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1. THE TECHNOLOGY OF ARTIFICIAL LIFT METHODS
Volume 1
Inflow Performance
Multiphase Flow In Pipes
The Flowing Well
Kermit E. Brown
H. Dale Beggs
2. Preface
Whenan earlierbook,βGasLift TheoryandPractice,βwas completed,Ipromisedmyself Iwouldneverwrite another
book.But here isbook#1 of a three-volume series.Althoughthisbookstandsalone asatextand engineeringsource
book,itis alsoa prelude tothe secondtextonβArtificial LiftMethodsβ.
In a nutshell,thisfirstbookprovidesall the material neededtodesignanartificial liftinstallation.
The firstchapter dealswithamuch-neglected areaof petroleumengineering-inflowperformance.IthankMike
Fetkovichforpermission fortouse muchof hismaterial inthischapter. Also,manyof my graduate students,including
Mansoor, Yousaf,andKadi,contributedtothischapter.
The secondchapter,dealingwith multiphase flow inpipes,wasco-authoredbyDale Beggs.IextendthanksalsotoJim
Brill forhismany contributions.Muchisyetto be learnedaboutmultiphaseflow andIamproud to be part of the
learnedaboutmultiphaseflow andIamproud to be part of the Universityof Tulsawhere researchiscontinuing inthis
area underthe supervisionof Dr.JamesBrill andDr. Dale Beggs.
Chapter1 and2 are usedto predictthe behaviorof aflowingwell asfoundinChapter3.Again,I thankmu graduate
studentfortheirassistance,includingCelioFonseca,GustavoLopez,PedroRegnauld, HugoMarin, VictorGomez,and
Harry Hong.
As isthe case inalmostany text,Ihave leanedheavilyonpublishedmaterial inpiecingvarioussectionstogether.
Finally,Ithankmyfamilyfortheirdedicationandencouragement.
3. Chapter 1. Inflow performance
1.11 INTRODUCTION
The inflow performance of awell representsthe abilityof thatwell togive upfluids.A typical plotisnotedin Figure
1.1. andshowshowthe shapesof the curvesmay differ.Forexample,flowingpressure vsrate maybe essentiallya
straightline (waterdrive and/orpressureabove saturationpressure)oritmay curve (solutiongasdrive andflowbelow
the bubble point).The abilityof awell togive upfluidsdependstoa greatextentuponthe type of reservoiranddrive
mechanism,andsuchvariablesasreservoir pressure,permeability,etc.itiscommonpractice toassume that inflowinto
a particularwell withconstantconditionsisdirectionallyproportional to(πΜ π ).Note curve A inFig.1.1 whichisa straight
line.Normallythisistrue onlyforflowingpressure above the bubble point.
For curve A the PI of the well isconstantandisrepresentedinFig.1.1.by the inverse of the slope of the straightline.
PI isdefinedasbarrelsof total productionperdayperpsi of pressure drop (
π/π
psi
)or symbolicallyππΌ = π½ =
ππ+ππ€
πΜ π βππ€π
where
ππ = oil flowrate, ππ€ =waterflowrate, πΜ π = average staticwell bore pressure, ππ€π = flowingwellbore pressure.
Whenthe value of thisslope isconstantthe well issaidto gave a single PI.
However,itisknownthatcurvature existsinthisline formanywells.Inthiscase a well cannotbe saidto have a linear
PI (straightline),because the slope varieswiththe variationindrawdown.(NotecurvesBandC inFig.1.1). Two field
casesare showninFigs 1.1). Two fieldcasesare showninFig.1.2 and1.3. Fig.1.2 showsthree flow teststakenona
solutiongasdrive well withflowingpressuresbelow the bubble point.Fig.1.3showsthree flow teststhatexhibita
straightline withflowbeingabove the bubblepoint.
For a veryactive waterdrive inwhichthe pressure remainsabove the bubblepoint,the PIremainsconstant,andfora
solutiongasdrive inwhichthe flowingpressuresare belowthe bubblepointthe PIchangesrapidly.If aninflowcurve is
constructedat any time inthe life of the reservoirof Fig.1.4 (say3,000,000 stk bbl oil) the q vs. ππ€π curve will probably
be a straightline forthe waterdrive reservoirandcurvedforthe solutiongasdrive reservoir.The PIwill probablybe
highfor the waterdrive withverylittle dropinaverage reservoirpressureatthe highrates.
5. 1.2 TYPES OF RESERVOIRS
In orderto properlydesignanartificial liftinstallation,anunderstandingof the reservoirdrive mechanismisimportant.
The type of reservoirwill materiallyinfluence the productionrate,hence the type of artificial liftinstallation.There are
generallyconcededtobe three basicreservoirtypeswithpossible two- andthree-waycombinationsothese three.A
brief discussionof eachfollows:
1.21 Solution gas drive (Fig. 1.5)
Thistype of reservoirmayalsobe referredtoasinternal gasdrive,depletiondrive,and/orvolumetricperformance.
Some of the associatedcharacteristicsare:
1) A constantvolume.Thismeansthatthere isnochange inthe initial size of the reservoir.There isnowater
encroachmentforthisparticulartype of drive mechanism.
2) There istwo-phase flowatpressure belowthe bubblepoint.Inotherwords,gascomingoutof solutionflows
alongwithoil.
3) The gas comesout of solutionbutdoesnotmove upwardtoform a gas cap. Gas bubblesformedinthe oil
phase remaininthe oil phase,resultinginsimultaneousflow of bothoil andgas.
4) Oil productionisthe resultof the volumetricexpansionof the solutiongasandthe volumetricexpulsionof oil.
5) Thistype of reservoirdrive mechanismapproachesaflashgas-liberationprocess.
In the earlyproductionlife of the reservoir,oil isreplacedbygason an equal volume basis,butaspressure declines,a
largergas phase develops.More gasexpansionisthenrequiredperunitvolume of oil producedbecause of the fre e-flow
abilityof the gasphase.
It isknownthat creatingexcessive drawdowninadepletiondrive reservoirresultsinanincreasedgasphase inthe
reservoir-anincreasedpermeabilitytogasanddecreasedpermeabilitytooil.
6. Fig.1.5 showstypical performance curvesforthistype of reservoir.Of particularimportance isthe rapiddeclinein
pressure,the rapiddeclineinPIandthe increase inthe gas-oil ratio.
In general,wellsinthistype of reservoircanexpecttobe low rate producersintheir latterlife unlesstheyhave long
pay intervalssuchasthe CookInlet inAlaska.Many wellsof thistype will be candidatestoproduce lessthan100 bpdor
perhapslessthan25 bpd.
1.22 Water drive (Fig. 1.6)
The water drive mechanismmayalsobe referredtoaswaterencroachmentorhydrauliccontrol.Some of the
associatedcharacteristicsare:
1) The reservoirvolume foroil doesnotremainconstant.Waterencroaches,changingthe initial volume of the
container.
2) There isa displacementof the oil bywater.
3) Thisreservoirtype couldalsohave agas phase,resultinginacombinationwaterdepletiondrive.
4) There will be anoptimumrate of productionforthisreservoirtype.
Fig.1.6 showstypical performance curvesforthistype of reservoir.Ina veryactive waterdrive the pressure decline
may be verysmall,andinfact, pressure mayremainconstant.Of greatimportance isthe trendof the PI to remain
constantoverthe life of the well.Inturnthe GOR alsoremainsconstant.
7. In a waterdrive reservoirthe PIβsof individualwellsare normallymore reliable thanthose of adepletiondrive
reservoir.Waterencroachmentmaybe suchthat there isverylittle lossinbottomhole pressure.Itisgenerally
concededthatPI informationmaybe extrapolatedlinearlyfordrawdownsnecessarytogive the desiredproduction.
Probably,inmostcases,the pressure remainsabove the bubble point.Mostartificialliftinstallationscanbe designed
withmore reliabilityandconfidence forwaterdrive thanforanyothertype of drive mechanism.However,there are
instanceswhere anincrease inwater-cutcausesadecrease inPI.
1.23 Gas cap expansion drive (Fig. 1.7)
Thistype of reservoirdrive mechanismmayalsobe referredtoassegregationorgravitydrainage.The rese rvoirisina
state of segregation βan oil zone overlainbyagas cap. The drive may be furtherclassifiedastowhetherornot gas
comingout of solutioninthe reservoirflowstothe gascap. A segregationdrive withcounterflow will have gascoming
out of solutionandmovingtothe gas cap. As productionproceedsthe gascap expandsandmovesdown,resultingin
gas cap expansiondrive.
Generally,the permeabilityof the formationdetermineswhetherornotcounterflow willoccur.Asan estimate,it
wouldbe expectedtooccur.Asan estimate,itwouldbe expectedtooccurfor permeabilitiesinexcessof 100
millidarcies.
The segregationdrive withcounterflowapproachesadifferential gasliberationprocess,definedasaprocessin which
the gasesliberatedfromsolutioninthe oil whenthe pressure isreducedare removedfromcontactwiththe oil as
rapidlyastheyare formed.
In Fig.1.7 the performance curvesappeartobe somewhere betweenthose forsolutiongasdrive andwaterdrive.In
general the pressure declinesfairlyrapidlyandthe PIfollowsthe same trend.
8. 1.24 Summary
There are manyreservoirshavingcombinationdrivemechanismandtheirperformance maydifferconsiderablyfrom
the typical curvesgivenforwaterdrive,solutiongasdrive, orgas-capexpansiondrive mechanism.Aneffortshouldbe
made to identifythe reservoirdrive mechanisminordertopermita betterdeterminationof the abilityof the wellto
give upfluids.There isnosubstitute forgooddatatakenduringthe earlylife of the reservoirinpredictingfuture
performance.
Several fieldcasesforreservoirsare showninFigs.1.8,1.9, 1.10 and 1.11. thesesexampleswere takenfromreference
(1).
9. 1.3 INFLOW PERFORMANCE RELATIONSHIPS
1.31 Introduction
In discussinginflow performancerelationshipswe mustkeepinmindthe type of reservoir,andthe shape of the IPR
curve,whichisa plotof flowingpressure vsrate (Fig.1.1).we mustalsorecall that the inflow performance of awell is
verylikelytochange withtime andcumulative production.
1.32 Productivity index
The commonly-usedterm, PI(productivityindex),representsone pointonthe inflowperformance curve.The PIis
definedas π/Ξπ inbpdper psi pressure dropfromstaticreservoirpressure toflowingbottomhole pressure.
In the designof artificial liftinstallationsthe productivityindexisexpressedin π/π of total liquid(oil `water)andis
definedas:
π½ =
ππ + ππ€
πΜ β ππ€π
(1.1)
where:
ππ = stock tanksbbl of oil perday
ππ€ = stock tanksbblsof waterperday
ππ€π = bottomhole flowingpressure,psi
πΜ π = staticpressure,psi
Fig.1.12 showsPI inan ideal case where astraightline istan π =
ππ΅
ππ΄
.
In Fig.1.13 we note a case where a straightline relationshipdoesnotexist,representingflow below the bubblepoint
pressure.
10. We note that
ππΌ = π½ = β
ππ
πππ€
(1.2)
where the negative signindicate adecreasingPIforanincreasingrate.
11. EXAMPLE PROBLEM #1 HOW TO DETERMINE PI (LINEAR CASE)
Given:
πΜ π = 2,400 psi
ππ = 200 bpd
ππ€ = 300 bpd
ππ€π = 2,200 psi
Findthe PI assumingittobe a linearrelationship (idealcase)
ππΌ = π» =
ππ + ππ€
πΜ π + ππ€π
= (200 + 300)/)2,400 β 2,200 = 2.5 πππ/ππ π
CLASS PROBLEM #1-A: TO DETERMINE PI (LINEAR CASE)
Given:
πΜ π = 2,800 psi
ππ€π = 2,200psi
ππ = 600 bpd
ππ€ = 1,000 bpd
Findthe PI average forthisflowtest.
It iscommon practice to measure one ortwo PIβsinthe earlylife of a well,thenuse thatsame PIto estimate
drawdownsnecessaryforgreaterproductionratesaswell asassumingthatthe same PI existslater inthe life of the well.
We are probablyfairlysafe indoingthisforwells inawaterdrive fieldwhere the flowingpressure isabove the bubble
point.Butthismas be inerror forwellsina solutiongasdrive reservoirorat flow below the bubblepoint.
12. AnothertermsometimesusedisspecificPI.Thisisthe productivityindex dividedbythe netfeetof pay.Itis commonly
usedto compare differentwellsinagivenfield.
π½π =
π½
β
=
ππ π‘π
β(πΜ :π β ππ€π)
= bpd/psi/ft (1.3)
EXAMPLE PROBLEM #2: HOW TO DETERMINE SPECIFIC PI
Given:
πΜ π = 2,500 psi
ππ€π = 2,200 psi
ππ = 200 bpd
β = 20 ft
Findthe specificPI:
π½π =
π½
β
=
ππ π‘π
β(πΜ π β ππ€π)
=
200
(20)(2,500 β 2,200)
= 0.0333
bpd/psi
ft
CLASS PROBLEM #2-A: TO DETERMINE SPECIFIC PI
Given:
πΜ π = 3,000 psi
ππ€π = 2,800 psi
ππ = 1,000 bpd
β = 50 ft
Findthe specificPI.
The PI referstoSurface productionof liquidandtosandface pressure differences.
In reservoirengineeringthe PIbasedonsurface oil productiononlyissometimesused.Forflowingandartificial lift
wells,itisnecessarytoliftthe total liquidfromthe well,andtherefore the PIequationshouldinclude the producing
water/oil ratio:
π½ =
ππ + πππΉ
π€π
πΜ π + ππ€π
(1.4)
where πΉ
π€π = wateroil ratio.Thisisthe same as equation1.1givenagainas π½ =
ππ+ππ€
πΜ π βππ€π
, and ismore commonlyusedin
thisform.
The variables whichaffectthe PIcan easilybe seenbyderivingthe PIequationusingDarcyβsradial flow equation.If
bothwater andoil are flowinginthe reservoir,the radial flow equationsforeachfluidmustbe used:
Oil
ππ =
7.08πΎπβ(πΜ π β ππ€π)
π΅πππ ln(ππ/ππ€)
(1.5)
Water
ππ€ =
7.08πΎπ€β(πΜ π β ππ€π)
π΅π€ππ€ ln(ππ/ππ€)
(1.6)
The PI basedon total liquidproductioncanthenbe calculatedasfollows:
13. π½ =
ππ + ππ€
πΜ π β ππ€π
=
7.08(πΜ π β ππ€π)β
(πΜ π β ππ€π)ln(ππ/ππ€)
[
πΎπ
π΅πππ
+
πΎπ€
π΅π€ππ€
] =
7.08β
ln(ππ/ππ€)
[
πΎπ
π΅πππ
+
πΎπ€
π΅π€ππ€
] (1.7)
Althoughthe pressure termsdropout,the PIremainsdependentonpressure sincethe viscositiesandformation
volume factorsare functionsof pressure.The effective permeabilitiesare alsofunctionsof the fluidsaturations.
1.321 Estimated productivity index
An estimate of the PIcan be obtainedfromthe followingapproximation:
Let
7.008
ln ππ/ππ€
= 1.0
For a well bore radiusof 2.5 in.thisrepresentsaradiusof drainage of 247 ft,and for3.5 in.it is346 ft.
then:
π½ = β [
πΎπ
π΅πππ
+
πΎπ€
π΅π€ππ€
] (1.8)
For the special case of negligible waterproduction(πΎπ€ = 0) the preedingequationbecomes:
π½ =
βπΎ
π΅πππ
(1.9)
EXAMPLE PROBLEM #3: HOW TO ESTIMATE PI
An 8,000 ft well on40 acre spacinghas a netpay thicknessof 20 ft. the following additional dataare known:
πΜ π = 4,000 psi
ππ€π = 3,800 psi
ππ€ = 3 in. (0.25 ft)
π΅π = 1.25
π΅π€ = 1.05
ππ = 10 cp
ππ€ = 0.8 cp
πΎπ = 2 darcies
πΎπ€ = 0.01 darcies
π0 = 550 bpd
ππ€ = 50 bpd
Find:(1) the estimatedproductivityindex neglectingwaterproduction;(2) the estimatedproductivityindex including
waterproduction;(3) the actual productivityindex of the well,basedonthe radial flow equationandaradiusof
drainage equivalenttoa40-acre circle;and (4) PIbasedon flow test.
Solution
(1) π½ =
βπΎ
π΅πππ
=
(20)(2)
(10)(1.25)
= 3.2 b/d/psi
(2) π½ = β [
πΎπ
π΅πππ
+
πΎπ€
π΅π€ππ€
] = (20) [
2
(10)(1.25)
+
0.01
(0.8)(1.05)
] = 3.44 b/d/psi
(3) π½ =
7.08
ln(ππ/ππ€)
β [
πΎπ
π΅πππ
+
πΎπ€
π΅π€ππ€
] =
7.08
ln(ππ/ππ€)
(3.44) =
24.37
ln(ππ/ππ€)
Determinationof ππ:
πππ
2 = (40 ππππ)(43,560 sq ft/acre)
ππ = 745 ft
14. π½ =
24.37
ln (
745
0.25
)
= 3.04 bpd/psi
(4) PI based on flow test =
550+50
4,000β3,800
= 3.0 bpd/psi
CLASS PROBLEM #3-A: TO ESTIMATE PI
Given:
Depth= 10,000 ft
Spacing= 60 acres
h= 30 ft
πΜ π = 3,800 psi
ππ€π = 3,765 psi
Well bore diameter=5 in.
π΅π = 1.23
π΅π€ = 1.04
ππ = 0.7 cp
ππ€ = 1.9 darcies
πΎπ = 1.9 darcies
πΎπ€ = .02 darcies
ππ = 500 bpd
ππ€ = 25 bpd
Find:
1) Estimate PIneglectingwaterproduction
2) EstimatedPIincludingwΓ‘terproduction
3) PI forthe well based onall available dataandassumingacircular drainage areafor 60 acres
4) PI basedonflowtest
Lewis,Horner,andStekoll showedthatthe productivityindex couldbe relatedbythe followingequations:
ππΌ = π½ = 5.9 Γ 10β4
πΎβ
πππ΅π
for πΎ = ππ (1.10)
Thiscan be approximatedas:
π½ =
0.6πΎβ
πππ΅π
(1.11)
πΎ = darcies
β = ft
ππ = centipoise
π΅π = formationvolume factor
LewisandHorner presentedacertainamountof data where theycomparedmeasuredvaluesof PIagainstcalculated
valuesof PI.The data for calculatedvs. measuredproductivityindicesare includedinTables1.1.and1.2.
In Fig.1.14 theyplotted
πΎβ
ππ΅
againstvaluesof measuredPI.Line A isdrawnthroughpointsthatrepresentwellsinwhich
the average reservoirpressure wasrelatively highatthe time of test,andflow wasprobablyabove the bubble point.
Line B representsdatafromwellswithalowbottomhole pressure withflowing pressure below the bubblepoint.An
average gas saturationof 11.5% wasestimatedanda relative permeabilitytooil of 51% wasdetermined.Applyingthese
15. values,Fig.1.15 was preparedandnowbothsetsof pointsare in agreement.Line A of fig.1.15 represents acorrelation
constantof 5.9 Γ 10β4 πΎβ
ππ΅
where πΎ = md,β = ft, π = cp, and ππΌ =
bpd
psi
. π΅ =
bbl
stk bbl
Fig.1.14 and 1.15 also howtwotheoretical linesascalculatedfromthe radial equation(equation1.5) forvaluesof
ππ
ππ€
= 4,000, and
ππ
ππ€
= 400. For and ππ€ value of 3 in. these valuescorrespondtodrainage radiiof 1,000 and 100 ft
respectively.The positionof Line A issomewhatabove the theorical lines,therefore,the measuredvaluesof PIare less
than wouldbe calculatedfromthe radial flowequation.Dependingupon ππ/ππ€ the correlationconstantmaydifferfrom
5.9 Γ 10β4.
TABLE 1.1
DATA FOR CALCULATING PRODUCTIVITY INDEXa
Well
no.
Average
permeability Γ
thickness, πΎβ,
millidarcy-ft
Measured
reservoir
viscosity, π
centipoises
Measured
formation
volume factor,
π΅π
Perforated or
open hole
Measured
productivity
index
Calculated
productivity indexb
1 16,100 0.614 1.33 P 19.0 11.7
2 15,540 0.614 1.33 P 15.4 11.2
3 14,400 0.614 1.33 P 11.9 10.4
4 11,545 0.44 1.48 P 9.5 10.5
5 7,800 0.614 1.33 P 6.3 5.6
6 8.,330 3.42 1.02 OH 0.81 0.7
7 6,910 3.42 1.02 OH 0.64 0.66
8 8,980 3.42 1.02 OH 0.62 0.77
9 5.200 3.42 1.02 OH 0.52 0.45
10 3,360 3.42 1.02 OH 0.28 0.29
11 2,500 3.42 1.02 OH 0.17 0.21
12 1,830 3.42 1.02 OH 0.12 0.16
13 575 3.42 1.02 OH 0.043 0.050
14 400 1.60 1.20 OH 0.105 0.126
a
Pointslistedare plottedascirclesonFigs.1.14 and 1.15
b
Calculatedbyusingline A onfig.1.15
c
Wells6 through13 had gas saturationof 11.5% of pore space whenmeasurementswere made.Effective
permeabilityof 51%was usedincalculations.
16. TABLE 1.2
ADDITIONAL PRODUCTIVITY INDEX DATA
Well
no.
Average
permeability Γ
thickness, πΎβ,
millidarcy-ft
Reservoir
viscosity, π
centipoises
Formation-volume
factor, π΅π
Measured
productivity index
Calculated
productivity indexc
15 11,800 0.80b
1-45 6.0 6.1
16 3,179 0.80b
1.45 2.1 1.62
17 2,516 0.80b
1.45 1.7 1.30
18 15,000b
1.5 1.2 5.0 4.92
19 28,000b
1.5 1.2 6.5 9.2
20 24,000b
1.5 1.2 5.4 7.87
21 150,000b
1.5 1.2 53.0 49.0
22 70,000b
1.5 1.2 29.0 23.0
23 23,000b
2.5 1.1 5.11 4.93
24 22,000 0.5b
1.26 16-19 20.6
25 2,630 0.53b
1.48b
1.22 1.97
a
Pointslistedare plottedas trianglesonFigs.1.14 and 1.15
b
Numerical value representsestimationrequiredbecause of incomplete core recoveryolackof bottom-hole sample
data. Othervalues, exceptcalculatedproductivityindex,were measured.
c
Calculatedbyusingline A onFig.1.15
17. EXAMPLE PROBLEM #4: HOW TO DETERMINE PI BY METHOD OF LEWIS AND HORNER
Lewisand Horner2
gave the followingfieldexample thatwascheckedagainstthe measuredPIfromaflowingpressure
tests:
πΎβ = 11,545 millidarcy-ft
ππ = 0.44 cp (reservoir conditions)
π΅π = 1.48
ππΌ =
(5.9 Γ 10β4)(11,545)
(0.44)(1.48)
= 10.5 bpd/psi
The PI of thiswell wasfoundtobe 9.5 by conductingatest onthe well
CLASS PROBLEM #4-A
Given:
πΎπ = 2 darcies
β = 40 ft
ππ = 0.8 cp
π΅π = 1.35
Determine the PIbythe methoof LewisandHorner.
We have definedPIasfollows:
π½ =
ππΏ
πΜ π β ππ€π
Thisrepresentsandaverage PI between the staticreservoirpressure andthe stabilizedflowingpressure forππΏ.This
definitionassumesthe PIisa straightline relationship,andinturnthat a plotof q vs. ππ€π will yieldastraightline
relationship.Althoughthismaybe true inmanycases forflow above the bubble point,thereare numeroussolutiongas
drive reservoirsinwhichthiswill notbe true.We mustkeepinmindthat thisalsorepresentsaPIteston one specific
day inthe life of the well.Thismayalsochange withtime and cumulative recoveryasdiscussedinthe followingsection.
1.322 Productivity index change with time
It isknownthat the PI changeswithcumulative recoveryasnotedinFig.1.4. There isa decidedchange insome cases
due to changesinpermeability tooil andwater,andwhenflow isbelow the bubble pointpressure.
A methodforpredictingthe PIinthe future3
wasnotedbymakinga semi-logplotof Fig.1.4 as showninFig.1.16. as
noted,the waterdrive plotremainsastraightline,andthe pointforthe solutiongasdrive alsogive essentiallyastraight
line.Some cautionisnecessaryintryingtoobtainmeaningful PIβs,thatis,teststakenatthe same drawdowninpressure.
The PIβsmay vary dependinguponthe drawdowninpressure,andhence the flowingpressure.
The gas cap expansionalsoapproachesastraightline relationship.Thisrepresentsone method thatcanbe utilizedin
predictingfuture PIβs.
Thisis furtherillustratedforasolutiongasdrive reservoirinFigs.1.17and 1.18. in Fig.1.17 the PI changesfrom2.0 to
0.4 withlessthan20,000 bblscumulative production,andthe extrapolationof thiscurvesbecome difficult.Whenthe
same three pointsare plottedonFig.1.18 (semi-logplot) astraightline results,givingameansof predictingPIβsat
future dateswithsome degree of confidence2
.Althoughthismethodworkedverywell forthisparticularwell,itmaynot
be applicable inothercases.Itisan empirical procedure,butmayverywell applytoothersolutiongasdrive wells.
Most of thischange inPI islikelycausedbyanincreasedfree gassaturationaroundthe well bore whichincreasesthe
permeabilitytogasand decreasesthe permeabilitytooil.Otherpossibilitiesare increasedoil viscositywithpressure
drop belowthe bubble point,andreductioninpermeabilitydue toformationcompressibility.
18.
19. 1.33 Some early discussion on PI
Some of the firstworkonPIβsandchange inPIβswasdone byEvingerandMuskat4
in1942. Thisworkwasalsodiscussed
by Calhoun5
inhis bookon reservoirengineering.Theynotedthata plotof flowingpressure againstrate wasnot always
a straight line.
Starting with the radial flow equation as follows:
ππ =
7.08πβπΎπππ
πππ΅πππ
(1.13)
Equation 1.13 can be rearranged as follows by defining πΎππ =
πΎπ
πΎ
:
ππ
7.08πΎβ
β«
ππ
π
ππ
ππ€
= β«
πΎππ
πππ΅π
ππ
ππ
ππ€
(1.14)
where
ππ
π
canbe integratedbetweenlimitsandthe term
πΎππ
πππ΅π
can be evaluatedas a function of pressure and integrated.
At anyone time we have a constant gas oil ratioas givenbythe followingequation:
π =
πΎπ
πΎπ
ππ
ππ
π΅π
π΅π
+ π π
(1.15)
where π = current producing gas oil ratio and π π = gas in solution at current pressure.
For a given π value a plot of oil saturation vs pressure can be made such as Fig. 1.19.
From Fig. 1.19 the values of
πΎπ
πΎπ
can be determined from the oil saturation and the values of
ππ
ππ
, π΅π, π΅π, and π π can be
determined by knowing the pressure, temperature, and fluid properties. Once we have the pressure vs. saturation
correlation,the πΎππ valuescan be determinedforthe saturationat a given pressure andthe term
πΎππ
πππ΅π
can be evaluated
and plotted against pressure as in fig. 1.20.
The right handside of the equation1.14 can now be integratedgraphicallybytaking the areabetweenvaluesof π1 and
π2under the curve as shown in Fig. 1.20. therefore, the equation then appears as follows:
ππ =
7.08πΎβ
ln
ππ
ππ€
(area under curve) (1.16)
EvingerandMuskat4
statedthat the PIcouldbe expressedintermsof three parameters;(1) the producinggasoil ratio,
(2) the reservoir pressure, and (3) the pressure gradient in the well system.
The following equation expresses oil flow in the reservoir:
ππ =
7.08πΎβ
ln
ππ
ππ€
β«
πΎππ
πππ΅π
ππ
ππ
ππ€
(1.17)
The integral canbe evaluatedasshownin Fig.1.21byfindingtheareaunderthecurve betweenanytwopressurepoints.
The PI can then be determined from the equation:
ππΌ =
ππ
πΜ π β ππ€π
=
7.08πΎβ(area under curve)
(πΜ π β ππ€π)ln
ππ
ππ€
(1.18)
Several factors are noted from an examination of equation 1.18, and Fig. 1.21:
1) The PI will notdouble if (πΜ π β ππ€π) doublesbecause the areaunder the area under the curve will not double.
20. 2) If a constant value of (πΜ π β ππ€π) is taken at a high pressure as compared to a low pressure, the area will be
greaterat the highpressure.Therefore the PIwill be greaterathigherreservoirpressuresandlowdrawdowns.
3) The PI value will depend on the producing gas oil ratio.Each gas-oil ratio value means a different steadystate
system for which a different curve will apply. Calhoun showed how the function changes with R values (Fig.
1.22).
Calhoun5
alsopreparedFig.1.23 whichshowsthe dependence of PIonreservoirpressure andpressure drawdownfor
one gas-oil ratioonly.A similarfigure wouldbe neededforeachdifferentgas-oil ratio.
The theoretical ProductivityIndex of aradial systemforsteadystate flow canbe expressedasfollows:
ππΌ =
7.08πΎβ
(πΜ π β ππ€π)(ln
ππ
ππ€
)
β«
πΎππ
πππ΅π
ππ
πΜ π
πΜ π€π
(1.19)
21. EvingerandMuskat4 suggestedthatinorderto use the PI value asa meansof comparison,the comparisonsbe made
for (πΜ π β ππ€π) equal tozero,that is, for ππ€π approachingπΜ π as a limit.Thenthe PIequationbecomes:
ππΌ =
7.08πΎβ
ln
ππ
ππ€
(
πΎππ
πππ΅π
)
πΜ π
(1.20)
In orderto compare PI valuesattwo differenttimeswe have:
(ππΌ)1
(ππΌ)2
=
(
πΎππ
πππ΅π
)
1
(
πΎππ
πππ΅π
)
2
(1.21)
It isthenpossible toevaluate the PIof a solutiongasdrive well.Inordertodo this we mustevaluate the oil saturationat
some future time.Byknowingthe pressure we canobtain π΅π,and ππ, andπΎππ isdefinedfromthe saturation.Calhoun5
showedatypical decline inPIbasedonhisanalysis(Fig.1.24).
Fig.1.25 showshowthe PIvariesat differentpressure forconstantsaturations.Therefore if the pressure and
saturationare knownwe couldobtain ππΌ/ππΌI or if the PI and pressure were know saturationcouldbe estimated.Fig.
1.25 was preparedfora particularreservoirhavingandoriginal oil saturationof 80% and an original pressure of 3,500
psi.
The methodproposedbyEvingerandMuskat has been usedextensivelyinfieldcasesandhasproventobe verygood
inmany cases.It isrecommendedthatthisprocedure be evaluatedalongwith otherproceduresdescribedinthis
chapter.It may verywell prove tobe sufficientlyaccurate forpredictionof PIβsandinflow curves. Calhoun5
givesan
excellentdiscussiononthisprocedure.
22. 1.34 Inflow performance curves
1.341 Introduction
The PI not onlychangeswithtime orcumulative productionbutisalsosubjecttochange withincreased drawdownat
any one specifictime inthe life of the well.If we measuredseveral PIβsinawell duringaspecifictime interval,a
relationshipwillbe obtainedbetweenrate andflowingpressurewhichnormallyisnotlinearforasolutiongasdrive
field.Thisphenomenonmaybe attributedtoone ormore of the followingfactors:
1) Increasedgassaturationwithsubsequentloweringof permeabilitytooil nearthe well bore asa resultof
reducedreservoirpressure nearthe well bore athigherproducingrates.
2) Changesfromlaminarto turbulentflow insome flow capillaries nearthe well bore atincreasedproducing
rates.
3) Exceedingcritical flow ratesthroughporesatformationface inthe well bore.These poresactas orificesand
whenthe critical rate is exceeded,increaseddrawdownshave adiminishedeffectonincreasingrates.
Thisplotof π vs. ππ€π iscalledinflowperformance andwasfirstusedbyGilbertindescribingwell performance6
.A
typical plotisnotedinFig.1.1 and differsdependinguponthe type of reservoir.
1.342 Vogelβs work
A publicationbyVogel inJanuary19687
offeredasolutionindetermininganinflow performancecurve fora solution
gas drive fieldforflowbelowthe bubble point.Bythe use of a computer,he calculatedIPRcurvesforwellsproducing
fromseveral fictitioussolution-gasdrivereservoirsthatcoveredawide range of oil PVTpropertiesandreservoirrelative
permeabilitycharacteristics.He made severalassumptionssuchascircular,radial uniformflow withaconstantwater
saturation. He neglectedgravitysegregationandhissolutionisvalidfortwo-phase flow inthe reservoironly.Vogel7
showedhowrate vs.flowingbottomhole pressureasa functionof cumulative recoverychangedinFig.1.26.as noted,
the resultisa progressive deteriorationof the IPRβsasdepletionproceedswithtime inasolution-gasdrive reservoir.
Vogel7
alsopresentedFig.1.27whichshowsthe effectof viscosityandgas-oil ratio(GOR). Curve Buseda crude oil
withaboutΒ½ the viscosityof the crude forCurve A.Also,the crude of Curve B useda GOR abouttwice thatof the crude
for Curve A.
23. He plottedall the IPRβsasβdimensionlessIPRβs.βThe pressure foreachpointonan IPR curve isdividedbythe
maximumorshut-inpressure forthatparticularcurve,andthe correspondingproductionrate is dividedbythe
maximum(100% drawdown) producingrate forthe same curve.Whenthisisdone,the curvesfromFig.1.26 can
replottedasshowninFig.1.28.
It isthenapparentthat withthistype of constructionthe curvesare remarkablysimilarthroughoutmostof the
producinglife of the reservoir.He alsonotedthatthe same dimensionlessplotof Fig.1.27 gave IPRβsthat were similar
as inFig. 1.29.
Before constructinghisfinal curve he made calculationsformore viscouscrudes,varyingGORs,varyingrelative
permeabilities, differentwell spacings,fracturedwells,andforwellswithskineffect.Insummary,hiscalculationsfor21
reservoirconditionsresultedinIPRβsgenerallyexhibitingasimilarshape.One exceptionwasawell forskineffectin
whichthe IPR approached a straightline.The more viscouscrudesandreservoirsabovethe bubble pointalsoshowed
significantdeviation.However,curvature wasapparent.
Vogelβswork7
resultedinhisconstructionof areference curve (Fig.1.30) whichisal that is neededfrom hispaperto
construct an IPRcurve from one flowingtestona well.Thiscurve shouldbe regardedasageneral solutionof the
solution-gasdrivereservoirflowequationswith the constantsforparticularsolutionsdependinguponanindividual
reservoir, andforflowingpressurebelowthe bubblepoint.
The equationof the curve of Fig.1.30 is:
ππ
(ππ)max
= 1 β 0.20 (
ππ€π
πΜ π
) β 0.80 (
ππ€π
πΜ π
)
2
(1.22)
where ππ isthe producingrate correspondingtoa givenwell intakepressure ππ€π,πΜ π the corresponding pressure,and
(ππ)max is the maximum(100%drawdown) producingrate. πΜ π isthe average reservoirpressure.
For comparison,the relationshipforastraight-line IPRis:
ππ
(ππ)max
= 1 β
ππ€π
πΜ π
(1.23)
24. Vogel comparedthe reference curve of Fig.1.30 withthose calculatedonthe computer.The curve matchesmore
closelythe IPRcurvesforearlystagesof depletionthanthe IPRcurvesforlaterstagesof depletion.Inthisway,the
percenterroris lesswhendealingwiththe higherproducingratesinthe earlystagesof depletion.The percentage error
becomesgreaterinthe laterstagesof depletion,but here productionratesare low and,as a consequence,absolute
errorswouldbe less.These comparisonscan be foundinFig.11 of hisoriginal paper.7
Maximumerrorinthe use of the
curve will occurwhenwell testsare made at verylow producingratesandlow drawdownsandthenan attemptismade
to extrapolate to100% drawdowns.Vogel statesthatmosterrors shouldnotexceed10%.Ihave personallyconsulted
withmanyengineersthathave usedthiscurve andtheyhave beenverypleasedwiththe results.
Vogel7
alsomade the followingobservations:
1) Thisprocedure wouldnotbe consideredcorrectwhere othertypesof drive exist.However,some engineers
have usedthisprocedure forothertypesandcombinationsof drive mechanismswithgoodresults.Forwells
producingwitha flowingpressure below the bubble point,hisworkmayverywell be good.
2) Since the reference curve isforthe two-phase flowof oil andgas only,itwouldnotbe consideredvalidwhen
three phases(oil,gas,andwater) are flowing.However,againsome engineershave noteditscontinued
accuracy for three-phase flow.
3) Its comparisontosingle-phase liquidflow andsingle-phase gasflow isnotedinFig.1.31.
4) The conclusionsare basedoncomputersolutionsinvolvingseveral simplifyingassumptionsandadditional
comparisonswithfielddataare needed.However,Ifeel thatthisrepresentsthe best solutiontodate andis
more accurate than assumingalinearrelationship.
5) The reference curve of Fig.1.30 isverysimple touse.All thatisneededisone flow testof flowingbottomhole
pressure vsrate and the static reservoirpressure.Fromthe ratio of ππ€π/πΜ π (ordinate) avalue of ππ/(ππ)max
can be foundand (ππ)max determined.Once (ππ)max hasbeendeterminedavalue of π orany ππ€πcan be
foundandthe constructionof an inflow performance curve isthenpossible.Solutionsare offeredinthe
followingexamples.
25. EXAMPLE PROBLEM #5: HOW TO DETERMINE MAXIMUM FLOW RATE FOR SOLUTION GAS DRIVE WELL
The followingtestwasconductedona solutiongasdrive well:
Givendata: πΜ = 2,000 βππ πβ
(AfterVogel7
) ππ€π = 1,500 psi
ππ = 65 bpd
Find:
a) (ππ)max
b) ππ for ππ€π = 500 psi
Solution:
a)
ππ€π
πΜ π
=
1,500
2,000
= 0.75
From Fig.1.30 we find
ππ
(ππ)max
= 0.40
Then
65
(ππ)max
= 0.40
(ππ)max =
65
0.40
= 162 bpd
The well makes162 bpdfor ππ€π = 0 whichisessentiallyimpossiblefromapractical pointof view.
b) Findππ for ππ€π = 500 psi;
ππ€π
πΜ π
= 0.25
From fig.1.30 we find
ππ
(ππ)max
= 0.90.we now know (ππ)max therefore
ππ
162
= 0.90
ππ = (0.90)(162) = 146 bpd
CLASS PROBLEM #5-A
Givendata: πΜ π = 2,500 psi, ππ = 2,500 psi
ππ€π = 1,800 psi
ππ = 100 bpd
Find: (a) (ππ)max
(b) π0 for ππ€π = 1,000 psi
CLASS PROBLEM #5-B
Givendata: πΜ π = 3,000 psi, ππ = 3,000 psi
ππ€π = 2,500 psi, π0 = 500 bpd
Find: (a) (ππ)max
(b) ππ for ππ€π = 1,200 psi
(c) Find(ππ)max if a linearrelationshipisassumed.
EXAMPLE PROBLEM #6: HOW TO CONSTRUCT IPR CURVE FROM ONE FLOW TEST
Constructa complete inflow Performance Curve forthe datagiveninExample Problem#5.
Thisproblemissolvedbyassumingvariousflowingpressureanddeterminingthe correspondingflow rates.We will
make use of the informationfoundpreviously.
Prepare a table as follows:
26. Assumedππ€π π Remarks
0 162 Previouslycalculated
1,500 65 Given
500 146 Previouslycalculated
2,000 0 Given
Additional valuesof ππ€π are assumedasnecessaryandthe correspondingflow ratesdetermined.Thisinformationis
thenplottedasinFig.1.32. It shouldbe kept inmindthat thisrepresentsaninflow performance curve forthe testdate
and will differata laterdate.However,itwill probablyretainthe same general shape butwill beginwithalowerstatic
reservoirpressure.A plotof thistype wasshown inFig. 1.26.
CLASS PROBLEM #6-A: TO FIND MAXIMUM FLOW RATE AND FLOW RATE FOR ANY FLOWING PRESSURE
Givendata: πΜ π = 2,300 psi
ππ€π = 1,400 psi
ππ = 95 bpd
Find: (a) (ππ)max
(b) ππ for ππ€π = 700 psi
(c) construct a complete IPRcurve of π vs. ππ€π
27. CLASS PROBLEM #6-B
(a) Constructan IPR curve for ClassProblem#5-B
(b) Constructan IPR curve for ClassProblem#5-Bassuminga linearrelationship.
1.343 Standingβs extension of Vogelβs work to account for damaged or improved wells
The initial workof Vogel assumedaflowefficiencyof 1.00 and didnot accountfor wellsthatwere damaged or
improved.Standingproposedacompanionchartto accountfor conditionswhere the flow efficiencywasnotequal to
1.00.
In Fig.1.33 flowefficiencyisdefinedas:
πΉπΈ =
Ideal drawdown
Actual drawdown
=
πΜ π β ππ€π
β²
πΜ π β ππ€π
where ππ€π
β²
= ππ€π + Ξππ πππ
substituting:
πΉπΈ = πΜ π β
ππ€π + Ξππ πππ
πΜ β ππ€π
= πΜ π β ππ€π β
Ξππ πππ
πΜ π β ππ€π
(1.24)
whichisthe ratio of useful pressure dropacrossthe systemtototal pressure drop.For a well draininga cylindrical
volume:
πΉπΈ = ln
0.47ππ
ππ€
/[ln
0.47ππ
ππ€
+ π] (1.25)
where π isthe dimensionlessskinfactor.Some confusionexistswhenwe thinkthat πΉπΈ alsoexpressesthe wellβsflow
rate withdamage tothe flowrate withoutdamage.Onlywhenthe IPRcurve isa straightline (undersaturatedliquids) is
πΉπΈ definedbyequation1.24alsoequal to the ratioof flow rates,where equation1.24 is:
πΉπΈ =
πΜ π β ππ€π β Ξππ πππ
πΜ π β ππ€π
(1.24)
As notedinFig.1.33, an undamagedwell wouldflowatrate π fora flowingpressure of ππ€π
β²
while the damagedwell
mustflowat the lowerpressure of ππ€π inorderto produce the same rate π.
28. The Ξππ πππ isthus seentobe the difference between ππ€π
β²
andππ€π.Fig.1.34 shows the regionof addedresistance to
flownearthe well-bore. There maybe manyfactorswhichcause or control thisaddedresistance includinginvasionof
the zone by mudor βkill-fluids,βswellingof shale,andothers.Thismayalsorepresentaregionof improvementaftera
stimulationtreatment.
The determinationof Ξππ πππ ismade byfirstdetermining π (skinfactor) fromastandardpressure build-uptestona
well asinFig.1.35