096 3 898

9 5.464 5 434 5 322 5.060 4 735 4.385 4 18410 5.767 5 739 5 598 5.319 5 000 4.647 4.443ii 5.964 5.935 5 829 5.561 5 240 4.884 468112 6.188 6158 6044 5.780 5463 5.107 490313 6.380 6.350 6 240 5.983 5 670 5.316 5 113

14 6.559 6.529 6421 6171 5.863 5.511 5.30915 6.725 6.694 6.589 6345 6.044 5.695 549516 6876 6.844 6.743 6506 6.213 5867 567117 7014 6983 6.885 6656 6.371 6030 5.838

18 7140 7113 7.019 6792 6523 6187 5.99919 7.261 7.240 7 140 6.913 6 663 6.334 6.153

20 7.376 7344 7.261 7.028 6.785 6.479 630222 7.518 7 507 7 451 7.227 6.982 6.691 6 524

24 7618 7.607 7.518 7.361 7149 6.870 6.714

26 7.697 7.685 7607 7473 7283 7.026 6.88128 7752 7.752 7674 7563 7.395 7.160 7.026

30 7 808 7 797 7 741 7.641 7.484 7.283 7 160

34 7.864 7864 7819 7.741 7.618 7.451 7.35038 7.909 7 909 7.875 7.808 7.7 19 7.585 749642 7.931 7 931 7.909 7.864 7.797 7.685 7.618

46 7.942 7 942 7 920 7.898 7.842 7.752 7 69750 7 954 7 954 7 942 7.920 7.875 7.808 7 764

60 7 968 7.968 7 965 7.954 7.931 7.898 7 864

70 7 976 7.976 7 976 7.968 7.965 7 942 7 920

80 7 982 7.982 7 987 7.976 7.976 7.965 7 954

90 7987 7987 7987 7984 7.983 7976 7965100 7.987 7 987 7.987 7 987 7 987 7.983 7 976

120 7.987 7.987 7.987 7987 7.987 7.987 7.987

322 Water Influx

TABLE 8.8.Dimensionless Influx, W00, for ri, =6 for bottom-water drive

Zi,

1D 005 0.1 0.3 05 0.7 09 JO

6 4780 4.762 4.597 4.285 3.953 3.611 34147 5309 5.289 5.114 4.779 4422 4.053 38378 5.799 5.778 5.595 5.256 4.875 4.478 4 247

9 6 252 6.229 6 041 5.712 5.310 4.888 4 642

10 6.750 6729 6498 6135 5.719 5.278 501911 7137 7.116 6916 6.548 6.110 5.648 537812 7569 7.545 7325 6945 6.491 6009 5.72813 7.967 7 916 7 719 7.329 6.858 6.359 6.06714 8.357 8.334 8.099 7.699 7.214 6697 639515 8.734 8.709 8.467 8 057 7.557 7.024 6.71316 9.093 9.067 8.819 8.398 7 884 7.336 7.017

17 9.442 9.416 9 160 8.730 8 204 7.641 7.31518 9.775 9.749 9 485 9.047 8.510 7.934 7.601

19 10.09 10.06 9.794 9.443 8 802 8.214 7.87420 10.40 10.37 10.10 9.646 9087 8.487 8.14222 1099 1096 10.67 10.21 9631 9.009 8.65324 ii 53 11 50 11.20 10.73 10.13 9.493 9.130

26 1206 12.03 11.72 11.23 1062 9.964 9.59428 12.52 12.49 12.17 11.68 11 06 10.39 10.0130 1295 12.92 12.59 12.09 1146 10.78 10.4035 1396 13.93 13.57 13.06 1241 11.70 11.3240 14.69 14.66 14.33 13.84 13 23 12.53 12.1545 1527 15.24 14.94 14.48 1390 13.23 12.8750 15 74 15.71 15.44 15.01 14.47 13.84 13.4960 16.40 16.38 1615 1581 1534 14.78 14.4770 16.87 16.85 1667 16.38 15.99 15.50 15.24

80 17.20 17.18 17.04 16 80 16.48 16.06 15.83

90 1743 17.42 17.30 17.10 16.85 16.50 1629100 17.58 17.58 17.49 17.34 17.12 16.83 16.66110 17.71 17.69 17.63 17.50 17.34 17.09 16.93120 17.78 17 78 17 73 17.63 17.49 17.29 17.17

130 17.84 1784 17.79 1773 17.62 17.45 17.34

140 17 88 17.88 17.85 17 79 17.71 17.57 17.48

150 17.92 17.91 17.88 17.84 17.77 17.66 17.58

175 17.95 17.95 17.94 17.92 17.87 17.81 1776200 17.97 17.97 17.96 17.95 17.93 17.88 17.86

225 17.97 17.97 17.97 17 96 17.95 17.93 17.91

250 17.98 17.98 17.98 17.97 17.96 17.95 17.95

300 17.98 17.98 17.98 17 98 17.98 17.97 17.97350 17.98 17.98 17.98 17.98 17.98 17.98 17 98

400 17.98 17.98 17.98 17.98 17.98 17.98 17.98

450 17.98 17.98 17.98 17.98 17 98 17.98 17 98

500 17.98 17.98 17.98 1798 17.98 17.98 1798

3. Models 323

TABLE 8.9.Dimensionless Influx, rt, =8 for bottom-water drive

1D 0.05 0.1 03 05 0.7 0.9 109 6.301 6.278 6.088 5.756 5.350 4.924 4.675

10 6.828 6.807 6.574 6.205 5.783 5.336 5.07211 7.250 7.229 7.026 6.650 6.204 5.732 5.45612 7.725 7 700 7.477 7.086 6.621 6.126 5 83613 8.173 8 149 7.919 7.515 7.029 6.514 6.21014 8.619 8.594 8.355 7.937 7.432 6.895 6.57815 9.058 9.032 8783 8.351 7828 7270 6.94016 9.485 9.458 9 202 8 755 8 213 7 634 7.29317 9.907 9.879 9613 9.153 8594 7997 7.64218 10.32 10.29 10.01 9 537 8 961 8.343 7.97919 10.72 10.69 10.41 9920 9.328 8.691 8.31520 11.12 11.08 10.80 10.30 9.687 9.031 864522 11.89 11.86 11.55 11.02 10.38 9.686 928024 12.63 1260 12.27 11.72 11.05 1032 9.896

26 1336 1332 1297 12.40 11.70 10.94 10.49

28 14.06 14.02 13.65 13.06 12.33 11.53 11.07

30 14.73 14.69 14.30 13.68 1293 12.10 11.62

34 16.01 15.97 15.54 14 88 14.07 13.18 12 67

38 17.21 17.17 16.70 15.99 15.13 14.18 13.65

40 1780 1775 1726 16.52 1564 14.66 14.12

45 19 15 19 10 18.56 17.76 16.83 15 77 15.21

50 20.42 2036 19.76 18.91 17.93 1680 16.24

55 21.46 21.39 20.80 19.96 18.97 17.83 17 24

60 22.40 22.34 21.75 20.91 19.93 18.78 18 19

70 23.97 23.92 23.36 2255 21.58 20.44 198680 25.29 25.23 24.71 2394 23.01 21.91 21.32

90 26.39 26.33 25.85 25 12 24.24 2.3.18 22.61

100 27.30 27 25 26.81 26.13 25.29 24.29 23 74

120 28 61 28.57 28.19 27.63 26.90 26.01 25 51

140 2955 2951 29.21 2874 28.12 27.33 2690

160 30.23 30.21 29.96 29 57 29.04 28.37 27 99

180 30.73 30.71 30.51 30 18 29.75 29.18 28 84

200 31.07 31.04 3090 3063 30.26 29.79 2951

240 31.50 31.49 31 39 31 22 30.98 30.65 30.45

280 31.72 31.71 31 66 31 56 31.39 3117 31.03

320 31.85 31 84 31.80 31.74 31.64 31 49 31.39

360 31 90 31 90 31.88 31 85 31.78 31 68 31.61

400 31.94 31.94 31.93 31.90 31.86 31.79 31.75

450 31 96 31.96 31 95 31.94 31.91 31.88 31 85

500 31 97 31.97 31 96 31 96 31.95 31 93 31.90550 31.97 31.97 31.91 3196 31.96 31.95 31.94

600 31.97 31.97 31.97 31.97 31.97 31.96 31.95700 31.97 31.97 31.97 31.97 31.97 31.97 31.97

800 31 97 31.97 31.97 31.97 31.97 31.97 31.97

324 Water Influx

TABLE 8.10.Dimensionless Influx, W,,,, for = 10 for bottom-water drive

tL) 0.05 0.1 03 05 07 - 0.9 1.0

22 1207 1204 1174 1121 1056 9.865 944924 1286 12.83 1252 11.97 11.29 10.55 10 12

26 1365 13.62 1329 12.72 1201 11.24 10.78

28 1442 1439 14.04 13.44 1270 11.90 11.42

30 15 17 15 13 14.77 14 15 13 38 12.55 12.0532 15.91 1587 15.49 14.85 1405 13.18 12.67

34 16.63 1659 16.20 1554 14.71 1381 13.28

36 17.33 17.29 1689 16.21 1535 14.42 13.87

38 18.03 17.99 17.57 16.86 1598 15.02 14.45

40 18.72 1868 1824 17.51 1660 1561 15.02

42 19.38 1933 1889 18.14 1721 16.19 15.58

44 20.03 1999 19.53 18.76 17.80 1675 16.14

46 20.67 20.62 20 15 19.36 18 38 17 30 16.67

48 21 30 21.25 20 76 19 95 18 95 17 84 17.20

50 2192 21.87 2136 2053 1951 18.38 1772

52 22.52 22.47 21 95 2110 20.05 18.89 18 22

54 2311 23.06 2253 2166 20.59 19.40 187256 2370 2.3.64 2309 2220 2111 19.89 19.21

58 24.26 24.21 23.65 22.74 21.63 2039 19.68

60 24 82 24.77 24 19 22 13 20 87 20.15

65 26.18 26 12 25 50 24 53 23.34 22.02 21 28

70 27.47 27 41 26.75 2.5 73 24.50 23.12 22.36

75 2871 2855 27.94 2688 2560 24.17 23.39

80 29.89 29 82 29.08 27.97 26.65 25 16 24.36

85 31.02 3095 30.17 2901 2765 2610 25.31

90 32.10 3203 31.20 3000 2860 2703 26.25

95 33.04 32.96 32.14 3095 2954 2793 27.10

100 33.94 33.85 33.03 31 85 30 44 28 82 27 98

110 35.55 3546 34.65 3349 3208 3047 2962

120 36.97 36.90 36.11 34 98 33 58 31 98 3114130 38.28 38.19 37.44 3633 3496 3338 3255

140 3944 39.37 3864 3756 3623 3467 3385

150 40 49 40.42 39.71 38 67 37 38 35.86 35 04

170 4221 42.15 4151 4054 39.33 37.89 3711

190 43 62 43.55 42.98 42 10 40.97 39.62 38 90210 4477 44.72 44.19 4340 42.36 41.11 40.42230 45 71 45 67 45.20 44 48 43.54 42.38 41.74

250 4648 46 44 46.01 45 38 44.53 43 47 42 87270 47.11 4706 4670 46.13 4536 4440 4384290 47.61 4758 4725 46.75 4607 45.19 44.68

310 48.03 48.00 47 72 47 26 46.66 45.87 45.41

330 4838 4835 4810 4771 47.16 4645 4603350 48 66 48 64 48.42 48.08 47 59 46.95 46.57

4. Pseudosteady-State Models 325

TABLE 8.10.(Cont'd)

z,005 0! 03 05 07 0.9 1.0

400 49.15 49.14 48.99 48.74 48.38 4789 4760450 49.46 49.45 49.35 49.17 4891 48.55 48.31500 49.65 49.64 49 58 49.45 49.26 48 98 48.82600 49.84 4984 49.81 49.74 49.65 49.50 49.41700 49.91 49.91 4990 4987 4982 4974 4969800 49.94 49 94 49.93 49.92 49 90 49.85 49.83900 49.96 4996 49.94 49.94 49 93 4991 49.90

1,000 49.96 4996 49.96 49.96 49.94 4993 49931.200 49.96 49.96 49 96 49.96 49 96 49.96 49.96

4. PSEUDOSTEADY-STATE MODELS

The edge-water and bottom-water, unsteady-state methods discussed in Sect.3 provide correct procedures for calculating water influx in nearly any reser-voir application. However, the calculations tend to be somewhat cumbersome.and therefore there have been various attempts to simplify the calculations.The most popular and seemingly accurate method is one developed by Fctko-vich using an aquifer material balance and an equation that describes the flowrate from the aquifer.15 The cquations for flow rate used by Feikovich arcsimilar to the productivity index equation defined in Chapter 7. The produc-tivity index required pseudosteady-state flow conditions. Thus, this methodneglects the effects of the transient period in the calculations of water influx,which will obviously introduce errors into the calculations. However, themethod has been found to give results similar to those of the van Everdingcnand Hurst model in many applications.

Fetkovich first wrote a material balance equation on the aquifer forconstant water and rock compressibilities as

p =— (i;) + p, (8.15)

where is the average pressure in the aquifer after the removal of bbl ofwater, p is the initial pressure of the aquifer, and W(J is the initial encroachabicwater in place at the initial pressure, Fetkovich next defined a generalized rateequation as

q1B,,=J(, (8.16)

where is the flow rate of water from the aquifer, I is the productivityindex of the aquifer and is a function of the aquifer geometry, PR is the

326 Water Influx

pressure at the reservoir-aquifer boundary, and ma is equal to 1 for Darcy flowduring the pscudostcady-state flow region. Equations (8.15) and (8.16) can becombined to yield the following equation (see References 15 and 16 for thecomplete derivation).

(8.17)

This equation was derived for constant pressures at both the reservoir-aquiferboundary, PR, and the average pressure in the aquifer, Al this point, toapply the equation to a typical reservoir application where both of thesepressures are changing with time, it would normally be required to use theprinciple of superposition. Fetkovich showed that by calculating the waterinflux for a short time period, with a corresponding average aquifer pres-sure, and an average boundary pressure, PR. and then starting the calcu-lation over again for a new period and new pressures, superposition was notneeded. The following equations are used in the calculation for water influxwith this method

f wri)(8.18)

= p,—

(8.19)

PRIX (8.20)

where n represents a particular interval, is the average aquifer pressureat the end of the n — l time interval, is the average reservoir-aquiferboundary pressure during interval n, and is the total, or cumulative, waterinflux and is given by

We = (8.21)

The productivity index, J, used in the calculation procedure is a function of thegeometry of the aquifer. Table 8 11 contains several aquifer productivityindexes as presented by Fctkovich.13 When you use the equations for thecondition of a constant pressure outer aquifer boundary, the average aquiferpressure in Eq. (8.18) will always be equal to the initial outer boundarypressure, which is usually p, Example 8.5 illustrates the use of the Fetkovichmethod.

Example 8.5. Repeat the water influx calculations for the reservoir inEx. 8.3 using the Fetkovich approach.

4 Models 327

— —FRn)

F,' i 3793(1

'- = 176.3(10)6)

TABLE 8.11.Productivity indices for radial and linear aquifers (taken from reference 15)Fetkovich

Type of Outer Aqwfrr

Finite—no flow

Radial Flow' Linear F1OWb

= 0.75J

0003381kwh

Finite—constant pressure—

/— J41fl(r,IrR)]

Units are in normal field units with * in millidarcies

w is width and L is length of linear aquifer

p.L

SOLUTION:

area of aquifer =

area of reservoir =

— ct

5.615

112

[250.000(43560)1 = 83,263 ftor0.Sir J

= 5807 ftor rR=10.5w j

— 58072)19.2(0.209)3793

we'5.615

0.00708k/i/ /180'

—

83,263\= 0.751— 39.08

Jp,A:

II)P'

39081793)(91 3)— 176.3(10)6 /

— 3793

176.3(10)6 bbl

(8.22)

(8.23)

328 Water Influx

Solving Eqs. (8 22) and (8.23), we get Table 8.12.

TABLE 8.12.

Time PR P,,- 1 PR., t&W, We -

0 3793 3793 0 0

-

0 37931 3788 3790.5 2.5 8,600 8,600 3792 82 3774 3781 11.8 40,500 49,100 3791 9

3 3748 3761 30.9 106,100 155,200 3789.7

4 3709 3728.5 61.2 210,000 365,300 3785.1

5 3680 3694.5 90.6 311,200 676,500 3778.4

6 3643 3661.5 116.9 401,600 1,078,100 3769.8

The water influx values calculated by the Fetkovich method agree fairlyclosely with those calculated by the van Everdingen and Hurst method usedin Ex. 8 3 The Fetkovich method consistently gives water influx values smallerthan the values calculated by the van Everdingen and Hurst method (or thisproblem (Fig. 8.14). This result could be because the Fetkovich method doesnot apply to an aquifer that remains in the transient time flow. It is apparentfrom observing the values of which are the avcrage pressure values in theaquifer, that the pressure in the aquifer is not dropping very fast, which wouldindicate that the aquifer is very large and that the water flow from it to thereservoir could be transient in nature.

1,200.000

'C

C

600,000

300.000

0

Fig. 8.14. Results of water influx calculations from Ex 8.5

900.000

Quarter

Problems 329

PROBLEMS

8.1 Assuming the Schilthuis steady-state water influx model, use the pressure drophistory for the Conroe Field given in Fig 8.15. and a water influx constant, k'.of 2170ft3/day/psi, to find the cumulative water encroachment at the end of thesecond and fourth periods by graphical integration for Table 6.1.

—200—

— — — — — — — — —

0 4 8 2 16 20 24 28 32 36 40TIME Ct) IN MONTHS

Fig. 8.15. Calculation of quantity of water that has encroached into Conroe Field.(After Schilthuis 5)

8.2 The pressurc history (or the Peoria Field is given in Fig. 8.16 Between 36 and48 months, production in the Peona Field remained substantially constant at8450 STI3/day, at a daily gas-oil ratio of 1052 SCF/STB, and 2550 STh of waterper day. The initial solution GOR was 720 SCF/STB The cumulative producedGOR at 36 months was 830 SCF1STB, and at 48 months it was 920 SCF/STBThe two-phase formation volume factor at 2500 psia was 9.050 ft3ISTB , and the

Fig. 8.16. Prcssurc decline in the Peoria Field.

I I I I I I I IWI 0

:3— _____cbt___.(g_I')____4.— k— .—, ——lb WI q e -1 -c.-

'2 ; :—

TiME IN MONTHS

330 Water Influx

gas volume factor at the same pressure, 0 00490 ft3ISCF. Calculate the cumu-lative water influx during the first 36 months.

8.3 Dunng a penod of production from a certain reservoir, the average reservoirpressure remained constant at 3200 psia Dunng the stabilized pressure, the oiland water producing rates were 30,000 SThIday and 5000 STh/day, respectively.Calculate the incremental water influx for a later penod when the pressure dropsfrom 3000 to 2800 psia. Assume the following relationship for pressure and timeholds:

= —0.003p. psia/month

Other data are the

p = 3500 psia

Rsoi = 750 SCF1STB

B,= 1.45 bbl/STB at 3200 psia

Hg = 0.002 bbl/STB at 3200 psia

R = 800 SCF/STB at 3200 psia

B.. = 1.04 bbl/STB at 3200 psia

8.4 The pressure decline in a reservoir from the initial pressure down to a certainpressure,p, was approximately linear at —0.500 psi/day Assuming the Schilthuissteady-state water influx model and a wafer influx constant of k', in ft3/day-psia,determine an expression for the water influx as a function of time in bbl.

8.5 An aquifer of 28,850 ac includes a reservoir of 451 ac. The formation has aporosity of 22%, thickness of 60 ft. a compressibility of psi', and apermeability of 100 md. The water has a viscosity of 0.30 cp and a compressi-bility of 3(10) The connate water saturation of the reservoir is 26%, andthe reservoir is approximately centered in this closed aquifer. It is exposed towater influx on all of its penphery.(a) Calculate the effective radii of the aquifer and the reservoir, and their ratio

(b) Calculate the volume of water the aquifer can supply to the reservoir by rockcompaction and water expansion per psi of pressure drop throughout theaquifer.

(c) Calculate the volume of the initial hydrocarbon contents of the reservoir.

(d) Calculate the pressure drop throughout the aquifer required to supply waterequivalent to the initial hydrocarbon contents of the reservoir.

(e) Calculate the theoretical time conversion constant for the aquifer.(f) Calculate the theoretical value of B' for the aquifer.

(g) Calculate the water influx at 100, 200, 400. and 800 days if the reservoirboundary pressure is lowered and maintained at 3450 psia from an initialpressure of 3500 psia.

(ii) If the boundary pressure were changed from 3450 psia to 3460 psia after 100days and maintained there, what would the influx be at 200, 400, and 800days as measured from the first pressure decrement at time zero?

Problems 331

(I) Calculate the cumulative water influx at 500 days from the following bound-ary pressure history:

r(days) 0 100 200 300 400 500p(psia) 3500 3490 3476 3458 3444 3420

U) Repeat part (i) assuming an infinite aquifer, and again assuming r,/rR 5.0

(k) At what time in days do the aquifer limits begin to affect the influx'(I) From the limiting value of W,L, for 8.0, find the maximum water influx

available per psi drop Compare this result with that calculated in part (b).

8.6 Find the cumulative water influx for the fifth and sixth periods in Ex 8.3 andTable 8.3.

8.7 The actual pressure history of a reservoir is simulated by the following data,which assume that the pressure at the onginal oil-water contact is changedinstantaneously by a finite amount, Ap.(a) Use the van Everdingen and Hurst method to calculate the total cumulative

water influx

(b) How much of this water influx occurred in the first two years'

Time in Years A!', psi

0 4005 601.0 941.5 1862.0 1102.5 1203.0 —

Other reservoir properties include the followingS

Reservoir area = 19,600,000 ft2Aquifer area = 686,900,000 ft'k=10.4md 4=25%

h =10 ft

8.8 An oil reservoir is located between two intersecting faults as shown in the arealview in Fig. 8.17. The reservoir shown is bounded by an aquifer estimated bygeologists to have an area of 26,400 ac. Other aquifer data are the following

4=0.21 k=275md

h=30ft c,=7(l0) 6psi'cp

332 Water Influx

Fig. 8.17. Rcscrvoir bctwccn interconnecting faults

The average reservoir pressure, measured at three-month intervals, is as follows:

Time in Days

0

p. psia

298791.3 2962

182 6 2927273 9 2882365.2 2837456 5 2793

Use both the van Everdingen and Hurst and the Fetkovich methods to calculatethe water influx that occurred during each of the three-month intervals. Assumethat the average reservoir pressure history approximatcs the oil reservoir-aquifer boundary pressure history

8.9 For the oil reservoir—aquifer boundary pressure relationship that follows, use thevan Everdingen and Hurst method to calculate the cumulative water influx ateach quarter (see Fig 8

4=020 k=200md

h 40ft

=080 cp

Area of oil reservoir = 1000 ac

Area of aquifer = 15.000 ac

Aquifer

Oil Reservoir1350 acres

Oil watercontact

Fault I

Fault II

References 333

4100 -.

—

4u00

3800-

I

31003678

3600 —.-'--————— i — -, — ——

0 2 4 5 6Quarter

Fig. 8.18. Boundary pressure relationship for Prob 8 9

8.10 Repeat Prob. 8.9 using the Fctkovich method, and compare the results with theresults of Prob. 8.9

REFERENCES

W 0. Moore and I. G. Truby. Jr., 'Pressure Performance of Five Fields Completedin a Common Aquifer." Trans AIME 195, 297.

2F. M Stewart, F' H. Callaway, and R. E. Gladfclter. "Comparison of Methods forAnalyzing a Water Drive Ficid, Torchlight Tenslcep Reservoir, Wyoming," TransAIME (1955), 204, 197

3D. Havicria and A S Odeb, "The Material Balance as an Equation of a StraightLine," Jour. of Petroleum Technology (August 1968), 846—900.

'D liaviena and A S Odeb, "The Material Balance as an Equation of a StraightLine. Part IT— Field Cases," Jour, of Petroleum lechnology (July 1964), 815—822

J. Schilthuis. "Active Oil and Reservoir Energy," Trans AIME (1936), 118, 37.b$ J. Pirson, Elements of Oil Reservoir Engineering, 2nd ed. New York. McGraw-i liii,1958, p. 608

7A. F van Everdingen and W. Hurst, "The Application of the Laplace Transformationto Flow Problems in Reservoirs," Trans. A1ME (1949), 186, 305.

334 Water Influx

A F. van Everdingcn, E 1-1. Timmerman, and J J McMahon. "Application of theMaterial Balance Equation to a I'artial Water-Drive Reservoir," Trans AIME (1953),198, 51

9A. T Chatas, Practical Treatment of Nonsteady-State Flow Problems in ReservoirSystems." Petroleum Lngineering (May 1953), 25, No 5, 8-42; No. 6. June. p B-38;No 8, August, p B-44.

'°M. J Edwardson, et al , "Calculation of Formation Temperature DisturbancesCaused by Mud Circulation," Jour of Petroleum lechnology (April 1962), 416-25;Trans. AIME, 225.

R Analytical Representation of the van Evcrdingcn-l lurst InfluenceFunctions for Reservoir Simulation," SPE Jour (June 1985), 405—06

12M A. Klins. A. 1. Bouchard, and C. L. Cable. "A Polynomial Approach to theEvcrdingcn-Hurst Dimensionless Variables for Water Encroachment," SPE ReservoirEngineering (reb 1988). 320—326.

K. H Coats, "A Mathematical Model for Water Movement about Bottom-Water-Drive Reservoirs," SPL' Jour. (March 1962) 44—52; Trans. AIME. 225

'4D. R Allard and S. M Chen, "Calculation of Water Influx for Bottomwater DriveReservoirs," SPE Reservoir Engineering (May 1988), 369—379

'5M. J Fetkovich. "A Simplified Approach to Water Influx Calculations—FiniteAquifer Systems," Jour of Petroleum Technology (July 1971), 814—828

16L. P Dake, Fundamentals of Reservoir Engineering. New York. Elsevier. 1978.

contact with J T. Smith

Chapter 9

The Displacement of Oiland Gas

1. INTRODUCTION

This chapter includes a discussion of the displacement of oil and gas both byexternal flooding processes and by internal displacement processes. It is notmeant to be an exhaustivc trcatise but only an introduction. Several goodbooks cover the material in this chapter more extensively.* The rescrvoirengineer should be exposed to these concepts because they form the basis forunderstanding secondary and tertiary flooding techniques as well as someprimary recovery mechanisms.

2. RECOVERY EFFICIENCY

The overall recovery efficiency E of any fluid displacement process is given bythe product of the macroscopic. or volumetric displacement, efficiency, E,.and the microscopic displacement efficiency, Ed:

E=E1Ed (9.1)

'References throughout thc text are given at the cnd of each chapter

336 The Displacement of Oil and Gas

The macroscopic displacement efficiency is a measure of how well the displac-ing fluid has contacted the oil-bearing parts of the reservoir. The microscopicdisplacement efficiency is a measure of how well the displacing fluid mobilizesthe residual oil once the fluid has contacted the oil.

'Ihe macroscopic displacement efficiency is made up of two other terms:the arcal, E1, sweep efficiency and the vertical, E1, sweep efficiency.

2.1. MicroscopIc Displacement Efficiency

The microscopic displacement efficiency is affected by the following factors:interfacial and surface tension forces, wcttability, capillary pressure, and rela-tive permeability.

When a drop of one immiscible fluid is immersed in another fluid andcomes to rest on a solid surface, the surface area of the drop will take aminimum value owing to the forces acting at the fluid-fluid and rock-fluidinterfaces. The forces per unit length acting at the fluid-fluid and rock-fluidinterfaces are referred to as interfacial tensions. The interfacial tension be-tween two fluids represents the amount of work required to create a new unitof surface area at the interface. The interfacial tension can also be thought ofas a measure of the immiscibility of tw& fluids. Typical values of oil-brineinterfacial tensions are on the order of 20 to 30 dynes/cm.

The tendency for a solid to prefer one fluid over another is called wet-zahiity. Wettability is a function of the chemical composition of both the fluidsand the rock. Surfaces can be either oil-wet or water-wet, depending on thechemical composition of the fluids. The degree to which a rock is either oil-wetor water-wet is strongly affected by the absorption or desorption of constitu-ents in the oil phase. Large, polar compounds in the oil phase can absorb ontothe solid surface, leaving an oil film that may alter the wettability of thesurface

The concept of wcttability leads to another significant factor in the re-covery of oil. This factor is capillary pressure. To illustrate capillary pressure,consider a capillary tube that contains both oil and brine, the oil having a lowerdensity than the brine. The pressure in the oil phase immediately above theoil-brine interface in the capillary tube will be slightly greater than the pres-sure in the water phase just below the interface. This difference in pressure iscalled the capillary pressure, of the system. The greater pressure willalways occur in the nonwetting phase. An expression relating the contactangle, 0, the radius, of the capillary in feet, the oil-brine interfacial tension,

in dynes/cm, and the capillary pressure in psi is given by:

cos 0(9.2)

This equation suggests that the capillary pressure in a porous medium is afunction of the chemical composition of the rock and fluids, the pore size

2 Recovery 337

distribution of the sand grains in the rock, and the saturation of the fluids inthe pores. Capillary pressures have also been found to be a function of thesaturation history, although this dependence is not reflected in Eq. (9.2). Forthis reason, different values of capillary pressure are obtained during thedrainage process (i.e , displacing the wetting phase by the noriwetting phase),then during the imbibition process (i.e., displacing the nonwetting phase withthe wetting phase) This historesis phenomenon is cxbibited in all rock-fluidsystems.

it has been shown that the pressure rcquircd to force a noriwetting phasethrough a small capillary can be very large. For instance, the pressure droprequired to force an oil drop through a tapering constriction that has a forwardradius of 0.00002 ft. a rcarward radius of 0.00005 ft, a contact angle of 00, andan interfacial tension of 25 dyn/cm is 0.71 psi. if the oil drop were 0.00035 ftlong, a pressure gradient of 2029 psi/ft would be required to move the dropthrough the constriction. Prcssurc gradients of this magnitude are not realiz-able in reservoirs. TypIcal pressure gradients obtained in reservoir systems areof thc order of 1 to 2 psi/ft.

Another factor affecting the microscopic displacement efficiency is thefact that when two or more fluid phases are present and flowing, the saturationof one phase affects the permeability of the other(s). The next section dis-cusses in detail the important concept of relative permeability.

2.2. Relative Permeability

Except for gases at low pressures, the permeability of a rock is a property ofthe rock and not of the fluid that flows through ii, provided that the fluid 100%saturates the pore space of the rock. This permeability at 100% saturation ofa single fluid is called the absolule permeability of the rock. If a core sample0.00215 It2 in cross section and 0.1 ft long flows a 1.0 cpbrine with a formationvolume factor of 1.0 bbl/STB at the rate of 0.30 STB/day under a 30 psipressure differential, it has an absolute pcrmeability of

k0.30(1.0)(0.t)

—413 d— 0.001 — 0.001 127(0.00215)(30)

— m

If the water is replaced by an oil of 3.0 cp viscosity and 1.2 bbl/STB formationvolume factor, under the same pressure differential the flow rate will be 0.0834STI.3/day, and again the absolute permeability is

k — — 0.0834(1 .2)(3.0)(O. 1)0.001 — 0.001 127(0.00215)(30)

— 13 md

If the same core is maintained at 70% water saturation (S., = 70%) and30% oil saturation =30%), and at these and only these saturations and


under the same pressure drop it flows 0.J8 STB/day of the brine and 0.01STB/day of the oil, then the effective permeability to water is

k— — 0.18(1.0)(L0)(O.l)_248

d— — 0.001 127(O.00215)(30)

— m

and the effective permeability to oil is

k_ —

mcio— 0.001 127(O.00215)(30)

The effective permeability, then, is the permeability of a rock to a particularfluid when that fluid has a pore saturation of less than 100%. As noted in theforegoing example, the sum of the effective permeabilities (i.e., 298 md) isalways less than the absolute permeability, 413 md.

When two fluids, such as oil and water, are present, their relative ratesof flow arc determined by their relative viscosities, their relative formationvolume factors, and their relative permeabilities. Relative permeability is theratio of effective permeability to the absolute permeability. For the previousexample, the relative permeabilities to water and to oil are

1. W

k 413

L. °"° k 413

The flowing water-oil ratio at reservoir conditions depends on the vis-cosity ratio and the effective permeability ratio (i.e., on the mobility ratio), or

0.001

- - — - - M0001127k0A4p — k0Ip.0 —

For the previous example

149q0B0 — k0Ip.,, — 50/3.0 —

At 70% water saturation and 30% oil saturation, the water is flowing at 14 9times the oil rate. Relative permeabilitics may be substituted for effectivepcrmeabilities in the previous calculatiop because the relative permeability

2 Recovery Efficiency 339

rano, kr,,/k,, equals the effective perincability ratio, The term relativepermeability ratio is more commonly used For the previous example:

km k11/k 248 0.60

k,0k,,Ik k0 50

Water flows at 14.9 times the oil rate because of a viscosity ratio of 3 and arelative permeability ratio of 5, both of which favor the water flow. Althoughthe relative permeability ratio varies with the water-oil saturation ratio, in thisexample 70/30. or 2 33, the relationship is unfortunately far from one ofsimple proportionality

Figure 9.1 shows a typical plot of oil and water relative permeabilitycurves for a particular rock as a function of water saturation. Starting at 100%water saturation, the curves show that a decrease in water saturation to 85%(a 15% increase in oil saturation) sharply reduces the relative permeability towater from 100% down to 60%, and at 15% oil saturation the relative pcrtne-ability to oil is essentially zero. This value of oil saturation, 15% in this case,is called the critical saturation, the saturation at which oil first begins to flowas the oil saturation increases. It is also called the residual saturation, the valuebelow which the oil saturation cannot be reduced in an oil-water system. Thisexplains why oil recovery by water drive is not 100% efficient. If the initial

Fig. 9.1. Water-oil relative permeability curves

0 0.2 0.6WATER SATURATION, FRACTION OF PORE SPACE


connate water saturation is 20% for this particular rock, then the recoveryfrom the portion of the reservoir invaded by high-pressure water influx is

initial-final 0.80 — 0.15Recovery=——j.1=--0.80

—=81%

Experiments show that essentially the same relative permeability curvesare obtained for a gas-water system as for the oil-water system, which alsomeans that the critical, or residual, gas saturation will be the same Further-more, it has been found that if both oil and free gas arc present, the residualhydrocarbon saturation (oil and gas) will be about the same, in this case 15%.Suppose, then, that the rock is invaded by water at a pressure below saturationpressure so that free gas is present. If, for example, the residual free gas satu-ration behind the flood front is 10%, then the oil saturation is 5%, and neglect-ing small changes in the formation volume factors of the oil, the recovery isincreased to:

0.80—0.05Rccovery=—

Returning to Fig. 9.1, as the water saturation decreases further, therelative permeability to water continues to decrease and the relative perme-ability to oil increases At 20% water saturation, the (connate) water is immo-bile, and the relative permeability to oil is quite high. This explains why somerocks may contain as much as 50% connate water and yet produce water-freeoil. Most reservoir rocks are preferentially water wet—that is, the water phaseand not the oil phase is next to the walls of the pore spaces. Because of this,at 20% water saturation the wateroccupies the least favorable portions of thepore spaces—that is, as thin layers about the sand grains, as thin layers on thewalls of the pore cavities, and in the smaller crevices and capillaries. The oil,which occupies 80% of the pore space, is in the most favorable portions of thepore spaces, which is indicated by a relative permeability of 93%. The curvesfurther indicate that about 10% of the pore spaces contribute nothing to thepcrmeability, for at 10% water saturation, the relative permeability to oil isessentially 100%. Conversely, on the other end of the curves, 15% of thepore spaces contribute 40% of the permeability, for an increase in oil satura-tion from zero to 15% reduces the relative permeability to water from 100%to 60%.

in describing two-phase flow mathematically, it is always the relativepermeability ratio that enters the equations. Figure 9.2 is a plot of the relativepermeability ratio, k0Ik,,,, versus water saturation for the same data of Fig 9.1.Because of the wide range of values, the relative permeability ratio isusually plotted on the log scale of semilog paper. Like many relative perme-ability ratio curves, the central or main portion of the curve is quite linear. As

2. Recovery Efficiency 341

a straight line or semilog paper, the relative permeability ratio may be ex-pressed as a function of the water saturation by:

Ic0 bSP1 (9.3)

The constants a and b may be determined from the graph shown in Fig. 9 2,or they may be determined from simultaneous equations. At S,. =0 30,k0/k,,, = 25, and at 5,,, 0.70, k,/k,,, 0.14. Then

25 ae -O30b and 0.14=aeO7Ob

Solving simultaneously, the intercept a = 1220, and the slope b = 13.0. Equa-tion (9.3) indicates that the relative permeability ratio for a rock is a functionof only the relative saturations of the fluids present. Although it is true thatthe viscosities, the intcrfacial tensions, and other factors have some effect on

Fig. 9.2. Scmilog plot of relative permeability ratio versus saturation

WATER SATURATION, FRACTiON OF PORE SPACE


the relative permeability ratio, for a given rock it is mainly a function of thcfluid saturations.

In many rocks there is a transition zone between the water and the oilzones. In the true water zone, the water saturation is essentially 100%, al-though in some reservoirs a small oil saturation may be found a considerabledistance vertically below the oil-water contact In the oil zone, there is usuallypresent connate water, which is essentially immobile. For the present ex-ample, the connate water saturation is 20% and the oil saturation is 80%.Only water will be produced from a well completed in the true water zone,and only oil will be produced from the true oil zone. In the transition zone(Figure 9 3), both oil and water will be produced, and the fraction that is waterwill depend on the oil and water saturations at the point of completion, ifthe well in Fig. 9 3 is completed in a uniform sand at a point where = 60%and 40%, the fraction of water or water cut may be calculated usingEq (7.19)'

qWBW

p. ln(r,/r

Since water cut, is defined as

(94)

Fig. 9.3. Ske'ch showing the variation in oil and water saturations in the transitionzone

q0B0

2 Recovery Efficiency

Combining these equations and cancelling common terms,

— k,../p.,.,

— k,,Ip.,, + k0/p.0

(9.5)

k,,,.p.0

Since is a function of saturation, Eq. (9.3) may be substituted inEq. (9.5) to give

/ 1(9.6)

I +\ Po/

Eq. (9.5) or (9.6) can be used with the data of Fig. 9.1 or 9.2, respec-tively, and with viscosity data to calculate the water cut. From Fig 9 1, at5,,, = 0.40, k,,, = 0.045 and km = 0.36. If p.,.. = 1.0 cp and p.0 = 3.0 cp,

0.36(1.0) =0.27

0.045(3.0)

If the calculations for the water-cut are repeated at several water satura-tions, and then the calculated values plotted versus water saturation, Fig. 9.4will be the result. This plot is referred to as the fractionalfiow curve The curveshows that the fractional flow of water ranges from 0 (for S,,, the connatewater saturation) to I (for S,, � one minus the residual oil saturation).

2.3. Macroscopic Displacement Efficiency

The following factors affect the macroscopic displacement efficiency: hetero-geneities and anisotropy, mobility of the displacing fluids compared with themobility of the displaced fluids, the physical arrangement of injection andproduction wells, and the type of rock matrix in which the oil or gas exists.

Heterogeneities and anisotropy of a hydrocarbon-bearing formationhave a significant effect on the macroscopic displacement efficiency Themovement of fluids through the reservoir will not be uniform if there arc largevariations in such properties as porosity, permeability, and clay cement. Lime-stone formations generally have wide fluctuations in porosity and perme-ability. Also, many formations have a system of microfractures or large macro-fractures. Any time a fracture occurs in a reservoir, fluids will try to travelthrough the fracture because of the high permeability of the fracture, whichmay lead to substantial bypassing of hydrocarbon.


S

0.8-————

S

06-—--—- -—

-

0.2-

00' , - - -,----0.0 02 0.4 06 08 10

SW

Fig. 9.4. Fractional flow curve for the relative permeability data of Fig 9.1

Many producing zones arc variable in permeability, both vertically andhorizontally, leading to reduced vertical, E1, and areal, sweep efficiencies.Zones or strata of higher or lower permeability often exhibit lateral continuitythroughout a reservoir or a portion thereof. Where such permeability strati-fication exists, the displacing water sweeps faster through the more permeablezones so that much of the oil in the less permeable zones must be producedover a long period of time at high water-oil ratios. The situation is the samewhether the water comes from natural influx or from injection systems, com-monly called pressure mainlenance when the reservoir pressure is fairly high,and secondary recovery when the reservoir pressure is low or depleted

The areal sweep efficiency is also affected by the type of flow geometryin a reservoir system. As an example, linear displacement Occurs in uniformbeds of constant cross section where the entire input and outflow ends areopen to flow. Under these conditions, the flood front advances as a plane(neglecting gravitational forces), and when it breaks through at the producingend, the sweep efficiency is 100%—that is, 100% of the bed volume has beencontacted by the displacing fluid. If the displacing and displaced fluids areinjected into and produced from wells located at the input and outflow endsof a uniform linear bed. such as the direct line-drive pattern arrangementshown in Fig. 9 5(a), the flood front is not a plane, and at breakthrough thesweep efficiency is far from 100%, as shown in Figure 9.5(b).

2. Recovery Efficiency 345

Mobility is a reJativc measure of how easily a fluid moves thorugh porousmedia. The apparent mobility has been defined as the ratio of effective perme-ability to fluid viscosity. Since the effective permeability is a function of fluidsaturations, several apparent mobilities can be defined. When a fluid is beinginjected into a porous medium containing both the injected fluid and a secondfluid, the apparent mobility of the displacing phase is usually measured at theaverage displacing phasc saturation when the displacing phase just begins tobreak through at the production site. The apparent mobility of the non-displacing phase is measured at the displacing phase saturation that occurs justbefore the beginning of the injection of the displacing phase Areal swcepefficiencies are a strong function of the mobility ratio. A phenomenon calledviscous fingering can take place if the mobility of the displacing phasc is muchgTeater than the mobility of the displaced phase. Viscous fmgering simplyrefers to the penetration of the much more mobile displacing phase into thephase that is being displaced.

o 0 0 0 0 0

A- --—--— A -- -- A — —--A—---— --A—--—---Ao pioducing wellA injection well

(a)

Fig. 9.5(a). Direct-line-drive flooding network.

Jig. 9.5(b). The photographic history of a direct-line-drive fluid-injection system,under steady-state conditions, as obtained with a blotting-paper clectrolytic model.(After Wyckoff, Botset, and Muskat.6)


Viscous fingering simply refers to the penetration of the much more mobiledisplacing phase into the phase that is being displaced.

Figure 9.6(b) shows thc effect of mobility ratio on areal sweep efficiencyat initial breakthrough for a five-spot network (shown in Fig. 9.6(a) obtainedusing the X-ray shadowgraph. The pattern at breakthrough for a mobility ratioof I obtained with an electrolytic model is included for comparison.

The arrangement of injection and production wells depends primarily onthe geology of the formation and the size (areal extent) of the reservoir. Fora given reservoir, an operator has the option of using the existing well arrange-

o

A well

Fig. 9.6(a). Five-spot flooding network

— X-RAY- —— ELECTROLYTIC MODEL

MOBIUTY RATIO=O 45

MOBILITYRATIO—i

MOBILITYRATIO-=1 67

Fig. 9.6(b). X-ray shadowgraph studies showing the effect of mobility ratio on arcalsweep efficiency at breakthrough (After Slobod and

LEGEND

3. Immiscible Displacement Processes 347

ment or drilling new wells in other locations. If the operator opts to use theexisting well arrangement, there may be a need to consider converting produc-tion wells to injection wells, or vice versa. An operator should also recognizethat, when a production well is converted to an injection well, the productioncapacity of the reservoir will have been reduced. This decision can often leadto major cost itcms in an overall project and should be given a great deal ofconsideration Knowledge of any directional permeability effects and otherhetcrogeneitics can aid in the consideration of well arrangements. The pres-ence of faults, fractures, and high-permeability streaks can dictate the shuttingin of a well near one of these heterogeneities. Directional permeability trendscan lead to a poor swecp efficiency in a developed pattern and can suggest thatthe pattern be altered in one direction or that a different pattern he used.

Sandstone formations arc characterized by a more uniform pore geome-try than limestone formations. Limestone formations have large holes (vugs)and can have fractures that are often connected. Limestone for-mations are associated with connate water that can have high levels of divalentions such as Ca2' and Vugular porosity and high divalent ion content intheir connate waters hinders the application of injection processes in lime-stone reservoirs. A sandstone formation can be composed of such small sandgrains and be so tightly packed that fluids will not readily flow through theformation.

3. IMMISCIBLE DISPLACEMENT PROCESSES

3.1. The DIsplacement Mechanism

Oil is displaced from a rock by water somewhat as fluid is displaced from acylinder by a leaky piston. Buckley and Leverett developed a theory ofdisplacement based on the relative permeability concept Their theory ispresented here.

Consider a linear bed containing oil and water (Fig. 9.7). Let the totalthroughput, q1' = + q0B0 in reservoir barrels be the same at all cross

Fig. 9.7.


sections. For the present, we will neglect gravitational and capillary forces thatmay be acting. Let be the water saturation in any element at time t (days).Then if oil is being displaced from the clement, at time (t + di) the watersaturation will be + If 4) is the total porosity fraction, is the crosssection in square feet, and dr is the thickness of the clement in feet, then therate of increase of water in the element at time t in barrels per day is

97di — 5.615 )

The subscript x on the derivative indicates that this derivative is different foreach element. If is the fraction of water in the total flow of q,' barrels perday, then is the rate of water entering the left-hand face of the elementdx. The oil saturation will be slightly higher at the right-hand face, so thefraction of water flowing there will be slightly less, or Then the rateof water leaving the element is (j, — The net rate of gain of water inthe clement at any time then is

= df,,,)q,' = (9.8)

Equating (9.7) and (9 8),

—9 9

'ÔX/, ( . )

Now for a given rock, the fraction of water is a function only of the watersaturation as indicated by Eq. (9 5), assuming constant oil and waterviscosities. The water saturation, however, is a function of both time andposition. x, which may be expressed as = and S., = G(I, x) Then

=di dx (9.10)

Now. there is interest in determining the rate of advance of a constant satura-tion plane, or front, i.e., where 5,., is constant. Then, from Eq. (9.10)

(öx\9( .11)

Substituting Eq (9.9) in Eq. (9.11),

(ax\ 5 615q9 12(


But

Eq. (9.12) then becomes

—

(aSp/ax), —(9.13)

_5.615q,'(9.14)

Because the porosity, area, and throughput are constant and because for anyvalue of S,,,, the derivative is a constant, then the rate dxldr is constant.This means that the distance a plane of constant saturation, S,,,, advances isdirectly proportional to time and to the value of the derivative at thatsaturation, or

4,A.(9 15)

We now apply Eq. (9.15) to a reservoir under active water drive whercthe walls are located in unifotrn rows along the stnkc on 40 ac spacing, asshown in Fig. 9.8. This gives rise to approximate linear flow, and if the dailyproduction of each of the three wells located along the dip is 200 STh of oilper day, then for an active water drive and an oil volume factor of 1 50bbl/STB, the total reservoir throughput, will be 900 bbl/day.

The cross-sectional area is the product of the width, 1320 ft, and the trueformation thickness, 20 ft, so that for a porosity of 25%, Eq. (9 15) becomes

5615X900x:X 0.25x

Fig. 9.8.

Srs/DAy


If we let x = 0 at the bottom of the transition zone, as indicated in Fig. 9 8,then the distances the various constant water saturation planes will travel in.say, 60, 120, and 240 days are given by:

= 46

= 92

x240= 184 (9.16)

The value of the derivative may be obtained for any value ofwater saturation, by plotting from Eq. (9.5) versus and graphicallytaking the slopes at values of This is shown in Fig. 9.9 at 40% watersaturation using the relative permeability ratio data of Tabie 9.1 and awater-oil viscosity ratio of 0.50. For example, at Si,, = 0.40, where k0/k,,, =5 50(Table 9 1),

50_0 267

The slope taken graphically at S,, = 0.40 and fW = 0.267 is 2.25, as shown inFig. 9.9.

The derivative (afW/aSW) may also be obtained mathematically using

11g. 9.9.

U)

U.'

Ui

3 Immiscible Displacement Processes 351

TABLE 9.1.frontal advance calculations

(1) (2)

=050

Eq. (95)

81w

85.Eq. (917)

(60 days)Eq (916)

(120 days)Eq. (9.16)

I84

(240 days)Eq (9.26)

0.20 inf. 0.000 0.00 0 0 0030 17.0 0.105 1.08 50 100 2000.40 5.50 0.267 2.25 1.04 208 4160.50 1.70 0541 2.86 131 262 524060 055 0.784 1.95 89 179 3580.70 0.17 0.922 0.83 38 76 153

0.80 0.0055 0973 0.30 14 28 550.90 0000 1.000 0.00 0 0 0

Eq. (9.3) to represent the relationship between the relative permeability ratioand thc water saturation. Differentiating Eq. (9.16), we obtain

af...(9 17)

35,. [1 + (p.,,./p,0)ae [1 +

For the k0/k,. data of Table 9.1, a = 540 and 1, 11.5. Then, at 0.40, forexample, by Eq. (9.17).

aJ. 0.SOx 11.5x5.50_225aS,,, — + 0.50 x 5.5012 —

Hgure 9.9 shows the fractional water cut, f,,,, and also the derivative (af,J8S.,)plotted against water saturation from the data of Table 9.1. Equation (9.17)was used to determine the values of the derivative. Since Eq. (9 3) does notbold for the very high and for the quite low water saturation ranges, someerror is introduced below 30% and above 80% water saturation. Since theseare in the regions of the lower values of the derivatives, the overall effect onthe calculation is small

The lowermost curve of Fig. 9.10 represents the initial distribution ofwater and oil in the linear sand body of Fig. 9.8. Above the transition zone,the connate water saturation is constant at 20%. Equation (9.16) may be usedwith the values of the derivatives, ca(culated in Table 9.1. and plotted in Fig.9.9. to construct the frontal advance curves shown in Fig. 9.10 at 60, 1.20, and240 days. For example, at 50% water saturation, the value of the derivative is


Fig. 9.10. Fluid distributions at initial conditions and at 60, 120, and 240 days.

2.86; so by Eq (9.16), at 60 days the 50% water saturation plane, or front, willadvance a distance of:

=46x2.86131 feet

This distance is plotted as shown in Fig. 9.10 along with the other distancesthat have been calculated in Table 9.1 for the other time values and otherwater saturations. These curves are characteristically double-valued or triple-valued. For example, Fig 9.10 indicates that the water saturation after 240days at 400 ft is 20, 36, and 60%. The saturation can be only one value at anyplace and time. The difficulty is resolved by dropping perpendiculars so thatthe areas to the right (A) equal the areas to the left (B), as shown in Fig. 9.10.

Figure 9.11 represents the initial water and oil distributions in the reser-voir unit and also the distributions after 240 days, provided the flood front hasnot reached the lowermost well. The area to the right of the flood front inFig. 9.11 is commonly called the oil bank and the area to the left is sometimescalled the drag zone. The area above the 240-day curve and below the 90%water saturation curve represents oil that may yet be recovered, or dragged outof the high-water saturation portion of the reservoir by flowing large volumesof water through it The area above the 90% water saturation represents un-recoverable oil since the critical oil saturation is 10%.

This presentation of the displacement mechanism has assumed that cap-illary and gravitational forces are negligible. These two forces account for the


100——- —'---—PERCENT UNRECOVERABLE OIL

- RECO'/ERABLE OIL I

TNIINITE

HRbUGHP(JT

Z— GRAVITYOIL RECOVERED

240 -

FRONT

I

0 100 200 300 400 500 600DISTANCE ALONG BED, FEET

Fig. 9.11.

initial distribution of oil and water in the reservoir unit, and they also act tomodify the sharp flood front in the manner indicated in Fig. 9.11. If produc-tion ceascs after 240 days, the oil-water distribution will approach one similarto the initial distribution, as shown by the dashed curve in Fig. 9.11.

Figure 9.11 also indicates that a well in this reservoir unit will producewater-free oil until the flood front approaches the well. Thereafter, in a rela-tively short period, the water cut will rise sharply and be followed by arelatively long period of production at high, and increasingly higher, watcrcuts. For example, just behind the flood front at 240 days, the water saturationrises from 20% to about 60%—that is, the water cut rises from zero to 78.4%(see Table 9.1). When a producing formation consists of two or more ratherdefinite strata, or stringers, of different permcabilities, the rates of advance inthe separate strata will be proportional to their permeabilities, and the overalleffect will be a combination of several separate displacements, such as de-scribed for a single homogeneous stratum.

3.2. The Displacement of Oil by Gas, with and withoutGravitational Segregation

The method discussed in the previous section also applies to the displacementof oil by gas drive. The treatment of oil displacement gas in this sectionconsiders only gravity drainage along dip. Richardson and Blackwell showedthat in some cases there can be a significant vertical component of drainage.9


Due to the high oil-gas viscosity ratios and the high gas-oil relativepermeability ratios at low gas saturations, the displacement efficiency by gasis generally much lower than that by water, unless the gas displacement isaccompanied by substantial gravitational segregation. This is basically thesame reason for the low recoveries from reservoirs produced under the dis-solved gas-drive mechanism. The effect of gravitational segregation in water-drive oil reservoirs is usually of much less concern because of the higher dis-placement efficiencies and the lower oil-water density differences, whereas theconverse is generally true for gas-oil systems. Welge showed that capillaryforces may generally be neglected in both, and he introduced a gravitationalterm in Eq. (9.5), as will be shown in the following equations.'° As with waterdisplacement, a linear systcm is assumed, and a constant gas pressure through-out the system is also assumed so that a constant throughput rate may be used.These assumptions also allow us to eliminate changes caused by gas density,oil density, oil volume factor, and the like. Equation (7.1) may be applied toboth the oil and gas flow, assuming the connate water is essentially immobile,so that the fraction of the flowing reservoir fluid volume, which is gas, is

1— II- I (9.18)U1 J

The total velocity is u,, which is the total throughput rate divided by thecross-sectional area The reservoir gas density, is in lb,,,/ft3. Whencapillary forces are neglected, as they arc in this application, the pressuregradients in the oil and gas phases are equal. Equation (7.1) may be solved forthe pressure gradient by applying it to the oil phase, or

— 27k+ 0.00694 Po cos a (9.19)

Substituting the pressure gradient of Eq. (9.19) in Eq. (9.18),

I =—

0 00694 (Po — pg) cos (9.20)

Expanding and multiplying through by

uo(921)

u1

But uju, is the fraction of oil flowing, which equals 1 minus the gas flowing,(1 —f). Then, finally.


— — pg) cosal

(9.22)

The relative permeability ratio (km/k,R) may be used for the effective perme-ability ratio in the denominator of Eq. (9.22); however, the permeability tooil, k0, in the numerator is the effective permeability and cannot be replacedby the relative permeability It may, however, be replaced with (k,0k), wherek is the absolute permeability. The total is the total throughputrate, q,', divided by the cross-sectional area, Inserting these equivalents,the fractional gas flow equation with gravitational segregation becomes

1 —[7 Pg) COS ctlfL P.O (923)

1 +kg

If the gravitational forces are small, Eq (9.23) reduces to the same type offractional flow equation as Eq. (9.5), or

fg = (9.24)I +

Although Eq. (9.24) is not rate sensitive (i.e., it does not depend on thethroughput rate), Eq. (9.23) includes the throughput velocity and istherefore rate sensitive. Since the total throughput rate, q,', is in the denoin-inator of the gravitational term of Eq. (9.23), rapid displacement (i.e., large

reduces the size of the gravitational term, and so causes an increasein the fraction of gas flowing. f1. A large value of fg implies low displacementefficiency. If the gravitational term is sufficiently large, fg becomes zero, oreven negative, which indicates counter-current flow of gas updip and oil down-dip, resulting in maximum displacement efficiency In the case of a gas capthat overlics most of an oil zone, the drainage is vertical, and cos a = 1.00; inaddition, the cross-sectional area is large. If the vertical effective permeabilityk0 is not reduced to a very low level by low permeability strata, gravitationaldrainage will substantially improve recovery

The use of Eq. (9.23) is illustrated using the data given by Welge for theMile Six Pool, Peru, where advantage was taken of good gravitational segrega-tion characteristics to improve recovery.'0 Pressure maintenance by gas injec-tion has been practiced since 1933 by returning produced gas and other gas tothe gas cap so that reservoir pressure has been maintained within 200 psi of its


initial value. Figure 9.12 shows the average relative permeability character-istics of the Mile Six Pool reservoir rock. As is common in gas-oil systems, thesaturations are expressed in percentages of the hydrocarbon porosity, and theconnate water, being immobile, is considered as part of the rock. The otherpertinent reservoir rock and fluid data are given in Table 9.2. Substitutingthese data in Eq. (9.25),

[7.821(106)(300)(L237(10)6)(48.7 — 5.)cos 72.5°1 ( k,0

1.32

k8(1.32)

— (9.25)I

The values of have been calculated in Table 9.2 for three conditions:(a) assuming negligible gravitational segregation by using (9.24); (b) usingthe gravitational term equal to 2.50 k,0 for the Mile Six Pool, Eq. (9.25); and(c) assuming the gravitational term equals 1.25 k,0, or half the value at MileSix Pool. The values of fg for these three conditions are shown plotted in Fig.9.13. The negative values of f, for the conditions that existed in the Mile SixPool indicate counter-current gas flow (i.e., gas updip and oil downdip) in therange of gas saturations between an assumed critical gas saturation of 5% andabout 17%.

The distance of advance of any gas saturation plane may be calculated

Fig. 9.12. Relative permeabilities for the Mile Six Pool, Peru.

GAS SATURATION, FRACTIONOF HYDROCARBON PORE SPACE

TA

BLE

9.2

.M

ile S

ix P

ool R

eser

voir

data

and

cal

cula

tions

Avg

. Abs

olut

e pe

rmea

bilit

y =

300

md

Res

ervo

ir o

il sp

gr

=0.

78(w

ater

=A

vg h

ydro

carb

on p

oros

ity =

0 16

25R

eser

voir

gas

sp

gr. =

0.08

(wat

er =

1)A

vg. c

onna

te w

ater

=0.

35R

eser

voir

tem

pera

ture

=11

4°F

Avg

dip

ang

le17

° 30

(a

=90

°—

17°3

0)A

vera

ge r

eser

voir

pre

ssur

e =

850

psia

Ave

cro

ss-s

ectio

nal a

rea

=1,

237,

000

sq f

tA

vera

ge th

roug

hput

11,6

00 r

eser

voir

bbl

per

day

Res

ervo

ir o

il vi

scos

ity =

1.32

cpO

il vo

lum

e fa

ctor

=1.

25bb

l/ST

BR

eser

voir

gas

vis

cosi

ty =

0 01

34 c

pSo

lutio

n ga

s at

850

psi

a =

400

SCF/

STB

Gas

dev

iatio

n fa

ctor

=0.

74

Sg00

50.

100.

1502

00.

2503

00.

3504

00.

4505

005

5-

0.60

kolk

glfl

f.38

8.80

3 10

1.40

0.72

0.36

4D

uO01

180.

072

M24

000

Gra

vity

term

=0

00.

720

0 91

809

690.

986

0.99

30.

996

0.99

80.

990

1 00

1 00

1 00

ôfR

/ÔS$

7.40

1.20

060

0.30

237

3819

10 Gra

vity

term

=2.

50 x

k,0

km0.

7705

90.

440.

340.

260.

1901

40.

1000

650.

040

0018

0.00

2 50

Xkm

1. 9

21

481.

100

850.

650.

480.

350.

250

160

0.10

0.04

50.

001-

2.5

-0.9

2—

0.48

—0

100.

150.

350.

520.

650.

750

840.

900

955

1 00

0—

0.29

—0

092

0 14

50

345

0.51

60.

647

0.74

90.

840

0 90

00

955

1.00

Of,I

ÔS

,3.

304

404.

303.

603.

002.

501.

951.

601.

200.

8032

8f,/a

S,

106

141

138

115

9680

6251

3826

-G

ravi

tyte

rm.=

i.25x

k,,,

1.25

k,0

0 96

0 74

0.55

0.42

50.

325

0 24

00.

175

0 12

50.

080

005

00.

023

0 00

1-1

25 k

,,,0

040.

260.

450

575

0 67

50.

760

0.82

50

875

0 92

00

950

0.97

71

00fg

0.19

004

130.

557

0666

0.75

50.

822

0873

0.92

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— 5.615(11,600)(100) (3j.'\X— 0.1625(1,237,000)

xas,

COUNTER CURRENT GASFLOW—NEGATIVE Fe


1.0

0.8

w

0.4

0.20

0.1 0.2GAS SATURATION. S1, FRACTION

Fig. 9.13. Fraction of gas in reservoir stream for the Mile Six Pool, Peru

for thc Mile Six Pool, using Eq. (9 15), replacing water as the displacing fluidby gas, or

5.615q,'t (aJ\X

4i1,

In 100 days, then,

(9.26)

The values of the derivatives given in Table 9.2 have been de-termined graphically from Fig. 9.13. Figure 9.14 shows the plots of Eq. (9.26)to obtain the gas-oil distributions and the positions of the gas front after 100days. The shape of the curves will not be altered for any other time. Thedistribution and fronts at 1000 days, for example, may be obtained by simplychanging the scale on the distance axis by a factor of 10.

Welge showed that the position of the front may be obtained by drawinga secant from the origin as shown in Fig. 9.13.10 For examplc, the secant istangent to the lower curve at 40% gas saturation. Then in Fig. 9.14, the frontmay be found by dropping a perpendicular from the 40% gas saturation asindicated. This will balance the areas of the S-shaped curve, which was doneby trial and error in Fig. 9.10 for water displacement. In the case of waterdisplacement, the secant should be drawn, not from the origin, but from the


1.00

0.80 UNRECOVERABLE OIL AT 5q060

TERM 2.504 IT TANGENT AT S '040S

(SEE FIG. 10)

TERM' 1.25 K10U)

TANGENT AT S5 '0.18

GRAVITY

J SECANT TANGENT AT

80 120 160 200 240DISTANCE IN IOODAYS,FEET

Fig. 9.14. 1-luid distributions in the Mile Six Pool after 100 days injection

connate watcr saturation, as indicated by the dashcd line in Fig. 9.9. This istangent at a watcr saturation of 60%. Referring to Fig. 9.10, the 240-day frontis seen to occur at 60% water saturation. Owing to the presence of an initialtransition zone, the fronts at 60 and 120 days occur at slightly lower values ofwater saturation.

The much greater displacement efficiency with gravity segregation thanwithout is apparent from Fig. 9.14. Since thc permeability to oil is essentiallyzero at 60% gas saturation, the maximum recovery by gas displacement andgravity drainage is 60% of the initial oil in place. Actually some small perme-ability to oil exists at even very low oil saturations, which explains why somefields may continue to produce at low rates for quite long periods after thepressure has been depicted. The displacement efficiency may be calculatedfrom Fig. 9.14 by the measurement of areas. For example, thc displacementefficiency at Mile Six Pool with full gravity segregation is in excess of

Area B 32.5Recovery= or 87.4%AreaA +AreaB 4.7+32.5

If the gravity segregation had been half as effective, the recovery would havebeen about 60%; without gravity segregation, the recovery would have beenonly 24%. These recoveries arc expressed as percentages of the recoverableoil. In terms of the initial oil in place, the recoveries are only 60% as large, or52 4, 36.0, and 14.4%, respectively. Welge, Shreve and Welch, Kern. andothers have extended the concepts presented here to the prediction of gas-oilratios, production rates, and cumulative recoveries, including the treatment ofproduction from wells behind the displacement front.'° 12 has used the


magnitude of the gravity term — p1) cos aJ as a criterion for deter-mining those reservoirs in which gravity segregation is likely to be of consid-erable importance.'3 The data of Table 9.3 indicate that this gravity term musthave a value above about 600 in the units used to he effective. An inspectionof Eq. (9.23), however, shows that the throughput velocity is also ofprimary importance -

One interesting application of gravity segregation is to the recovery ofupdip or "attic" oil in active water-drive reservoirs possessing good gravitysegregation characteristics. When the structurally highest well(s) has gone towater production, highpressure gas is injected for a period. This gas migratesupdip and displaces the oil downdip, where it may be produced from thesame well in which the gas was injected The injected gas is of course, unre-coverable.

it appears from the previous discussions and examples that water isgenerally more efficient than gas in displacing oil from reservoir rocks, mainlybecause (a) the water viscosity is of the order of 50 times the gas viscosity and(b) the water occupies the less conductive portions of the pore spaces, whereasthe gas occupies the more conductive portions. Thus, in water displacementthe oil is left to the central and more conductive portions of the pore channels,whereas in gas displacement the gas invades and occupies the more conductiveportions first, leaving the oil and water to the less conductive portions. Whathas been said of water displacement is true for preferentially water wet (hydro-philic) rock, which is the case of most reservoir rocks. When the rock ispreferentially oil wet (hydrophobic), the dispiacing water will invade the morcconductive portions first, just as gas does, resulting in lower displacementefficiencies In this case, the efficiency by water still exceeds that by gasbecause of the viscosity advantage the water has over the gas.

3.3 Oil Recovery by Internal Gas Drive

Oil is produced from volumetric, undersaturated reservoirs by expansion ofthe reservoir fluids. Down to the bubble-point pressure, the production iscaused by liquid (oil and connate water) expansion and rock compressibility(see Chapter 5, Sect. 6). Below the bubble point, the expansion of the connatewater and the rock compressibility are negligible, and as the oil phase con-tracts owing to release of gas from solution, production is a result of expansionof the gas phase. When the gas saturation reaches the critical value, free gasbegins to flow. At fairly low gas saturations, the gas mobility, becomeslarge and the oil mobility, small, resulting in high gas-oil ratios and inlow oil recoveries, usually in the range of 5 to 25%

Because the gas orginates internally within the oil, the method describedin the previous section for the displacement of oil by external gas drive is notapplicable. In addition, constant pressure was assumed in the external dis-placement so that the gas and oil viscosities and voiume factors remainedconstant during the displacement With internal gas drive, the pressure drops

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as production proceeds, and the gas and oil viscosities and volume factors con-tinually change, further complicating the mechanism.

Because of the complexity of the internal gas drive mechanism, a num-ber of simplifying assumptions must be made to keep the mathematical formsreasonably simple. The following assumptions, generally made, do reduce theaccuracy of the methods, but in most cases not appreciably:

1. Uniformity of the reservoir at all times regarding porosity, fluid satu-rations, and relative permeabilities. Studies have shown that the gas and oilsaturations about wells are surprisingly uniform at all stages of depletion.

2. Uniform pressure throughout the reservoir in both the gas and oil zones.This means the gas and oil volume factors, the gas and oil viscosities, andthe solution gas will be the same throughout the reservoir.

3. Negligible gravity segregation forces.4. Equilibrium at all times between the gas and the oil phases.5. A gas liberation mechanism which is the same as that used to determine the

fluid properties.6. No water encroachment and negligible water production.

Several methods appear in the literature for predicting the performanceof internal gas drive reservoirs from their rock and fluid properties. Three arediscussed in this chapter: (1) Muskat's method, (2) Schilthuis' method, and(3) Tamer's IS These methods relate the pressure decline to the oilrecovery and the gas-oil ratio.

You will recall that the material balance is successful in predicting theperformance of volumetric reservoirs down to pressures at which free gasbegins to flow. In the study of North Snyder, for example, in Chapter 5, Sect.4, the produced gas-oil ratio was assumed to be equal to the dissolved gas-oilratio down to the pressure at which the gas saturation reached 10%, the criti-cal gas saturation assumed for that reservoir. Below this pressure (i.e., athigher gas saturations), both gas and oil flow to the weilbores, their relativerates being controlled by their change with pressure and bytheir relative permeabilities, which change with their saturations. It is notsurprising, then, that the material balance principle (static) is combined withthe producing gas-oil ratio equation (dynamic) to predict the performance atpressures at which the gas saturation exceeds the critical value.

In the Muskat method, the values of the many variables that affect theproduction of gas and oil and the values of the rates of changes of these vari-ables with pressure are evaluated at any stage of depletion (pressure). As-suming these values hold for a small drop in pressure, the incremental gas andoil production can be calculated for the small pressure drop. These variablesare recalculated at the lower pressure, and the process is continued to anydesired abandonment pressure. To derive the Muskat equation, let be the


reservoir pore volume in barrels. Then, the stock tank barrels of oil remainingN, at any pressure is given by:

N,=

stock tank barrels (9.27)

Differentiating with respect to pressure,

dN, (1 dS0 S0 dB0928

dp dp dp

The gas remaining in thc reservoir, both free and dissolved, at the samepressure, in standard cubic feet, is

= +—

5,—(9.29)

Differentiating with respect to pressure,

dG,dp

v dS0 S0 dR30 R,0S0 dB0 — (I — till8 — I dS.930

"1.80 dp + B0 dp dp B dp B,, dp

if reservoir pressure is dropping at the rate dpldi, then the current or produc-ing gas-oil ratio at this pressure is

931dN,Idp

Substituting Eqs. (9 28) and (9.30) in Eq. (9 31),

R,0 dS0 So dR,0 R,050 d80 (1 — dB8 1

dp ÷ B0 dp tip - tip B8 dpR =

- I dS0 S0dB0 (9.32)

B0dp

Equation (9.32) is simply an expression of the material balance for volumetric.undersaturated reservoirs in differential form The producing gas-oil ratio mayalso be written as

D_D gPo 01%

KOjJ.81)8


Eq. (9.33) applies both to the flowing free gas and to the solution gas whichflows to the weilbore in the oil. These two types of gas make up the totalsurface producing gas-oil ratio, R in SCF/STB. Equation (9.33) may beequated to Eq. (9.32) and solved for dS0Idp to give:

So8g dR50 + So kg P'o dB0 (1—dS0 — B0 dp B0 k0 dp dp_

dp — 1+&Eo(9.34)

k0

To simplify the handling of Eq. (9.34), the terms in the numerator which arefunctions of pressure only may be grouped together and given thc groupsymbols X(p), Y(p), and Z(p) as follows:

X(p) = 0 Y(p) = J.. Z(p)= (9.35)

Using these group symbols and placing Eq. (9.34) in an incremental form,

1(9.36)

k0

Equation (9 36) gives the change in oil saturation that accompanies a pressuredrop, The functions X(p), Y(p), and Z(p) are obtained from the reser-voir fluid properties using Eqs. (9.35). The values of the derivatives dR50/dp,dB0Idp, and arc found graphically from the plots of B0, andversus pressure. It has been found that when determining dBg/dp the numbersarc more accurately obtained by plotting l/Bg versus pressure. When this isdone, the following substitution is used:

d(1/B8) -dp —

dp dp

or

(937)

In calculating iXS0 for any pressure drop Ap, the values of X(p),Y(p), Z(p), k1/k0, and at the beginning of the interval may be used.


Better results will be obtained, however, if values at the middle of the pressuredrop interval are used. The value of S0 at the middle of the interval can beclosely estimated from the AS0 value for the previous interval, and the valueof used corresponding to the estimated midinterval value of the oil satu-ration. In addition to Eq. (9.36), the total oil saturation must he calculated.This is done by simply multiplying the value of AS0 lAp by the pressure drop,Ap, and then subtracting the AS0 from the oil saturation value that corre-sponds to the pressure at the beginning of the pressure drop interval, as shownin the following equation:

S01S0(, 1)—Ap (9.38)

where, j corresponds to the pressure at the end of the pressure increment andj — 1 corresponds to the pressure at the beginning of the pressure increment.

The following procedure is used to solve for the AS0 for a given pressuredrop Ap:

1. Plot R,0, B0, and B, or 1/B, versus pressure and determine the slope of eachplot.

2. Solve Eq. (9.36) for the AS0 lAp using the oil saturation that correspondsto the initial pressure of the given 4p.

3. Estimate S0, using Eq. (9.38).4. Solve Eq. (9.36) using the oil saturation from Step 3.5. Determine an average value for AS0 lAp from the two values calculated in

Steps 2 and 4.6. Using solve for S0, using Eq. (9.38). This value of becomes

S0(1_l) for the next pressure drop interval.7. Repeat Steps 2 through 6 for all pressure drops of interest.

The Schilthuis method begins with the general material balance equa-tion, which reduces to the following for a volumetric, undersaturated reser-voir, using the single-phase formation volume factor:

N— (939)

Notice that this equation contains variables that are a function of only thereservoir pressure. B,, and B0, and the unknown variables, Ri,, and

of course, is the ratio of cumulative oil production, to cumulative gasproduction. To use this equation as a predictive tool a method mustbe developed to estimate R,,. The Schilthuis method uses the total surface pro-

R01.17

SL=SW+(l N iB01

+!Sg(RpR,o)j

Tb;;— A0;+ — R0) —

(940)

The Displacement of Oil and Gas

ducing gas-oil ratio or the instantaneous gas-oil ratio, R. defined previously inEq. (9.33) as

(9.33)

The first term on the nght-band side of Eq (9.33) accounts for the pro-duction of solution gas, and the second term accounts for the production offree gas in the reservoir. The second term is a ratio of the gas to oil flowequations discussed in Chapter 7. To calculate R with Eq. (9 33), informationabout the permeabilities to gas and oil is required This information is usuallyknown from laboratory measurements as a function of fluid saturations and isoften available in graphic form (see Fig 9.15). Thc fluid saturation equationis also needed:

where

SL is the total liquid saturation (i.e., SL = Se,, + S0. which also equals 1 — Sg)

The solution of this set of equations to obtain production values requiresa trial-and-error procedure. First, the material balance equation is rearrangedto yield the following'

1=0 (9.41)

fluid Saturation

Fig. 9.15. Permeability ratio versus fluid saturation.


All the parameters in Eq. (9.41) are known as functions of pressure fromlaboratory studies except N,1N and When the correct values of these twovariables are used in Eq. (9.41) at a given pressure, then the side ofthe equation equals zero. The trial-and-error procedure follows this sequenceof steps:

1. Guess a value for an incremental oil production IN) that occurs duringa small drop in the average reservoir pressure (Ap)

2. Determine thc cumulative oil production to pressure p, Pr' — Ap byadding all the previous incremental oil productions to the guess duringthe current pressure drop. The subscript, j — refers to the conditions atthe beginning of the pressure drop and j to the conditions at the end of thepressure drop.

(9.42)

3. Solve the total liquid saturation equation, Eq. (9.40), for 5,. at the currentpressure of interest.

4. Knowing S,., determine a value for k,/k0 from permeability ratio versussaturation information, and then solve Eq. (9.33) for R1 at the currentpressure.

5. Calculate the incremental gas production using an average value of thegas-oil ratio over the current pressure drop:

R,. = R,,+ R,(943)

944N N

6. Determine the cumulative gas production by adding all previous in-cremental gas productions in a similar manner to Step 2 in which thecumulative oil was determined.

G AG(945)

7. Calculate a value for with the cumulative oil and gas amounts.

(9.46)

8. With the cumulative oil recovery from Step 2 and the from Step 7, solve


Eq (9.41) to determine if the left-hand side equals zero. If the left-handside does not equal zero, then a new incremental recovery should beguessed and the procedure repeated until Eq. (9.41) is satisfied.

Any one of a number of iteration techniques can be used to assist in thetrial and error procedure. One that is frequently used is the secant method,'7which has the following iteration formula:

rxn—xn_'iI (947)

Un J, I J

To apply the secant method to the foregoing procedure, the left-handside of Eq. (9.41) becomes the function, f, and the cumulative oil recoverybecomes x. The secant method provides the new guess for oil recovery, andthe sequence of steps is repeated until the function, f, is zero or within aspecified tolerance (e.g., ± 10 4) The solution procedure described earlier isfairly easy to program on a computer.

The Tamer method for predicting reservoir performance by internal gasdrive is presented in a form proposed by Tracy.'8 Neglecting the formation andwater compressibility terms, the general material balance in terms of thesingle-phase oil formation volume factor may be written as foUows:

- — RsoBgI + GpBg — (W. -- (.)

B0 — + (Riot — + —

Tracy suggested writing

- B0 - R50Bg- (.)

B0 — B01 + — Rso)Bg + — Bg,)

(9.50)— T3oi + (R50, Rso)Bg + — B82)

mB0(9.51)

B0 — + (R501 — R30)B8 + -j_(B8 —

where cD,,, and are simply a convenient collection of many terms, all ofwhich arc functions of pressure, except the ratio m, the initial free gas-to-oilvolume. ilie general material balance equation may now bc written as

N = + — (We — (9.52)

3 Immiscible Displacement Processes

Applying this equation to the case of a volumetric, undersaturated reservoir,

N + (9.53)

in progressing from the conditions atany pressure, P,-i' to a lowerpressure, p,, Tracy suggested the estimation of thc producing gas-oil ratio, R,at the lower pressure rather than estimating the production during theinterval, as we did in the Schilthuis method. The value of R may be estimatedby extrapolating the plot of R versus pressure, as calculated at the higherpressure. Then, The estimated averag: gas-oil ratio between the two pressuresis given by Eq. (9.43):

(9.43)

From this estimated average gas-oil ratio for the Ap interval, the estimatedproduction, AN,,, for the interval is made using Eq. (9.53) in the followingform:

N — 1) + ÷ fl+ (9.54)

From the value of AN,, in Eq. (9 58), the value of is found:

(9.55)

In addition to these equations, the total liquid saturation equation is required,Eq. (9 40). The solution procedure becomes as follows:

I. Calculate the values of cI,,, and cI's as a function of pressure.

2. Estimate a value for R, in order to calculate an for a pressure drop ofinterest, Ap.

3. Calculate AN,, by rearranging Eq. (9.54) to give:

AN 956( . )

4. Calculate the total oil recovery from Eq. (9.55).5. Determine by calculating the total liquid saturation, SL from Eq.

(9.40) and using k,/k0 versus saturation information.

6. Calculate a value of R• by using Eq (9 33), and compare it with theassumed value in Step 2. If these two values agree within some tolerance.then the calculated in Step 3 is correct for this pressure drop interval.If the value of R, does not agree with the assumed value in Step 2, then the


TABLE 9.4.Fluid property data for Example 9 1

Pressure,psa

-

bhl/S TB SCMSTBB,,

bbIISCFIL.,cp

p.,,- cp

2500 1.498 72] 0.001048 0.488 0.0170

2300 1 463 669 0.001155 0.539 0.01662100 1 429 617 0.001280 0.595 0.01621900 1 395 565 0.001440 0 658 0.01581700 1.361 513 0001634 0726 001541500 1.327 461 0.001884 0802 0.01501300 1.292 409 0.002206 0.887 0.01461100 1.258 357 0.002654 0.981 0.0142900 1.224 305 0.003300 1.085 0 0138700 1.190 253 0004315 1199 00134500 1156 201 0.006163 1.324 00130300 1.121 149 0.010469 1 464 0.0126

tOO 1.087 97 0.032032 1617 0.0122

calculatedrepeated.

value should be used as the new guess and Steps 2 through 6

As a further check, the value of Rave can be recalculated and Eq. (9.56) solvedfor Again, if the new value agrees with what was previously calculatedin Step 3 within some tolerance, it can be assumed that the oil recovery iscorrect.

The three methods are illustrated in Ex Prob. 9.1.

Example 9.1. Calculating oil recovery as a function of pressure using the(a) Muskat, (b) Schilthuis, and (c) Tamer methods for an undersaturated,volumetric reservoir. The recovery is calculated for the first 200 psi pressureincrement from the initial pressure down to a pressure of 2300 psia

Given'

Initial reservoir pressure 2500 psiaInitial reservoir temperature = 180°FInitial oil in place =56 x 106 STBinitial water saturation = 0.20Fluid property data are given in Table 9.4.Permeability ratio data are plotted in Fig. 9.16.

steps:SoLuTioN: (a) The Muskat method involves the following sequence of


1. R40, B,, and were plotted versus pressure to determine the slopesAlthough the plots are not shown, the following values were

0.26 = 0.000171 = 0.433dp dp dp

The values of X(p), Y(p). and Z(p) arc tabulated as follows as a functionof pressure.

Pressure X(p) Y(p) Z(p)

2500 psia 0 000182 0 003277 0 0004542300 psia 0000205 0003795 0.000500

2 Calculate using X(p). Y(p), and Z(p) at 2500 psia.

EtS0 — 0 20(0.000182) 0 + 00 ooo 146

1+0 -

3 Estimate S01:

0.80 — 200(0 000146) 0.7709

4. Calculate using the S01 from Step 3 and X(p), and Y(p), and Z(p)at 2300 psia

0.7709(0.000205) ÷ 0.7709(0.00001)0 003795(1 0—0 2 — 0.7709)0.000500

— I +i€(000001)

=0000173'ftp

5. Calculate the average

— 0.000146 + 0.000173—0 000159

2

6. Calculate using from Step 5•

S0, = 0.8 — 0.000159(200) = 0.7682


This value of S0 can now be used to calculate the oil recovery that has occurreddown to a pressure of 2300 psia:

(957)

[ STB

(b) Because the Schilthuis method involves a trail-and-error procedure, thesecant iteration formula is used to assist in the solution. For this problem, xin the secant formula becomes Ne/N and f becomes the value of the left-handside of Eq. (9.41). The secant method requires two values of IN to beginthe iteration process

1. Assume incremental oil recovery:

—=0.01 and —-0.02

2. Calculate cumulative oil recovery'

— —001 and = =002NN N N

3. Calculate 5,'

5L1 = 0.2 + (1 — 0.2)(1 — .9735

9657

4 Determine from Fig. 9.16 and calculate R:

k01k02

R1 1=R 2721

R2 1 = R22 = 669 + 0.00001 = 669.4

where

R1 k = instantaneous gas-oil ratioj = current pressure counterk guess or iteration counter


kglko kg/ko

001 10

005 -1 -05

002 H 02

- -1- - - -0130 40 50 60 70 80 90 100

lotol liquid saturation, %

Fig. 9.16. Permeability ratio relationship for Example Problem 9 1

5 Calculate incremental gas

= Rave2 = = 695.2

= 0.01(695.2) = 6.952


6. Calculate cumulative gas recovery:

and

7. Calculate

and 695.2

8. Solve Eq. (9.41) to determine if the left-hand side equals hero or is

within some chosen tolerance, say, ± 10

— 0 01[1.523 + 0.001155(695.2 — 721)1— 27ft

— (1.523 — 1.498)— 1 — — .40

02[J .523 ÷ 0.001155(695.2 — 721)] —

12— (1.523— 1.498)—1—.1946

Neither value is within the chosen tolerance, so the two initial guesses of theincremental oil recovery for the first pressure increment are wrong The secantiteration formula is used to determine a new guess for the incremental oil

19=0.0167

With the new guess, Steps 2 to 8 are repeated to determine if the tolerance ismet for Eq. (9.41). The left-hand side of Eq. (9.41) is found to be —0.0025,which is within the tolerance. Thus, the correct value of fractional oil recoverydown to 2300 psia is 0.0167.

lo compare with the Muskat method, the recovery ratio must be multipliedby the initial oil in placeS 56 M STB, to yield the total cumulative recovery:

= 56(10)6(0.0167) = 935,200 STB

(c) The Tamer method requires the following steps:

I Calculate ct,, and cI's at 2300 psia:

— 1 463—669(0.001155)— 1.463 — 1.498 (721 — 669)0.001155

— 27 546

0.0011551.463—I 498+(721 —669)0001155

004609

4. Introduction to Watertlooding 375

2. Assume R, = 670 SCF/STB, which is just slightly larger then sug-gesting that only a very small amount of gas is flowing to the weliborcand is being produced:

3. Calculate AN,,:

56.000,000—0—000

— 27.546 + 0.04609(695.9)— 939,3 SIB

4. Caiculatc

= = 939,300 STB

5 Determine kg/ko

SL=O.2+(l

From this value of SL, the permeability ratio, can be obtained from Fig.9.16 Since the curve is off the plot, a very small value of = 0.00001 isestimated.

6. Calculate R1 and compare it with the assumed value in Step 2:

R, = ÷ =6694

This value agrees very well with the value of 670 that was assumed in Step 2.For the numbers of Ex. Prob. 9.1, the three methods of calculation yieldedvalues of that are within 0.5%. This suggests that any one of the threemcthods may he used to predict oil and gas recovery, especially consideringthat many of the parameters used in the equations could be in error more than0.5%.

4. INTRODUCTION TO WATERFL000ING

Waterflooding is the use of injection water to produce reservoir oil. Theprocess was discovered quite by accident more than 100 years ago when waterfrom a shaliow water-bearing horizon leaked around a packer and entered anoil column in a well. The oil production from the well was curtailed, but pro-duction from surrounding wells increased. Over the years. the use of water-


flooding has slowly grown until it is now the dominant fluid injection recoverytechnique. The next several sections provide an overview of this process. Youare referred to several good books on the subject that provide detailed designcritena.12 in the following sections. the characteristics of good watcrfloodcandidates, the location of injectors and producers in a waterflood, and waysto estimate the recovery of a waterflood are hnefly discussed

4.1. Waterflood Candidates

Several factors lend an oil reservoir to a successful waterflood. They can begenCrali7ed in two categories: reservoir characteristics and fluid character-istics.

The main reservoir characteristics that affect a watcrflood are depth,structure, homogeneity, and petrophysical properties such as porosity, satura-tion, and average permeability. The depth of the reservoir affects the water-flood in two ways. First, investment and operating costs generally increase asthe depth increases as & result of the increase in drilling and lifting costs.Second, the reservoir must be deep enough for the injection pressure to be lessthan the fracture pressure of the reservoir. Otherwise, fractures induced byhigh water injection rates could lead to poor sweep efficiencies if the injectedwater channels through the reservoir to the producing wells. If the reservoirhas a dipped structure, gravity effects can often be used to increase sweepefficiencies. The homogeneity of a reservoir plays an important role in theeffectiveness of a waterflood. The presence of faults, permeability trends, andthe like effect the location of new injection wells because good communicationis required between injection and production weils However, if serious chan-neling exists, as in some reservoirs that arc significantly heterogeneous, thenmuch of the reservoir oil will be bypassed and the water injection will he ren-dered useless. If a reservoir has insufficient porosity and oil saturation, thena waterflood may not he economically justified on the basis that not enoughoil will be produced to offset investment and operating costs. The averagereservoir permeability should be high enough to allow sufficient fluid injectionwithout parting or fracturing the reservoir.

The principal fluid characteristic is the viscosity of the oil compared tothat of the injected water. The important variable to consider is actually themobility ratio, which was defined earlier in Chapter 7 and includes not onlythe viscosity ratio 'ut a ratio of the relative permeabilities to each fluid phaseS

k0

A good waterflood has a mobility ratio around 1. if the reservoir oil is ex-tremely viscous, then the mobility ratio will likely be much greater than IWhen this is the case, the wLter will "finger" through the reservoir and bypassmuch of the oil.

4 Introduction to Waterflooding 377

4.2. Location of Injectors and Producers

The injection and production wells in a waterflood should be placed to accom-plish the following: (a) provide the desired oil productivity and the necessarywater injection rate to yield this oil productivity; and (b) take advantage of thereservoir characteristics such as dip, faults, fractures, and permeability trends.In general, two kinds of flooding patterns are used: peripheral flooding andpattern flooding.

Pattern flooding is used in reservoirs having a small dip and a large sur-face area. Some of the more common patterns are shown in Fig. 9.17. Table9.5 lists the ratio of producing wells to injection wells in the patterns shownin Fig 9.17. If the reservoir characteristics yield lower injection rates thanthose desired, the operator should consider using either a seven- or a nine-spotpattern where there are more injection wells per pattern than producing wells.A similar argument can be made for using a four-spot pattern in a reservoirwith low flow rates in the production wells

TABLE 9.5.Ratio of producing wells to injection wells for several pattern arrangements

Ratio

Patternof Producing Wellsto Injection Wells - -—

Four-spot 2Five-spot 1

Sevcn-spot . 1/2

Nine-spot 1/3

Direct-line-drive I

Staggered-line-drive 1

The direct-line-drive and staggered-line-drive patterns are frequentlyused because they usually involve the lowest investment. Some of the eco-nomic factors to consider include the cost of drilling new wells, the cost ofswitching existing wells to a different type (i.e., a producer to an injector), andthe loss of revenue from the production when making a switch from a pro-ducer to an injector

In peripheral flooding, the injectors arc grouped together unlike inpattern floods where the injectors are interspersed with the producers. Figure9.18 illustrates two cases in which peripheraL floods arc sometimes used. InFig 9 18(a), a schematic of an anticlinal reservoir with an underlying aquiferis shown. The injectors are placed so that the injected water either enters theaquifer or is near the aquifer-reservoir interface. The pattern of wells on thesurface, shown in Fig. 9.18(a), is a ring of injectors surrounding the producersA monoclinal reservoir with an underlying aquifer is shown in Fig. 9.18(b).In this case, the injectors arc again placed so that the injected water either

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There arc several methods of obtaining estimates for the microscopic dis-placement efficiency Ihe basis for one method was presented in Sect. 3.1 inthis chapter. The area! displacement, or sweep, efficiency is largely a functionof pattern type and mobility ratio. The vertical displacement efficiency is pri-marily a function of reservoir heterogeneities and thickness of the reservoirformation.

Waterflooding is an important process for the reservoir engineer tounderstand It has made and will continue to make large contributions to therecovery of reservoir oil.

5. INTRODUCTION TO ENHANCED OIL RECOVERYPROCESSES

Enhanced oil recovery (EOR) refers to the process of producing liquid hydro-carbons by methods other than the conventional use of reservoir energy andreservoir rcpressurizing schemes with gas or water On the average, conven-tional methods of production produce about one-third of the initial oil in placein a given reservoir. The remaining oil, nearly two-thirds of the initial re-source, is a large and attractive target for EOR methods. The next few sec-tions provide an introduction to this important topic in reservoir engineering.

5.1. MobilIzation of Residual Oil

During the early stages of a waterflood in a water-wet reservoir system. thebrine exists as a film around the sand grains and the oil fills the remaining porespace. At a time intermediate during the flood, the oil saturation has been de-creased and exists partly as a continuous phase in some pore channels but asdiscontinuous droplets in other channels.. At the end of the flood, when theoil has been reduced to residual oil saturation, S0,, the oil exists primarily asa discontinuous phase of droplets or globules that have been isolated andtrapped by the displacing brine.

The watcrflooding of oil in an oil-wet system yields a different fluid dis-tribution at S0,. Early in the waterfLood, the brine forms continuous flow pathsthrough the center portions of some of the pore channels. The brine entersmore and more of the pore channels as the waterflood progresses. At residualoil saturation, the brine has entered a sufficient number of pore channels toshut off the oil flow. The residual oil exists as a film around the sand grainsIn the smaller flow channels, this film may occupy the entire void space.

The mobilization of the residual oil saturation in a water-wet systemrequires that the discontinuous globuics be connected to form a continuousflow channel that leads to a producing well. In an oil-wet porous medium, thefilm of oil around the sand grains must be displaced to large pore channels andbe connected in a continuous phase before it can be mobilized The mobiliza-tion of oil is governed by the viscous forces (pressure gradients) and the inter-facial tension forces that exist in the sand grain-oil-water system.

5 Introduction to Enhanced Oil Recovery Processes 381

'Ihere have been several investigations of the effect of viscous forces andinterfacial tension forces on the trapping and mobilization of residual 20 21

From these studies, corrclations between a dimensionless parameter called thecapillary number, and fraction of oil recovered have been developed Thecapillary number is the ratio of viscous force to interfacial tension force andiS defined by Eq. (9.58).

= (constant) = (constant) (9.58)

where v is velocity, is the viscosity of the displacing fluid, is the inter-facial tension between the displaced and displacing fluids, k0 is the effectivepermeability to the displaced phase, 4, is the porosity, and is the pressuredrop associated with the velocity.

Figure 9.19 is a schematic representation of the capillary number cor-relation in which the capillary number is plotted on the abscissa, and the ratioof residual oil saturation (after conducting an EOR process to the residual oilsaturation before the EOR process) is plotted as the vertical coordinate. Thecapillary number increases as the viscous force increases or as the interfacialtension force decreases The EOR methods that have been developed andapplied to reservoir situations are designed either to increase the viscous forceassociated with the injected fluid or to decrease the interfacial tension forcebetween the injected fluid and the reservoir oil. The next three sections discuss

10

CS,, I after05

IS,,,) belore

0010' 106 10'

Capillary Number

Fig. 9.19. Capillary number correlation


the three general types of EOR processes: miscible flooding, chemical flood-ing, and thermal flooding.

5.2. MiscIble Injection Processes

In Sect. 2.1 of this chapter, it was noted that microscopic displacement effi-ciency is largely a function of interfacial forces acting among the oil, rock, anddisplacing fluid. If the interfacial tension between the trapped oil and dis-placing fluid could be lowered to 10-2 to dyn/cm, the oil droplets couldbe deformed so that they would squeeze through the pore constrictions andcombine with other droplets to yield a continuous oil phase. A miscible pro-cess is one in which the interfacial tension is zero; that is, the displacing fluidand residual oil mix to form one phase. If the interfacial tension is zero, thecapillary number becomes infinite and the microscopic displacement effi-ciency is maximized.

Figure 9.20 is a schematic of a miscible process. Fluid A is injected intothe formation. When the fluid contacts residual oil droplets, an oil bank isformed at the leading edge of the injected material. A mixing zone developsbetween fluid A and the oil bank and grows as a result of molecular diffusionand other processes. Fluid A is then followed by fluid B, which is miscible withfluid A but not generally miscible with the oil and which is much cheaper thanfluid A. A mixing zone is also created at the fluid A—fluid B interface. It isimportant that the amount of fluid A injected be large enough that the twomixing zones do not come in contact but still be small enough to avoid largecosts for injected chemicals.

Consider a miscible process with n-decane as the residual oil, propaneas fluid A. and methane as fluid B. The system pressure and temperature are

Injection ProductionWell Well

OilBank

Connate Water

Fig. 9.20. Schematic of a miscible injcction process

(

5. Introduction to Enhanced Oil Recovery Processes 383

2000 psia and 100°F, respectively. At these conditions, both n-decane andpropane are liquids and are therefore miscible in all proportions. The systemtemperature and pressure indicate that any mixture of methane and propanewould be in the gas state; therefore, the methane and propane would be misci-ble in all proportions. However, the methane and n-decanc would not be mis-cible in all proportions because one is a liquid and thc other a gas Notice that,in this example, the propane appears to act as a liquid when in the presenceof n -decane and as a gas when in contact with methane. It is this uniquecapacity of propane and other intermediate gases that leads to the miscibleprocess.

There are, in general, two types of miscible processes. One is referredto as the single-contact miscible process and involves such injection fluids asliquified petroleum gases (LPG) and alcohols. Thc injected fluids are misciblewith residual oil immediately on contact. The second type is the multiple-contact, or dynamic, miscible process. The injected fluids in this case are usu-ally methane, inert fluids, or an enriched methane gas supplemented with aCrC6 fraction; this fraction of alkanes has the unique ability of behaving likea liquid or a gas at many reservoir conditions. The injected fluid and oil areusually not miscible on first contact but rely on a process of chemical exchangeof the intermediate hydrocarbons between phases to achieve miscibility.These processes are discussed in great detail elsewhere.4'5 22

5.3. Chemical Injection Processes

Chemical flooding relies on the addition of one or more chemical compoundsto an injected fluid either to reduce the interfacial tension between the reser-voir oil and injected fluid or to improve the sweep efficiency of the injectedfluid by making it more viscous, thereby improving the mobility ratio. Bothmechanisms are designed to increase the capillary number.

Three general methods are used in chemicaL flooding technology.4first is polymer flooding, in which a large macromolecuic is used to increasethe displacing fluid viscosity. This process leads to improved sweep efficiencyin the reservoir of the injected fluid. The remaining two methods, micellar-polymer flooding and alkaline flooding, make use of chemicals that reducc theinterfacial tension between an oil and a displacing fluid.

The addition of molecules of large molecular weight (i.e., polymers) toan injected water can often increase the effectiveness of a conventional water-flood. Polymers are usually added to the water in concentrations ranging from250 to 2000 parts per million (ppm). A polymer solution is more viscous thana brine without polymer. in a flooding application, the increased viscosityalters the mobility ratio between the injected fluid and the reservoir oil. Theimproved mobility ratio leads to better vertical and areal sweep efficienciesand thus higher oil recoveries. Polymers have also been used to alter grosspermeability variations in some reservoirs, in this application, polymers forma gel-like material by cross-linking with other chemical species. The polymer


gel sets up in high permeability streaks and diverts the flow of subsequentlyinjected fluids to a different location

The improvement in oil recovery when using polymers is a result of animproved sweep efficiency over what is obtained during a conventional water-flood. Typical oil recoveries from polymer flooding are in the range of 1 to 5%of the initial oil in place. It has been found that a polymer flood is more likelyto be successful if it is started early in the producing life of the reservoir.

The miccllar-polymcr process uses a surfactant to lower the interfacialtension between the injected fluid and the reservoir oil. A surfactant is asurface-active agent that contains a hydrophobic ("dislikes" water) part to themolecule and a hydrophilic ("likes" water) part. The surfactant migrates tothe interface between the oil and water phases and helps make the two phasesmore miscible. interfacial tensions can be reduced from -'-30 dyn/cm, foundin typical waterflooding applications, to dyn/cm with the addition of aslittle as 0 1 to 5.0 weight % surfactant to water-oil systems Soaps and deter-gents used in the cleaning industry are examples of surfactants. The sameprinciples involved in washing soiled linen or greasy hands are used in "wash-ing" residual oil off of rock formations. As the interfacial tension between anoil phase and a water phase is reduced, the capacity of the aqueous phase todisplace the trapped oil phase from the pores of the rock matrix increases. Thereduction of interfacial tension results in a shifting of the relative permeabilitycurves such that the oil will flow more readily at lower oil saturations.

When an alkaline solution is mixed with certain crude oils, surfactantmolecules are formed by chemical reactions between the alkaline solution andthe oil When the formation of surfactant molecules occurs in situ, the inter-facial tension between the brine and oil phases can be significantly reduced.The reduction of interfacial tension causes the microscopic displacement effi-ciency to increase, thereby increasing oil recovery

5.4. Thermal Processes

Primary and secondary production from reservoirs containing heavy, low-gravity crude oils is usually a very small fraction of the initial oil in placeThese types of oils are very thick and viscous and as a result do not migratereadily to producing wells, it is not uncommon for viscosities of certain heavycrudes to decrease by several orders of magnitude with an increase of tem-perature of 100 to 200°F. This suggests that if the temperature of a crude oilin the reservoir can be raised by 100 to 200°F over the normal reservoirtemperature, the oil viscosity will be reduced significantly and will flow muchmore easily to a producing well. The temperature of a reservoir can be raisedby injecting a hot fluid or by generating thermal energy in situ by combustingthe oil. hot water or steam can be injected as the hot fluid. Three types ofprocesses are generally used in the industry: (1) the continuous injection ofhot fluids, such as hot water or steam; (2) the intermittent injection of steam,

5. Introduction to Enhanced Oil Recovery Processes 385

referred steam cycling; and (3) the injection of air or oxygen-enriched airto aid in the combustion of reservoir 5 23

The continuous injection of hot fluids is usually accomplished by inject-ing either hot water or steam and is much like a conventional waterflood.When steam is injected into the formation, the thermal energy is used to heatthe reservoir oil. Unfortunately, the energy also heats the entire environment,such as formation rock and water. Some energy is also lost to the underburdenand overburden. Once the oil viscosity is reduced by the increased tem-perature, the oil can flow more readily to the producing wells. The steammoves through the reservoir and comes in contact with cold oil, rock, andwater. As the steam contacts the cold environment, it condcnses. A ho' waterbank is formed and acts as a waterflood, pushing additional oil to the produc-ing wells.

Several mechanisms responsible for the production of oil from a steaminjection process have been identified. These include thermal expansion of thecrude oil, viscosity reduction of the crude oil, changes in surface forces as thereservoir temperature increases, and steam distillation of the lighter portionsof the crude oil

The intermittent injection of steam, referred to as the steam stimulationprocess or the cyclic steam process, begins with the injection of steam for aperiod of days to weeks. The well is then shut in, and the steam is allowed tosoak the area around the injection well. This soak period is fairly short, usuallyfrom one to five days. The injection well is then placed on production. Thelength of the production period is dictated by the oil production rate, and itcan last from several months to a year or more. The cycle is repeated as manytimes as economically feasible. The oil production decreases with each newcycle.

Mechanisms of oil recovery that result from this proccss include (I)reduction of flow resistance near the wellbore by reducing the crude oil viscos-ity and (2) enchancement of the solution gas drive mechanism by decreasingthe gas solubility in an oil as temperature increases.

In heavy oil reservoirs, the steam stimulation process is often applied todevelop injectivity around an injection well so that a continuous steam injec-tion proccss can be conducted.

The injection of air or oxygen-enriched air is referred to as the in situcombustion process. Early attempts to apply the combustion process involvedwhat is called the forward dry combustion process. The crude oil was igniteddownhole, and then a stream of air or oxygen-enriched air was injected intothe well where the combustion was originated. The flame front was then prop-agated through the reservoir. Large portions of heat energy were lost to thesurroundings with this process. To reduce the heat losses, a reverse combus-tion process was conceived In reverse combustion, the oil is ignited as in for-ward combustion, but the air stream is injected into a different well. The airis then "pushed" through the flame front as the flame front moves in the

3S6 The Displacement of Oil and Gas

opposite direction. Researchers found the process to work in the iatoratory,but when it was tried in the field on a pilot scale, it was never successful. Inthe field, the flame would shut off because there was no oxygen supply. Whenoxygen was injected, the oil would often scif-ignite. The whole process wouldthen revert to a forward combustion process.

When the reverse combustion process failed, a new technique called theforward wet combustion process was introduced. This process begins as aforward dry combustion, but once the flame front has been established, theoxygen stream is replaced with water. As the water comes in contact with thehot zone left by the combustion front, it flashes to steam, using energy thatotherwise would have been wasted. The steam moves through the reservoirand aids in the displacement of oil. The wet combustion process has becomethe primary method of conducting combustion processes.

Not all crude oils arc amenable to the combustion process. For the com-bustion process to function properly, the crude oil must contain enough heavycomponents to serve as the source of fuel for the combustion. This usuallyrequires an oil of low API gravity.

Most of the oil that has been produced by EOR methods to date hasbeen a result of applying thermal processes. There is a practical reason forthis, as well as several technical reasons. To produce more than I to 2% of theinitial oil in place from a heavy oil reservoir, thermal methods had to be em-ployed. As a result, thermal methods were investigated much earlier thaneither miscible or chemical methods, and the resulting technology was devel-oped much more rapidly.

PROBLEMS9.1 (a) A rock 10 cm long and 2 sq cm in cross section flows 0.0080 Cu Cm/SeC of a

25 cp oil under a 1.5 atm pressure drop If the oil saturates the rock 100%,what is its absolute permeability')

(b) What will be the rate of 0 75 cp brine in the same core under a 2.5 atmpressure drop if the brine saturates the rock 100%?

(c) Is the rock more permeable to the oil at 100% oil saturation or to the bnneat 100% bnne saturation')

(d) The same core is maintained at 40% water saturation and 60% oil satur-ation. Under a 2.0 atm pressure drop, the oil flow is 0.0030 cu cm/sec andthe water flow is 0 004 cu cm/sec. What arc the effective permeabilities towater and to oil at these saturations')

(e) Explain why the sum of the two effective permeabilities is less than theabsolute permeability.

(1) What are the relative permeabilities to oil and water at 40% water satu-ration?

(g) What is the relative permeability ratio at 40% water saturation')

Problems 387

(h) Show that the effective permeability ratio is equal to the relative perme-ability ratio.

9.2 The following permeability data were measured on a sandstone as a function ofits water saturation

S.., 0 10 20 30* 40 50 60 70 7580 90100k,0 1.0 1.0 1.0 0.94 0.80 044 0.16 0045 0 0 0 0k.,, 0 0 0 0 0.04 0.11 0.20 0.30 0.36 0.44 0.68L0Cri tical saturations for oil and water

(a) Plot the relative permeabilities to oil and water versus water saturation onCartesian coordinate paper.

(b) Plot the relative permeability ratio versus water saturation on semilog paper.

(c) Find the constants a and h in Eq. (9.3) from the slope and intercept of yourgraph. Also find a and b by substituting two sets of data in Eq. (9.3) andsolving simultaneous equations.

(d) If 3.4 cp. cp. B0= 1.50 bbl/STB, and 1.05 bbl/STB,what is the surface water cut of a well completed in the transition zone wherethc water saturation is

(c) What is the reservoir cut in part (d)"(f) What percentage of recovery will be realized from this sandstone under

high-pressure water dnve from that portion of the reservoir above the tran-sition zone invaded by water? The initial water saturation above the transi-tion zone is 30%.

(g) If water drive occurs at a pressure below saturation pressure such that theaverage gas saturation is 15% in the invaded portion, what percentage ofrecovery will be realized? The average oil volume factor at the lower pres-sure is 1 35 bbl/STh and the initial oil volume factor is 1.50 bbl/STB

(h) What fraction of the absolute permeability of this sandstone is due to theleast permeable pore channels that make up 20% of the pore volume? Whatfraction is due to the most permeable pore channels that make up 25% ofthe pore volume?

9.3 Given the following reservoir data

Throughput rate = 1000 bbl/dayAverage porosity = 18%Initial water saturation = 20%Cross-sectional area = 50,000 ft2

Water viscosity = 0.62 cpOil viscosity = 2.48 cp

versus S.., data in Fig 9 1 and 92

Assume zero transition zone'

(a) Calculate I,, and plot versus SW.

(b) Graphically determine a number of points, and plot versus S.


(C) Calculate at several values of using Eq (9.17), and compare withthe graphical values of part (b).

(d) Calculate the distances of advance of the constant saturation fronts at 100,200, and 400 days. Plot on Cartesian coordinate paper versus Equalizethe areas within and without the flood front lines to locate the position ofthe flood fronts.

(e) Draw a secant line from S,. — 0.20 tangent to the versus S.,, curve in part(b), arid show that the value of SW at the point of tangency is also the pointat which the flood front lines arc drawn

(f) Calculate the fractional recovery when the flood front first intercepts a well,using the areas of the graph of part (d) Express the recovery in terms of (1)the initial oil in place and (2) the recoverable oil in place (i.e., recoverableafter infinite throughput)

(g) To what surface water cut will a well rather suddenly rise when it is just en-veloped by the flood fronts' Use B. = 1 50 bbl/STB and BW = 1 05 bbl/STB.

(h) Do the answers to parts (1) and (g) depend on how far the front has trav-elled? Explain.

9.4 Show that for radial displacement where r

r is the distance a constant saturation front has travelled.9.5 Given the following reservoir data:

5, 10' 15 20 25 30 35 40 45 50 62*

k,k, 1) 0.08 0 20 0.40 0.85 1.60 3.00 5 50 10.0k, 0.70 0.52 0.38 0.28 0.20 0.14 0 11 0.07 0.04 0

• Cntical for gas and oil

Absolute permeability 400 mdhydrocarbon porosity = 15%Connate water = 28%Dip angle = 20°

Cross-sectional area — 750,000 ft2

Oil viscosity = 1 42 cp(ia.s viscosity =0 015 cpReservoir oil specific gravity =0 75Reservoir gas specific gravity = 0.15 (water = I)Reservoir throughput at constant pressure = 10,000 bbl/day

(a) Calculate and plot the fraction of gas, ), versus gas saturation similar to Fig9.13 both with and without the gravity segregation term.

(b) Plot the gas saturation versus distance after 100 days of gas injection bothwith and without the gravity segregation term

(c) Using the areas of part (b). calculate the recovenes behind the flood frontwith and without gravity segregation in terms of both initial oil and recov-erable oil

References 389

9.6 Derive an cquation including a gravity term similar to Eq (9.23) for waterdisplacing oil.

9.7 Rework the water displacement calculation of Table 9.1, and include a gravitysegregation term. Assume an absolute permeability of 500 md, a dip angle of450, a density difference of 20% between the reservoir oil and water and an oilviscosity of 1.6 cp Plot water saturation versus distance after 240 days. andcompare with Figure 9 11.

S,, 20 30 40 50 60 70 80 90093 0.60 035 0.22 0.12 0.05 001 0

9.8 Continue the calculations of Ex. Piob. 9 1 down to a reservoir pressure of 100psia using:

(a) Muskat method(b) Schilthuis method(c) Tamer method

REFERENCES

'C. R Smith, Mechanics of Secondary Oil Recovery Huntington, NY Robert EKneger Publishing, 196620. P Wilihite, Wa:erflooding. Dallac, TX: Society of Petroleum Engineers, 1986

'F F. Craig, '1 he Reservoir Engineering Aspects of Waterflooding," Monograph 3,Society of Petroleum Engineers, Dallas. TX, 1971

4H. K. van Poollen and Associates, Inc., Enhanced Oil Recovery Tulsa, OK Penn-Well Publishing, 1980.

W. Lake, Enhanced Oil Recovery. Englewood Cliffs, NJ: Prentice Hall, 1989

6R D. Wycoff, H. (i. Botset, and M Muskat, Trans AIME (1933), 103, 219.R I. Slobod and B. H Candle. Trans AIME (1952). 195, 265

as E. Buckley and M. C. Leverett. "Mechanism of Fluid Displacement in Sands,"Trans. AIME (1942), 146, 107.9j• U. Richardson and R J. Blackwell, "Usc of Simple Mathematical Models forPredicting Reservoir Behavior." Jour Of Petroleum Technology, (September 1971),1145.

'°H J Welge, 'A Simplified Method for Computing Oil Recoveries by Gas or WaterDrive," Trans AIME (1952). 195, 91

" 1). R Shrcvc and 1.. W. Welch, Jr., "Gas Dnve and Gravity Drainage Analysis forPressure Maintenance Operations," Trans. AIME (1956). 207, J 36.

'2L. R. Kern, "Displacement Mechanism in Multi-Well Systems," Trans. AIME(1952), 195, 39

"R H Smith reported by J. A. Klotz, "The Gravity Drainage Mechanism," Jour. ofPetroleum Technology. (April 1953), V, 19

'4M. Muskat, Production Histories of Oil Producing Gas-drive Reservoirs."Jour of Applied Physics (1945), 16, 147


'50il and Gas Journal, "Practical Reservoir Engineering." No 2, (Petroleum Publish-ing Co ), Tulsa, OK

Tamer, "How Different Si7e Gas Caps and Pressure Maintenance ProgramsAffect Amount of Recoverable Oil," Oil Weekly, June 12, 1944, 144, 32—34.

"1 13. Riggs, An Introduction to Numerical Methods for Chemical Engineers. Lub-bock. TX. Texas Tcch. University Press, 1988.

ING. W Tracy, "Simplified Form of the Material Balance Equation," Trans AIME(1955). 204, 243

19R B Terry, "Enhanced Oil Recovery." Encyclopedia of Physical Science and Tech-nology, Vol. 5 New York: Academic Press, 1987.

3. Taher, "Dynamic and Static Forces Required to Remove a Discontinuous OilPhase from Porous Media Containing Both Oil and Water," Soc Pet Engr. Jour.(March 1969), 3.

21 0. L. Stcgcmeier, "Mechanisms of Entrapment and Mobilization of Oil in PorousMedia," in Improved Oil Recovery by Surfacian: and Polymer Flooding, D. 0. Shahand R. S Schechtcr. cd. New York Academic Press, 1977

22F I Stalkup Jr, "Miscible Displacement." Monograph 8. Society of PetroleumEnginecrs. Dallas, TX, 1983

M, Prats, "Thermal Recovery," Monograph 7, Society of Petroleum Engineers,Dallas, TX, 1983.

Chapter 10

History Matching

1. INTRODUCTION

One of the most important job functions of the reservoir engineer is theprcdiction of future production rates from a given reservoir or specific well.Over the years, enginecrs have developed several methods to accomplish thistask The methods range from simple decline curve analysis techniques tosophisticated multidimensional. multiflow reservoir simulators *.l 2 3

a simple or complex method is used, the general approach taken to predictproduction rates is first to calculate producing rates for a period for which theengineer already has production information If the calculated rates match theactual rates, the calculation is assumed to be correct and can then be used tomake future predictions, if the calculated rates do not match the existing pro#duction data, some of the process parameters are modified and the calculationrepeated. The process of modifying thesc parameters to match the calculatedrates with the actual observed rates is referred to as history matching.

The calculational method along with the nccessary data used to conductthe history match is often referred to as a mathematical model or simulator.When decline curve analysis is used as the calculational method, the engineer

'Refercnccs throughout the text are givcn at the end of each chapter

101

392 History Matching

is doing littie more than curve fitting. and the only data that are necessary arethe existing production data However, when the calculational technique in-volvcs multidimensional mass and energy balance equations and multiflowequations, a large amount of data is required as well as a computer to conductthe calculations With this complex model, the reservoir is usually divided intoa grid. This allows the engineer to use varying input data, such as porosity,permeability, and saturation, in different grid blocks. This often requires esti-mating much of the data since the engineer usually knows data only at specificcoring sites that occur much less frequently then the grid blocks used in thecalculational procedure.

Since the purpose of this chapter is to discuss the concept of historymatching, a model of complexity in between that of the simple decline curveanalysis and the complex multidimensional, multiflow simulator is used. Themodel presented here uses the simple zero-dimensional Schilthuis materialbalance equation discussed in earlier chapters.

2. DEVELOPMENT OF THE MODEL

The material balance equations presented in Chapters 2 to 6 and Sect. 3.3 inChapter 9 do not yield information on future production rates because theequations do not have a time dimension associated with them. These equa-tions simply relate average reservoir pressure to cumulative production Toobtain rate information, a method is needed whereby time can be related toeither the average reservoir pressure or cumulative production. In Chapter 7,single-phase flow in porous media was discussed and equations were devel-oped for several situations that related flow rate to average reservoir pressure.It should be possible, then, to combine the material balance equations ofChapters 2 to 6 and 9 with the flow equations from Chapter 7 in a model orsimulator that would provide a relationship for flow rates as a function of time.The model will require accurate fluid and rock property data and past produc-tion data Once a model has been tested for a particular well or reservoirsystem and found to reproduce actual past production data, it can he used topredict future production rates The importance of the data used in the modelcannot be over emphasiicd. If the data are correct, the prediction of pro-duction rates will be fairly accurate.

2.1. The Material Balance Part of the Model

The problem we consider in this chapter involves a volumetric, internal gas-drive reservoir In Sect. 3 3 of Chapter 9, several different methods to calcu-late the oil recovery asa function of reservoirpressure forthis typeofreservoirwere presented. For the example in this chapter, the Schilthuis method isused You will remember that the Schilthuis method requires permeability

2. Development of the Model 393

ratio versus saturation information and the solution of Eqs. (9.33), (9.40),(9.41) (written with the two-phase formation volume factor):

(9.33)

S, = S,. + (1 — [i (9.40)

—1=0 (9.41)

2.2. Incorporating a Flow Equation Into the Model

The procedure mentioned in the previous section yields oil and gas productionas a function of the average reservoir pressure, but it does not give any indi-cation of the time requ'red to produce the oil and gas To calculate the timeand rate at which the oil and gas are produced, a flow equation is needed. Itwas found in Chapter 7 that most wells reach the pseudosteady-state afterflowing for a few hours to a few days. An assumption will he made that the wellin this chapter's problem has been produced for a time long enough for thepseudosteady-state flow to have been reached. For this case, Eq. (7 50) can beused to describe the oil flow rate into the weilbore:

0.00708k0hq0=—

1i0B0(7.50)

This equation assumes pseudosteady-state, radial geometry for an incom-pressible fluid The subscnpt, o, refers to oil, and the average reservoir pres-sure, is the pressure used to determine the production, in the Schilthuismaterial balance equation. The incremental time required to produce anincrement of oil for a given pressure drop is found by simply dividing theincremental oil recovery by the rate computed from Eq. (7.50) at the corre-sponding average pressure:

=q0

The total time that corresponds to a particular average reservoir pressure canbe determined by summing the incremental times for each of the incrementalpressure drops until the average reservoir pressure of interest is reached

Since Eq (10.1) requires and the Schilthuis equation determines


N, the initial oil in place, must be estimated. In Sect. 2 of Chapter 5,it was shown that the initial oil in place could be estimated from the volumetricapproach by the use of Eq (10.2):

N7758Ah4(l—S,,,j

B01(10.2)

Combining these equations with the solution of the Schilthuis material balanceequation yields the necessary production rates of both oil and gas.

3. AN EXAMPLE OF A HISTORY MATCH

We now apply the reservoir model developed in the previous two sections tohistory match production data from a well in a volumetric, internal gas-drivereservoir. The data for the problem were obtained from personnel at theUniversity of Kansas and is used here by permission.4

The well is located in a reservoir that is a sandstone and is produced fromtwo zones separated by a thin shale section of approximately 1102 ft in thick-ncss. The reservoir is classified as a stratigraphic trap. The two producingzoncs decrease in thickness and permeability in directions where it is bet leved

Fig. 10.1. Cii and gas viscosity for history matching problem.

012

011U

U)0UU)>Cl)

010 0

OIL AND GAS VISCOSITY3.0

20

300 400

PRESSURE, PSIA

-I

U)Ui

U.UU,

0

w

-I0>U)

C,U.0-jU0

UUi

112

3 An Example of a History Match 395

that a pinch-out occurs. Permeability and porosity decrease to unproductivelimits both above and below the producing formation The initial reservoirpressure was 620 psia The average porosity and initial water saturation valueswere 21.5% and 37%, respectively The area drained by the well is 40 ac.Average thicknesses and absolute permeabilities were reported to be 17 ft and9.6 md for zone 1 and 14 ft and 7 2 md for zone 2. Laboratory data for fluidviscosities, formation volume factors, solution gas-oil ratio, oil relative perme-ability, and the gas-to-oil permeability ratio are plotted in Figs. 10 1 to 10.5.

240

200

160

120

80

40

111

1.10

109z

U)U)

1.060

U.Lii

1 07

106

105

1.04

100 200 300 400 500 600 700

PRESSURE, PSIA

Fig. 10.2. Oil and gas formation volume factor for history matching problem


.

160'

- I ISOLUTION GAS-OIL RATIO R_,

--./

60 /

PRESSURE. PSIA

Fig. 10.3. Solution gas-oil ratio for history matching problem

Actual oil production and instantaneous gas-oil ratios for the first three yearsof the life of the well are plotted in Fig. 10 6.

3.1. Solution Procedure

One of the first steps in attempting to perform the history match is to convertthe fluid property data provided in Figs 10.1 to 10.3 to a more usuable form.This is done by simply regressing the data to obtain equations for each param-eter—for example, oil and gas viscosity, as a function of pressure. The result-ing equations can be found in the program listing in Tabie JO. 1. The two per-nieability relationships also nccd to be regressed to be used in the program.

-jU)

z

1-U)LI.UU,

-J0zU)4L$.0

I

100 200 300 400 500 700


08

07

06

0

$1

0

0

0.3

0

0

11g. 10.4. Oil relative permeability for history matching problem

Both the relative permeability to oil and the permeability ratio can be ex-pressed as functions of gas saturation. These equations are input into the pro-gram in a different manner from the fluid property equations. The constantsfor the regressed equations are placed in a data file to be read into the pro-gram at the time of program execution General expressions arc written in theprogram which use these constants These relationships are handled this wayto facilitate modifications to the equations used in the program if necessary

10

09

010 015 0.20 025

GAS SATURATION, FRACTION PORE SPACE

E EE E E

— —

.-

—

- — /- --— -

- — - r- - --- — —

7i i I I I IS

4

3.

7.

—I

005 010 015 020 025 030 035GAS SATURAT1ON. FRACTION PORE SPACE

10 c. nitin fnr hictnrv matching' rnnhlent

U—7.6—

S

I

10

S.

7

1,ao0 00 0 15

7-

0

7—5-S

10,

50 a:

1-

-H

10'

101

S

3. An Example of a History Match

CCC

CCC

WELL 15 1

399

U,I-

800 U,

LI.0U,

0

1200

1000

60

-I-- ——— —

p.U,

Uip. ——-—- -x.-J0

20 - --x

— —--——--200V GOR

— OIL PRODUC1 ION

1 2 3 4 5 6 7 8

TIME, YEARS

Fig. 10.6. Actual production and instantaneous GOR for history matching problem

TABLE 10.1.FORTRAN hsting of program for history matching problem

CC THIS PROGRAM CONDUCTS A HISTORY MATCHC THE PROGRAM IS UNIQUE FOR THE PROBLEM IN CHAPTER 10C OF THE REVISED CRAFT AND HAWKINS RESERVOIR ENGINEERINGC TEXT BY RONALD E. TERRYC

CHARACTER'6 DUMMYFREAL Ki .K2.KRO(60).N.NGP(60),NOP(60).NP(60)REAL BG(60),BO(60),BT(60).TOP(60),TGP(60).

I DT(60).DMB(1 00),R(60).RP(60).RS(BO).QO(60).2 VG(60),VO(50),P(60),TOT(60),A0(5),A1(5).3 SGL(5),SGU(5),C0(I 0),Cl(1 0).GASL(1 0),GASU(1 0)

iNPUT DATA

DATA PI,DP.PH1.SW.ACRE/620 .5., 215.37,40/DATA DOP,RW,RE,EPS,PW/1 .E-4..25,660 ,1 E-5,25.0/

GET INPUT VARIABLES FROM FILE NAMED BY USER


TABLE 10.1.(Continued)

WRITE(,598)598 FORMAT(IX'ENTER DATA FILE NAME')

READ(*,599)DUMMYF

599 FORMAT(A6)OPEN(5,F1LE= DUMMYF,STATUS= 'OLD')

CC NKGKO = NUMBER OF KG/KO CURVESC SGL(I) = LOWER GAS SATURATION FOR Rh CURVEC SGU(I) = UPPER GAS SATURATION FOR Ith CURVEC A0 (I) AND Al (I) = CONSTANTS IN REGRESSED EQUATIONC log (KGKO) = A0 + A?SGC

READ(5,600)NKGKO600 FORMATQ5)

DO 4 I=l,NKGKOREAD(5,605)SGL(I),SGU(I),AO(I)Al(I)

4 CONTINUE605 FORMAT(5F1 0.0)

CC NKRO= NUMBER OF KRO CURVESC GASL(I) GASU(I) = LOWER AND UPPER GAS SATURATION FORC Ith CURVEC C0(I) AND C1(I) - CONSTANTS IN REGRESSED EQUATIONC KRO=COCI'SGC

READ(5,600)NKROD041 I-l,NKRO

READ(5,605)GASL(I),GASU(I),C0(I),Cl(I)41 CONTINUE

READ(5,605) Hl,H2READ(5,605), KI ,K2CLOSE(5)

CC OPEN OUTPUT FILE NAMED MATCHC

OPEN(6,FILE= 'MATCH')CC CALCULATE FLUID PROPERTiESC FROM REGRESSED EQUATIONSC

DO 10 1=1.60

IF(P(I).LT.200)GOTO 11VO(I)=2.354—0 001 134'P(I)VG(I)=009684+.3267E—5'P(I)+ 11 16E—9P(i)'2RS(I)-=52 45 $ .2475*P(i)_ 2097E—4P(I)"2



062+ 9257E—4P(l)BG(I)=i 446+.3625P(I)-I .4532E_4*P(l)**2BG(I)=1JBG(I)GOTO 10

11 VO(I)=2.670— 004566P(I)+.9461 E—5'P(I)"2VG(J)= .007580+ .2497E—4P(I)— 5622E—7'P(I)'2RS(I)=16 994- 6565*P(I)_.001 198P(I)'2BO(I)= 1.0334BG(l)=8.874+ 3676*P(l)BG(I)=1 IBG(1)

10 CONTINUECC BEGiN SCHILTHUIS MATERIAL BALANCE CALCULATIONC

NP(1)=0.R(1)=RS(1)BT(1)= BO(1)NOP(1)=0.NGP(1)=0.TOT(1)=0.DO 201=2,60

BT(J)= BO(l)+(RS(1 )—RS(I))'BG(I)TOP(1)=NOP(l—1)DMB(1)=0.01

DOPDO 30 J=2,100

SG'=(l —(1 _TOP(J))*BO(I)/BO(1 ))'(l —SW)IF(SGLT.0 0)SG:=0.0

CC DETERMINE PROPER KG/KO VALUEC

0050 KK=1NKGKOIF(SG GT SGU(KK)) GOTO 50A=A0(KK)B=A1(KK)GO 1055

50 CONTINUE55 CONTINUE

CC DETERMINE PROPER KRO VALUEC

DO 42 KI-=1,NKROIF(SG.GT GASIJ(KI)) GO 1042AZ=C0(KI)BZ=C1(Kl)GO 1043



42 CONTINUE43 CONTINUE

KRO(I)=AZ+ BZ*SGIF(KRO(I).LT.0.0) KRO (I)=0 0IF(KRO(I).GT.1 .0) KRO(I)= 1 0RKGKO= I 0."(A+ B'SG)R(I)=RS(1)+ VO(I)'BO(I)/VG(I)/BG(I)'RKGKORAV=(R(1 — 1)+ R(l))/2DGP=DOP'RAVTGP(J)=NGP(I- 1)+DGPRP(I)=TGP(J)ITOP(J)DMB(J)=TOP(J)'(BT(I) + BG(1)'(RP(I)— RS(1 )))/

I (BT(I)—BT(1))—1TOP(J i 1)=TOP(J)—DMB(J)'(TOP(J)—TOP(J— 1))!

1 (DMB(J)—DMB(J— 1))1F(ABS(DMB(J)).GT EPS) GO TO 35NOP(I)=TOP(J)NGP(I)-TOP(J)'RP(I)GOTO 20

35 DOP=TOP(J+ 1)-NOP(I- 1)30 CONTINUE20 CONTINUE

CC WRITE OUT RESULTS FOR NP/N AND PRESSUREC

WRITE(6,1 01)WRITE(6,102) P(1),NOP(1),R(1)DO 202 1=2,402

WRITE(6,1 02) P(I),NOP(I)R(1)202 CONTINUE

CC CALCULATE THE INITIAL OIL-IN-PLACEC

HTOT-H1 +H2CAPAC=H1 'KI +H2K2N=7758 'ACRE'HTOTPHI(l .-SW)/BO(1)WRITE(6,501)

501 FORMAT ('1',3X, THE HISTORY MATCH FOR WELL I')WRITE(6,502)

502 FORMAT(ll,IX,' PRESSURE.PSIA ',2X,'OIL PROD RATE, BPD,1 2X. 'INSTANT GOR, SCF/STB,2X,' TOTAL TIME, YEARS')

CC CALCULATED FLOW RATE AND TIMEC

DO 40 1=2,40NP(I)= NOP(I)'N



00708/BO(l)NO(I)/(ALOG(RE/RW)— 75)(P(I)— PW)1 KRO(I)CAPAC

DT(I)=(NP(I)—NP(I—1))/QO(I)/365.TOT(I)=TOT(I - 1)i- DT(I)IF(MOD(I,2) EQ 0)THEN

CC WRITE OUT PRESSURE. FLOW RATE, iNSTANTANEOUS GOR, AND TIMEC

WRITE(6.503) P(I),QO(l),R(I),TOT(I)503 FORMAT(1X,F18 6,2X.F18 6,2X,F20 6,2X,F16 6)

ENDIF40 CONTINUE

101 FORMAT('T ',5X,'PRESSURE'.1 OX,'NP/N',l OX,'INSTANT GOR')102 FORMAT(8X.F5 0.1 OX.F7.612X.F6 0)

CLOSE(6)STOPEND

With thc data of Figs. 10.1 to 10.5 expressed in equation format, theprogram is now ready to be executed. A listing of the FORTRAN code forthe program is presented in Table 10.1 This version of the program was com-piled and executed on a personal computer using a commercially availableFORTRAN compiler. The program begins by declaring variables and initial-izing some of the data with DATA statements. Next, the program prompts theuser for a name of an input file that contains the constants for the permeabilityrelationships, the zone thicknesses, and absolute permeabilities. An exampleof an input file that, contains the initial data given in the problem statementis listed in Table 10.2. The first data entry in the file, the number 4, refers tothe number of straight-line segments used to describe the original perme-ability ratio data in Fig. 10.5. The next four lines of data contain the lower andupper gas saturation limits and the constants for the four equations that arein the following general format:

log A 0+ A 1 *SG (10 3)

The next several lines of data provide similar information for the relativepermeability to oil curve in Fig. 10.4. Seven straight-line segments are used todescribe this curve, and the saturation and constant values are then listed Thegeneral format for the equations representing the relative permeability to oildata is

KJO= C0+ C1*SG (10.4)


TABLE 10.2.Example of data input file for history matching problem

40000 0430 —4.7981 54.65200430 .0810 —3.8450 32.01 700810 1090 —2.6308 16.98901090 3500 —2.1119 122410

7.0000 .0156 1.0000 —8.24000156 .0450 1.0980 —14.4000.0450 .0688 .7500 —661330688 1008 5427 —3.54671008 .1520 .3583 —1 73331520 .2088 .2215 —.8300.2088 .3325 .1262 —.3760

17.0 14096 7.2

The last two lines of input represent the thicknesses and the absolutepermeabilities, respectively.

Returning to the main program, the program now opens up an outputfile called MATCH. Fluid properties are then calculated as a function ofsure, beginning at the initial pressure of 620 psia and at subsequent pressuresdetermined by deincremcnting the pressure by 5 psi. Once the fluid propertiesarc calculated, the program is ready to begin the Scbilthuis material balancecalculation. After the material balance calculation has generated fractional oilrecovery and instantaneous GOR values, these are written to the output file.An example of the output file is shown in Table 10.3 The initial oil in placeis then calculated so that oil production values can be determined from thefractional oil recovery numbers Oil production rates and corresponding timesare then calculated, and the average pressures, oil production rates,neous GOR values, and times are written to the output file. Programcution is terminated at this point

3.2. Discussion of History Matching Results

When the program is executed using the original data given in the problemstatement as input, the oil production rate and R, or instantaneous GOR,values obtained result in the plots shown in Fig 10 7. The oil production rateis plotted in Figure 10 7(a) and compared with the actual data. Notice that thecalculated rates begin higher than the actual rates and decrease faster withtime or with a greater slope. The calculated instantaneous GOR values arecompared with the actual 00K values in Fig 10.7(b) and found to be muchlower than the actual values


TABLE 10.3.Example of output file for history matching problem

PRESSURE NP/N INSTANT GOR620 .000000 198615 .003574 198.605. .010883 196.595. .018410 195585 026131 196575 033986 200565. 041858 209.555. .049524 226545. 056676 264535 063061 311525. 068856 353.515. 074112 405505. 078835 465495 083060 535.485 .086836 611.

475 .090217 695465. .093254 786455. 095994 882.445 098478 983435. .100740 1089425 .102811 1200.

THE HISTORY MATCH FOR WELL 1

PRESSURE, OiL PROD INSTANT. GOR, TOTAL TIME,PSIA RATE, BPD SCF/STB YEARS

615000000 81.731461 197 584778 .139468605.000000 76.426636 196124374 439406595.000000 71.218536 195383209 .770589585 000000 65003090 196.084351 1139370575.000000 57519310 199.663467 1.561816565 000000 50337765 208.837708 2 043982555 000000 43595257 228.218506 2.584777545.000000 37472816 264.157867 3.170556535 000000 32096237 310 664001 3780574525 000000 29040087 353.154480 4403254515.000000 26670746 404.841736 5018327505.000000 24536453 465.493835 5619408495.000000 22.615231 534 584167 6203073485.000000 20882790 611 443237 6768256475 000000 19315432 695 369751 7315524465 000000 17891640 785 692627 7.846456455 000000 16592621 881.798767 8.363103445000000 15842712 983.164307 8.856463435000000 15035938 1089 271851 9 330001425 000000 14276489 1199706299 9.786566


90-

___--

80 .70-

z 60—a

5°

30a

20 - -__Actual GOR

10 — •—Calculatcd GOR ---_______ -

0 01 2 3 4

TIMF(YEARS)

(a) Oil production rate

I-700 0 Actual 0

S Calculated 0600 CI—

- -___

___-

co(n —- --

• •—.—100- ——-— —-- --

0-0 1 2 3 4

TIME(YEARS)

(b) Instantaneous GOR

Fig. 10.7. 1 match using onginal data.

At this point. it is necessary to ask how the calculated instantaneousGOR values could be raised in order for them to match the actual values. Anexamination of Eq (9.33) suggests that R is a function of fluid property dataand the ratio of gas-to-oil permcabilities. To calculate higher values for R.either the fluid property data or the permeability ratio data must be modified.


Because fluid property data are readily and accurately obtained and the per-meabilitics could change significantly in the rcscrvoir owing to diffcrent rockenvironments, it seems justified to modify the permeability ratio data. It isoften the case when conducting a history match, an engineer will find differ-ences between laboratory measured permeability ratios and field measuredpermeability ratios. Mueller, Warren, and West showed that one of the mainreasons for the discrepancy between laboratory k8(k0 values and field-mea-sured values can be explained by the unequal stagcs of depletion in the res-ervoir.5 For the same reason, field instantaneous GOR valuesseldom show theslight decline predicted in the early stages of depiction and conversely usuallyshow a rise in gas-oil ratio at an earlier stage of depletion than the prediction.Whereas the theoretical predictions assumc a negligible (actually zero) pres-sure drawdown, so that the saturations are therefore uniform thoughout thereservoir, actual well pressure drawdowns will deplete the reservoir in the vi-cinity of the welibore in advance of areas further removed. In developmentprograms, too, some wells are often completed years before other wells, anddepletion is naturally further advanced in the area of the older wells, whichwill have gas-oil ratios considerably higher than the newer wells. And evenwhen all wells are completed within a short period, when the formation thick-ness varies and all wells produce at the same rate, the reservoir will be de-pleted faster when the formation is thinner. Finally, when the reservoir com-prises two or more strata of different specific permeabilities, even if theirrelative permeability characteristics are the same, the strata with higher per-meabilitics will be depleted before those with lower permeabilitics. Since allof thesc effects arc minimized in high-capacity formations, closer agreementbetween field and laboratory data can he expected for higher capacity for-mations On the other hand, high capacity formations tend to favor gravitysegregation. When gravity segregation occurs and advantage is taken of it byshutting in the high-ratio wells or working over wells to reduce their ratios, thefield-measured kg/k0 values will be lower than the laboratory values. Thus thelaboratory kg/ko values may apply at every point in a reservoir without gravitysegregation, and yet the field k8 1k0 values will be higher owing to the unequaldepletion of the various portions of the reservoir

'l'he following procedure is used to generate new permeability ratiovalues from the actual production data.

I. Plot the actual R values versus time and determine a relationship betweenR and time

2. Choose a pressure and determine the fluid property data at that pressure.From the chosen pressure and the output data in Table 10.3. find the timethat corresponds with the chosen prcssurc.

3. From the relationship found in Step 1, calculate R for the time found inStep 2


4. With the value of R found in Step 3 and the fluiG property data found inStep 2, rearrange Eq. (9.33) and calculatc a value of the permeability ratio.

S. From the pressure chosen in Step 2 and from the values in the outputdata in Table 10.3, calculate the value of the gas saturation that corre-sponds with the calculated value of the permeability ratio.

6. Repeat Steps 2 through 5 for several pressures. The result will be a new per-meability ratio—gas saturation relationship.

You should rcalize that in Steps 2 and 5 the original permeability ratio wasused to generate the data of Table 10.3. This suggests that the new perme-ability ratio-gas saturation relationship could he in error because it is basedon the data of Table 10.3 and that to have a more correct relationship, it mighthe necessary to repeat the procedure. The quality of the history match ob-tained with the new permeability ratio values will dictate whether this iterativeprocedure should he used in generating the new permeability ratio-gas satu-ration relationship. The new permeability ratios determined from the fore-going six-step procedure are plotted with the original permeability ratios inFig 10.8.

It is now necessary to regress the new permeability ratio—gas saturationrelationship and input the new data into the program before the program canbe executed again to obtain a new history match. When this is done, the pro-gram yields the results plotted in Fig. 10.9.

The new permeability ratio data has significantly improved the match ofthe instantaneous GOR values, as can be seen in Fig. 10.9(b) However, theoil production rates are still not a good match. In fact, the new permeabilityratio data have yielded a steeper slope for the calculated oil rates, as shownin Fig. 10.9(a), than what is observed in Fig. 10.7(a) from the original data.A look at the calculation scheme helps to understand the effect of the newpermeability ratio data.

Because the new values of instantaneous GOR were calculated with thenew permeability ratio data, which in turn were determined by using Eq.(9 33) and the actual GOR values, it should be expected that the calculated(JOR values will match the actual GOR values. The flow rate calculation,which involves Eq. (7.50), does not use the permeability ratio, so the mag-nitude of the flow rates would not be expected to he affected by the newpermeability ratio data however, the time calculation does involve whichis a function of the permeability ratio in the Schilthuis material balance calcu-lation. Therefore, the rate at which the flow rates decline will be altered withthe new permeability ratio data.

To obtain a more accurate match of oil production rates, it is necessaryto modify additional data. This raises the question: What other data can bejustifiably changed? We have argued that it was not justifiable to modify thefluid property data However, the fluid property data and/or equations shouldbe carefully checked for possible errors. In this case, the equations were

010'

10'

10

ozGAS SATURATLON, FRACTION PORE SPACE

Fin. 10.8. Eircl -

History Matching

-

80. -

70-—— •— —— -- —. --

60 —:-i_'—5:

a1— 40 .-j0

20' —- - —1 '

CalculatedQ

0---—-- --0 I 2 3 4

TIMECYEARS)


--

- DActualGOR• Calculated GOR

600 - - -—a--.-o —

SI-400 .

(ncn S300--

S200

100-—

0-—---0 1 2

TIME(YEARS)


Fig. 10.9. 1 listory match after modifying permeability ratio values.

checked by calculating values of B0, A1, and p.1 at several pressuresand comparing them with the original data. The fluid property equations werefound to be correct and accurate. Other assumed reservoir properties thatcould be in error include the 'one thicknesses and absolute permeabilities.


The thicknesses are determined from logging and coring operations fromwhich an isopach map is created. Absolute permeabilities are measured froma small sample of a core taken from a limited number of locations in the reser-voir The number of coring locations is limited largely because of the costsinvolved in performing the coring operations. Although the actual measure-ment of both the thickness and permeability from coring material is highlyaccurate, errors are introduced when one tries to extrapolate the measuredinformation to the entire drainage area of a particular well. For instance, whenconstructing the isopach map for the zone thickness, you need to make as-sumptions regarding the continuity of the zone in between coring locationsThese assumptions may or may not be correct Because of the possible errorsintroduced in determining average values for the thickness and permeabilityfor the well drainage area, varying these parameters and observing the effectof out history match IS justified. In the remainder of this section, the effect ofchanging these parameters on the history matching process is examined. Table10.4 the cases that are discussed.

TABLE 10.4.Description of cases

case Number Puramezer Varied From Original Data

I None2 Pcrmcabulity ratio3 Same as Case 2 plus zone thickness4 Same as Case 2 plus absolute permeability5 Same as Case 2 pIus iune thickness and absolute

permeability6 A second iteration on the permeability ratio data plus

ione thickness and absolute permeability

In Case 3, the thicknesses of both £OflCS were adjusted to determine theeffect on the history match Since the calculated flow rates arc higher than theactual (low rates, the thicknesses were reduced. Figurc 10.10 shows the effecton oil producing rate and instantaneous GOR when the thicknesses are re-duced by about 2.0%.

By reducing the thicknesses, the calculated oil production rates areshifted downward, as shown in Fig. 10.10(a). This yields a good match with theearly data but not with the later data because the calculated values decline ata much more rapid rate than do the actual data. The calculated instantaneousGOR values still closely match the actual GOR values. These observations canbe supported by noting that the zone thickness enters into the calculationschemc in two places. One is in the calculation for N, the initial oil in place,which is performed by using EEq. (10.2). Then N is multiplied by each of the

values determined in the Schilthuis balance. The second place thethickness is used is in the flow equation. Eq. (7.50), which is used to calculate


/0 ———

_________

—________

60

S

30- --——-- .-QCfl-j0 20-

10Actual GOR -

. Calculated GOR

0 I

0 1 2 3 4

TIME(YEARS)


700 Actual 0 —

S Calculated 0600- --—-

Q_500_ S

-___-S

U) Cl) - -.CD

S

200 •

100

0-0 1 2 3 4

TIME(YEARS)

(b) Instantaneous (JOR

Fig. 10.10. History match of Case 3 Case 3 used the permeability ratio data andreduced zone thicknesses

q0. The instantaneous GOR values are not affected because neither the calcu-lation for N nor the calculation for q0 is used in the calculation for instantane-ous GOR or R However, the oil flow rate is directly proportional to thethickness, so as the thickness is reduced, the flow rate is also reduced At firstglanceS it appears that the decline rate of the flow rate would be altered But

3. An Example of a History Match 413

upon further study, it is found that although the flow rate is obviously a

function of the thickness, the time is not. To calculate the time, an incrementalis divided by the flow rate corresponding with that incremental produc-

tion. Since both and q0 arc directly proportional to the thickness, the thick-ness cancels out, thereby making the time independent of the thickness Insummary, the net result of reducing the thickness is as follows: (1) the mag-nitude of the oil flow rate is reduced, (2) the slopes of the oil production andinstantaneous GOR curves are not altered, and (3) the instantaneous GORvalues are not altered.

To determine the effect on the history match of varying the absolute per-meabilities, the permeabilities were reduced by about 20% in Case 4. Figure10.11 shows the oil production rates and the instantaneous GOR. plots for thisnew case. The quality of the match of oil production rates has improved, butthe quality of the match of the instantaneous GOR values has decreased.Again, if the equations involved arc examined, an understanding of howchanging the absolute permeabilities have affected the history match can beobtained.

Equation (7.50) suggests that the oil flow rate is directly proportional tothe effective permeability to oil, k0. Equation (10.5)

k0=k,0k (10.5)

shows the relationship between the effective permeability to oil and the abso-lute permeability, k. Combining Eqs (7 50) and (10.5), it can be seen that theoil flow rate is directly proportional to the absolute permeability. Therefore,when the absolute permeability is reduced, the oil flow rate is also reduced.Since the time values are a function of q0. the time values are also affected.The magnitude of the instantaneous GOR values arc not a function of the ab-solute permeability since neither the effective nor the absolutc pcrmeabilitiesare used in the Schilthuis material balance calculation. However, the timevalues are modified, so the slope of both the oil production rate and theinstantaneous GOR curves arc altered. This is exactly what should happen inorder to obtain a better history match of the oil production values. However,although it has improved the oil production history match, the instantaneousGOR match has been made worse. By reducing the absolute pcrmeahilitics,it has been found that (1) the magnitude of the oil flow rates are reduced, (2)the magnitude of the instantaneous GOR va(ues ate not changed, and (3) theslopes of both the oil production and instantaneous GOR curves are altered.

By modifying the zone thicknesses and absolute permeabilities, the mag-nitude of the oil flow rates and the slope of the oil flow rate curve can be modi-fied. Also, while adjusting the oiJ flow rate, slight changes in the slope of theinstantaneous GOR curve arc obtained. In Case 5, both the zone thicknessand the absolute permeability are changed in addition to using the new perme-ability ratio data. Figure 10.12 contains the history match for Case 5. As canbe seen in Figure 10.12(a), the calculated oil flow rates are an excellent match


70

a1— 30

-jO 20

10Acli.aI 0

• Calculated 0

0 I

O 1 2 3 4

1 IME(YEARS)


700• ..flActualGOR• Calculated GOR

Q ——

____--

500 -----.__-__

__—

So 300 —-——I

.200

100 - -

___-

0 -—

0 1 2 3 4

TIME(YEARS)

(h) Instantaneous

Fig. 10.11. History match of Case 4 Case 4 used the new permeability ratio data andreduced absolute pcrmcabilities

to the actual field oil production values The match of instantaneous GORvalues has worsened from Cases 2 to 4 but is still much improved over thematch in Case 1, which was obtained by using the original permeability ratiodata.

We have said that a second iteration of the permeability ratio values maybe necessary, depending on the quality of the final history match that is ob-tained. This is because the procedure used to obtain the new permeability

3 An Example of a History Match


4.—

(b) Instantaneous 00k

415

Fig. 10.12. History match of Case 5 Case 5 used the new permeability ratio data andmodified ione thicknesses and absolute permeabilities

ratio data involves using the old permeability ratio data. The calculated instan-taneous GOR values do not match the actual field GOR values very well, soa second iteration of the permeability ratio values is warranted. Following theprocedure of obtaining new permeability ratio data in conjunction with theresults of Case 5, a second set of new permeability ratios is obtained. Thissecond set is plotted in Fig. 10.13 along with the original data and the first setused in Cases 2 to 5

70

z0U4

cr1-Q.U)-J0

30--—-——- —

- Actual Q• Calculated Q

.

0

20

10

2 4

TIME(YEARS)

800

700

600

500

400-

0cr1-.jce

5U) U)4—0

U Actual GOR• Calculated GOR

300 ——a)- Do. S

.200

100

0-

.

0 1 2 3 4

1 IME(YEARS)

S • —. —— —

O( 010 015 0.25 030 035

1r

= ==

-- Otiginal Data

- First IterationI I I

• Second Iteration

H H - H

i i Ii I' I i0 005 010 015 020 025 0.30 035

GAS SATURATION, FRACTION PORE SPACE

3. An Example of a History Match 417

By using the permeability ratio data from the second iteration and byadjusting the zone thicknesses and absolute permeabilities as needed, theresults shown in Fig. 10.14 are obtained. It can be sccn that the quality of thehistory match for both the oil production rate and the instantaneous GORvalues is very good. When a history match is obtained that matches both the


200

100

0

0

U)C1)4—


Fig. 10.14. history match of Case 6 Case 6 used permeability ratio data from a seconditeration and modified zone thicknesses and absolute perrncabilitics.

z0

cxI-a-U)-J0

11_U

:

--

:10 ActualO -

• Calculated Q

--0

0 1 2 3

TIME(YEARS)

4

800

700

600

500

400

0 1 2 3 4

TIME(YEARS)

History Matching

oil production and instantaneous GOR curves this well, the model can be usedwith confidence to predict future production information.

3.3. Summary Comments Concerning HistoryMatching Example

Now that a model has been obtained that matches the available productiondata and that can be used to predict future oil and gas production rates, anassessment of the modifications to the data that were performed during thehistory match should be made. Table 10.5 contains information concerning thedata that were varied in the six cases we have discussed.

TABLE 10.5.Input data for history matching example

CasePermeabilityRatio Data

AbsolutePermeability, md

ZoThickn

neess, ft-

Zone 1 Zone 2 Zone F Zone 2

1 Original data 9 6 7.2 17.0 14 02 First iteration 9.6 7.2 17.0 14.03 First iteration 9.6 72 13.6 11 24 First iteration 7.7 5 7 17 0 14.05 First iteration 5 9 4 4 22.0 18.26 Second iteration 5 9 44 22.0 18.2

All other input data were held constant at the original values for all sixcases The first and second iterations of the gas-to-oil permeability ratio dataled to values higher than the original data, as can be seen in Fig. 10.13. In thelast section, several reasons for a discrepancy between Laboratory-measuredpermeability ratio values and field-measured values were discussed. Thesereasons included greater drawdown in areas closer to the weilbore than inareas some distance away from the welibore, completion and placing on pro-duction of some wells before others, two or more strata of varying perme-abilities, and gravity drainage effects. All these phenomena lead to unequalstages of depletion within the reservoir. The unequal stages of depletion causevarying saturations throughout the reservoir and hence varying effective per-meabilities. The data plotted in Fig. 10.13 suggest that the discrepancy be-tween the laboratory-measured permeability ratio values and the field-determined values is not great and that the use of the modified permeabilityratio values in the matching process is justified.

For the final match in Case 6, thc zone thicknesses were increased byabout 30% over the original data of Case 1, and the absolute permeabilitieswere reduced by about 39%. At first glance, these modifications in zone thick-ness and absolute permeability might seem excessive. however, rememberthat the values of zone thickness and absolute permeability were determined

Problems 419

in the laboratory on a core sample. approximately 6 in. in diameter Althoughthe techniques used in the laboratory are very accurate in the actual measure-mcnt of these parameters, to perform the history match, it was necessary toassume that the measured values would be used as avcrage values over the en-tire 40 ac of the drainage area of the well This is a very large extrapolation,and there could be significant error in this assumption. if the small magnitudesof the original values arc considered, the adjustments made during the historymatching process arc not large in magnitude. The adjustments were only 2.8to 3.7 md and 4.2 to 5 0 ft. These numbers are large relative to the initialvalues hut are certainly not large in magnitude.

We can conclude that the model developed to perform the history matchfor the well in question is reasonable and defendable. More sophisticatedequations could have been developed, but for this particular example theSchilthuis material balance coupled with a flow equation were quite adequate.As long as the simple approach meets the objectives, there is great merit inkeeping things simple. However, you should realize that the principles we havediscussed about history matching are applicable no matter what degree ofmodel sophistication is used

PROBLEMS

10.1 The following data arc taken from a volumetric. undersaturated reservoir Cal-culate the relative permeability ratio k, at each pressure, and plot versus totalliquid saturation.

Connate water, S,, = 25%

Initial oil in place 150 MMS1B

B0— 1.552 bbl/STB

ppsia

RSCF/S1B MMS1B

B0bbl/STB bbl/SCF

R,,SCF/STB

4000 903 3 75 1 500 0.000796 820 3113500 1410 13 50 1 430 0 000857 660 37.13000 2230 20.70 1 385 0 000930 580 42.52500 3162 2700 1348 000115 520 5082000 3620 32.30 1.310 0.00145 450 61.21500 3990 37.50 1 272 0.00216 - 380 77.3

10.2 Discuss the effect of the following on the relative permeability ratios calculatedfrom production data

(a) Error in the calculated value of initial oil in place.

(b) Error in the value of the connate water


(c) Effect of a small but unaccounted for water dnvc.

(d) Effect of gravitational segregation both where the high gas-oil ratio wells areshut in and where they arc not.

(e) Unequal reservoir depletion

(f) Presence of a gas cap.

10.3 For the data that follow and that are given in Figs 10.15 to 10.17 and the fluidproperty data presented in the chapter, perform a history match on the produc-tion data in Figures 10.18 to 10 21, using the computer program in Table 10 1Use the new permeability ratio data plottcd in Figures 10 8 and 10 13 to fine-tune the match.

Laboratory core permeability measurements

ability to Air(md)Average Absolute PermeWell Zonel Zone2

5—6 5.1 4.08—16 8.3 689—13 11.1 60

14-12 8.1 76

12 11 10- 500 .....—..--—————-— a • . . .

a-¶4

2

"1

-600

13

/a

': . . S S

$13 III $5

I •'3 SI. 9.15 —

. a • . S .

24 23

Fig. 10.15. Structural map of well locations for Problem 10 3.

Fig. 10.17. Isopach map of Zonc 2 for Problem 10 3

13

24

13

24

History Matching

- - - - — 1200

1000

I-U)

U-C-)(I)

• -400

- -—— • ZOO

422

60

50

40

3°

20

10

HOIL PRODUCTIONx 1GOFI

2 3 4 5 6 7 8

lIME. YEARS

Fig. 10.18. Actual oil production and instantaneous GOR for Wells 5—6 for Prob-lem 10.3.

1200

I—800 w

ILUCl)

Wv 0

400

200

1 2 4

TIME, YEARS

Fig. 10.19. Actual oil production and instantaneous GOR for Wells 8—16 for Prob-lem 10 3.

Problems 423

70' 1200

60 1000

800

40

30' 400

20GOR

.

OIL PRODUC1 ION

1 2 3 45678TIME. YEARS

Fig. 10.20. Actual oil production and instantaneous GOR for Wells 9—13 for Prob-lem 10.3.

60 OIL PRODUCTION -— 1200

GOR

50-- -----—————-----———-—-1000

1-———------ -- --—800I- U-U,

1 2 3 4 5 6 7 8

TIME. YEARS

Fig. 10.21. Actual oil production and instantaneous GOR for Wells 14—12 for Prob-1cm 10 3.


10.4 Write a computer program that uses the Muskat method discussed in Chapter9 in place of the Schilthuis method used in Chapter 10 to perform the historymatch on the data in Chapter 10

10.5 Write a computer program that uses the Tamer method discussed in Chapter 9in place of the Schilthuis method used in Chapter 10 to perform the historymatch on the data in Chapter 10

REFERENCES

'A. W. McCray, Petroleum Evaluations and Economic Decisions Englewood Cliffs,NJ: Prentice Hall, 1975.

211 B. Crichiow, Modern Reservoir Engineering—A Simulation Approach EnglewoodCliffs, NJ: Prentice Hall, 1977.

'P. H. Yang and A. T. Watson, "Automatic History Matching With Variable-MetricMethods," Society of Petroleum Engineering Reservoir Engineering Jour., (August1988), 995.

4Pcrsonal contact with D. W. Green

'T. 1) Mueller, 3. E. Warren, and W. J. West, "Analysis of Reservoir PerformanceCurves and a Laboratory K,IK. Curve Measured on a Core Sample," Trans(1955), 204, 128.

Index

Abon-Kassern, see DranchukAgarwai, Al-Hussainy, and Ramcy, 82Al-liussamy and Rantey, 241AI.llussainy, Ramey, and Crawford, 240Allard and Chen, 299, 300Allen and Roe, 130, 131, 132Allen, 107Amencan Society for Tcaiing and Materials

(ASTM), 12Anschutz Ranch East Unit, 140Aquifcr, 4

definition of, 273Arcaro and Bauiouni, 83Arps, 151 13Ashman, see JogiAziz, see Mauar

Bacon Lime Zone, 130, 134Barnes, see FancherBassiouni, see ArcaroBedding planes

bottom-water drive, 150, 152. See alsoVolumetric oil reseivoirs

edge-water drive, 150.5cc aLsoVohimctnc oil reservoirs

Beggs and Robinson, 41Bcggs, 31 See also VasquezBdll gas ficld, 16, 22, 24, 78, 79, 81Bernard, 97, 98Berry, see JacobyBierwang field, 83

Big Sandy reservoir, 32, 53, 34, 36, 52BlackwclI, see RichardsonBlasingarne, 27, 28Bobrowaki, see CookBorshcel, see SiunhaBotset, see WycoffBottom-hole pressure gauge, 2Bourgoyne, Hawkins, Lavaquasl, and

Wickcnhauser, 98Boyd, see McCarthyBrady, see LutesBrar, see MattarBruce and Welge, 73Bniakotter, see RusselBubble-point prcssurc, 2, 8. 9, 32, 127 See

also Saturated oil reservoirs, Volumetric oilreservoirs, Single-phase fluid flow

Buckley and Leverett, 3, 347, 351.5cc alsoDisplacement, oil and gas

Buckley, 153 See also Craze, TamerBurrows, see Carr

C

Calculation (initial), gas and oil, 110Callaway, see StewartCalvin, see KleinsteibcrCanyon Reef reservoir (Kelly-Snyder field,

Texas), 158Carpenter, Schroeder, and Cook, 194Carr, Kobayashi, and Burrows, 29, 30Caudle, see SlobodChatas, 298Chen, see AllardChiang, see Lutes

425

426 index

Chin, see CookClai* and Wesseiy, 4Coats, 299, 300Coleman, Wdde, and Moore, 3Connate water saturation, 2, 3Connate water, 73Conroc Field (rexas), 187-188, 190-193,

205, 271-278, 329Cook, Spencer, and Bobrowski, 201Cook, Spencer, Bobrowaki, and Chin, 148,

198Core Laboratories, Inc, 147, 197, 198, 199Crawford, see AJ-HussainyCraze and Buckley, 151, 152Crrcondentherm, 6, 7, 9, i 18Crude oil properties, 31

formation volume factor, 34isothermal compressibility, 38solution gas-oil ratio, 32, 125viscosity, 40

Dirty's equation, 232-233, 235Dirty's law, 211, 221, 276Daxy, 2, 210Darcy, as unit of measure, 212

niillidarcy, 212. 229, 230Davis, seeDelta pressure, 96Dew-point reservoir, 9, 118, 127, 128Displacement efficiency, 135Displacement, oil and gas, 335

Ruckley-Levereit displacemantmechanism, 347, 351

drag zone, 352flood front, 352oil bank, 352

enhanced oil recovery processes (EOR),380

in oil-wet system, 380alkaline flooding, 383-384capillary number, 381chcrntcai injection processes, 383dynamic miscible process, 383forward dry combustion process, 385forward wet combustion process, 386hot Water or steam stimulation

process, 385in situ combustion, 385in water-Wet system, 380micellar-polymer flooding. 383-384miscible injection processes. 382multiple Contact miscible process,

383polymer flooding, 383-384residual oil, mobilization of, 380single contact miscible process, 383thermal processes. 384

flooding techniques, 335external, 335internal dIsplacement, 335

immiscible displacement processes, 347frontal advance curve, 351

macroscopic displacement efficiency, 343anisotropy of hydro-carbon-bcarmg

formation, effect on, 343areal sweep efficiency, 344, 346heterogeneities of hydro-carbcn-

bearing formation, effect on, 343,341 21

injection well, change to, 347limestome formations, 347pressure maintenance, 344sandstone formations, 347secondary recovery, 344vertical sweep efficiency, 344viscous fingenng, 345-346

microscopic displacement efficiency, 336absolute pcmieabdity, 337-342capillary pressure. 336critical saturation, 339fractional flow curve, 343hydrocarbon saturation, 340intcrfacial tensions between fluids,

336relative permeability, 337-342residual saturation, 341transition zone, 342wettability, 336

oil recovery by internal gas drive, 360bubble point, 360gas-oil ratios, 360, 364, 369iteration techniques, 368secant method, 368, 372, 374

recovery efficiency, 335waterfloodzng, 315

direct-line-dnve, 377discovery of method, 375factors in, 376pattern flooding, 377peripheral flooding, 377placcment of wells, 377recovery efficiency, estimating, 378staggered-line-drive, 377

Displacement, oil by gas, 353downdip oil, 360gravity segregation, 360oil wet rock, 360updip ('attic") oil, 360water wet rock, 360

Distillate, 107Downdip water wells, 8Dranchuk and Abou-Kassem, 19, 20, 26Dry gas reservoir, 69, 88Dry gas, 7 5cc aLso Lean gas

Index 427

Eakin, see LeeEarlougher, 244-245, 249Earlougber, Matthews, and Russdll, 253East Tc,ias field, 64Echo Lake field, 100, 101Economics, in relation to gas, 95Egbogah, 40Ederta, 21.See also MuskatElk Basin field (Wyoming and Montana).

148Elk City field (Oklahoma), 201Ellenburger formation (West Texas), 274Equthbnum ratios, 125, 126, 127, 128, 131,

148

Pancher, Lewis, and Bamcs, 2larshad, see Ramagostlatt and Davis, 219Ietkovicb, 3, 325-328, 332, 333Flash proccss, 133 See also Saturated oil

reservoiisFluid flow, single phase, see Single-phase

fluid flowfluids, I

Formation volume factor and density, 23FORTRAN, 399, 403Free gas volune, 60, 66

Gas and oil (initial) calculation, 110Gas cap, 4Gas comprcssibtlity factor, 24Gas deviation factor, 114Gas displacement. 335Gas properties, 12

gas deviation factor, 15ideal gas law, 12ma] gas law, 14

supercompressibility factor, 14specific gravity, 13, 14

two-phase, 129Gas reservoirs

abnormally pressurcd, 97as storage reservoirs, 95

Gas reservoirs, single-phase, 69Gas volume factors, 23Gas-condensate reservoirs, 107, 128, 136Gas-condensate, 4Gas-distillate, 4Gas-oil ratios, see Equilibnum ratios,

Volumetnc gas reservoirs, Voluinciric oilreservoirs, Saturated oil rcscrvoirs

Gashydrates, IGecrtsma, 11

Geffen, Parish, Haynes. and Morse, 80Gcneral mstenal balance cquation, see

Material balance equationGladfelter, see StewartGlen Rose Formation, 130Gloyd-Mitchell Zone (Rodessa field), 163

average monthly production data, 166dcvelopment, production, and reservoir

pressure curves, 165gas capansion, 163liquid expansion, 163production history versus

cumulative produocd oil, 168time, 168

solution gas-dnve reservoir, 203, 204Gold, McCa.n, and Jennings. 93, 94Gonzalez, see LeeGoodrich, see RusselGray, see Jogi

H

Hall, 170Harrison, see Rodgers}lazvillc arid Hawkins, 97Haviena and Odeh, 56, 191, 192. 193, 208,

275. See also OdehHawkins, see Bourgoyne, HarvdleHaynes, see GeffenHigh crudes, IHistory matching, 391

definition of, 391development of model, 392

mcorporating flow equation, 393matenal part of model, 392

discussion of results, 404example of, 394gas production rate, 418gas-oil ratios, 407, 418introduction to, 391muLtidimcnsional, multiflow reservoir

simulators, 391simple decline curve analysis, 391oil production rate, 417, 418permeability ratio, 408, 413-415, 417,

420permeability ratio gas saturation

relationship, 408solution procedure, 396

fluid property data, conversion of,396

summary, 418Ilolden field, 103Holland, see SiunhalIollis, 96Hurst, 279, 280Hydraulic gradients, 211, 213

428 . Index

Tces, 1Ikoku, 95Improved zecovcry, 6Initial unit reserve, 77Interstitial water, 73ira Rinehart': Yearbook,, 109, 110Isopachous map, 70, 72, 87Isothermal compressibility, 24

Jackson, see MatthcsJacoby and Berry, 201Jacoby, Koefler, and Beriy, 127Jennings, see GoldJogi, Gray, Ashman, and Thompson, 98Jones sand, 73, 75Juan County, Utah, 134

Katz and Tek, 9Katz, see Matthews, StandingKaveler, 74Kelly-Snyder field, Texas (Canyon Reef

reservoir), 158, 159, 161, 162, 178Kennedy, see WielandKenneyly and Reudelhuber, 147Kennerly, 147, 197, 198, 199Kern, 359)Cleinstelber, Wendschlag, and Calvin, 140Koeller, see JacobyKovayashi, see Cur

Laminar flow, 211and Darcy's law, 211

LaSalle OL1 field, SiLavaguad, see BourgoyneLean gas, 108, 138

cycling, 134-136, 139Lce, 253Lee, Gonzalez and Eakin, 30[everett and Lewis, 2Leverett, 3. See also BuckleyLewis, see Fancher, [everettLiquificd petroleum gas (LPG), 120Louisiana Gulf Coast Eugene Island Block

305 reservoir, 83Lutcs, Chiang, Brady, and Rossen, 83

M Sand reservoir, 101, 102Marudiak, see MatthesMaterial balance equation, 56

denvation, 56Havlena-Odeh mcthod, 66

Matenal balance method, 83uses and lunitations, 64

Manar, Brat, and Aziz, 26-27Matthes, Jackson, Schuler, and Marudiak, 83Matthews and Russell, 234, 235Matthews, Roland, and Katz, 115, 117Matthews, see EarlougherMcCain, 45, 46, 47, 48. See also GoldMcCarthy, Boyd, and Reid, 93McCord, 148McMahon, see van EvcrdingcnMercury, 119Mdc Six pool (Peru), 204, 355, 356, 357,

358, 359, 361Millikan and Sidweil, 2Moore and Truby, 274Moore, see ColemanMorsc, see GeffenMoscnp, see WoodyMueller, Warren, and West, 407Muskat, 3, 362, 370, 374, 424 See also

WycoifMusket, Standing, Thornton, and Eulerts, 110

N

Net isopachous map, see I.sopachous mapNewman, 11Newton-Raphson, 20Nitrogen, for pressure maintenance, 139North Sea gas field, see Rough gas fieldNorth Snyder, 362

0

Odeh and Havlena, 3Odeh, see HavienaOil and Gas Journal, The, 159Oil displacement, 335Oil volume, 60Oil zone, 5

p

P Sand reservoir, 103Paradox limestone formation, 133, 134Paraffin, IPansh, see GeffenPeona ficld, 329Perry, see RussclPetro-physics, 2Petroleum reservoirs

production from, 4undersaturated, 146

Pirson, 63, 187Poiseuille's law, 225Pore volume compressibility, Ii

Index 429

Piessure maintenance program, 5Pressure maintenance, using nitrogen, 139Pressure transient testing, 253

buildup testing, 259Homer plot, 260, 262pseudosteady-state time region in, 259shut-bi pressure, 260skm factor in, 260supeiposiuon, use of1 259

drawdown testing. 253damage xone in, 255pseudosteady-state period in, 259skin factor in, 255-256, 257transient time period in, 259

Pressure-temperature phase, 6Pressure-volume-temperature relations, 2Production

primary production (hydrocarbons), 5secondary recovery operation, Stertiary recovery processes, 5

Productivity index (P1). 246mjectivity index, 247

as used with salt water disposal andinjection wells, 247

productivity ratio (PR), 248Pseudosteady flow systems, 242

radial flowcompressible fluids, 246slightly ccanprcssible fluids, 242

wrcell, 120, 141, 138, 144, 179, 180, 197,

202. 205, 278properties, 178, 201studies, 112, 200

Quantities of gas liberated, 2

R Sand reservoir, 179Radial diffemenriation equation, development

of, 231Ramagost and Farshid, 98Ranrey, see Agarwal, Al-llussainy,

WancnbargcrRangely Field (Colorado), 148Recoverable gross gas, 127Redlich-Kwong, equation of stale, 140Reed, see Wycoft'Regier, see RodgersReid, see McCarthyReserves, SReservoir

definition, 4Reservoir engineenng

definition, 4history, I

Reservoir flow systems, classification, 213geometries

linear, 215radial, 215spherical, 215

phasessingle vs. muki, 215

tune dependencelate transient, 217pseudosteady, 217steady-state, 215. See also Steady-

state flow systemstransient, 216

type of fluid, 233compressible, 214incompressible, 213-214slightly compressible, 214

Reservoir mathematical modeling, 3Reservoir prcssure. 2Reservoir simulation, 3Reservoir types defined, 6Reservoir water properties, 45

formation volume factor, 45isothermal compressibility, 46solution gas-water ratio, 45viscosity, 47

Residual gas saturation, 80Resouive (hydrocarbons), 5Retrograde condensation, 7, 120, 133, 135,

139Retrograde liquid

flow, 133saturation, 119

Reudelhuber and I finds, 201Rcudelhubcr, see 1(errnerlyRichardson and Blaekwell, 351Robinson, see BeggsRock collapse theory, 97Rock properties, 9

fluid saturations, 11isothermal compressibility, 10porosity, 9

Rodessa field, see Cloyd-Mitchcll /LoncRodgers, Harrison, and Regier, 125, 127,

133, 134Roe, see AllenRoland, see MatthewsRossen, see LutesRough gas field, 96Russel, Goodrich, Perry, and Bruskottcr, 240Russell, see F.arloughecRussell, see Matthews

S

Sabine gas field, 102Sabine oil field, 50Saturated oil reservoirs

bubble-point pressure, 172-177, 195, 20!

430 Index

differential vaporization and separatortests, 200

formation volume factor, 200gas liberation process

differcntial, 194-195, 199, 200flash, 194-195. 199, 202

gas-oil ratios, 197, 198, 200. 201, 204introduction to, 184

continuous uniform formations, 185gas cap, 185, 186, 191expansion, 191gravitational segregation

characteristics, 185hydraulic control, 185

matcrial balance as straight line, 191material balance in, 186maxunum efficiency rate (MER), 203

rate sensitivity, 203recovery mechanism, 191volatile oil reservoirs, 201water drive, 185, 191

bottom-water drive, 185edge-water, 185

Saturation pressure, see Bubble-pointpressure

Scbatz, see SiimhaSchilthuis, 2, 3, 56. 186, 188, 275, 217,

279, 329, 362, 365, 369, 370, 372, 411,419, 424

Schuler field, 73Schuler, see MatthesSciater and Stephenson, 2Scurry Reef field (Texas), 148, 198Separator system, 92

gas, 114-118liquid, 114-118

Shieve and Welch, 359Shnnkage of oil, 2Sidwell, see MillikanSimpson. rule, 241Single-phase fluid flow, 210

bubble point, 246oil-gas ratios, 246

Siunha, holland, Borshcel, and Schatz, 98Slaughter field, 64Slobod and Candle, 346Slurries, ISmith, 359Smith, R. II., 361Society of Petroleum Fnginecrs (SPE), 5Spencer, see CookSt. John oil field, 102Standing and Katz, 18, 19Standing, 31, 125, 126. See also MuskatSteady-statc flow systcm:,

cross flow, 224flow through capilianes and fractures, 225linear flow, incompressible fluids, 218

linear flow, slightly compressible fluids,219

parallel flow, 224, 226perineabiiity variations in linear systcms

222radial flow, 230radial flow, compressible fluids, 227radial flow, slightly compressible fluids,

227permeability variations, 229

radii, 227external, 227welibore, 227

viscous flow, 225Stephenson, see SclatcrStewart, Callaway, and Gladfclter, 274Straugraphic trap, 394Substances

Crude oil, INatural gas, IWater, I

Subsurface contour map, 70Summit County, Utah, 140Superposition, 249

in bounded or partially boundedreservoirs, 251

images method, 252Sutton, 16, 11

T

Tamer and Buckley, 3Tamer, 362, 370, 374, 424Tek, see KatzThompson, see JogiThornton, see MuskatTimmerm.n, see van EverdingenTorchlight Tcnsleep reservoir, 274Total quantity of gas dissolved in oil, 2Transient flow systems, 233

line source solution, 235radial flow, compressible fluids, 239radial flow, slightly compressible fluids,

234Traps, 4Truby, see Moore

U

U S Bureau of Mines, 194Unsta County, Wyoming, 140Undersaturated oil reservoirs, 146.5cc also

Volumetric oil reservoirsUnit recovery, see Volurnctnc gas reservoirsUniversity of Kansas, 394Updip water wells, 83

V

van der Knaap, 10

Index

van Bverdmgen and Hurst, 3,255, 280. Seealso Water influx

van Everdingen, Tunmeiman, and McMahcn,66

Vapor, IVaporization, 8Vasqucz and Beggs, 39, 41, 42Villena-Lanzi, 39Viscosity, 29Vogel and Yarborough, 139Volumetric gas reservoirs

calculation of depletion performance,120, 127, 136

calculation of unit recovery from, 76comparison between predicted and actual

performance of, 130dchnicion of, 84net cumulative produced gas-oil ratio,

156, 157, 159, 160, 163-165, 168.112performance of, 118, 127under water drive, 78, 80, 81, 99

Volumetric method (for calculating gas inplace), 70

Volumetric method, liznitauons of equationsand eriors, 99

Volumetric oil reservoirsbedding planes

bottom-water drive, ISO, 152edge-water dnve, 150

bubble-point pcessure,154, 157, 158calculation of initial oil in place, 148

material balance studies, 149volwnetric method, 149

calculations including formation and watercompressibduties, 170

effective fluid compressibility, Illestimation of oil recoveries, 148free gas phase, 154, 155, 156hydraulic control, 150material balance in undersaturated, 153net cumulative produced gas-oil ratio.

156, 157, 159, 160, 163-165, 168-172

sandstone, 152

Volumetric oil reservoirs, 148artificial gas cap, 155, 159

Water influx, 273pseudo-steady state, 325

bottom-water method, 325Darcy flow, 326

w

edge-water method, 325steady-state, 215

current gas-oil ratios, 276free gas rite, daily, 276net gas-oil ratio, daily, 276reservoir voidage rate, 276volumetric withdrawal rate, 276water influx constant, 278

unsteady-state, 280bottom-water drive model, 299constant terminal pressure case, 281constant terminal rate case, 281edge-water drive model, 280

van Evcrdingen and Hurst, 280, 281, 282,298. 299, 300, 325, 331, 332

Warren, see MuellerWater volume, 60Water-drive gas reservoir, 87Water-drive oil reservoir, 275Waterflooduig, 5Wattenbarger and Ramey, 221Welch, see ShreveWelge, 354, 359.See also BruceWendschlag, see KleinsteiberWessely, see ClarkWest, see MuellerWestern Ovenhrust Belt, 140Wet gas, 108, 139Wichert and Az.; 22-23Wickenbauser, see BourgoyneWieland and Kennedy, 64Wildcat reservoir, 182Wilde, see ColemanWoody and Moscnp, 57Wycoff and Botset, 2Wycoff, Bolset, and Muskat, 345Wycoff, Basset, Musket, and Reed, 2

V

Yarborough, see Vogel

z

Z-factor, 20

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